Achieving Gold-Medal-Level Olympiad Reasoning via Simple and Unified Scaling

# Achieving Gold-Medal-Level Olympiad Reasoning via Simple and Unified Scaling
Source: [https://arxiv.org/html/2605.13301](https://arxiv.org/html/2605.13301)
Yafu Li1,2,Runzhe Zhan111footnotemark:1,Haoran Zhang1,411footnotemark:1,Shunkai Zhang1,511footnotemark:1,Yizhuo Li111footnotemark:1, Zhilin Wang1,Jiacheng Chen2,Futing Wang1,Xuyang Hu1,Yuchen Fan1, Bangjie Xu3,Yucheng Su3,Xinmiao Han3,Chenxi Li1,Haodi Lei1,Yufeng Zhao1, Zejin Lin3,Qianjia Cheng1,Tong Zhu1,Xiaoye Qu1,Ganqu Cui1,Peng Ye122footnotemark:2, Yun Luo122footnotemark:2,Zhouchen Lin5,Yu Qiao1,Bowen Zhou1,322footnotemark:2,Ning Ding3,122footnotemark:2,Yu Cheng2,122footnotemark:2 1Shanghai AI Laboratory2The Chinese University of Hong Kong3Tsinghua University 4Shanghai Jiao Tong University5Peking UniversityCore contributors\. Yafu Li is the project lead\. Corresponding authors\. Contact:yafuly@gmail\.comandchengyu@cse\.cuhk\.edu\.hk\.

###### Abstract

Recent progress in reasoning models has substantially advanced long\-horizon mathematical and scientific problem solving, with several systems now reaching gold\-medal\-level performance on International Mathematical Olympiad \(IMO\) and International Physics Olympiad \(IPhO\) problems\. In this paper, we introduce a simple and unified recipe for converting a post\-trained reasoning backbone into a rigorous olympiad\-level solver\. The recipe first uses a reverse\-perplexity curriculum for SFT to instill rigorous proof\-search and self\-checking behaviors, then scales these behaviors through a two\-stage RL pipeline that progresses from RL with verifiable rewards to more delicate proof\-level RL, and finally boosts solving performance with test\-time scaling\. Applying this recipe, we train a30B\-A3Bbackbone with SFT on around340K sub\-8K\-tokentrajectories followed by200RL steps\. The resulting model,SU\-01, supports stable reasoning on difficult problems with trajectories exceeding100Ktokens, while achieving gold\-medal\-level performance on mathematical and physical olympiad competitions, includingIMO 2025/USAMO 2026andIPhO 2024/2025\. It also demonstrates strong generalization of scientific reasoning to domains beyond mathematics and physics\.

![[Uncaptioned image]](https://arxiv.org/html/2605.13301v1/x3.png)

Figure 1:Overall comparison on IMO\-ProofBench\.⋆\\stardenotes results reported in the original paper, and†\\daggerdenotes results with test\-time scaling\. Blue bars trace the evolution of our pipeline from the 30B\-A3B backbone through rigorous SFT \([Section2](https://arxiv.org/html/2605.13301#S2)\), coarse RL, refined RL \([Section3](https://arxiv.org/html/2605.13301#S3)\), and test\-time scaling \([Section4](https://arxiv.org/html/2605.13301#S4)\), culminating in gold\-medal\-level olympiad reasoning \([Section5](https://arxiv.org/html/2605.13301#S5)\)\.
## 1Introduction

Olympiad competitions provide one of the clearest stress tests for long\-horizon reasoning\. Unlike many standard benchmarks, these problems require a model to search over many possible solution paths, control assumptions precisely, verify intermediate claims, and present a final argument that can survive strict grading across mathematical and scientific settings\. Recent systems have made rapid progress in this direction: AlphaGeometry combined neural guidance with symbolic search for olympiad geometry\(Trinhet al\.,[2024](https://arxiv.org/html/2605.13301#bib.bib106)\), while AlphaProof, AlphaGeometry 2, and Gemini Deep Think reached silver\- or gold\-medal standards on International Mathematical Olympiad problems with larger search and verification budgets\(Google DeepMind,[2024](https://arxiv.org/html/2605.13301#bib.bib107);[2025](https://arxiv.org/html/2605.13301#bib.bib108)\)\. At the same time, general reasoning models have improved through chain\-of\-thought prompting, math\-specialized post\-training, and reinforcement learning with verifiable rewards\(Weiet al\.,[2022](https://arxiv.org/html/2605.13301#bib.bib47); Shaoet al\.,[2024](https://arxiv.org/html/2605.13301#bib.bib41); Yanget al\.,[2024](https://arxiv.org/html/2605.13301#bib.bib52); Guoet al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib35); Yanet al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib51); Zhanet al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib87)\), while scientific olympiad benchmarks test transfer to modeling, derivation, and competition\-style justification\(Heet al\.,[2024](https://arxiv.org/html/2605.13301#bib.bib32); Chenet al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib83); Luoet al\.,[2026](https://arxiv.org/html/2605.13301#bib.bib111)\)\.

A central question is therefore whethera reasoning backbone can be pushed to olympiad\-level performance with a compact, domain\-unified recipethat applies the same reasoning\-centric pipeline across mathematical and scientific problems\. Using a 30B\-A3B model, we build a modular pipeline: SFT reshapes reasoning behavior, RL scales solving capability, and TTS allocates additional inference compute to the hardest proof\-search problems\. Together, these stages align behavior shaping, reward design, experience replay, and self\-verification into a compact recipe for rigorous mathematical and scientific reasoning\. The desgin follows a specializable\-generalist view: rather than building a narrow olympiad solver, we specialize a broadly capable post\-trained model toward expert\-level proof reasoning while preserving transfer across scientific domains\.

The first stage aims to instill a more disciplined proof\-search pattern\.Starting from a post\-trained model that is already competitive on scientific reasoning tasks, we curate long\-form solution, self\-verification, and self\-refinement trajectories from mathematical, scientific, coding, and instruction\-following sources\. After filtering, the SFT mixture contains 338K trajectories with responses shorter than 8K tokens\. SFT on this rigorous proof data instills reasoning behaviors centered on proof search, self\-checking, and repair\. We then order the examples by reverse perplexity so that each pass starts with trajectories most mismatched to the initial policy before consolidating on more familiar examples\. This curriculum helps preserve and recover the capability of the post\-trained model with its reasoning behavior reshaped\.

The second stage scales this behavior through two levels of RL\.Coarse RL uses verifiable prompts and efficient outcome checking to scale the reasoning behaviors introduced by SFT under reliable binary rewards, following the broader RLVR paradigm for efficient reasoning improvement\(Guoet al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib35); Shaoet al\.,[2024](https://arxiv.org/html/2605.13301#bib.bib41)\)\. Refined RL then shifts the target from answer correctness to proof quality\. It combines a proof\-level generative reward model for scoring complete proofs, self\-refinement prompts for training critique\-and\-repair behavior, and experience replay for preserving rare successful trajectories on hard problems\. Finally, we apply test\-time scaling through a self\-verification\-and\-refinement loop to elevate the trained model to olympiad\-level reasoning\(Huang and Yang,[2025](https://arxiv.org/html/2605.13301#bib.bib89)\)\.

On answer\-verifiable benchmarks, the resulting model,SU\-01, nearly matches the strongest similar\-size baseline, Qwen3\.6\-35B\-A3B, across AnswerBench, AMO\-Bench, AIME 2025/2026, and FrontierScience\-Olympiad\. On proof\-oriented evaluation, SU\-01 reaches 57\.6% on IMO\-ProofBench with direct generation and 70\.2% with TTS, substantially outperforming similar\-size models and approaching competitive commercial systems such as Gemini 3\.1 Pro Thinking\. Beyond solving competition problems, SU\-01 obtains the best similar\-size overall score on FrontierScience\-Research, suggesting that the recipe generalizes scientific reasoning toward research\-style problems beyond olympiad benchmarks\.

On official competition problems, SU\-01 shows strong end\-to\-end reasoning beyond benchmark\-style evaluation\. Direct SU\-01 already exceeds the IPhO gold lines for both 2024 and 2025, and clears the bronze\-medal lines on IMO 2025 and USAMO 2026\. With test\-time scaling, it reaches35 pointson both mathematical olympiads, meeting theIMO 2025 goldline and exceeding theUSAMO 2026 goldline by 10 points\. Notably, on USAMO 2026, this matches the highest reported human total among 340 competitors, indicating that the overall recipe can elicit top\-level human\-like olympiad reasoning from a compact 30B\-A3B model\. The TTS traces further show how this capability emerges at inference time: SU\-01 can sustain reasoning trajectories beyond 100K tokens, condition on its own drafts and error analyses, and repeatedly verify and repair candidate proofs\. Overall, these results support a specializable\-generalist view of compact reasoning models: with the right training and inference recipe, a broadly capable backbone can be driven toward expert\-level proof reasoning while retaining meaningful scientific transfer\.

![Refer to caption](https://arxiv.org/html/2605.13301v1/x4.png)Figure 2:Overview of the SU\-01 training and inference pipeline\. The recipe first reshapes the backbone with rigorous long\-form SFT, then scales the resulting behavior through coarse and refined RL, and finally applies test\-time verification and refinement for olympiad\-level problem solving\.
## 2Instilling Rigorous Reasoning via SFT

The first stage of the SU\-01 pipeline uses supervised fine\-tuning to reshape the model’s reasoning behavior\. We choose P1\-30B\-A3B\(Chenet al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib83)\)as the initial model because it already shows competitive performance in scientific reasoning, including both mathematics and physics\. Despite its strong results on verifiable tasks, we observe that its solutions are not always organized around rigorous proof\-search patterns\. The purpose of SFT is therefore to reshape its reasoning behavior toward more explicit, disciplined, and proof\-oriented long\-form reasoning while preserving as much of its existing capability as possible\.

We empirically find that applying SFT to a post\-trained backbone is more efficient than training the same reasoning behavior from a base model\. A post\-trained model already contains useful instruction\-following behavior, problem\-solving ability, and broad scientific competence\. Starting from that checkpoint allows SFT to focus on changing the reasoning pattern rather than rebuilding these capabilities from scratch\. In this framing, SFT specializes the generalist backbone toward rigorous proof\-search behavior while preserving its broad scientific competence, providing a stronger starting policy for subsequent RL to scale\. The launch configuration and optimization hyperparameters for this stage are summarized in[AppendixC](https://arxiv.org/html/2605.13301#A3)\.

### 2\.1SFT Data Curation

We curate SFT prompts from a broad mixture of mathematical, scientific, instruction\-following, and coding sources\. The mathematical subset includes problems from Evan Chen’s olympiad materials111Evan Chen’s olympiad materials:[https://web\.evanchen\.cc/](https://web.evanchen.cc/)\., the Shuzhimi Forum222The Shuzhimi Forum is an online Chinese mathematical problem\-solving community\., AoPS \(Art of Problem Solving\)333AoPS:[https://artofproblemsolving\.com/](https://artofproblemsolving.com/)\., online mathematical competition training books444The book subset is curated from publicly available online mathematical competition training materials\., and DeepMath problems with difficulty at least 6\(Heet al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib79)\)\. For scientific reasoning, we include prompts from NaturalReasoning\(Yuanet al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib80)\)\. To improve the generalization of the SFT model beyond narrow olympiad\-style mathematics, we also include chat prompts from Nemotron\-Instruction\-Following\-Chat\-v1555Nemotron\-Instruction\-Following\-Chat\-v1 Hugging Face dataset card:[link](https://huggingface.co/datasets/nvidia/Nemotron-Instruction-Following-Chat-v1)\.and coding prompts from Eurus\-2\-RL\-Data\(Cuiet al\.,[2025a](https://arxiv.org/html/2605.13301#bib.bib60)\)and OpenCodeReasoning\-2666OpenCodeReasoning\-2 Hugging Face dataset card:[link](https://huggingface.co/datasets/nvidia/OpenCodeReasoning-2)\.; the latter extends the OpenCodeReasoning data\-distillation line for competitive coding\(Ahmadet al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib81)\)\.

![Refer to caption](https://arxiv.org/html/2605.13301v1/x5.png)Figure 3:Composition of the SFT data after filtering\. Math, STEM, Code, and IF form the direct\-generation group; Self\-Verify and Self\-Refine form the self\-improvement group\.Before generation, we first filter contaminated problems from the prompt pool\. For each remaining prompt, we use DeepSeek\-V3\.2\-Speciale\(DeepSeek\-AI,[2025a](https://arxiv.org/html/2605.13301#bib.bib82)\)to generate high\-quality long\-form reasoning trajectories\. We then filter noisy generations and remove trajectories longer than 8,192 tokens\. This filtering step keeps the supervised signal focused on rigorous and usable reasoning traces, while avoiding extremely long outputs that are more likely to introduce truncation or unstable optimization\.

In addition to direct solution trajectories, we further equip the model with self\-verification and self\-refinement behaviors\. For the mathematical subset, we ask DeepSeek\-V3\.2\-Speciale to generate verification traces for the generated solutions, followed by refinement traces that address issues identified during verification\. These examples expose the model to the behaviors that are especially important for olympiad\-level reasoning: checking whether a proof is actually justified and improving an argument when a flaw is found\. Finally, we obtain a filtered SFT mixture of 338K trajectories, as shown in[Figure3](https://arxiv.org/html/2605.13301#S2.F3)\.

### 2\.2Reverse\-Perplexity Curriculum for SFT

Long\-CoT SFT on a post\-trained reasoning model is a delicate optimization problem\. The model already contains a strong instruction\-following and reasoning policy, so SFT is not simply adding a new capability to an empty backbone; it is modifying an existing policy while trying to preserve its original competence\. If the supervised signal is too narrow or the training is stopped too early, performance can degrade substantially even when the model starts to imitate more explicit long\-form reasoning\. This tension is consistent with the long\-CoT degradation phenomenon studied byLuoet al\.\([2025](https://arxiv.org/html/2605.13301#bib.bib77)\): a post\-trained model often needs sufficient data scale and enough SFT epochs to absorb the new reasoning style without overwriting the useful competence installed by previous post\-training stages\.

In our setting, recovery depends strongly on both training duration and the length behavior of the resulting model\(Renet al\.,[2026](https://arxiv.org/html/2605.13301#bib.bib78)\)\. For trajectories capped at 8,192 tokens, we empirically find that four epochs are usually sufficient to recover most of the model capability after the initial behavioral shift, provided that the data mixture and learning rate are well controlled\. We also treat validation truncation rate as an operational indicator of SFT sufficiency\. A post\-trained model that has not been sufficiently adapted to rigorous long\-CoT supervision often exhibits shallow reasoning behaviors: it circles around local heuristics, repeats intermediate claims, and continues reasoning without making decisive progress\. These repetitive and endless\-reasoning patterns naturally increase truncation\. In practice, we find that a truncation rate below 5% is a useful sign that the model has largely adapted to the target reasoning style\.

To make long\-CoT SFT more stable, we use a reverse\-perplexity training curriculum\. Let𝒟=\{\(xi,yi\)\}i=1N\\mathcal\{D\}=\\\{\(x\_\{i\},y\_\{i\}\)\\\}\_\{i=1\}^\{N\}be the SFT set, wherexix\_\{i\}is the prompt andyi=\(yi,1,…,yi,Ti\)y\_\{i\}=\(y\_\{i,1\},\\ldots,y\_\{i,T\_\{i\}\}\)is the teacher trajectory\. Given the initial policyπ0\\pi\_\{0\}, we score each example by its length\-normalized perplexity,PPL​\(xi,yi\)=exp⁡\(−1Ti​∑t=1Tilog⁡π0​\(yi,t∣xi,yi,<t\)\)\\mathrm\{PPL\}\(x\_\{i\},y\_\{i\}\)=\\exp\\left\(\-\\frac\{1\}\{T\_\{i\}\}\\sum\_\{t=1\}^\{T\_\{i\}\}\\log\\pi\_\{0\}\(y\_\{i,t\}\\mid x\_\{i\},y\_\{i,<t\}\)\\right\)\. Instead of presenting examples in random order or in ascending perplexity, we sort the data in descending perplexity and train from high\-PPL examples to low\-PPL examples within each epoch\. This order repeatedly starts each pass from teacher trajectories that are most mismatched with the current policy, using unfamiliar proof\-search patterns for behavioral adaptation before consolidating on more familiar examples\. We discuss the empirical effect of this ordering in[Section6\.3](https://arxiv.org/html/2605.13301#S6.SS3)\.

## 3Boosting Reasoning Capability with RL

Once the model has acquired a stronger long\-form reasoning pattern, reinforcement learning provides the scalable feedback mechanism for turning this pattern into stronger expert behavior\. We split this stage into two levels\.Coarse RLconverts the SFT reasoning pattern into stronger answer\-seeking behavior under reliable, mostly verifiable reward signals, improving search, coverage, and direct solving performance on hard tasks\.Refined RLthen specializes the policy toward complete, auditable proof construction, using more fine\-grained feedback to encourage proof rigor and self\-refinement\. The shared RL launch configuration and stage\-specific hyperparameters are summarized in[AppendixD](https://arxiv.org/html/2605.13301#A4)\.

### 3\.1RL Data Curation

RL training uses a separate prompt pool from SFT, curated to support both answer\-verifiable optimization and proof\-quality refinement\. The physics subset is derived from olympiad\-level physics data associated with P1\(Chenet al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib83)\)\. The mathematical subset follows the same source families as our SFT data, including AoPS, online competition training books, Evan Chen’s olympiad materials, and the Shuzhimi Forum\. We refer readers to §[2\.1](https://arxiv.org/html/2605.13301#S2.SS1)for source attribution\. We additionally include OPC777OPC dataset card:[link](https://huggingface.co/datasets/INSAIT-Institute/OPC)\., a human\-evaluated corpus of advanced mathematical proofs\(Dekonincket al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib84)\), to increase coverage of proof\-oriented prompts\.

We split the resulting RL pool into a verifiable set and a non\-verifiable set\. The verifiable set contains prompts whose final answers or structured outputs can be checked reliably, while the non\-verifiable set includes proof\-oriented or open\-ended reasoning prompts that require softer judgment, e\.g\., generative reward\. Before training, we first deduplicate and decontaminate the prompt pool\. We then apply rejection sampling to remove examples that are already too easy or too hard for the current policy, and further filter noisy prompts that are poorly formatted or otherwise unreliable\. The final RL pool contains 8,967 verifiable prompts and 16,287 non\-verifiable prompts\.

### 3\.2Coarse RL

Coarse RL trains the SFT model on the 8,967 verifiable prompts described above\. We formulate this stage as reinforcement learning with verifiable rewards \(RLVR;Lambertet al\.[2024](https://arxiv.org/html/2605.13301#bib.bib16); Guoet al\.[2025](https://arxiv.org/html/2605.13301#bib.bib35)\), using Group Sequence Policy Optimization \(GSPO;Zhenget al\.[2025](https://arxiv.org/html/2605.13301#bib.bib85)\)\. GSPO is better aligned with outcome\-reward training than token\-level GRPO because both reward assignment and policy clipping operate at the complete\-response level\. For each promptq∈𝒟verq\\in\\mathcal\{D\}\_\{\\mathrm\{ver\}\}\(the verifiable prompt set\), the rollout policyπθold\\pi\_\{\\theta\_\{\\mathrm\{old\}\}\}samples a group ofKKcandidate solutions𝒢q=\{oi\}i=1K\\mathcal\{G\}\_\{q\}=\\\{o\_\{i\}\\\}\_\{i=1\}^\{K\}\. The verifier converts each final answer into a binary outcome rewardr​\(q,o\)=1r\(q,o\)=1if the extracted final answer is verified as correct, andr​\(q,o\)=0r\(q,o\)=0otherwise\. The group\-relative advantage is computed from the within\-prompt reward baseline\. We use the unnormalized form without group standard\-deviation normalization,A^i=r​\(q,oi\)−μ𝒢q\\widehat\{A\}\_\{i\}=r\(q,o\_\{i\}\)\-\\mu\_\{\\mathcal\{G\}\_\{q\}\}, whereμ𝒢q=1K​∑j=1Kr​\(q,oj\)\\mu\_\{\\mathcal\{G\}\_\{q\}\}=\\frac\{1\}\{K\}\\sum\_\{j=1\}^\{K\}r\(q,o\_\{j\}\)\. The key GSPO quantity is the length\-normalized sequence\-level importance ratiosi​\(θ\)=exp⁡\{1\|oi\|​∑t=1\|oi\|log⁡πθ​\(oi,t∣q,oi,<t\)πθold​\(oi,t∣q,oi,<t\)\}s\_\{i\}\(\\theta\)=\\exp\\\{\\frac\{1\}\{\|o\_\{i\}\|\}\\sum\_\{t=1\}^\{\|o\_\{i\}\|\}\\log\\frac\{\\pi\_\{\\theta\}\(o\_\{i,t\}\\mid q,o\_\{i,<t\}\)\}\{\\pi\_\{\\theta\_\{\\mathrm\{old\}\}\}\(o\_\{i,t\}\\mid q,o\_\{i,<t\}\)\}\\\}\. The policy is updated with the clipped sequence\-level surrogate

𝒥GSPO​\(θ\)=𝔼q,\{oi\}​\[1K​∑i=1Kmin⁡\(si​\(θ\)​A^i,clip⁡\(si​\(θ\),1−ϵ,1\+ϵ\)​A^i\)\]\.\\mathcal\{J\}\_\{\\mathrm\{GSPO\}\}\(\\theta\)=\\mathbb\{E\}\_\{q,\\\{o\_\{i\}\\\}\}\\left\[\\frac\{1\}\{K\}\\sum\_\{i=1\}^\{K\}\\min\\left\(s\_\{i\}\(\\theta\)\\widehat\{A\}\_\{i\},\\operatorname\{clip\}\(s\_\{i\}\(\\theta\),1\-\\epsilon,1\+\\epsilon\)\\widehat\{A\}\_\{i\}\\right\)\\right\]\.\(1\)These definitions are also the interface used by the experience replay variants in the subsequent subsection: replayed trajectories can reuse the same reward, advantage, and sequence\-ratio notation while changing the source policy in the denominator ofsi​\(θ\)s\_\{i\}\(\\theta\)\. Following the routing\-replay motivation in GSPO\(Zhenget al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib85)\), we freeze the MoE router during RL so replayed trajectories are evaluated under stable expert\-routing decisions, which reduces replay\-induced instability\.

The reward system is intentionally layered to keep high\-precision automatic checks before more expensive model\-based judgments\. We first extract the final answer and apply canonicalized text matching\. Unresolved cases are then checked by Math\-Verify888Math\-Verify repository:[link](https://github.com/huggingface/Math-Verify)\., a rule\-based mathematical\-expression evaluation pipeline for LLM outputs\. Samples that still fail these rule\-based checks are sent to gpt\-oss\-120b999gpt\-oss\-120b model card:[link](https://huggingface.co/openai/gpt-oss-120b)\.\(OpenAI,[2025](https://arxiv.org/html/2605.13301#bib.bib88)\)for generative verification\. This ordering makes the reward conservative by default, while still recovering correct solutions whose final answers are equivalent but difficult to normalize with rule\-based parsers alone\.

### 3\.3Refined RL

After coarse RL has established strong search behavior, refined RL shifts the optimization target from answer correctness to proof quality\. The central issue is that many olympiad solutions can reach a correct final answer while still containing hidden gaps, unjustified transformations, or incomplete case analysis\. Refined RL therefore uses a stronger process\-level reward and adds two memory mechanisms: self\-refinement, which turns recent failures into repair tasks, and experience replay, which preserves rare successful proofs long enough for the policy to learn from them\.

#### Generative proof reward\.

We use DeepSeekMath\-V2 as a generative reward model for refined RL\(DeepSeek\-AI,[2025b](https://arxiv.org/html/2605.13301#bib.bib86)\), except for physics prompts\. For every rollout from both the verifiable and non\-verifiable subsets, the reward model reads the problem and the complete solution or proof, then outputs a binary scorerproof​\(q,o\)∈\{0,1\}r\_\{\\mathrm\{proof\}\}\(q,o\)\\in\\\{0,1\\\}\. Unlike the coarse verifier in §[3\.2](https://arxiv.org/html/2605.13301#S3.SS2), this score is not restricted to checking whether the final answer matches a reference answer\. It evaluates whether the full reasoning path is mathematically valid, sufficiently rigorous, and complete\. This makes the reward more aligned with the final goal of olympiad reasoning, but also more expensive and more vulnerable to judge artifacts\. We therefore apply anti\-hack preprocessing before sending a response to the reward model: malformed generations with leaked chat\-template tokens, unbalanced thinking delimiters, or severe repetition are replaced by a safe fallback answer\. This prevents the policy from receiving reward by exploiting formatting or verifier\-input pathologies rather than by improving the proof\. The reward\-model serving configuration is summarized in[AppendixE](https://arxiv.org/html/2605.13301#A5)\.

#### Self\-refinement\.

Self\-refinement exposes the policy to the same repair pattern that we use at test time: propose a solution, inspect it, locate gaps, and produce a corrected proof\. After each rollout, responses are grouped by query\. If a query group has average proof reward below a thresholdτref=0\.5\\tau\_\{\\mathrm\{ref\}\}=0\.5, failed responses from that group are converted into refinement prompts\. Each prompt contains the original problem, the previous incorrect solution, and an instruction to critique the argument, fix proof errors, fill missing justifications, and output a complete final solution\. These prompts are stored in a self\-refinement buffer and mixed into subsequent batches with target ratioηref=0\.2\\eta\_\{\\mathrm\{ref\}\}=0\.2\. Normal samples displaced by refinement queries are returned to a buffer, so refinement does not silently discard fresh training data\. We also do not recursively enqueue failed refinement attempts, which avoids spending repeated updates on examples that remain outside the current policy’s learnable region\.

#### Experience replay\.

On difficult proof problems, the policy may occasionally discover a valid solution trajectory even though it usually fails on the same query\. Immediately discarding such a trajectory wastes a high\-value training signal\. Following ExGRPO\(Zhanet al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib87)\), we keep a replay bufferℰ\\mathcal\{E\}keyed by query, but our implementation is simpler: it uses the same GSPO\-style update and does not apply the policy\-shaping transform introduced in ExGRPO\. After each rollout, a query is admitted to the replay buffer only when it is hard but solvable, operationalized as0<n\+​\(q\)<20<n\_\{\+\}\(q\)<2, wheren\+​\(q\)n\_\{\+\}\(q\)is the number of successful trajectories in the current group\. In answer\-only RLVR, such a unique success can be a lucky hit: a trajectory may end with the correct final answer while still containing brittle or invalid reasoning\(Zhanet al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib87)\)\. In our refined RL setting, however, success is assigned by the DeepSeekMath\-V2 proof reward, which inspects the full solution rather than only the final answer\. This does not eliminate reward\-model noise, but it makes a rare successful rollout substantially more likely to encode a reusable proof pattern and therefore a safer replay target\. Stored trajectories are deduplicated, and a query is retired once fresh on\-policy rollouts solve it often enough, with thresholdn\+​\(q\)≥4n\_\{\+\}\(q\)\\geq 4\.

Replay is mixed with fresh proof\-reward training rather than run as a separate mode\. A replay rollout injects one stored successful trajectory for the selected query, and the replay ratio is controlled byρ=0\.25\\rho=0\.25over the non\-refinement portion of the batch\. When multiple successful trajectories are stored for a query, we select the lowest\-entropy one,o∗=arg⁡mino∈ℰ​\(q\)⁡H​\(o;πθ\)o^\{\*\}=\\arg\\min\_\{o\\in\\mathcal\{E\}\(q\)\}H\(o;\\pi\_\{\\theta\}\), using rollout\-side top\-kklog probabilities as an efficient entropy estimate, following the trajectory\-selection principle in ExGRPO\(Zhanet al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib87)\)\. The resulting refined objective is

𝒥refined​\(θ\)=\(1−ρ\)​𝔼ℬfresh​\[𝒥GSPO​\(q,𝒢q;θ,πθold\)\]\+ρ​𝔼ℬexp​\[𝒥GSPO​\(q∗,\{o∗\}∪𝒢q∗;θ,πθsrc\)\],\\mathcal\{J\}\_\{\\mathrm\{refined\}\}\(\\theta\)=\(1\-\\rho\)\\,\\mathbb\{E\}\_\{\\mathcal\{B\}\_\{\\mathrm\{fresh\}\}\}\\left\[\\mathcal\{J\}\_\{\\mathrm\{GSPO\}\}\(q,\\mathcal\{G\}\_\{q\};\\theta,\\pi\_\{\\theta\_\{\\mathrm\{old\}\}\}\)\\right\]\+\\rho\\,\\mathbb\{E\}\_\{\\mathcal\{B\}\_\{\\mathrm\{exp\}\}\}\\left\[\\mathcal\{J\}\_\{\\mathrm\{GSPO\}\}\(q^\{\*\},\\\{o^\{\*\}\\\}\\cup\\mathcal\{G\}\_\{q^\{\*\}\};\\theta,\\pi\_\{\\theta\_\{\\mathrm\{src\}\}\}\)\\right\],\(2\)whereπθsrc=πθpast\\pi\_\{\\theta\_\{\\mathrm\{src\}\}\}=\\pi\_\{\\theta\_\{\\mathrm\{past\}\}\}for the replayed trajectory andπθsrc=πθold\\pi\_\{\\theta\_\{\\mathrm\{src\}\}\}=\\pi\_\{\\theta\_\{\\mathrm\{old\}\}\}for fresh rollouts\. This replay design is targeted rather than exhaustive: it stores rare valid proofs, prefers the most stable stored trajectory, replays it at a controlled ratio, and removes it once the current policy can reliably reproduce the behavior on\-policy\.

## 4Achieving Gold\-Medal\-Level Reasoning via Test\-time Scaling

Even with a strong reasoning policy, the hardest problems often require substantial search and revision at inference time\. This is not merely a matter of sampling more answers\. IMO\-style tasks demand complete and rigorous proofs, and a solution with the right final conclusion can still fail if it contains a hidden gap or a logical fallacy\. Recent work on IMO 2025 makes this point explicit: strong frontier models already contain significant mathematical capability, but their single\-pass outputs and even best\-of\-many selection can remain far below the level obtained by a structured verification\-and\-refinement pipeline\(Huang and Yang,[2025](https://arxiv.org/html/2605.13301#bib.bib89)\)\.

The need for test\-time scaling is also tied to reasoning budget\. A single generation has a finite context and thinking budget, while an olympiad proof may require several rounds of exploration, lemma checking, counterexample search, and exposition repair\. A model can spend most of its budget discovering a promising approach and still fail to fully close the proof\. Breaking inference into repeated solve–verify–refine stages effectively allocates additional computation to the same problem while keeping each step focused and auditable\. This extra budget is useful only when the model can remain coherent across repeated drafts, critiques, and repairs\. After the full training pipeline, SU\-01 is able to use this budget productively on difficult problems, sustaining coherent reasoning trajectories longer than 100K tokens during inference\.

Our TTS procedure follows the verification\-and\-refinement paradigm ofHuang and Yang \([2025](https://arxiv.org/html/2605.13301#bib.bib89)\)as a self\-verification and refinement loop\. The model first produces an initial solution under a solver prompt that prioritizes proof rigor rather than merely reaching a final answer\. It then enters refinement, where it revisits the draft, repairs weak steps, and tries to turn a promising argument into a complete proof\. The refined candidate is next checked through a verification prompt: the model inspects the full solution and writes a structured bug report, identifying issues such as critical errors, unjustified claims, or missing cases\. A verdict step interprets this report and decides whether the candidate should be accepted, rejected, or sent back for another refinement round\. This loop is repeated until the solution consistently passes self\-verification or the refinement budget is exhausted\. Multiple independent runs can be executed in parallel or serial, and accepted candidates are selected only after the proof is stable under repeated verification\. The corresponding inference setting is summarized in[AppendixE](https://arxiv.org/html/2605.13301#A5)\.

## 5Experimental Results

The experimental section is organized around three complementary evaluation views: answer\-verifiable reasoning tasks, non\-verifiable or proof\-oriented tasks, and official olympiad competition problems\.

### 5\.1Benchmarks

We organize evaluation into three benchmark families\. The first family contains answer\-verifiable reasoning tasks, where correctness can be checked by a final answer or a high\-confidence automatic verifier\. It includes AMO\-Bench\(Anet al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib105)\), AIME 2025 and AIME 2026101010AIME official competition page:[https://maa\.org/math\-competitions/aime](https://maa.org/math-competitions/aime)\., AnswerBench from the IMO\-Bench evaluation suite\(Luonget al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib109)\), and FrontierScience\-Olympiad, the Olympiad subset of FrontierScience\(Wanget al\.,[2026](https://arxiv.org/html/2605.13301#bib.bib110)\)\. These benchmarks mainly test whether the model can produce correct final answers under single\-pass or fixed\-budget inference\.

The second family contains non\-verifiable or proof\-oriented tasks\. We include ProofBench from IMO\-Bench\(Luonget al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib109)\), which emphasizes proof quality rather than only final\-answer matching, and FrontierScience\-Research, the research subset of FrontierScience\(Wanget al\.,[2026](https://arxiv.org/html/2605.13301#bib.bib110)\)\. These tasks are used to probe whether training improves rigorous reasoning and scientific problem solving beyond answer\-checkable settings\.

The third family contains official olympiad competition problems, including IMO 2025111111International Mathematical Olympiad official archive:[https://www\.imo\-official\.org/](https://www.imo-official.org/)\., USAMO 2026121212USA Mathematical Olympiad:[https://maa\.org/math\-competitions/usamo](https://maa.org/math-competitions/usamo)\.and IPhO \(2024, 2025\)131313International Physics Olympiad official site:[https://www\.ipho\-new\.org/](https://www.ipho-new.org/)\.\. Detailed grading and verifier settings are summarized in[AppendixG](https://arxiv.org/html/2605.13301#A7)\.

Table 1:Performance on answer\-verifiable reasoning tasks\. Results for AnswerBench, AMO\-Bench, AIME 25/26, and FrontierScience\-Olympiad are averaged over 4, 8, 8, and 4 runs, respectively\. FrontierScience\-Olympiad abbreviates the Olympiad subset of FrontierScience\. Avg\. is the mean of AnswerBench, AMO\-Bench, AIME 2025, AIME 2026, and FrontierScience\-Olympiad\. Within each comparison block,boldmarks the best score and underline marks the second best\.
### 5\.2Verifiable Problems

As shown in[Table1](https://arxiv.org/html/2605.13301#S5.T1), SU\-01 reaches a 77\.3% average score across AnswerBench, AMO\-Bench, AIME 2025, AIME 2026, and FrontierScience\-Olympiad, nearly matching the strongest similar\-size baseline, Qwen3\.6\-35B\-A3B \(77\.4%\)\. Importantly, SU\-01 achieves this level of performance with a simpler unified post\-training recipe and substantially lower training cost, highlighting the efficiency of our approach\. The mathematical benchmarks show where this improvement is most pronounced\. SU\-01 achieves the best similar\-size results on AMO\-Bench \(59\.8%\) and AIME 2025/2026 \(94\.6%/93\.3%\), which are closer to competition\-style problem solving than routine answer extraction\. On AnswerBench, SU\-01 remains competitive at 77\.5%, close to Qwen3\.6\-35B\-A3B \(78\.0%\) and behind only Nemotron\-Cascade\-2 \(80\.5%\)\. FrontierScience\-Olympiad tests whether this behavior transfers beyond pure mathematics\. Although the RL stages use only math and physics signals, SU\-01 reaches 61\.5% overall and shows strong transfer to untrained STEM domains, including 69\.4% on Chemistry and 25\.0% on Biology\. This cross\-domain transfer supports the specializable\-generalist framing: the model is specialized through math and physics reasoning signals, yet the resulting capability does not collapse into a narrow contest solver\.

### 5\.3Non\-verifiable Problems

Non\-verifiable benchmarks test whether the training recipe improves the quality of full reasoning traces, rather than only optimizing final\-answer rewards\. On IMO\-ProofBench,[Table3](https://arxiv.org/html/2605.13301#S5.T3)shows that SU\-01 reaches 57\.6% overall in direct generation, already the strongest result among similar\-size models\. Test\-time scaling further raises the score to 70\.2%, including 91\.0% on the basic split and 49\.5% on the advanced split, bringing a 30B\-A3B model close to much larger frontier systems such as Gemini 3\.1 Pro\. This improvement indicates that self\-verification and refinement are especially useful when correctness depends on the complete proof, not merely on producing the right final answer\.

Table 2:Performance on physics olympiad problems, averaged over 8 runs\. Gold lines for IPhO 2024/2025 are 20\.8/19\.7 points\. For SU\-01,x/yx/yreports scores without and with test\-time scaling\.FrontierScience\-Research is a substantially harder research\-oriented subset of FrontierScience, covering physics, chemistry, and biology problems that require scientific modeling and multi\-step reasoning beyond standard contest formats\. Absolute scores remain low even for frontier systems, but SU\-01 obtains the best similar\-size overall score at 11\.7%\. It also leads the similar\-size block on Physics, ties for the best Chemistry score, and ranks second on Biology, despite our RL stages using only mathematics and physics signals\. This cross\-domain pattern suggests that the recipe learns a more general scientific reasoning behavior rather than only specializing to the training domains, providing early evidence of transferable research\-level reasoning in a compact model\.

Table 3:Performance on non\-verifiable benchmarks\. FrontierScience\-Research refers to the research subset of FrontierScience\. For SU\-01,x/yx/yreports scores without and with TTS on ProofBench\.
### 5\.4Olympiad Competition Problems

Table 4:Performance on mathematical olympiad competition problems\. Medal lines for IMO 2025 are 35/28/19 points, and medal lines for USAMO 2026 are 25/18/11 points\.⋆indicates that TTS results are evaluated by human experts, while direct generation results are evaluated automatically \([AppendixG](https://arxiv.org/html/2605.13301#A7)\)\.Even without TTS, SU\-01 averages 23\.5 points on IPhO 2024 and 20\.3 points on IPhO 2025, exceeding the corresponding gold lines of 20\.8 and 19\.7 points, as shown in[Table2](https://arxiv.org/html/2605.13301#S5.T2)\. TTS further raises the scores to 25\.3 and 21\.7 points, making SU\-01 the strongest similar\-size model in both years among models with available scores\.

[Table4](https://arxiv.org/html/2605.13301#S5.T4)reports the final competition\-style mathematics results\. In direct generation, SU\-01 reaches 21 points on IMO 2025 and 15 points on USAMO 2026, already clearing the bronze\-medal lines for both competitions\. The direct model obtains full credit on IMO 2025 P2 and USAMO 2026 P1/P4, near\-complete solutions on IMO 2025 P4/P5, and still fails several harder problems in a single pass\. This indicates that the base model has acquired substantial olympiad reasoning ability, but still benefits from additional search and self\-correction on the most brittle proof attempts\.

With test\-time scaling, SU\-01 reaches35 pointson both IMO 2025 and USAMO 2026, meeting theIMO goldline exactly and exceeding theUSAMO goldline by 10 points\. TTS upgrades five IMO 2025 problems to full credit while P6 remains unsolved, and recovers full\-credit solutions on five of six USAMO 2026 problems while P2 remains unresolved\. The USAMO 2026 score summary reports 340 competitors, a median score of 6, a top\-12 cutoff of 26, and a maximum score of 35141414https://web\.evanchen\.cc/exams/posted\-usamo\-statistics\.pdf\. SU\-01 therefore matches thehighestreported human total on this contest, while still exposing a concrete failure mode on P2\. This result suggests that our overall recipe can elicit top\-level human\-like olympiad reasoning from a compact 30B\-A3B model\.

#### Case study\.

We include the corresponding model\-generated solutions and expert verdicts in[AppendixH](https://arxiv.org/html/2605.13301#A8)\. Across the twelve IMO 2025 and USAMO 2026 problems, the model gives full\-credit solutions to ten problems, with failures on IMO 2025 P6 and USAMO 2026 P2\. Its main strength is translating olympiad problems into formal frameworks: coordinates or complex numbers for geometry, modular classifications for number theory, recurrences for functional equations, and automata\-based dynamic programming for digit problems\. A particularly striking example is USAMO 2026 P3: rather than following the standard synthetic geometry route, the model elegantly uses complex numbers to unify the unit circle, equilateral\-triangle rotations, chord relations, and tangent conditions within a single algebraic framework\. This yields an ingenious analytic reformulation of a configuration that olympiad solvers would typically approach through angle chasing and carefully chosen auxiliary constructions\. IMO 2025 P2 shows a complementary strength: the model reduces a configuration involving two intersecting circles, an orthocenter, and a tangency claim to coordinate and distance computations\. Other strong examples include the carry\-state dynamic programming approach for USAMO P4 and the number\-theoretic proof using totients, congruences, Vieta jumping, and Fibonacci structure in USAMO P6\. However, the failures show a clear limitation: the model can miss subtle structural constraints, as in the invalid column\-permutation reduction in IMO P6, or leave gaps in delicate global strategy arguments, as in USAMO P2\. Overall, the model performs well when a problem admits a rigid formal representation, but is less reliable when the core challenge is preserving combinatorial structure or proving a finely tuned process invariant\.

## 6Analysis and Discussion

### 6\.1Progressive Rigorous Reasoning

![Refer to caption](https://arxiv.org/html/2605.13301v1/x8.png)Figure 4:Progressive reasoning performance across training stages\.[Figure4](https://arxiv.org/html/2605.13301#S6.F4)separates two kinds of reasoning progress\. AnswerBench measures whether the model can recover a correct final answer under verifiable evaluation, whereas ProofBench grades the total solution and therefore exposes gaps in rigor, justification, and proof completion\. The starting P1\-30B\-A3B model is already strong on AnswerBench, but its ProofBench scores are much lower, especially on the Advanced split, indicating that answer\-seeking ability alone does not imply olympiad\-style proof reliability\.

The staged trend matches the intended role of each method component\.SFTlowers AnswerBench from 69\.2 to 59\.8, but raises ProofBench\-Basic from 33\.8 to 57\.6 and ProofBench\-Advanced from 6\.2 to 14\.8\. This is consistent with behavior shaping: the model is moved away from short answer recovery and toward longer proof\-search, self\-checking, and refinement patterns\.Coarse RLthen uses verifiable rewards to recover and improve direct solving ability, lifting AnswerBench to 77\.2 while also improving ProofBench\-Basic to 76\.7 and ProofBench\-Advanced to 25\.2\. This suggests that RLVR scales the rigorous reasoning behavior introduced by SFT into stronger problem\-solving capability\.

The final SU\-01 model keeps AnswerBench essentially saturated at 77\.5 and ProofBench\-Basic nearly unchanged at 77\.1, but improves ProofBench\-Advanced from 25\.2 to 38\.1\. The gain is therefore concentrated on harder non\-verifiable proof problems, matching the role ofRefined RL: proof\-level generative rewards provide supervision beyond final\-answer correctness, self\-refinement prompts train the model to critique and repair its own solutions, and experience replay keeps rare successful hard\-problem trajectories available long enough for the policy to learn more robust proof construction\.Test\-time scalingthen further lifts the proof\-oriented evaluation, reaching 91\.0 on ProofBench\-Basic and 49\.5 on ProofBench\-Advanced, showing that the trained self\-verification and refinement behavior remains useful when additional inference compute is spent on checking and repairing candidate proofs\.

### 6\.2Characterizing Inference Scaling

We further inspect the TTS traces from USAMO 2026 to understand where inference compute is spent during difficult proof search\. The key distinction is not only between short and long responses, but between qualitatively different reasoning contexts: initial generation starts from the problem statement, whereas refinement must condition on an existing solution together with verifier feedback or a bug report and then produce a revised proof\.

![Refer to caption](https://arxiv.org/html/2605.13301v1/x9.png)Figure 5:Generation\-length distribution of actions in the TTS pipeline on USAMO 2026\.[Figure5](https://arxiv.org/html/2605.13301#S6.F5)shows a clear allocation of computation across TTS actions\. Initial solution generation is the longest stage, with a median length of 106K tokens, reflecting broad proof search and candidate construction\. Refinement remains length\-intensive, with a median of 83K tokens and a heavier upper tail, consistent with substantial proof repair\. Verification is shorter but still substantive, with a median of 28\.7K tokens, reflecting its role in auditing complete arguments for hidden gaps\. Verdict parsing is lightweight, with a median of only 404 tokens\.

The long refinement traces indicate that the model can reason over complicated conditioning contexts rather than merely produce long first\-pass solutions: given a candidate proof and a structured critique, it often sustains another long reasoning trajectory to localize the flaw, preserve useful parts of the argument, and synthesize a corrected proof\. The pattern therefore suggests that the training recipe enables the 30B model to generalize beyond direct solution generation, sustaining complex reasoning beyond 100K tokens and repeatedly verifying and refining its own solutions toward stronger candidates\.

### 6\.3Reverse\-Perplexity Ordering

![Refer to caption](https://arxiv.org/html/2605.13301v1/x10.png)Figure 6:Validation results for SFT data ordering\.We compare the effect of different SFT data orderings on validation performance, as shown in[Figure6](https://arxiv.org/html/2605.13301#S6.F6)\. The results come from validation experiments whose SFT training data differ from the final training mixture used for SU\-01\. Data ordering has a large effect on both score recovery and generation stability\. Random ordering substantially under\-recovers the P1\-30B baseline, reaching 39\.5 on AnswerBench and 31\.0 on AMO\-Bench, with truncation rates of 7\.3% and 8\.0%, respectively\. Descending\-PPL ordering recovers much more of the original capability, reaching 55\.8 on AnswerBench and 40\.0 on AMO\-Bench, while reducing truncation to 0\.3% and 0\.0%\. The low\-PPL\-first setting is the weakest curriculum, degrading to 24\.3 on AnswerBench and 15\.0 on AMO\-Bench\. These results suggest that descending\-PPL ordering helps preserve the capability of the post\-trained model while reshaping its reasoning behavior, and also prevents training from falling into a superficial long\-generation regime with high truncation rates\.

### 6\.4Cost Analysis

SU\-01 uses a compact and transparent post\-training setup\. Starting from a 30B\-A3B backbone, we train on 338K SFT trajectories shorter than 8K tokens for four epochs with batch size 128\. The RL stage uses 25K prompts for 200 steps, with batch size 128, 8 rollouts per prompt, and a 160K\-token maximum response length\. For reference, DeepSeek\-V3\.2\(DeepSeek\-AI,[2025a](https://arxiv.org/html/2605.13301#bib.bib82)\)reports continued pre\-training with 1,000 indexer warm\-up steps over 2\.1B tokens and 15,000 sparse\-training steps over 943\.7B tokens, followed by post\-training with specialist distillation and mixed RL over thousands of continued\-RL steps; its high\-compute Speciale variant further trains on reasoning data with DeepSeekMath\-V2\-style proof rewards\. Nemotron\-Cascade 2\(Yanget al\.,[2026](https://arxiv.org/html/2605.13301#bib.bib102)\), a same\-size 30B\-A3B reference model, reports a substantially broader SFT mixture\. Summing the disclosed category counts gives roughly 26\.6M SFT samples across math, proof, code, science, long\-context, chat, instruction\-following, tool\-use, and software\-engineering data, trained with packed 256K\-token sequences for 33K steps\. Its post\-training then continues with a multi\-stage Cascade RL pipeline covering IF\-RL, multi\-domain RL, on\-policy distillation, RLHF, long\-context RL, CodeRL, and SWE RL\. These comparisons highlight the main design point of SU\-01: a simple and unified recipe can elicit strong olympiad\-level reasoning from a compact 30B\-A3B model while preserving scientific transfer\.

## 7Related Work

#### Post\-training for Large Reasoning Models\.

Post\-training has become the main mechanism for turning strong pretrained language models into reliable reasoning systems\. Early self\-improvement work showed that models can bootstrap rationales from their own successful attempts\(Zelikmanet al\.,[2022](https://arxiv.org/html/2605.13301#bib.bib14)\), while open post\-training recipes combine instruction tuning, preference optimization, and reinforcement learning to improve general assistant behavior\(Lambertet al\.,[2024](https://arxiv.org/html/2605.13301#bib.bib16)\)\. For mathematical and long\-CoT reasoning, DeepSeekMath introduced large\-scale mathematical pretraining together with GRPO\(Shaoet al\.,[2024](https://arxiv.org/html/2605.13301#bib.bib41)\), Qwen2\.5\-Math emphasized math\-specific self\-improvement and tool\-augmented data construction\(Yanget al\.,[2024](https://arxiv.org/html/2605.13301#bib.bib52)\), DeepSeek\-R1 demonstrated that large\-scale RL can induce long reasoning traces and self\-correction behaviors\(Guoet al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib35)\), and Kimi k1\.5 highlighted long\-context RL, curriculum design, and sampling\-based inference\(Kimi Team,[2025](https://arxiv.org/html/2605.13301#bib.bib15)\)\. Recent work further studies how to stabilize and reuse learning signal: off\-policy guidance and experience replay improve sample reuse for reasoning policies\(Yanet al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib51); Zhanet al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib87)\), entropy analyses explain exploration, entropy collapse, and selective entropy regularization in RLVR\(Cuiet al\.,[2025b](https://arxiv.org/html/2605.13301#bib.bib4); Jianget al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib5)\), and GSPO optimizes at the sequence level for MoE reasoning models\(Zhenget al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib85)\)\. Contemporary technical reports such as MiniMax\-M2\.5, Kimi\-K2\.5, and GLM\-5 also show the growing importance of agentic post\-training, efficient reasoning, long\-context execution, and tool\-oriented RL\(MiniMax AI,[2026](https://arxiv.org/html/2605.13301#bib.bib92); Moonshot AI,[2026](https://arxiv.org/html/2605.13301#bib.bib96); Z\.AI,[2026b](https://arxiv.org/html/2605.13301#bib.bib98)\)\.

#### Toward Olympiad\-Level Reasoning\.

Olympiad reasoning pushes beyond benchmark math accuracy because solutions must be complete, rigorous, and robust to hidden gaps\. One line of work addresses this challenge with specialized symbolic or neuro\-symbolic systems: AlphaGeometry solves geometry problems by combining a neural language model with symbolic deduction\(Trinhet al\.,[2024](https://arxiv.org/html/2605.13301#bib.bib106)\), and AlphaProof/AlphaGeometry 2 reached silver\-medal\-level IMO performance through formal reasoning and search\(Google DeepMind,[2024](https://arxiv.org/html/2605.13301#bib.bib107)\)\. More recent frontier systems have moved toward broader natural\-language reasoning and test\-time search, including Gemini Deep Think’s gold\-medal\-level IMO result\(Google DeepMind,[2025](https://arxiv.org/html/2605.13301#bib.bib108)\)\. In parallel, model\-agnostic verification\-and\-refinement pipelines show that repeated generation, critique, repair, and acceptance decisions can substantially improve proof quality without relying on a single\-pass answer\(Huang and Yang,[2025](https://arxiv.org/html/2605.13301#bib.bib89)\), and DeepSeekMath\-V2 studies self\-verifiable mathematical reasoning as a training and inference target\(DeepSeek\-AI,[2025b](https://arxiv.org/html/2605.13301#bib.bib86)\)\. Nemotron\-Cascade 2 provides another recent example of a compact MoE reasoning model approaching frontier mathematical and olympiad performance through cascade RL and multi\-domain on\-policy distillation\(Yanget al\.,[2026](https://arxiv.org/html/2605.13301#bib.bib102)\)\. Our contribution is a simple and unified recipe that enables a 30B\-A3B model to develop rigorous proof behavior through post\-training and reach olympiad\-level performance with self\-verification, refinement, and test\-time scaling\.

## 8Conclusion

This report presents a simple and unified recipe for turning a compact post\-trained reasoning model into a stronger mathematical and scientific reasoner\. Starting from a broadly capable 30B\-A3B backbone, SU\-01 combines reverse\-perplexity curriculum SFT, efficient coarse RL with outcome verification, refined RL with proof\-level rewards, self\-refinement and experience replay, and test\-time scaling through self\-verification and refinement\. Together, these stages decompose rigorous reasoning improvement into behavior shaping, scalable reward feedback, proof\-level specialization, and inference\-time repair\. The resulting model reaches gold\-medal\-level performance on mathematical and physical olympiad competitions, sustains reasoning trajectories beyond 100K tokens during inference, and shows transfer to scientific domains beyond the main math and physics training signals\. In summary, SU\-01 supports a specializable\-generalist view of compact reasoning models: with the right training and inference recipe, a broadly capable backbone can be driven toward expert\-level proof reasoning while retaining meaningful scientific transfer\.

## Acknowledgments

This work was supported by the Shanghai Artificial Intelligence Laboratory\. We thank the authors and maintainers of prior open research and infrastructure that made this work possible\. In particular, we are grateful to DeepSeek for open\-sourcing strong reasoning policies and generative reward models, which provided an important reference point for our work\(DeepSeek\-AI,[2025a](https://arxiv.org/html/2605.13301#bib.bib82);[b](https://arxiv.org/html/2605.13301#bib.bib86)\)\. IMO\-Bench, AMO\-Bench, and FrontierScience helped guide the overall system optimization by offering challenging mathematical and scientific reasoning benchmarks and evaluation protocols\(Luonget al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib109); Anet al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib105); Wanget al\.,[2026](https://arxiv.org/html/2605.13301#bib.bib110)\)\. We also thank prior data efforts that supported our SFT and RL data curation, including DeepMath, NaturalReasoning, Eurus, OpenCodeReasoning, P1, and OPC\(Heet al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib79); Yuanet al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib80); Cuiet al\.,[2025a](https://arxiv.org/html/2605.13301#bib.bib60); Ahmadet al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib81); Chenet al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib83); Dekonincket al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib84)\), as well as the many public problem sources and communities that cannot all be listed here\. We further acknowledge the broader open\-source infrastructure ecosystem, including slime for training and SGLang for efficient inference and serving\(THUDM,[2026](https://arxiv.org/html/2605.13301#bib.bib112); SGLang Team,[2026](https://arxiv.org/html/2605.13301#bib.bib113)\)\.

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## Appendix

### Contents

## Appendix AImplementation and Evaluation Details

This appendix collects the implementation details needed to interpret and reproduce the reported training, inference, reward\-model serving, and evaluation procedures\. We summarize the effective modeling, optimization, decoding, serving, and grading settings extracted from the corresponding launch scripts and evaluation protocols, while omitting infrastructure\-only commands that do not affect the training objective, data semantics, inference behavior, reward definition, or evaluation criteria\.

## Appendix BProblem\-Solving Prompt

Unless otherwise specified, all RL training and inference stages use the following fixed problem\-solving prompt for SU\-01:

Please solve the following olympiad problem\. Show your complete reasoning and proof\. 1\. Please use LaTeX format to represent the variables and formulas used in the solution process and results\. 2\. If the problem asks you to find specific values, please put the final answer\(s\) in \\boxed\{\}\. 3\. If the problem requires a proof, present a clear and rigorous argument\.

## Appendix CSFT Training Details

The SFT stage is implemented with slime\(THUDM,[2026](https://arxiv.org/html/2605.13301#bib.bib112)\), trained on 8 GPUs, initializes from P1\-30B, and optimizes the reverse\-perplexity\-ordered mixture described in[Section2\.1](https://arxiv.org/html/2605.13301#S2.SS1); rollout shuffling is disabled so that the curriculum order is preserved throughout training\. We train for four epochs with batch size 128, using Adam with learning rate1×10−51\\times 10^\{\-5\}, cosine decay to a minimum learning rate of1×10−61\\times 10^\{\-6\}, warmup fraction 0\.1, weight decay 0\.1, and momentum parametersβ1=0\.9\\beta\_\{1\}=0\.9andβ2=0\.95\\beta\_\{2\}=0\.95\. This configuration treats SFT as a controlled behavioral adaptation step: the objective is not to maximize benchmark performance directly, but to expose the post\-trained backbone to a stable sequence of rigorous long\-form trajectories while preserving as much of its existing mathematical and scientific competence as possible\.

## Appendix DRL Training Details

The RL stage is implemented with slime\(THUDM,[2026](https://arxiv.org/html/2605.13301#bib.bib112)\), trained on 64 GPUs, continues from the SFT checkpoint, and trains with a GSPO objective over a balanced mixture of verifiable prompts, proof\-reward prompts, self\-refinement prompts, and replayed experience\. In total, the model is trained for 200 RL steps, consisting of 96 coarse\-RL steps followed by 104 refined\-RL steps\. Each rollout uses prompt batch size 128, 8 samples per prompt, maximum response length 160k tokens, temperature 1\.0, and 4 policy\-update steps per rollout\. The data pipeline applies dynamic sampling with a non\-zero standard\-deviation reward requirement, partial rollout, and oversampling batch size 160; the oversampling and partial\-rollout mechanism increases the candidate prompt pool under filtering while recycling completed partial generations, and replay filtering prevents repeatedly training on queries that have become too easy for the current policy\. The policy objective uses the GSPO advantage estimator with KL coefficient 0, entropy coefficient 0, symmetric clip range10−310^\{\-3\}, and trajectory importance sampling enabled\.

The self\-refinement mechanism uses a refinement prompt template that includes the original problem and the previous failed answer\. The self\-refinement ratio is set to20%20\\%, which denotes the fraction of training queries sampled from refinement data, and the group reward threshold selects failed or partially failed rollout groups with an average reward below0\.50\.5for refinement\. Experience replay largely follows the ExGRPO setting without reward shaping and is re\-implemented in theslimetraining framework\. We use an experience admission threshold of online rollout correctness below25%25\\%, one replayed trajectory per replay query, and entropy\-based selection estimated from SGLang top\-1616log probabilities\. A query is admitted to the experience pool if only one sampled rollout succeeds, making the successful trajectory a hard positive example\. Experiences are retired once online rollout correctness reaches50%50\\%, indicating improved model performance on the corresponding query\. Optimization is performed using Adam with a constant learning rate of1×10−61\\times 10^\{\-6\}, weight decay of 0\.1, and momentum parametersβ1=0\.9\\beta\_\{1\}=0\.9andβ2=0\.98\\beta\_\{2\}=0\.98\.

## Appendix EInference and Reward\-Model Serving Details

For test\-time scaling, inference is served with SGLang\(SGLang Team,[2026](https://arxiv.org/html/2605.13301#bib.bib113)\)by default; the Nemotron\-Cascade\-2 and DeepSeek\-V3\.2\-Speciale models are served with vLLM\(vLLM Team,[2026](https://arxiv.org/html/2605.13301#bib.bib114)\)151515Nemotron\-Cascade\-2:vLLM\-0\.17\.2rc1\.dev148\+g47b7af0d8\.cu128 ; DeepSeek\-V3\.2\-Speciale: vLLM\-v0\.20\-CUDA12\.9\.\. We follow the same model\-agnostic verification\-and\-refinement setting asHuang and Yang \([2025](https://arxiv.org/html/2605.13301#bib.bib89)\)\.

Concretely, inference is organized as an iterative solve–verify–refine procedure: the solver first produces a candidate proof, the candidate is inspected by a verifier that returns a structured critique or bug report, and the solver then revises the proof conditioned on this feedback until the candidate is accepted or the refinement budget is exhausted\. Unless otherwise specified, each generation uses a maximum length of 160,000 tokens, temperature 1\.0, and top\-p 0\.95\. We use the following exceptions for model\- or benchmark\-specific constraints\. For Nemotron model, we set top\-p to 1\.0\. The maximum generation length is set to 131K tokens for AIME25/26, AMOBench, IPhO, and FrontierScience\-Olympic, and to 256K tokens for IMO\-AnswerBench, IMO\-ProofBench, and FrontierScience\-Research subset\. For API\-limited models, we set the maximum decoding completion length to the largest value allowed by the API request constraints: 128,000 tokens for GPT\-5\.5 API and 65,535 tokens for Gemini 3\.1\-Pro\.

The TTS loop uses explicit stopping rules following the verification\-and\-refinement protocol\. Within a single run,MAX\_VERIFICATION\_TRUE\_ROUNDS=5means that a candidate is accepted once it passes verification for five consecutive rounds, whileMAX\_VERIFICATION\_FALSE\_ROUNDS=10terminates the run early after ten consecutive failed verification rounds\. Each run allows at mostMAX\_EXPLORATION\_ROUNDS=30solve–verify–refine cycles, corresponding to the repeated exploration and correction loop in the TTS pipeline\. For each problem, we setMAX\_RUNS=10, so at most ten independent runs are launched\. This setting is intentionally aligned with the self\-refinement behavior used during refined RL, so the model is evaluated under the same type of proof\-repair workflow that it sees during training\.

For proof\-reward evaluation during refined RL, the DeepSeekMath\-V2 reward server is deployed on 32 GPUs with maximum context length 32k and data\-parallel degree 64\. Speculative decoding is enabled during server\-side inference using an MTP\-based configuration\. We adopt a33\-step speculative decoding scheme with asingletop\-ranked EAGLE draft candidate and44draft tokens per step\. This setting improves verifier efficiency while maintaining the evaluation correctness required by the refined\-RL training\.

## Appendix FCompared Models

The comparison tables use a mixture of public technical reports, official model cards, and official benchmark reports\. For the answer\-verifiable comparisons in[Table1](https://arxiv.org/html/2605.13301#S5.T1), the larger\-model group contains DeepSeek\-V3\.2\(DeepSeek\-AI,[2025a](https://arxiv.org/html/2605.13301#bib.bib82)\), GPT\-5\.5\(OpenAI,[2026](https://arxiv.org/html/2605.13301#bib.bib100)\), and Gemini 3\.1 Pro Thinking\(Google DeepMind,[2026](https://arxiv.org/html/2605.13301#bib.bib101)\)\. The similar\-size group contains GLM\-4\.7\-Flash\(Z\.AI,[2026a](https://arxiv.org/html/2605.13301#bib.bib99)\), Nemotron\-Cascade\-2\-30B\-A3B\(Yanget al\.,[2026](https://arxiv.org/html/2605.13301#bib.bib102)\), Qwen3\.6\-35B\-A3B\(Qwen,[2026](https://arxiv.org/html/2605.13301#bib.bib103)\), and Gemma\-4\-31B\(Google,[2026](https://arxiv.org/html/2605.13301#bib.bib104)\), together with ourSU\-01\. These same active comparison families are reused in the non\-verifiable benchmark table in[Table3](https://arxiv.org/html/2605.13301#S5.T3), with missing benchmark entries marked by –\. GPT\-5\.5\-High is treated as the high\-reasoning variant of GPT\-5\.5, Gemini 3\.1 Pro Thinking is drawn from the Gemini 3\.1 Pro family, and DeepSeek V3\.2 uses the same DeepSeek\-V3\.2 source as above\. When no paper\-style technical report is available, we cite the corresponding official model card, repository, or product page\.

## Appendix GEvaluation Details

For answer\-verifiable tasks, we use a layered automatic grading pipeline\. We first apply rule\-based verification, including canonicalized answer matching and symbolic or expression\-level checks with Math\-Verify161616Math\-Verify repository:[https://github\.com/huggingface/Math\-Verify](https://github.com/huggingface/Math-Verify)\.\. If a response is not resolved by these checks, we send it to gpt\-oss\-120b for generative verification\(OpenAI,[2025](https://arxiv.org/html/2605.13301#bib.bib88)\)\. For FrontierScience\-O, we use the same verifiable grading pipeline, but adopt the dedicated prompt described in the original FrontierScience paper\(Wanget al\.,[2026](https://arxiv.org/html/2605.13301#bib.bib110)\)\.

For non\-verifiable and proof\-oriented tasks, we follow each benchmark’s official evaluation protocol\. ProofBench is evaluated following the IMO\-Bench framework\(Luonget al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib109)\): solutions are graded with the four\-level score set\{0,1,6,7\}\\\{0,1,6,7\\\}, and we use Gemini\-2\.5\-Pro as the grading model to align with the official reported setting\. FrontierScience\-R is evaluated with the official FrontierScience procedure using GPT\-5\-high as the judge\(Wanget al\.,[2026](https://arxiv.org/html/2605.13301#bib.bib110)\)\.

For IPhO, we follow the evaluation paradigms used in P1 and P1\-VL\(Chenet al\.,[2025](https://arxiv.org/html/2605.13301#bib.bib83); Luoet al\.,[2026](https://arxiv.org/html/2605.13301#bib.bib111)\), which combine physics\-specific grading prompts with competition\-style scoring criteria\. For IMO and USAMO, we construct a ProofBench\-style automatic grading framework from annotated reference solutions and grading guidelines prepared by expert annotators who have won gold medals in mathematical olympiads, such as IMO or CMO\. For the final verdict, three gold\-medal experts independently assign scores from\{0,1,6,7\}\\\{0,1,6,7\\\}, and we report the worst of the three scores as a conservative final score\.

## Appendix HModel Solutions for IMO 2025 and USAMO 2026

This section lists the model\-generated solutions used for the human\-expert scoring reported in[Table4](https://arxiv.org/html/2605.13301#S5.T4)\. Each solution is followed by its final expert verdict\. For USAMO 2026 Problem 2, we show only an excerpt because the original response is substantially longer than the other solutions\.

### H\.1IMO 2025

#### IMO 2025 Problem 1\.

A line in the plane is called*sunny*if it is not parallel to any of thexx\-axis, theyy\-axis, or the linex\+y=0x\+y=0\.Letn≥3n\\geq 3be a given integer\. Determine all nonnegative integerskksuch that there existnndistinct lines in the plane satisfying both of the following:•for all positive integersaaandbbwitha\+b≤n\+1a\+b\\leq n\+1, the point\(a,b\)\(a,b\)lies on at least one of the lines; and•exactlykkof thennlines are sunny\.

Solution

Definition\.For an integern≥3n\\geq 3let

Pn=\{\(a,b\)∈ℕ2∣a≥1,b≥1,a\+b≤n\+1\}\.P\_\{n\}=\\\{\(a,b\)\\in\\mathbb\{N\}^\{2\}\\mid a\\geq 1,\\;b\\geq 1,\\;a\+b\\leq n\+1\\\}\.We are asked for which nonnegative integerskkthere existnndistinct lines in the plane such that

⊳\\trianglerightevery point ofPnP\_\{n\}lies on at least one of the lines, and

⊳\\trianglerightexactlykkof the lines are*sunny*\(i\.e\. not parallel to thexx\-axis, theyy\-axis, or the linex\+y=0x\+y=0\)\.

We will prove that for everyn≥3n\\geq 3the only possible values ofkkare0,1,30,1,3\.

1\. Constructions fork=0,1,3k=0,1,3

We give explicit families ofnndistinct lines that satisfy the two conditions\.

k=0k=0

Take thennhorizontal lines

y=1,y=2,…,y=n\.y=1,\\;y=2,\\;\\dots,\\;y=n\.Each point\(a,b\)∈Pn\(a,b\)\\in P\_\{n\}lies on the liney=by=b\. No line is sunny \(all are parallel to thexx\-axis\)\. Hencek=0k=0is attainable\.

k=1k=1

Take then−1n\-1horizontal lines

y=1,y=2,…,y=n−1\.y=1,\\;y=2,\\;\\dots,\\;y=n\-1\.These cover all points withb≤n−1b\\leq n\-1\. The only point not covered is\(1,n\)\(1,n\)\(sincea\+n≤n\+1a\+n\\leq n\+1forcesa=1a=1\)\. Add one extra line, for instance

L:y=x\+\(n−1\)\.L:\\;y=x\+\(n\-1\)\.LLhas slope11, so it is sunny\. It passes through\(1,n\)\(1,n\)\. The total number of lines is\(n−1\)\+1=n\(n\-1\)\+1=n, and exactly one line is sunny\. Thusk=1k=1is attainable\.

k=3k=3

*Ifn=3n=3:*Use the three sunny lines

L1\\displaystyle L\_\{1\}:y=x,\\displaystyle:y=x,L2\\displaystyle L\_\{2\}:y=−2​x\+5,\\displaystyle:y=\-2x\+5,L3\\displaystyle L\_\{3\}:y=−12​x\+52\.\\displaystyle:y=\-\\tfrac\{1\}\{2\}x\+\\tfrac\{5\}\{2\}\.Check:L1L\_\{1\}contains\(1,1\)\(1,1\)and\(2,2\)\(2,2\);L2L\_\{2\}contains\(1,3\)\(1,3\)and\(2,1\)\(2,1\);L3L\_\{3\}contains\(1,2\)\(1,2\)and\(3,1\)\(3,1\)\. The six points ofP3P\_\{3\}are all covered\. All three lines are sunny, sok=3k=3\.

*Ifn\>3n\>3:*First take then−3n\-3horizontal lines

y=1,y=2,…,y=n−3\.y=1,\\;y=2,\\;\\dots,\\;y=n\-3\.These cover all points withb≤n−3b\\leq n\-3\. The remaining points are those withb=n−2,n−1,nb=n\-2,\\,n\-1,\\,n; there are exactly3\+2\+1=63\+2\+1=6such points\. Translate the situation by\(0,−\(n−3\)\)\(0,\\,\-\(n\-3\)\): the map

\(x,y\)⟼\(x,y−\(n−3\)\)\(x,\\,y\)\\;\\longmapsto\\;\(x,\\,y\-\(n\-3\)\)sends these six points bijectively onto the setP3P\_\{3\}\(becausea\+b≤n\+1a\+b\\leq n\+1becomesa\+\(b′\+\(n−3\)\)≤n\+1⇔a\+b′≤4a\+\(b^\{\\prime\}\+\(n\-3\)\)\\leq n\+1\\iff a\+b^\{\\prime\}\\leq 4\)\. Therefore, if we take the three sunny lines that coverP3P\_\{3\}in the translated coordinates and then translate them back, we obtain three sunny lines in the original plane:

S1\\displaystyle S\_\{1\}:y=x\+\(n−3\),\\displaystyle:y=x\+\(n\-3\),S2\\displaystyle S\_\{2\}:y=−2​x\+\(n\+2\),\\displaystyle:y=\-2x\+\(n\+2\),S3\\displaystyle S\_\{3\}:y=−12​x\+2​n−12\.\\displaystyle:y=\-\\tfrac\{1\}\{2\}x\+\\tfrac\{2n\-1\}\{2\}\.Together with then−3n\-3horizontal lines we havenndistinct lines, exactly three of which are sunny\. Hencek=3k=3is attainable for everyn≥3n\\geq 3\.

2\. Upper bound on points per sunny line

Lemma 1\.For any sunny lineLL, we have\|L∩Pn\|≤⌊n\+12⌋\|L\\cap P\_\{n\}\|\\leq\\left\\lfloor\\dfrac\{n\+1\}\{2\}\\right\\rfloor\.

*Proof\.*IfLLcontains no integer point, then obviously\|L∩Pn\|≤1≤⌊\(n\+1\)/2⌋\|L\\cap P\_\{n\}\|\\leq 1\\leq\\lfloor\(n\+1\)/2\\rfloor\(sincen≥3n\\geq 3\)\. So assumeLLcontains at least one integer point\. SinceLLis sunny, it is not vertical, so we can write its equation asy=m​x\+dy=mx\+d\.

⊳\\trianglerightIfmmis irrational, thenLLcan contain at most one integer point \(two integer points would give a rational slope\)\. Hence the bound holds\. Thus we may assumemmis rational\.

Writem=p/qm=p/qin lowest terms,q\>0q\>0,gcd⁡\(p,q\)=1\\gcd\(p,q\)=1\. Choose an integer point\(x0,y0\)\(x\_\{0\},y\_\{0\}\)onLL\. Then all integer points onLLare

\(x,y\)=\(x0\+q​t,y0\+p​t\),t∈ℤ\.\(x,y\)=\(x\_\{0\}\+qt,\\;y\_\{0\}\+pt\),\\qquad t\\in\\mathbb\{Z\}\.Let the set ofttfor which the point lies inPnP\_\{n\}be a consecutive integer intervaltmin,…,tmaxt\_\{\\min\},\\dots,t\_\{\\max\}; letNNbe the number of suchtt\. Thentmax−tmin=N−1t\_\{\\max\}\-t\_\{\\min\}=N\-1\.

For any such point we have the constraints

1≤x≤n,1≤y≤n,2≤x\+y≤n\+1\.1\\leq x\\leq n,\\qquad 1\\leq y\\leq n,\\qquad 2\\leq x\+y\\leq n\+1\.In particular,

q​\(N−1\)\\displaystyle q\\,\(N\-1\)=\|xmax−xmin\|≤n−1,\(A\)\\displaystyle=\|x\_\{\\max\}\-x\_\{\\min\}\|\\leq n\-1,\\quad\\text\{\(A\)\}\|p\|​\(N−1\)\\displaystyle\|p\|\\,\(N\-1\)=\|ymax−ymin\|≤n−1,\(B\)\\displaystyle=\|y\_\{\\max\}\-y\_\{\\min\}\|\\leq n\-1,\\quad\\text\{\(B\)\}\|p\+q\|​\(N−1\)\\displaystyle\|p\+q\|\\,\(N\-1\)=\|\(x\+y\)max−\(x\+y\)min\|≤n−1\.\(C\)\\displaystyle=\|\(x\+y\)\_\{\\max\}\-\(x\+y\)\_\{\\min\}\|\\leq n\-1\.\\quad\\text\{\(C\)\}SetM=max⁡\{q,\|p\|,\|p\+q\|\}M=\\max\\\{q,\\ \|p\|,\\ \|p\+q\|\\\}\. Then \(A\)\-\(C\) give

N−1≤n−1M\.N\-1\\leq\\frac\{n\-1\}\{M\}\.Now, becauseLLis sunny, we havep≠0p\\neq 0\(otherwise horizontal\) andp\+q≠0p\+q\\neq 0\(otherwise slope−1\-1\)\. We claim thatM≥2M\\geq 2\. SupposeM=1M=1\. Thenq=1q=1\(sinceq\>0q\>0\),\|p\|≤1\|p\|\\leq 1,\|p\+q\|≤1\|p\+q\|\\leq 1\. Asp≠0p\\neq 0, eitherp=1p=1orp=−1p=\-1\.

⊳\\trianglerightIfp=1p=1, then\|p\+q\|=\|1\+1\|=2\>1\|p\+q\|=\|1\+1\|=2\>1, contradiction\.

⊳\\trianglerightIfp=−1p=\-1, thenp\+q=0p\+q=0, which is forbidden\.

HenceM≥2M\\geq 2\. Consequently,

N−1≤n−12⟹N≤n\+12\.N\-1\\leq\\frac\{n\-1\}\{2\}\\quad\\Longrightarrow\\quad N\\leq\\frac\{n\+1\}\{2\}\.SinceNNis an integer,N≤⌊n\+12⌋N\\leq\\left\\lfloor\\frac\{n\+1\}\{2\}\\right\\rfloor\.□\\square

3\. No covering by all sunny lines forn≥4n\\geq 4

Lemma 2\.Forn≥4n\\geq 4there is no set ofnndistinct sunny lines that coversPnP\_\{n\}\.

*Proof\.*

*Evennn:*Writen=2​mn=2m\. By Lemma 1 each sunny line contains at mostmmpoints\. Therefore the total number of incidences \(counting each point as many times as lines through it\) is at mostn⋅m=2​m2n\\cdot m=2m^\{2\}\. But

\|Pn\|=n​\(n\+1\)2=m​\(2​m\+1\)=2​m2\+m\>2​m2\(m≥1\),\|P\_\{n\}\|=\\frac\{n\(n\+1\)\}\{2\}=m\(2m\+1\)=2m^\{2\}\+m\>2m^\{2\}\\qquad\(m\\geq 1\),so it is impossible to cover all points\.

*Oddnn:*Writen=2​m\+1n=2m\+1withm≥2m\\geq 2\(i\.e\.n≥5n\\geq 5\)\. Lemma 1 gives each sunny line at mostm\+1m\+1points\. Hence

∑i=1n\|Li∩Pn\|≤n​\(m\+1\)=\(2​m\+1\)​\(m\+1\)\.\\sum\_\{i=1\}^\{n\}\|L\_\{i\}\\cap P\_\{n\}\|\\leq n\(m\+1\)=\(2m\+1\)\(m\+1\)\.But

\|Pn\|=\(2​m\+1\)​\(2​m\+2\)2=\(2​m\+1\)​\(m\+1\)\.\|P\_\{n\}\|=\\frac\{\(2m\+1\)\(2m\+2\)\}\{2\}=\(2m\+1\)\(m\+1\)\.Thus the inequality must be an equality\. In particular, each line contains exactlym\+1m\+1points, and the sum of the sizes equals\|Pn\|\|P\_\{n\}\|\. Since every point is covered at least once, equality forces each point to be covered exactly once, and therefore thennlines are pairwise disjoint\.

We now classify all sunny lines that can contain exactlym\+1m\+1points inPnP\_\{n\}\.

Lemma 3\.Letn=2​m\+1≥3n=2m\+1\\geq 3\. If a sunny lineLLsatisfies\|L∩Pn\|=m\+1\|L\\cap P\_\{n\}\|=m\+1, thenLLis one of the three lines

LA:y=x,LB:y=−2​x\+\(n\+2\),LC:y=−12​x\+n\+22\.L\_\{A\}:y=x,\\qquad L\_\{B\}:y=\-2x\+\(n\+2\),\\qquad L\_\{C\}:y=\-\\tfrac\{1\}\{2\}x\+\\tfrac\{n\+2\}\{2\}\.*Proof\.*WriteLLin reduced formy=pq​x\+dy=\\frac\{p\}\{q\}x\+dwithq\>0q\>0,gcd⁡\(p,q\)=1\\gcd\(p,q\)=1,p≠0p\\neq 0,p\+q≠0p\+q\\neq 0\. LetN=m\+1N=m\+1\.

From \(A\)\-\(C\) we have \(withN−1=mN\-1=mandn−1=2​mn\-1=2m\)

q​m≤2​m,\|p\|​m≤2​m,\|p\+q\|​m≤2​m,q\\,m\\leq 2m,\\quad\|p\|\\,m\\leq 2m,\\quad\|p\+q\|\\,m\\leq 2m,hence

q≤2,\|p\|≤2,\|p\+q\|≤2\.q\\leq 2,\\quad\|p\|\\leq 2,\\quad\|p\+q\|\\leq 2\.We examine the possibilities\.

Caseq=1q=1\.Then\|p\|≤2\|p\|\\leq 2and\|p\+1\|≤2\|p\+1\|\\leq 2\. Sincep≠0p\\neq 0andp\+1≠0p\+1\\neq 0\(i\.e\.p≠−1p\\neq\-1\), the admissible integers arep=1p=1andp=−2p=\-2\.

Caseq=2q=2\.Thengcd⁡\(p,2\)=1\\gcd\(p,2\)=1, soppis odd\.\|p\|≤2\|p\|\\leq 2forcesp=±1p=\\pm 1\.\|p\+2\|≤2\|p\+2\|\\leq 2eliminatesp=1p=1because\|3\|\>2\|3\|\>2; thus onlyp=−1p=\-1remains\.

Now we determine the interceptddso that exactlym\+1m\+1integer points ofPnP\_\{n\}lie on the line\.

Subcase \(1, 1\):L:y=x\+dL:y=x\+d\.

Let the integer points be\(a\+i,a\+i\+d\)\(a\+i,\\ a\+i\+d\)fori=0,1,…,mi=0,1,\\dots,m\(sincexxincreases byq=1q=1\)\. Thexx\-coordinates range fromaatoa\+ma\+mand must satisfy1≤a\+i≤2​m\+11\\leq a\+i\\leq 2m\+1\. Hence1≤a≤m\+11\\leq a\\leq m\+1\. The constraints from the bounds onyyand onx\+yx\+yyield:

\{a\+d≥1,a\+m\+d≤2​m\+1,2​a\+2​m\+d≤2​m\+2\.\\begin\{cases\}a\+d\\geq 1,\\\\ a\+m\+d\\leq 2m\+1,\\\\ 2a\+2m\+d\\leq 2m\+2\.\\end\{cases\}The third inequality givesd≤2−2​ad\\leq 2\-2a; the second givesd≤m\+1−ad\\leq m\+1\-a; the first givesd≥1−ad\\geq 1\-a\.

For consistency we need1−a≤2−2​a1\-a\\leq 2\-2a, i\.e\.a≤1a\\leq 1\. Thusa=1a=1\. Then the bounds become0≤d≤min⁡\{m,0\}=00\\leq d\\leq\\min\\\{m,\\,0\\\}=0, sod=0d=0\. HenceL:y=xL:y=x\.

Subcase \(1, \-2\):L:y=−2​x\+dL:y=\-2x\+d\.

Again the points are\(a\+i,−2​\(a\+i\)\+d\)\(a\+i,\\ \-2\(a\+i\)\+d\)withi=0,…,mi=0,\\dots,mand1≤a≤m\+11\\leq a\\leq m\+1\. The tightest constraints come fromy≥1y\\geq 1\(ati=mi=m\),y≤2​m\+1y\\leq 2m\+1\(ati=0i=0\), andx\+y≤2​m\+2x\+y\\leq 2m\+2\(ati=0i=0\):

−2​\(a\+m\)\+d≥1\\displaystyle\-2\(a\+m\)\+d\\geq 1⟹d≥1\+2​a\+2​m,\\displaystyle\\Longrightarrow\\;d\\geq 1\+2a\+2m,−2​a\+d≤2​m\+1\\displaystyle\-2a\+d\\leq 2m\+1⟹d≤2​m\+1\+2​a,\\displaystyle\\Longrightarrow\\;d\\leq 2m\+1\+2a,−a\+d≤2​m\+2\\displaystyle\-a\+d\\leq 2m\+2⟹d≤2​m\+2\+a\.\\displaystyle\\Longrightarrow\\;d\\leq 2m\+2\+a\.The first two give

1\+2​a\+2​m≤d≤2​m\+1\+2​a\.1\+2a\+2m\\leq d\\leq 2m\+1\+2a\.Henceddmust equal2​m\+1\+2​a2m\+1\+2a\(the only integer satisfying both\)\. Substituting into the third inequality yields

2​m\+1\+2​a≤2​m\+2\+a⟹a≤1\.2m\+1\+2a\\leq 2m\+2\+a\\;\\Longrightarrow\\;a\\leq 1\.Sincea≥1a\\geq 1, we havea=1a=1\. Thend=2​m\+1\+2=2​m\+3=n\+2d=2m\+1\+2=2m\+3=n\+2\. SoL:y=−2​x\+\(n\+2\)L:y=\-2x\+\(n\+2\)\.

Subcase \(2, \-1\):L:y=−12​x\+dL:y=\-\\tfrac\{1\}\{2\}x\+d\.

Thexx\-coordinates differ byq=2q=2\. Let the smallest beaa\. Then the points are\(a\+2​i,−12​\(a\+2​i\)\+d\)\(a\+2i,\\ \-\\tfrac\{1\}\{2\}\(a\+2i\)\+d\)fori=0,…,mi=0,\\dots,m\. To stay within the bounds1≤x≤2​m\+11\\leq x\\leq 2m\+1we need

a≥1anda\+2​m≤2​m\+1⟹a≤1,a\\geq 1\\quad\\text\{and\}\\quad a\+2m\\leq 2m\+1\\;\\Longrightarrow\\;a\\leq 1,soa=1a=1\. Thus thexx\-coordinates are1,3,5,…,2​m\+11,3,5,\\dots,2m\+1\.

For theyy\-coordinates to be integers,ddmust have the formd=d′\+12d=d^\{\\prime\}\+\\tfrac\{1\}\{2\}withd′∈ℤd^\{\\prime\}\\in\\mathbb\{Z\}\. Then

yi=−12​\(2​i\+1\)\+d=−i−12\+d′\+12=d′−i\.y\_\{i\}=\-\\tfrac\{1\}\{2\}\(2i\+1\)\+d=\-i\-\\tfrac\{1\}\{2\}\+d^\{\\prime\}\+\\tfrac\{1\}\{2\}=d^\{\\prime\}\-i\.Now apply the constraints:

⊳\\trianglerightyi≥1y\_\{i\}\\geq 1for allii: the smallestyyoccurs ati=mi=m\(since slope negative\), sod′−m≥1⇒d′≥m\+1d^\{\\prime\}\-m\\geq 1\\;\\Rightarrow\\;d^\{\\prime\}\\geq m\+1\.

⊳\\trianglerightyi≤2​m\+1y\_\{i\}\\leq 2m\+1for allii: the largestyyoccurs ati=0i=0, givingd′≤2​m\+1d^\{\\prime\}\\leq 2m\+1\.

⊳\\trianglerightxi\+yi=\(2​i\+1\)\+\(d′−i\)=d′\+i\+1≤2​m\+2x\_\{i\}\+y\_\{i\}=\(2i\+1\)\+\(d^\{\\prime\}\-i\)=d^\{\\prime\}\+i\+1\\leq 2m\+2: the most restrictive is ati=mi=m, givingd′\+m\+1≤2​m\+2⇒d′≤m\+1d^\{\\prime\}\+m\+1\\leq 2m\+2\\;\\Rightarrow\\;d^\{\\prime\}\\leq m\+1\.

Thusd′≥m\+1d^\{\\prime\}\\geq m\+1andd′≤m\+1d^\{\\prime\}\\leq m\+1, sod′=m\+1d^\{\\prime\}=m\+1\. Thend=m\+1\+12=2​m\+32=n\+22d=m\+1\+\\tfrac\{1\}\{2\}=\\tfrac\{2m\+3\}\{2\}=\\tfrac\{n\+2\}\{2\}\. HenceL:y=−12​x\+n\+22L:y=\-\\tfrac\{1\}\{2\}x\+\\tfrac\{n\+2\}\{2\}\.

We have exhausted all possibilities; therefore any sunny line with exactlym\+1m\+1points must be one of the three listed\.□\\square

Returning to the proof of Lemma 2 for oddn≥5n\\geq 5: we havenndistinct sunny lines, each must contain exactlym\+1m\+1points, hence each must be one ofLA,LB,LCL\_\{A\},L\_\{B\},L\_\{C\}\. But there are only three distinct such lines, whilen≥5n\\geq 5gives a contradiction\. Hence no covering exists for oddn≥5n\\geq 5either\. \(Forn=3n=3the three lines are exactly these three, and that case is covered by the construction\.\) This completes the proof of Lemma 2\.□\\square

4\. Induction onnn

We now prove by strong induction that for everyn≥3n\\geq 3the numberkkof sunny lines in any covering ofPnP\_\{n\}bynndistinct lines can only be0,1,30,1,3\.

Base casen=3n=3\.Constructions in 1 show thatk=0,1,3k=0,1,3are attainable\. We must show thatk=2k=2is impossible\.

Assume three lines coverP3P\_\{3\}with exactly two sunny\. Let the sunny lines beL1,L2L\_\{1\},L\_\{2\}and the third lineL3L\_\{3\}be non\-sunny\.

Lemma 1 implies that each sunny line contains at most two points ofP3P\_\{3\}\. The only pairs of points inP3P\_\{3\}that are collinear with a sunny line \(i\.e\. with slope different from0, undefined, or−1\-1\) are exactly the three pairs:

\{\(1,1\),\(2,2\)\}​\(slope​1\),\{\(1,3\),\(2,1\)\}​\(slope−2\),\{\(1,2\),\(3,1\)\}​\(slope−1/2\)\.\\\{\(1,1\),\(2,2\)\\\}\\ \(\\text\{slope \}1\),\\quad\\\{\(1,3\),\(2,1\)\\\}\\ \(\\text\{slope \}\-2\),\\quad\\\{\(1,2\),\(3,1\)\\\}\\ \(\\text\{slope \}\-1/2\)\.Hence any sunny line that contains two points must be one of the three lines

ℓ1:y=x,ℓ2:y=−2​x\+5,ℓ3:y=−12​x\+52\.\\ell\_\{1\}:y=x,\\quad\\ell\_\{2\}:y=\-2x\+5,\\quad\\ell\_\{3\}:y=\-\\tfrac\{1\}\{2\}x\+\\tfrac\{5\}\{2\}\.Now consider the possibilities\.

*If bothL1L\_\{1\}andL2L\_\{2\}contain two points and are disjoint\.*Then they must be two of the three linesℓ1,ℓ2,ℓ3\\ell\_\{1\},\\ell\_\{2\},\\ell\_\{3\}\. Those three lines are pairwise disjoint, so they cover four points\. The two remaining points lie on the thirdℓ\\ell, which is sunny\. The third lineL3L\_\{3\}is non\-sunny, so it cannot cover both of those points \(the unique line through them is sunny\)\. Contradiction\.

*If at least one ofL1,L2L\_\{1\},L\_\{2\}contains at most one point\.*Then the two sunny lines together cover at most three points\. Consequently, the non\-sunny lineL3L\_\{3\}must cover at least three points\. The only lines that contain three points ofP3P\_\{3\}are the three ”boundary” lines:

y=1,x=1,x\+y=4\.y=1,\\quad x=1,\\quad x\+y=4\.\(One verifies that any other line contains at most two points\.\) SoL3L\_\{3\}is one of these three\.

⊳\\trianglerightL3=y=1L\_\{3\}=y=1: covers\(1,1\),\(2,1\),\(3,1\)\(1,1\),\(2,1\),\(3,1\)\. The uncovered points are\(1,2\),\(1,3\),\(2,2\)\(1,2\),\(1,3\),\(2,2\)\. Each of these three points lies on a different one ofℓ1,ℓ2,ℓ3\\ell\_\{1\},\\ell\_\{2\},\\ell\_\{3\}\(namely\(2,2\)∈ℓ1\(2,2\)\\in\\ell\_\{1\},\(1,3\)∈ℓ2\(1,3\)\\in\\ell\_\{2\},\(1,2\)∈ℓ3\(1,2\)\\in\\ell\_\{3\}\)\. Two sunny lines can cover at most two of them \(eachℓi\\ell\_\{i\}contains exactly one\)\. Hence impossible\.

⊳\\trianglerightL3=x=1L\_\{3\}=x=1: covers\(1,1\),\(1,2\),\(1,3\)\(1,1\),\(1,2\),\(1,3\)\. Uncovered:\(2,1\),\(2,2\),\(3,1\)\(2,1\),\(2,2\),\(3,1\)\. Again these are one perℓi\\ell\_\{i\}; two sunny lines cannot cover all three\. Contradiction\.

⊳\\trianglerightL3=x\+y=4L\_\{3\}=x\+y=4: covers\(1,3\),\(2,2\),\(3,1\)\(1,3\),\(2,2\),\(3,1\)\. Uncovered:\(1,1\),\(1,2\),\(2,1\)\(1,1\),\(1,2\),\(2,1\), again one perℓi\\ell\_\{i\}\. Contradiction\.

Thusk=2k=2is impossible, so the only possible values forn=3n=3are0,1,30,1,3\.

Inductive step\.Assume the statement holds for allmmwith3≤m<n3\\leq m<n, wheren≥4n\\geq 4\. Consider any covering ofPnP\_\{n\}bynndistinct lines\.

Define three special lines:

R:y=1,C:x=1,D:x\+y=n\+1\.R:\\ y=1,\\qquad C:\\ x=1,\\qquad D:\\ x\+y=n\+1\.Note thatR,C,DR,C,Dare not sunny \(horizontal, vertical, slope−1\-1respectively\)\.

Claim\.At least one ofR,C,DR,C,Dbelongs to the covering\.

*Proof of claim\.*Suppose none of them is present\. Then:

⊳\\trianglerightThe liney=1y=1containsnndistinct points\(a,1\)\(a,1\)\. To cover a point\(a,1\)\(a,1\), the line must intersecty=1y=1at that point\. A horizontal line different fromy=1y=1does not intersecty=1y=1\. Hence every line that covers a point ony=1y=1must be non\-horizontal\. Moreover, a non\-horizontal line meetsy=1y=1in at most one point\. Therefore we need at leastnnnon\-horizontal lines; since there are exactlynnlines,*all*lines are non\-horizontal, and each covers exactly one point ony=1y=1\.

⊳\\trianglerightThe linex=1x=1containsnnpoints\(1,b\)\(1,b\)\. SinceCCis absent, a vertical line different fromx=1x=1does not contain any point withx=1x=1\. Hence every line covering a point onx=1x=1must be non\-vertical\. Since all lines are already non\-horizontal, they must all be non\-vertical to cover those points \(a vertical line would missx=1x=1unless it isx=1x=1itself\)\. Thus all lines are non\-vertical\.

⊳\\trianglerightThe linex\+y=n\+1x\+y=n\+1containsnnpoints\. SinceDDis absent, a line with slope−1\-1other thanDDis parallel to it and does not intersect it\. Therefore every line covering a point onDDmust have slope different from−1\-1\. Hence all lines have slope≠−1\\neq\-1\.

Therefore every line is non\-horizontal, non\-vertical, and slope≠−1\\neq\-1; i\.e\., every line is sunny\. But Lemma 2 states that forn≥4n\\geq 4there is no covering ofPnP\_\{n\}bynnsunny lines\. Contradiction\. Hence the claim holds\.□\\square

Now we examine which ofR,C,DR,C,Dis present\. \(If more than one is present, we may choose any one; the argument works in each case\.\)

*IfRRis present:*RemoveRRfrom the set\. The remainingn−1n\-1lines still cover all points ofPnP\_\{n\}withy≥2y\\geq 2\(becauseRRonly covered points withy=1y=1\)\. Apply the translationϕR:\(x,y\)↦\(x,y−1\)\\phi\_\{R\}:\(x,y\)\\mapsto\(x,y\-1\)\. This map sends the set\{\(x,y\)∈Pn∣y≥2\}\\\{\(x,y\)\\in P\_\{n\}\\mid y\\geq 2\\\}bijectively ontoPn−1P\_\{n\-1\}\. Moreover,ϕR\\phi\_\{R\}sends each line to a line \(translations preserve lines\)\. The resulting family consists ofn−1n\-1distinct lines \(translation preserves distinctness\) and coversPn−1P\_\{n\-1\}\(each point ofPn−1P\_\{n\-1\}corresponds to a point withy≥2y\\geq 2that was covered\)\. SinceRRis not sunny, the number of sunny lines among then−1n\-1lines remainskk\.

Thus we have a covering ofPn−1P\_\{n\-1\}byn−1n\-1distinct lines with exactlykksunny lines\. By the induction hypothesis \(applied tom=n−1m=n\-1\), we must havek∈\{0,1,3\}k\\in\\\{0,1,3\\\}\.

*IfCCis present:*RemoveCCand useϕC:\(x,y\)↦\(x−1,y\)\\phi\_\{C\}:\(x,y\)\\mapsto\(x\-1,y\)\. It maps\{\(x,y\)∈Pn∣x≥2\}\\\{\(x,y\)\\in P\_\{n\}\\mid x\\geq 2\\\}ontoPn−1P\_\{n\-1\}\.CCis non\-sunny, sokkunchanged\. Induction givesk∈\{0,1,3\}k\\in\\\{0,1,3\\\}\.

*IfDDis present:*RemoveDD\. The remainingn−1n\-1lines cover all points ofPnP\_\{n\}that are not onDD\. Observe thatPn−1⊆PnP\_\{n\-1\}\\subseteq P\_\{n\}andPn−1∩D=∅P\_\{n\-1\}\\cap D=\\emptyset\(since points inPn−1P\_\{n\-1\}satisfya\+b≤na\+b\\leq n, whileDDrequiresa\+b=n\+1a\+b=n\+1\. Hence they coverPn−1P\_\{n\-1\}\.DDis not sunny, sokkstays the same\. Induction yieldsk∈\{0,1,3\}k\\in\\\{0,1,3\\\}\.

In every possible case we concludek∈\{0,1,3\}k\\in\\\{0,1,3\\\}, which completes the inductive step\.

By strong induction, the statement holds for all integersn≥3n\\geq 3\.

5\. Conclusion

We have shown that for everyn≥3n\\geq 3the only possible numbers of sunny lines in a covering ofPnP\_\{n\}bynndistinct lines are0,11, and33\. Explicit constructions demonstrate that each of these values is indeed attainable\.

k∈\{0,1,3\}for alln≥3\.\\boxed\{k\\in\\\{0,\\,1,\\,3\\\}\\ \\text\{ for all \}n\\geq 3\.\}

#### IMO 2025 Problem 2\.

LetΩ\\OmegaandΓ\\Gammabe circles with centresMMandNN, respectively, such that the radius ofΩ\\Omegais less than the radius ofΓ\\Gamma\. SupposeΩ\\OmegaandΓ\\Gammaintersect at two distinct pointsAAandBB\. LineM​NMNintersectsΩ\\OmegaatCCandΓ\\GammaatDD, so thatCC,MM,NN,DDlie onM​NMNin that order\. LetPPbe the circumcenter of triangleA​C​DACD\. LineA​PAPmeetsΩ\\Omegaagain atE≠AE\\neq Aand meetsΓ\\Gammaagain atF≠AF\\neq A\. LetHHbe the orthocenter of triangleP​M​NPMN\.Prove that the line throughHHparallel toA​PAPis tangent to the circumcircle of triangleB​E​FBEF\.

Solution

We will prove the statement using analytic geometry\. The configuration is invariant under similarity transformations, so we may fix convenient coordinates\.

1\. Coordinate setup

PlaceMMandNNon thexx\-axis\. Without loss of generality, setM​N=1MN=1\(scale the figure appropriately\)\. Let

M=\(0,0\),N=\(1,0\)\.M=\(0,0\),\\qquad N=\(1,0\)\.Denote the radius ofΩ\\Omega\(with centerMM\) byrrand the radius ofΓ\\Gamma\(with centerNN\) byRR, with0<r<R0<r<R\. Since the circles intersect in two distinct points, we have

2\. Intersection pointsAAandBB

The equations of the circles are

\{x2\+y2=r2,\(Ω\)\(x−1\)2\+y2=R2,\(Γ\)\.\\begin\{cases\}x^\{2\}\+y^\{2\}=r^\{2\},&\\text\{\($\\Omega$\)\}\\\\\[5\.69054pt\] \(x\-1\)^\{2\}\+y^\{2\}=R^\{2\},&\\text\{\($\\Gamma$\)\}\.\\end\{cases\}Subtracting gives2​x−1=R2−r22x\-1=R^\{2\}\-r^\{2\}, so

x=a=1\+r2−R22\.x=a=\\frac\{1\+r^\{2\}\-R^\{2\}\}\{2\}\.Leth\>0h\>0be such thata2\+h2=r2a^\{2\}\+h^\{2\}=r^\{2\}\. The two intersections are symmetric about thexx\-axis; we take

A=\(a,h\),B=\(a,−h\)\.A=\(a,\\,h\),\\qquad B=\(a,\\,\-h\)\.
3\. PointsCCandDDon lineM​NMN

The lineM​NMNis thexx\-axis\. Intersection ofΩ\\Omegawith thexx\-axis:x2=r2⇒x=±rx^\{2\}=r^\{2\}\\Rightarrow x=\\pm r\. Because the order on the line isCC,MM,NN,DD, pointCCmust lie left ofMMandDDright ofNN\. Hence

C=\(−r,0\),D=\(1\+R,0\)\.C=\(\-r,\\,0\),\\qquad D=\(1\+R,\\,0\)\.
4\. CircumcenterPPof△​A​C​D\\triangle ACD

PointsCCandDDlie on thexx\-axis, so the perpendicular bisector ofC​DCDis the vertical line through the midpoint ofC​DCD:

midpoint=\(−r\+1\+R2,0\)=\(1\+R−r2,0\)\.\\text\{midpoint\}=\\left\(\\frac\{\-r\+1\+R\}\{2\},\\,0\\right\)=\\left\(\\frac\{1\+R\-r\}\{2\},\\,0\\right\)\.Thus

xP=1\+R−r2\.x\_\{P\}=\\frac\{1\+R\-r\}\{2\}\.To findyPy\_\{P\}, useP​A=P​CPA=PC:

\(xP−a\)2\+\(yP−h\)2=\(xP\+r\)2\+yP2\.\(x\_\{P\}\-a\)^\{2\}\+\(y\_\{P\}\-h\)^\{2\}=\(x\_\{P\}\+r\)^\{2\}\+y\_\{P\}^\{2\}\.Expanding and usinga2\+h2=r2a^\{2\}\+h^\{2\}=r^\{2\}yields

−2​h​yP=2​\(r\+a\)​xP⟹yP=−\(r\+a\)​xPh\.\-2hy\_\{P\}=2\(r\+a\)x\_\{P\}\\quad\\Longrightarrow\\quad y\_\{P\}=\-\\frac\{\(r\+a\)x\_\{P\}\}\{h\}\.Hence

P=\(1\+R−r2,−\(r\+a\)​xPh\)\.P=\\left\(\\frac\{1\+R\-r\}\{2\},\\;\-\\frac\{\(r\+a\)x\_\{P\}\}\{h\}\\right\)\.
5\. Simplifying notation with auxiliary parameters

Introduce

p=R\+r,q=R−r,S=1\+R\+r=1\+p\.p=R\+r,\\qquad q=R\-r,\\qquad S=1\+R\+r=1\+p\.Then

r=p−q2,R=p\+q2,a=1−p​q2,xP=1\+q2\.r=\\frac\{p\-q\}\{2\},\\quad R=\\frac\{p\+q\}\{2\},\\quad a=\\frac\{1\-pq\}\{2\},\\quad x\_\{P\}=\\frac\{1\+q\}\{2\}\.Fromh2=r2−a2h^\{2\}=r^\{2\}\-a^\{2\}we obtain

h2=\(p2−1\)​\(1−q2\)4\.h^\{2\}=\\frac\{\(p^\{2\}\-1\)\(1\-q^\{2\}\)\}\{4\}\.Also

r\+a=S​\(1−q\)2\.r\+a=\\frac\{S\(1\-q\)\}\{2\}\.
6\. VectorA​P→\\overrightarrow\{AP\}

Setv→=A​P→=\(v1,v2\)\\vec\{v\}=\\overrightarrow\{AP\}=\(v\_\{1\},\\,v\_\{2\}\)\.

v1=xP−a=1\+q2−1−p​q2=q\+p​q2=q​S2\.v\_\{1\}=x\_\{P\}\-a=\\frac\{1\+q\}\{2\}\-\\frac\{1\-pq\}\{2\}=\\frac\{q\+pq\}\{2\}=\\frac\{qS\}\{2\}\.Forv2v\_\{2\}:

v2=yP−h=−\(r\+a\)​xPh−h=−\(r\+a\)​xP\+h2h\.v\_\{2\}=y\_\{P\}\-h=\-\\frac\{\(r\+a\)x\_\{P\}\}\{h\}\-h=\-\\frac\{\(r\+a\)x\_\{P\}\+h^\{2\}\}\{h\}\.Sinceh2=\(r−a\)​\(r\+a\)h^\{2\}=\(r\-a\)\(r\+a\),

\(r\+a\)​xP\+h2=\(r\+a\)​\(xP\+r−a\)\.\(r\+a\)x\_\{P\}\+h^\{2\}=\(r\+a\)\(x\_\{P\}\+r\-a\)\.ComputexP\+r−ax\_\{P\}\+r\-a:

xP\+r−a=1\+q2\+r−1−p​q2=q\+2​r\+p​q2\.x\_\{P\}\+r\-a=\\frac\{1\+q\}\{2\}\+r\-\\frac\{1\-pq\}\{2\}=\\frac\{q\+2r\+pq\}\{2\}\.But2​r=p−q2r=p\-q, so

q\+2​r\+p​q=q\+\(p−q\)\+p​q=p\+p​q=p​\(1\+q\)\.q\+2r\+pq=q\+\(p\-q\)\+pq=p\+pq=p\(1\+q\)\.Thus

xP\+r−a=p​\(1\+q\)2\.x\_\{P\}\+r\-a=\\frac\{p\(1\+q\)\}\{2\}\.Now\(r\+a\)=S​\(1−q\)2\(r\+a\)=\\dfrac\{S\(1\-q\)\}\{2\}, so

\(r\+a\)​\(xP\+r−a\)=S​\(1−q\)2⋅p​\(1\+q\)2=p​S​\(1−q2\)4\.\(r\+a\)\(x\_\{P\}\+r\-a\)=\\frac\{S\(1\-q\)\}\{2\}\\cdot\\frac\{p\(1\+q\)\}\{2\}=\\frac\{pS\(1\-q^\{2\}\)\}\{4\}\.Therefore

v2=−1h⋅p​S​\(1−q2\)4=−p​S​\(1−q2\)4​h\.v\_\{2\}=\-\\frac\{1\}\{h\}\\cdot\\frac\{pS\(1\-q^\{2\}\)\}\{4\}=\-\\frac\{pS\(1\-q^\{2\}\)\}\{4h\}\.
7\. PointsEEandFF\(second intersections ofA​PAPwith the circles\)

Parameterize lineA​PAPasA\+λ​v→A\+\\lambda\\vec\{v\}\. Substitute intoΩ\\Omega:\|A\+λ​v→\|2=r2\|A\+\\lambda\\vec\{v\}\|^\{2\}=r^\{2\}\. Because\|A\|2=r2\|A\|^\{2\}=r^\{2\}, we get

2​λ​\(v→⋅A\)\+λ2​\|v→\|2=0\.2\\lambda\(\\vec\{v\}\\cdot A\)\+\\lambda^\{2\}\|\\vec\{v\}\|^\{2\}=0\.Henceλ=0\\lambda=0\(pointAA\) or

λE=−2​v→⋅A\|v→\|2\.\\lambda\_\{E\}=\-\\frac\{2\\,\\vec\{v\}\\cdot A\}\{\|\\vec\{v\}\|^\{2\}\}\.Similarly, forΓ\\Gammasubstitute into\|\(A\+λ​v→\)−N\|2=R2\|\(A\+\\lambda\\vec\{v\}\)\-N\|^\{2\}=R^\{2\}\(sinceN=\(1,0\)N=\(1,0\)\)\.\|A−N\|2=R2\|A\-N\|^\{2\}=R^\{2\}, so

λF=−2​v→⋅\(A−N\)\|v→\|2\.\\lambda\_\{F\}=\-\\frac\{2\\,\\vec\{v\}\\cdot\(A\-N\)\}\{\|\\vec\{v\}\|^\{2\}\}\.Now compute the needed dot products\.

v→⋅A=a​v1\+h​v2=a⋅q​S2\+h​\(−p​S​\(1−q2\)4​h\)=a​q​S2−p​S​\(1−q2\)4=S4​\(2​a​q−p​\(1−q2\)\)\.\\vec\{v\}\\cdot A=av\_\{1\}\+hv\_\{2\}=a\\cdot\\frac\{qS\}\{2\}\+h\\left\(\-\\frac\{pS\(1\-q^\{2\}\)\}\{4h\}\\right\)=\\frac\{aqS\}\{2\}\-\\frac\{pS\(1\-q^\{2\}\)\}\{4\}=\\frac\{S\}\{4\}\\bigl\(2aq\-p\(1\-q^\{2\}\)\\bigr\)\.But2​a​q=q​\(1−p​q\)=q−p​q22aq=q\(1\-pq\)=q\-pq^\{2\}, so

2​a​q−p​\(1−q2\)=q−p​q2−p\+p​q2=q−p\.2aq\-p\(1\-q^\{2\}\)=q\-pq^\{2\}\-p\+pq^\{2\}=q\-p\.Thus

v→⋅A=S4​\(q−p\)=−S​\(p−q\)4=−S⋅2​r4=−r​S2\.\\vec\{v\}\\cdot A=\\frac\{S\}\{4\}\(q\-p\)=\-\\frac\{S\(p\-q\)\}\{4\}=\-\\frac\{S\\cdot 2r\}\{4\}=\-\\frac\{rS\}\{2\}\.Next,

v→⋅\(A−N\)=v→⋅A−v1=−r​S2−q​S2=−S​\(r\+q\)2=−R​S2,\\vec\{v\}\\cdot\(A\-N\)=\\vec\{v\}\\cdot A\-v\_\{1\}=\-\\frac\{rS\}\{2\}\-\\frac\{qS\}\{2\}=\-\\frac\{S\(r\+q\)\}\{2\}=\-\\frac\{RS\}\{2\},becauser\+q=Rr\+q=R\.

Consequently,

λE=−2​\(−r​S2\)\|v→\|2=r​S\|v→\|2,λF=−2​\(−R​S2\)\|v→\|2=R​S\|v→\|2\.\\lambda\_\{E\}=\-\\frac\{2\\bigl\(\-\\frac\{rS\}\{2\}\\bigr\)\}\{\|\\vec\{v\}\|^\{2\}\}=\\frac\{rS\}\{\|\\vec\{v\}\|^\{2\}\},\\qquad\\lambda\_\{F\}=\-\\frac\{2\\bigl\(\-\\frac\{RS\}\{2\}\\bigr\)\}\{\|\\vec\{v\}\|^\{2\}\}=\\frac\{RS\}\{\|\\vec\{v\}\|^\{2\}\}\.Hence

E=A\+r​S\|v→\|2​v→,F=A\+R​S\|v→\|2​v→\.E=A\+\\frac\{rS\}\{\|\\vec\{v\}\|^\{2\}\}\\,\\vec\{v\},\\qquad F=A\+\\frac\{RS\}\{\|\\vec\{v\}\|^\{2\}\}\\,\\vec\{v\}\.
8\. OrthocenterHHof△​P​M​N\\triangle PMN

Vertices:M=\(0,0\)M=\(0,0\),N=\(1,0\)N=\(1,0\),P=\(xP,yP\)P=\(x\_\{P\},y\_\{P\}\)\.

SideM​NMNis horizontal, so the altitude fromPPis the vertical linex=xPx=x\_\{P\}\.

Altitude fromMMis the line throughMMperpendicular toP​NPN\. VectorP​N→=\(1−xP,−yP\)\\overrightarrow\{PN\}=\(1\-x\_\{P\},\\,\-y\_\{P\}\)\. A vector perpendicular toP​NPNis\(yP,1−xP\)\(y\_\{P\},\\,1\-x\_\{P\}\)\(since\(1−xP\)​yP\+\(−yP\)​\(1−xP\)=0\(1\-x\_\{P\}\)y\_\{P\}\+\(\-y\_\{P\}\)\(1\-x\_\{P\}\)=0\)\. Parametric equation:t​\(yP,1−xP\)t\\,\(y\_\{P\},\\,1\-x\_\{P\}\)\.

The orthocenterHHis the intersection ofx=xPx=x\_\{P\}with this altitude\. So set

t​yP=xP⟹t=xPyP\(yP≠0\)\.ty\_\{P\}=x\_\{P\}\\quad\\Longrightarrow\\quad t=\\frac\{x\_\{P\}\}\{y\_\{P\}\}\\ \\ \(y\_\{P\}\\neq 0\)\.Then

H=\(xP,t​\(1−xP\)\)=\(xP,xP​\(1−xP\)yP\)\.H=\\bigl\(x\_\{P\},\\ t\(1\-x\_\{P\}\)\\bigr\)=\\left\(x\_\{P\},\\ \\frac\{x\_\{P\}\(1\-x\_\{P\}\)\}\{y\_\{P\}\}\\right\)\.Now substituteyP=−\(r\+a\)​xPhy\_\{P\}=\-\\dfrac\{\(r\+a\)x\_\{P\}\}\{h\}:

xP​\(1−xP\)yP=xP​\(1−xP\)−\(r\+a\)​xPh=−h​\(1−xP\)r\+a\.\\frac\{x\_\{P\}\(1\-x\_\{P\}\)\}\{y\_\{P\}\}=\\frac\{x\_\{P\}\(1\-x\_\{P\}\)\}\{\-\\frac\{\(r\+a\)x\_\{P\}\}\{h\}\}=\-\\frac\{h\(1\-x\_\{P\}\)\}\{r\+a\}\.Compute1−xP=1−1\+q2=1−q21\-x\_\{P\}=1\-\\dfrac\{1\+q\}\{2\}=\\dfrac\{1\-q\}\{2\}\. Andr\+a=S​\(1−q\)2r\+a=\\dfrac\{S\(1\-q\)\}\{2\}\. Therefore

−h​\(1−xP\)r\+a=−h⋅1−q2S​\(1−q\)2=−hS\.\-\\frac\{h\(1\-x\_\{P\}\)\}\{r\+a\}=\-\\frac\{h\\cdot\\frac\{1\-q\}\{2\}\}\{\\frac\{S\(1\-q\)\}\{2\}\}=\-\\frac\{h\}\{S\}\.Thus

H=\(1\+q2,−hS\)\.H=\\left\(\\frac\{1\+q\}\{2\},\\;\-\\frac\{h\}\{S\}\\right\)\.
9\. CircumcenterO′O^\{\\prime\}of△​B​E​F\\triangle BEF

We will determineO′=\(x,y\)O^\{\\prime\}=\(x,y\)that is equidistant fromBB,EE,FF\.

BecauseEEandFFlie on lineA​PAP, the segmentE​FEFis collinear withv→\\vec\{v\}\. The perpendicular bisector ofE​FEFconsists of pointsXXsatisfyingv→⋅\(X−ME​F\)=0\\vec\{v\}\\cdot\(X\-M\_\{EF\}\)=0, whereME​FM\_\{EF\}is the midpoint ofE​FEF\.

The midpoint:

ME​F=E\+F2=A\+λE\+λF2​v→\.M\_\{EF\}=\\frac\{E\+F\}\{2\}=A\+\\frac\{\\lambda\_\{E\}\+\\lambda\_\{F\}\}\{2\}\\,\\vec\{v\}\.SinceλE\+λF=\(r\+R\)​S\|v→\|2=p​S\|v→\|2\\lambda\_\{E\}\+\\lambda\_\{F\}=\\dfrac\{\(r\+R\)S\}\{\|\\vec\{v\}\|^\{2\}\}=\\dfrac\{pS\}\{\|\\vec\{v\}\|^\{2\}\},

ME​F=A\+p​S2​\|v→\|2​v→\.M\_\{EF\}=A\+\\frac\{pS\}\{2\|\\vec\{v\}\|^\{2\}\}\\,\\vec\{v\}\.Hence the conditionv→⋅X=v→⋅ME​F\\vec\{v\}\\cdot X=\\vec\{v\}\\cdot M\_\{EF\}gives

v→⋅O′=v→⋅A\+p​S2\.\\vec\{v\}\\cdot O^\{\\prime\}=\\vec\{v\}\\cdot A\+\\frac\{pS\}\{2\}\.Butv→⋅A=−r​S2\\vec\{v\}\\cdot A=\-\\dfrac\{rS\}\{2\}, so

v→⋅O′=−r​S2\+p​S2=\(p−r\)​S2=R​S2\.\(1\)\\vec\{v\}\\cdot O^\{\\prime\}=\-\\frac\{rS\}\{2\}\+\\frac\{pS\}\{2\}=\\frac\{\(p\-r\)S\}\{2\}=\\frac\{RS\}\{2\}\.\\quad\\text\{\(1\)\}Next impose\|O′−B\|2=\|O′−E\|2\|O^\{\\prime\}\-B\|^\{2\}=\|O^\{\\prime\}\-E\|^\{2\}\. WriteU=O′−AU=O^\{\\prime\}\-A\. Then

O′−B=U−w,with​w=B−A=\(0,−2​h\),O^\{\\prime\}\-B=U\-w,\\quad\\text\{with \}w=B\-A=\(0,\\,\-2h\),and

O′−E=U−λE​v→\.O^\{\\prime\}\-E=U\-\\lambda\_\{E\}\\vec\{v\}\.Then

\|U−w\|2=\|U−λE​v→\|2\.\|U\-w\|^\{2\}=\|U\-\\lambda\_\{E\}\\vec\{v\}\|^\{2\}\.Expand:

\|U\|2−2​U⋅w\+\|w\|2=\|U\|2−2​λE​v→⋅U\+λE2​\|v→\|2\.\|U\|^\{2\}\-2U\\cdot w\+\|w\|^\{2\}=\|U\|^\{2\}\-2\\lambda\_\{E\}\\vec\{v\}\\cdot U\+\\lambda\_\{E\}^\{2\}\|\\vec\{v\}\|^\{2\}\.Cancel\|U\|2\|U\|^\{2\}:

−2U⋅w\+\|w\|2=−2λEv→⋅U\+λE2\|v→\|2\.\(2\)\-2U\\cdot w\+\|w\|^\{2\}=\-2\\lambda\_\{E\}\\vec\{v\}\\cdot U\+\\lambda\_\{E\}^\{2\}\|\\vec\{v\}\|^\{2\}\.\\quad\\text\{\(2\)\}NowU⋅w=\(x−a\)⋅0\+\(y−h\)​\(−2​h\)=−2​h​\(y−h\)U\\cdot w=\(x\-a\)\\cdot 0\+\(y\-h\)\(\-2h\)=\-2h\(y\-h\)\. Thus

−2​U⋅w\+\|w\|2=−2​\[−2​h​\(y−h\)\]\+4​h2=4​h​\(y−h\)\+4​h2=4​h​y\.\-2U\\cdot w\+\|w\|^\{2\}=\-2\[\-2h\(y\-h\)\]\+4h^\{2\}=4h\(y\-h\)\+4h^\{2\}=4hy\.So \(2\) becomes

4hy=−2λEv→⋅U\+λE2\|v→\|2\.\(3\)4hy=\-2\\lambda\_\{E\}\\vec\{v\}\\cdot U\+\\lambda\_\{E\}^\{2\}\|\\vec\{v\}\|^\{2\}\.\\quad\\text\{\(3\)\}We have

v→⋅U=v→⋅\(O′−A\)=v→⋅O′−v→⋅A=R​S2−\(−r​S2\)=\(R\+r\)​S2=p​S2\.\\vec\{v\}\\cdot U=\\vec\{v\}\\cdot\(O^\{\\prime\}\-A\)=\\vec\{v\}\\cdot O^\{\\prime\}\-\\vec\{v\}\\cdot A=\\frac\{RS\}\{2\}\-\\left\(\-\\frac\{rS\}\{2\}\\right\)=\\frac\{\(R\+r\)S\}\{2\}=\\frac\{pS\}\{2\}\.RecallλE=r​S\|v→\|2\\lambda\_\{E\}=\\dfrac\{rS\}\{\|\\vec\{v\}\|^\{2\}\}\. Substitute into \(3\):

4​h​y=−2​\(r​S\|v→\|2\)⋅p​S2\+\(r​S\|v→\|2\)2​\|v→\|2=−r​p​S2\|v→\|2\+r2​S2\|v→\|2=r​S2\|v→\|2​\(r−p\)=−r​p​S2\|v→\|2\+r2​S2\|v→\|2=−r​R​S2\|v→\|2\(since​p−r=R\)\.4hy=\-2\\left\(\\frac\{rS\}\{\|\\vec\{v\}\|^\{2\}\}\\right\)\\cdot\\frac\{pS\}\{2\}\+\\left\(\\frac\{rS\}\{\|\\vec\{v\}\|^\{2\}\}\\right\)^\{2\}\|\\vec\{v\}\|^\{2\}=\-\\frac\{rpS^\{2\}\}\{\|\\vec\{v\}\|^\{2\}\}\+\\frac\{r^\{2\}S^\{2\}\}\{\|\\vec\{v\}\|^\{2\}\}=\\frac\{rS^\{2\}\}\{\|\\vec\{v\}\|^\{2\}\}\(r\-p\)=\-\\frac\{rpS^\{2\}\}\{\|\\vec\{v\}\|^\{2\}\}\+\\frac\{r^\{2\}S^\{2\}\}\{\|\\vec\{v\}\|^\{2\}\}=\-\\frac\{rRS^\{2\}\}\{\|\\vec\{v\}\|^\{2\}\}\\qquad\(\\text\{since \}p\-r=R\)\.Thus

4hy=−r​R​S2\|v→\|2,soy=−r​R​S24​h​\|v→\|2\.\(4\)4hy=\-\\frac\{rRS^\{2\}\}\{\|\\vec\{v\}\|^\{2\}\},\\qquad\\text\{so\}\\qquad y=\-\\frac\{rRS^\{2\}\}\{4h\|\\vec\{v\}\|^\{2\}\}\.\\quad\\text\{\(4\)\}
10\. Computation of\|v→\|2\|\\vec\{v\}\|^\{2\}

We have

v1=q​S2,v2=−p​S​\(1−q2\)4​h\.v\_\{1\}=\\frac\{qS\}\{2\},\\qquad v\_\{2\}=\-\\frac\{pS\(1\-q^\{2\}\)\}\{4h\}\.Hence

v12=q2​S24,v\_\{1\}^\{2\}=\\frac\{q^\{2\}S^\{2\}\}\{4\},v22=p2​S2​\(1−q2\)216​h2\.v\_\{2\}^\{2\}=\\frac\{p^\{2\}S^\{2\}\(1\-q^\{2\}\)^\{2\}\}\{16h^\{2\}\}\.Usingh2=\(p2−1\)​\(1−q2\)4h^\{2\}=\\dfrac\{\(p^\{2\}\-1\)\(1\-q^\{2\}\)\}\{4\},

v22=p2​S2​\(1−q2\)4​\(p2−1\)\.v\_\{2\}^\{2\}=\\frac\{p^\{2\}S^\{2\}\(1\-q^\{2\}\)\}\{4\(p^\{2\}\-1\)\}\.Therefore

\|v→\|2=S24​\(q2\+p2​\(1−q2\)p2−1\)=S24⋅q2​\(p2−1\)\+p2​\(1−q2\)p2−1\.\|\\vec\{v\}\|^\{2\}=\\frac\{S^\{2\}\}\{4\}\\left\(q^\{2\}\+\\frac\{p^\{2\}\(1\-q^\{2\}\)\}\{p^\{2\}\-1\}\\right\)=\\frac\{S^\{2\}\}\{4\}\\cdot\\frac\{q^\{2\}\(p^\{2\}\-1\)\+p^\{2\}\(1\-q^\{2\}\)\}\{p^\{2\}\-1\}\.The numerator simplifies:

q2​\(p2−1\)\+p2​\(1−q2\)=p2−q2\.q^\{2\}\(p^\{2\}\-1\)\+p^\{2\}\(1\-q^\{2\}\)=p^\{2\}\-q^\{2\}\.Thus

\|v→\|2=S24⋅p2−q2p2−1\.\|\\vec\{v\}\|^\{2\}=\\frac\{S^\{2\}\}\{4\}\\cdot\\frac\{p^\{2\}\-q^\{2\}\}\{p^\{2\}\-1\}\.Nowp2−q2=\(R\+r\)2−\(R−r\)2=4​R​r\.p^\{2\}\-q^\{2\}=\(R\+r\)^\{2\}\-\(R\-r\)^\{2\}=4Rr\.Andp2−1=\(p−1\)​\(p\+1\)=\(R\+r−1\)​\(R\+r\+1\)=\(R\+r−1\)​S\.p^\{2\}\-1=\(p\-1\)\(p\+1\)=\(R\+r\-1\)\(R\+r\+1\)=\(R\+r\-1\)S\.

Hence

\|v→\|2=S24⋅4​R​r\(R\+r−1\)​S=S​R​rR\+r−1\.\(5\)\|\\vec\{v\}\|^\{2\}=\\frac\{S^\{2\}\}\{4\}\\cdot\\frac\{4Rr\}\{\(R\+r\-1\)S\}=\\frac\{SRr\}\{R\+r\-1\}\.\\quad\\text\{\(5\)\}
11\. Simplifyyyfrom \(4\)

Insert \(5\) into \(4\):

y=−r​R​S24​h⋅R\+r−1S​R​r=−S​\(R\+r−1\)4​h\.\(6\)y=\-\\frac\{rRS^\{2\}\}\{4h\}\\cdot\\frac\{R\+r\-1\}\{SRr\}=\-\\frac\{S\(R\+r\-1\)\}\{4h\}\.\\quad\\text\{\(6\)\}
12\. Determinexxfrom equation \(1\)

Equation \(1\):v→⋅O′=v1​x\+v2​y=R​S2\\vec\{v\}\\cdot O^\{\\prime\}=v\_\{1\}x\+v\_\{2\}y=\\dfrac\{RS\}\{2\}\.

We knowv1=q​S2v\_\{1\}=\\dfrac\{qS\}\{2\}, and from \(6\) we haveyy\. Computev2​yv\_\{2\}y\.

v2=−p​S​\(1−q2\)4​h,y=−S​\(R\+r−1\)4​h\.v\_\{2\}=\-\\frac\{pS\(1\-q^\{2\}\)\}\{4h\},\\qquad y=\-\\frac\{S\(R\+r\-1\)\}\{4h\}\.Thus

v2​y=p​S2​\(1−q2\)​\(R\+r−1\)16​h2\.v\_\{2\}y=\\frac\{pS^\{2\}\(1\-q^\{2\}\)\(R\+r\-1\)\}\{16h^\{2\}\}\.Recall4​h2=\(R\+r−1\)​S​\(1−q2\)4h^\{2\}=\(R\+r\-1\)S\(1\-q^\{2\}\)\. Then16​h2=4​\(R\+r−1\)​S​\(1−q2\)16h^\{2\}=4\(R\+r\-1\)S\(1\-q^\{2\}\)\.

v2​y=p​S2​\(1−q2\)​\(R\+r−1\)4​\(R\+r−1\)​S​\(1−q2\)=p​S4\.v\_\{2\}y=\\frac\{pS^\{2\}\(1\-q^\{2\}\)\(R\+r\-1\)\}\{4\(R\+r\-1\)S\(1\-q^\{2\}\)\}=\\frac\{pS\}\{4\}\.Now \(1\) becomes

q​S2​x\+p​S4=R​S2\.\\frac\{qS\}\{2\}\\,x\+\\frac\{pS\}\{4\}=\\frac\{RS\}\{2\}\.Divide byS≠0S\\neq 0:

q2​x\+p4=R2\.\\frac\{q\}\{2\}\\,x\+\\frac\{p\}\{4\}=\\frac\{R\}\{2\}\.Multiply by44:

Nowq=R−rq=R\-r,p=R\+rp=R\+r, so

2​\(R−r\)​x\+\(R\+r\)=2​R⟹2​\(R−r\)​x=2​R−\(R\+r\)=R−r\.2\(R\-r\)x\+\(R\+r\)=2R\\quad\\Longrightarrow\\quad 2\(R\-r\)x=2R\-\(R\+r\)=R\-r\.SinceR≠rR\\neq r\(strict inequality\), we obtain

x=R−r2​\(R−r\)=12\.x=\\frac\{R\-r\}\{2\(R\-r\)\}=\\frac\{1\}\{2\}\.Therefore the circumcenter of△​B​E​F\\triangle BEFis

O′=\(12,−S​\(R\+r−1\)4​h\)\.\(7\)O^\{\\prime\}=\\left\(\\frac\{1\}\{2\},\\;\-\\frac\{S\(R\+r\-1\)\}\{4h\}\\right\)\.\\quad\\text\{\(7\)\}
13\. Distance fromO′O^\{\\prime\}to the line throughHHparallel tov→\\vec\{v\}

Letℓ\\ellbe the line throughHHwith directionv→\\vec\{v\}\.

The distance from a point to a line with directionv→\\vec\{v\}throughHHis

d=\|\(O′−H\)×v→\|\|v→\|,d=\\frac\{\|\(O^\{\\prime\}\-H\)\\times\\vec\{v\}\|\}\{\|\\vec\{v\}\|\},where the cross product \(in the plane\) is taken as the scalarΔ​x⋅v2−Δ​y⋅v1\\Delta x\\cdot v\_\{2\}\-\\Delta y\\cdot v\_\{1\}\.

Coordinates:

H=\(1\+q2,−hS\)\.H=\\left\(\\frac\{1\+q\}\{2\},\\;\-\\frac\{h\}\{S\}\\right\)\.Compute differences:

Δ​x=12−xH=12−1\+q2=−q2\.\\Delta x=\\frac\{1\}\{2\}\-x\_\{H\}=\\frac\{1\}\{2\}\-\\frac\{1\+q\}\{2\}=\-\\frac\{q\}\{2\}\.Δ​y=yO′−yH=−S​\(R\+r−1\)4​h−\(−hS\)=−S​\(R\+r−1\)4​h\+hS\.\\Delta y=y\_\{O^\{\\prime\}\}\-y\_\{H\}=\-\\frac\{S\(R\+r\-1\)\}\{4h\}\-\\left\(\-\\frac\{h\}\{S\}\\right\)=\-\\frac\{S\(R\+r\-1\)\}\{4h\}\+\\frac\{h\}\{S\}\.Write as a single fraction:

Δ​y=−S2​\(R\+r−1\)\+4​h24​h​S\.\\Delta y=\\frac\{\-S^\{2\}\(R\+r\-1\)\+4h^\{2\}\}\{4hS\}\.But4​h2=\(R\+r−1\)​S​\(1−q2\)4h^\{2\}=\(R\+r\-1\)S\(1\-q^\{2\}\)\. Hence

Δ​y=\(R\+r−1\)​S​\(1−q2\)−S2​\(R\+r−1\)4​h​S=\(R\+r−1\)​S​\(\(1−q2\)−S\)4​h​S=\(R\+r−1\)​\(\(1−q2\)−S\)4​h\.\\Delta y=\\frac\{\(R\+r\-1\)S\(1\-q^\{2\}\)\-S^\{2\}\(R\+r\-1\)\}\{4hS\}=\\frac\{\(R\+r\-1\)S\\bigl\(\(1\-q^\{2\}\)\-S\\bigr\)\}\{4hS\}=\\frac\{\(R\+r\-1\)\\bigl\(\(1\-q^\{2\}\)\-S\\bigr\)\}\{4h\}\.NowS=1\+pS=1\+p, so

\(1−q2\)−S=1−q2−1−p=−p−q2\.\(1\-q^\{2\}\)\-S=1\-q^\{2\}\-1\-p=\-p\-q^\{2\}\.Thus

Δy=−\(R\+r−1\)​\(p\+q2\)4​h\.\(8\)\\Delta y=\-\\frac\{\(R\+r\-1\)\(p\+q^\{2\}\)\}\{4h\}\.\\quad\\text\{\(8\)\}Now compute the cross product:

Δ×v→=Δ​x⋅v2−Δ​y⋅v1\.\\Delta\\times\\vec\{v\}=\\Delta x\\cdot v\_\{2\}\-\\Delta y\\cdot v\_\{1\}\.Δ​x⋅v2=\(−q2\)​\(−p​S​\(1−q2\)4​h\)=q​p​S​\(1−q2\)8​h\.\\Delta x\\cdot v\_\{2\}=\\left\(\-\\frac\{q\}\{2\}\\right\)\\left\(\-\\frac\{pS\(1\-q^\{2\}\)\}\{4h\}\\right\)=\\frac\{qpS\(1\-q^\{2\}\)\}\{8h\}\.Δ​y⋅v1=\(−\(R\+r−1\)​\(p\+q2\)4​h\)​\(q​S2\)=−q​S​\(R\+r−1\)​\(p\+q2\)8​h\.\\Delta y\\cdot v\_\{1\}=\\left\(\-\\frac\{\(R\+r\-1\)\(p\+q^\{2\}\)\}\{4h\}\\right\)\\left\(\\frac\{qS\}\{2\}\\right\)=\-\\frac\{qS\(R\+r\-1\)\(p\+q^\{2\}\)\}\{8h\}\.Therefore

Δ×v→=q​p​S​\(1−q2\)8​h−\(−q​S​\(R\+r−1\)​\(p\+q2\)8​h\)=q​S8​h​\(p​\(1−q2\)\+\(R\+r−1\)​\(p\+q2\)\)\.\\Delta\\times\\vec\{v\}=\\frac\{qpS\(1\-q^\{2\}\)\}\{8h\}\-\\left\(\-\\frac\{qS\(R\+r\-1\)\(p\+q^\{2\}\)\}\{8h\}\\right\)=\\frac\{qS\}\{8h\}\\Bigl\(p\(1\-q^\{2\}\)\+\(R\+r\-1\)\(p\+q^\{2\}\)\\Bigr\)\.ButR\+r−1=p−1R\+r\-1=p\-1\. So the bracket becomes

p​\(1−q2\)\+\(p−1\)​\(p\+q2\)\.p\(1\-q^\{2\}\)\+\(p\-1\)\(p\+q^\{2\}\)\.Expand:

p​\(1−q2\)=p−p​q2,p\(1\-q^\{2\}\)=p\-pq^\{2\},\(p−1\)​\(p\+q2\)=p​\(p−1\)\+\(p−1\)​q2=p2−p\+p​q2−q2\.\(p\-1\)\(p\+q^\{2\}\)=p\(p\-1\)\+\(p\-1\)q^\{2\}=p^\{2\}\-p\+pq^\{2\}\-q^\{2\}\.Sum:

\(p−p​q2\)\+\(p2−p\+p​q2−q2\)=p2−q2\.\(p\-pq^\{2\}\)\+\(p^\{2\}\-p\+pq^\{2\}\-q^\{2\}\)=p^\{2\}\-q^\{2\}\.Thus

Δ×v→=q​S8​h​\(p2−q2\)=q​S8​h⋅4​R​r=q​S​R​r2​h\.\\Delta\\times\\vec\{v\}=\\frac\{qS\}\{8h\}\(p^\{2\}\-q^\{2\}\)=\\frac\{qS\}\{8h\}\\cdot 4Rr=\\frac\{qSRr\}\{2h\}\.Sinceq=R−rq=R\-r,

Δ×v→=\(R−r\)​S​R​r2​h\.\(9\)\\Delta\\times\\vec\{v\}=\\frac\{\(R\-r\)SRr\}\{2h\}\.\\quad\\text\{\(9\)\}
14\. Squared distance fromO′O^\{\\prime\}toℓ\\ell

d2=\(\(R−r\)​S​R​r2​h\)2\|v→\|2=\(R−r\)2​S2​R2​r24​h2⋅1\|v→\|2\.d^\{2\}=\\frac\{\\left\(\\dfrac\{\(R\-r\)SRr\}\{2h\}\\right\)^\{2\}\}\{\|\\vec\{v\}\|^\{2\}\}=\\frac\{\(R\-r\)^\{2\}S^\{2\}R^\{2\}r^\{2\}\}\{4h^\{2\}\}\\cdot\\frac\{1\}\{\|\\vec\{v\}\|^\{2\}\}\.Substitute\|v→\|2=S​R​rR\+r−1\|\\vec\{v\}\|^\{2\}=\\dfrac\{SRr\}\{R\+r\-1\}:

d2=\(R−r\)2​S2​R2​r24​h2⋅R\+r−1S​R​r=\(R−r\)2​S​R​r​\(R\+r−1\)4​h2\.\(10\)d^\{2\}=\\frac\{\(R\-r\)^\{2\}S^\{2\}R^\{2\}r^\{2\}\}\{4h^\{2\}\}\\cdot\\frac\{R\+r\-1\}\{SRr\}=\\frac\{\(R\-r\)^\{2\}SRr\(R\+r\-1\)\}\{4h^\{2\}\}\.\\quad\\text\{\(10\)\}Now use4​h2=\(R\+r−1\)​S​\(1−q2\)4h^\{2\}=\(R\+r\-1\)S\(1\-q^\{2\}\):

d2=\(R−r\)2​S​R​r​\(R\+r−1\)\(R\+r−1\)​S​\(1−q2\)=\(R−r\)2​R​r1−\(R−r\)2\.\(11\)d^\{2\}=\\frac\{\(R\-r\)^\{2\}SRr\(R\+r\-1\)\}\{\(R\+r\-1\)S\(1\-q^\{2\}\)\}=\\frac\{\(R\-r\)^\{2\}Rr\}\{1\-\(R\-r\)^\{2\}\}\.\\quad\\text\{\(11\)\}
15\. Radius of the circumcircle of△​B​E​F\\triangle BEF

ComputeRc=\|O′−B\|R\_\{c\}=\|O^\{\\prime\}\-B\|, withB=\(a,−h\)B=\(a,\\,\-h\)andO′O^\{\\prime\}as in \(7\)\.

Δ​x′=12−a=12−1−p​q2=p​q2\.\\Delta x^\{\\prime\}=\\frac\{1\}\{2\}\-a=\\frac\{1\}\{2\}\-\\frac\{1\-pq\}\{2\}=\\frac\{pq\}\{2\}\.Δ​y′=yO′\+h=−S​\(R\+r−1\)4​h\+h=4​h2−S​\(R\+r−1\)4​h\.\\Delta y^\{\\prime\}=y\_\{O^\{\\prime\}\}\+h=\-\\frac\{S\(R\+r\-1\)\}\{4h\}\+h=\\frac\{4h^\{2\}\-S\(R\+r\-1\)\}\{4h\}\.Again,4​h2=\(R\+r−1\)​S​\(1−q2\)4h^\{2\}=\(R\+r\-1\)S\(1\-q^\{2\}\),

Δ​y′=\(R\+r−1\)​S​\(1−q2\)−S​\(R\+r−1\)4​h=S​\(R\+r−1\)​\(1−q2−1\)4​h=−S​\(R\+r−1\)​q24​h\.\\Delta y^\{\\prime\}=\\frac\{\(R\+r\-1\)S\(1\-q^\{2\}\)\-S\(R\+r\-1\)\}\{4h\}=\\frac\{S\(R\+r\-1\)\(1\-q^\{2\}\-1\)\}\{4h\}=\-\\frac\{S\(R\+r\-1\)q^\{2\}\}\{4h\}\.Thus

Rc2=\(p​q2\)2\+\(−S​\(R\+r−1\)​q24​h\)2=p2​q24\+S2​\(R\+r−1\)2​q416​h2\.R\_\{c\}^\{2\}=\\left\(\\frac\{pq\}\{2\}\\right\)^\{2\}\+\\left\(\-\\frac\{S\(R\+r\-1\)q^\{2\}\}\{4h\}\\right\)^\{2\}=\\frac\{p^\{2\}q^\{2\}\}\{4\}\+\\frac\{S^\{2\}\(R\+r\-1\)^\{2\}q^\{4\}\}\{16h^\{2\}\}\.Substitute16​h2=4​\(R\+r−1\)​S​\(1−q2\)16h^\{2\}=4\(R\+r\-1\)S\(1\-q^\{2\}\):

Rc2=p2​q24\+S2​\(R\+r−1\)2​q44​\(R\+r−1\)​S​\(1−q2\)=p2​q24\+S​\(R\+r−1\)​q44​\(1−q2\)\.R\_\{c\}^\{2\}=\\frac\{p^\{2\}q^\{2\}\}\{4\}\+\\frac\{S^\{2\}\(R\+r\-1\)^\{2\}q^\{4\}\}\{4\(R\+r\-1\)S\(1\-q^\{2\}\)\}=\\frac\{p^\{2\}q^\{2\}\}\{4\}\+\\frac\{S\(R\+r\-1\)q^\{4\}\}\{4\(1\-q^\{2\}\)\}\.Write as

Rc2=q24​\(p2\+S​\(R\+r−1\)​q21−q2\)\.R\_\{c\}^\{2\}=\\frac\{q^\{2\}\}\{4\}\\left\(p^\{2\}\+\\frac\{S\(R\+r\-1\)q^\{2\}\}\{1\-q^\{2\}\}\\right\)\.NowS​\(R\+r−1\)=\(p\+1\)​\(p−1\)=p2−1\.S\(R\+r\-1\)=\(p\+1\)\(p\-1\)=p^\{2\}\-1\.Therefore

Rc2=q24⋅p2​\(1−q2\)\+\(p2−1\)​q21−q2=q24⋅p2−p2​q2\+p2​q2−q21−q2=q24⋅p2−q21−q2\.R\_\{c\}^\{2\}=\\frac\{q^\{2\}\}\{4\}\\cdot\\frac\{p^\{2\}\(1\-q^\{2\}\)\+\(p^\{2\}\-1\)q^\{2\}\}\{1\-q^\{2\}\}=\\frac\{q^\{2\}\}\{4\}\\cdot\\frac\{p^\{2\}\-p^\{2\}q^\{2\}\+p^\{2\}q^\{2\}\-q^\{2\}\}\{1\-q^\{2\}\}=\\frac\{q^\{2\}\}\{4\}\\cdot\\frac\{p^\{2\}\-q^\{2\}\}\{1\-q^\{2\}\}\.Sincep2−q2=4​R​rp^\{2\}\-q^\{2\}=4Rrandq2=\(R−r\)2q^\{2\}=\(R\-r\)^\{2\},

Rc2=\(R−r\)24⋅4​R​r1−\(R−r\)2=\(R−r\)2​R​r1−\(R−r\)2\.\(12\)R\_\{c\}^\{2\}=\\frac\{\(R\-r\)^\{2\}\}\{4\}\\cdot\\frac\{4Rr\}\{1\-\(R\-r\)^\{2\}\}=\\frac\{\(R\-r\)^\{2\}Rr\}\{1\-\(R\-r\)^\{2\}\}\.\\quad\\text\{\(12\)\}
16\. Conclusion

Comparing \(11\) and \(12\) we obtain

sod=Rcd=R\_\{c\}\(all quantities are nonnegative\)\.

Hence the distance from the circumcenterO′O^\{\\prime\}of△​B​E​F\\triangle BEFto the lineℓ\\ell\(throughHHparallel toA​PAP\) equals the radius of its circumcircle\. Thereforeℓ\\ellis tangent to the circumcircle of△​B​E​F\\triangle BEF\.

#### IMO 2025 Problem 3\.

A functionf:ℕ→ℕf\\colon\\mathbb\{N\}\\to\\mathbb\{N\}is said to be*bonza*iff​\(a\)dividesba−f​\(b\)f​\(a\)f\(a\)\\quad\\text\{divides\}\\quad b^\{a\}\-f\(b\)^\{f\(a\)\}for all positive integersaaandbb\.Determine the smallest real constantccsuch thatf​\(n\)≤c​nf\(n\)\\leq cnfor all bonza functionsffand all positive integersnn\.

Solution

1\. Preliminary lemmas

Lemma 1\.f​\(1\)=1f\(1\)=1\.

*Proof\.*Takinga=b=1a=b=1givesf​\(1\)∣11−f​\(1\)f​\(1\)=1−f​\(1\)f​\(1\)f\(1\)\\mid 1^\{1\}\-f\(1\)^\{f\(1\)\}=1\-f\(1\)^\{f\(1\)\}\.

Sincef​\(1\)∣f​\(1\)f​\(1\)f\(1\)\\mid f\(1\)^\{f\(1\)\}, we havef​\(1\)∣1f\(1\)\\mid 1\. Hencef​\(1\)=1f\(1\)=1\.

Lemma 2\.For every positive integeraa,f​\(a\)∣aaf\(a\)\\mid a^\{a\}\.

*Proof\.*Putb=ab=ain the definition:f​\(a\)∣aa−f​\(a\)f​\(a\)f\(a\)\\mid a^\{a\}\-f\(a\)^\{f\(a\)\}\.

Becausef​\(a\)∣f​\(a\)f​\(a\)f\(a\)\\mid f\(a\)^\{f\(a\)\}, it follows thatf​\(a\)∣aaf\(a\)\\mid a^\{a\}\.

Corollary\.If a primeppdividesf​\(a\)f\(a\), thenppdividesaa\.

*Proof\.*Fromf​\(a\)∣aaf\(a\)\\mid a^\{a\}, any prime divisor off​\(a\)f\(a\)must divideaaa^\{a\}, henceaa\.

2\. Determination off​\(2\)f\(2\),f​\(3\)f\(3\)andf​\(5\)f\(5\)

From Lemma 2 we have:

f​\(2\)∣22=4⟹f​\(2\)∈\{1,2,4\},f\(2\)\\mid 2^\{2\}=4\\quad\\Longrightarrow\\quad f\(2\)\\in\\\{1,2,4\\\},f​\(3\)∣33=27⟹f​\(3\)∈\{1,3,9,27\}\.f\(3\)\\mid 3^\{3\}=27\\quad\\Longrightarrow\\quad f\(3\)\\in\\\{1,3,9,27\\\}\.f​\(5\)∣55=3125⟹f​\(5\)=5δ​with​0≤δ≤5\.f\(5\)\\mid 5^\{5\}=3125\\quad\\Longrightarrow\\quad f\(5\)=5^\{\\delta\}\\ \\text\{with\}\\ 0\\leq\\delta\\leq 5\.Now use the mixed conditions\(a,b\)=\(2,3\)\(a,b\)=\(2,3\)and\(3,2\)\(3,2\):

\(2,3\)\\displaystyle\(2,3\):f​\(2\)∣32−f​\(3\)f​\(2\)=9−f​\(3\)f​\(2\),\\displaystyle:\\ f\(2\)\\mid 3^\{2\}\-f\(3\)^\{f\(2\)\}=9\-f\(3\)^\{f\(2\)\},\(3,2\)\\displaystyle\(3,2\):f​\(3\)∣23−f​\(2\)f​\(3\)=8−f​\(2\)f​\(3\)\.\\displaystyle:\\ f\(3\)\\mid 2^\{3\}\-f\(2\)^\{f\(3\)\}=8\-f\(2\)^\{f\(3\)\}\.Letx=f​\(2\),y=f​\(3\)x=f\(2\),\\ y=f\(3\)\. We test allx∈\{1,2,4\},y∈\{1,3,9,27\}x\\in\\\{1,2,4\\\},\\ y\\in\\\{1,3,9,27\\\}\(note allyyare odd\)\.

⊳\\trianglerightx=1x=1:\(2,3\)\(2,3\)automatic\.\(3,2\)\(3,2\)givesy∣8−1y=7y\\mid 8\-1^\{y\}=7\. Onlyy=1y=1divides 7\. So\(1,1\)\(1,1\)works\.

⊳\\trianglerightx=2x=2:

\(2,3\)\(2,3\):2∣9−y22\\mid 9\-y^\{2\}\. Sinceyyis odd,y2y^\{2\}is odd,9−odd9\-\\text\{odd\}is even \- holds for all oddyy\.

\(3,2\)\(3,2\):y∣8−2yy\\mid 8\-2^\{y\}\.

⊳\\trianglerighty=1y=1:1∣61\\mid 6\- true\.

⊳\\trianglerighty=3y=3:3∣8−8=03\\mid 8\-8=0\- true\.

⊳\\trianglerighty=9y=9:9∣8−512=−5049\\mid 8\-512=\-504\- true \(−504/9=−56\-504/9=\-56\)\.

⊳\\trianglerighty=27y=27: compute227mod272^\{27\}\\bmod 27\.29≡512≡26\(mod27\)2^\{9\}\\equiv 512\\equiv 26\\pmod\{27\}, then227=\(29\)3≡263≡\(−1\)3=−1≡26\(mod27\)2^\{27\}=\(2^\{9\}\)^\{3\}\\equiv 26^\{3\}\\equiv\(\-1\)^\{3\}=\-1\\equiv 26\\pmod\{27\}\. So8−26=−18≢0\(mod27\)8\-26=\-18\\not\\equiv 0\\pmod\{27\}\- false\.

Thus forx=2x=2we have\(2,1\),\(2,3\),\(2,9\)\(2,1\),\\ \(2,3\),\\ \(2,9\)\.

⊳\\trianglerightx=4x=4:

\(2,3\)\(2,3\):4∣9−y44\\mid 9\-y^\{4\}\. For oddyy,y4≡1\(mod4\)y^\{4\}\\equiv 1\\pmod\{4\}, so9−1=89\-1=8divisible by 4 \- holds for all oddyy\.\(3,2\)\(3,2\):y∣8−4yy\\mid 8\-4^\{y\}\.

⊳\\trianglerighty=1y=1:1∣8−4=41\\mid 8\-4=4\- true\.

⊳\\trianglerighty=3y=3:3∣8−64=−563\\mid 8\-64=\-56\- false\.

⊳\\trianglerighty=9y=9:43≡64≡1\(mod9\)4^\{3\}\\equiv 64\\equiv 1\\pmod\{9\}, so49≡1\(mod9\)4^\{9\}\\equiv 1\\pmod\{9\}, then8−1=7≢0\(mod9\)8\-1=7\\not\\equiv 0\\pmod\{9\}\- false\.

⊳\\trianglerighty=27y=27:43≡64≡10\(mod27\)4^\{3\}\\equiv 64\\equiv 10\\pmod\{27\}, so49≡103=1000≡1\(mod27\)4^\{9\}\\equiv 10^\{3\}=1000\\equiv 1\\pmod\{27\}, thus427≡1\(mod27\)4^\{27\}\\equiv 1\\pmod\{27\},8−1=7≢0\(mod27\)8\-1=7\\not\\equiv 0\\pmod\{27\}\- false\.

Hence only\(4,1\)\(4,1\)works\.

Therefore after\(2,3\)\(2,3\)and\(3,2\)\(3,2\)the possible pairs are

\(1,1\),\(2,1\),\(2,3\),\(2,9\),\(4,1\)\.\(1\)\(1,1\),\\ \(2,1\),\\ \(2,3\),\\ \(2,9\),\\ \(4,1\)\.\\quad\\text\{\(1\)\}Now incorporatef​\(5\)=5δf\(5\)=5^\{\\delta\}\. For each candidate pair we must satisfy

\(2,5\)\\displaystyle\(2,5\):x∣25−5δ​x,\\displaystyle:\\ x\\mid 5\-5^\{\\delta x\},\(3,5\)\\displaystyle\(3,5\):y∣125−5δ​y,\\displaystyle:\\ y\\mid 25\-5^\{\\delta y\},\(5,2\)\\displaystyle\(5,2\):5δ∣32−x5δ,\\displaystyle:5^\{\\delta\}\\mid 2\-x^\{5^\{\\delta\}\},\(5,3\)\\displaystyle\(5,3\):5δ∣243−y5δ\.\\displaystyle:5^\{\\delta\}\\mid 43\-y^\{5^\{\\delta\}\}\.We testδ=0,1,2,3,4,5\\delta=0,1,2,3,4,5for each pair\.

1\. Pair\(x,y\)=\(1,1\)\(x,y\)=\(1,1\)\.

⊳\\triangleright\(2,5\)\(2,5\)automatic\.

⊳\\triangleright\(3,5\)\(3,5\)automatic\.

⊳\\triangleright\(5,2\)\(5,2\):5δ∣32−15δ=315^\{\\delta\}\\mid 32\-1^\{5^\{\\delta\}\}=31\. So5δ∣315^\{\\delta\}\\mid 31forcesδ=0\\delta=0\.

⊳\\triangleright\(5,3\)\(5,3\):5δ∣243−15δ=2425^\{\\delta\}\\mid 243\-1^\{5^\{\\delta\}\}=242\. Again onlyδ=0\\delta=0works\.

⇒\(f​\(2\),f​\(3\),f​\(5\)\)=\(1,1,1\)\\Rightarrow\(f\(2\),f\(3\),f\(5\)\)=\(1,1,1\)\.

2\. Pair\(x,y\)=\(2,1\)\(x,y\)=\(2,1\)\.

⊳\\triangleright\(2,5\)\(2,5\):2∣25−52​δ2\\mid 25\-5^\{2\\delta\}\- difference even, automatic\.

⊳\\triangleright\(3,5\)\(3,5\):11divides everything\.

⊳\\triangleright\(5,3\)\(5,3\):5δ∣243−1=2425^\{\\delta\}\\mid 243\-1=242\. Since5∤2425\\nmid 242, onlyδ=0\\delta=0\.

⊳\\triangleright\(5,2\)\(5,2\)withδ=0\\delta=0:1∣32−2=301\\mid 32\-2=30\- true\.

⇒\(2,1,1\)\\Rightarrow\(2,1,1\)\.

3\. Pair\(x,y\)=\(4,1\)\(x,y\)=\(4,1\)\.

⊳\\triangleright\(2,5\)\(2,5\):4∣25−52​δ4\\mid 25\-5^\{2\\delta\}\. Mod 4:25≡125\\equiv 1,52​δ≡15^\{2\\delta\}\\equiv 1, difference divisible by 4 \- automatic\.

⊳\\triangleright\(3,5\)\(3,5\): automatic\.

⊳\\triangleright\(5,3\)\(5,3\):5δ∣2425^\{\\delta\}\\mid 242\- forcesδ=0\\delta=0\.

⊳\\triangleright\(5,2\)\(5,2\)withδ=0\\delta=0:1∣32−4=281\\mid 32\-4=28\- true\.

⇒\(4,1,1\)\\Rightarrow\(4,1,1\)\.

4\. Pair\(x,y\)=\(2,3\)\(x,y\)=\(2,3\)\.

⊳\\triangleright\(2,5\)\(2,5\): automatic \(even difference\)\.

⊳\\triangleright\(3,5\)\(3,5\):3∣125−53​δ3\\mid 125\-5^\{3\\delta\}\.

⊳\\triangleright\(5,2\)\(5,2\):5δ∣32−25δ5^\{\\delta\}\\mid 32\-2^\{5^\{\\delta\}\}\.

⊳\\triangleright\(5,3\)\(5,3\):5δ∣243−35δ5^\{\\delta\}\\mid 243\-3^\{5^\{\\delta\}\}\.

Testδ\\delta:

⊳\\trianglerightδ=0\\delta=0:f​\(5\)=1f\(5\)=1\. Then3∣125−1=1243\\mid 125\-1=124\- false\.

⊳\\trianglerightδ=1\\delta=1:f​\(5\)=5f\(5\)=5\.

⊳\\triangleright3∣125−53=03\\mid 125\-5^\{3\}=0\- true\.

⊳\\triangleright5∣32−25=05\\mid 32\-2^\{5\}=0\- true\.

⊳\\triangleright5∣243−35=05\\mid 243\-3^\{5\}=0\- true\.

Soδ=1\\delta=1works\.

⊳\\trianglerightδ=2\\delta=2:f​\(5\)=25f\(5\)=25\.3∣125−563\\mid 125\-5^\{6\}\. Mod 3:125≡2125\\equiv 2,56≡\(53\)2≡22=4≡15^\{6\}\\equiv\(5^\{3\}\)^\{2\}\\equiv 2^\{2\}=4\\equiv 1, so difference2−1=12\-1=1\- false\.

⊳\\trianglerightδ=3\\delta=3:f​\(5\)=125f\(5\)=125\.

⊳\\triangleright3∣125−593\\mid 125\-5^\{9\}:59≡53≡125≡2\(mod3\)5^\{9\}\\equiv 5^\{3\}\\equiv 125\\equiv 2\\pmod\{3\}, difference0\- true\.

⊳\\trianglerightCheck\(5,2\)\(5,2\):125∣32−2125125\\mid 32\-2^\{125\}\. Compute2125mod1252^\{125\}\\bmod 125:φ​\(125\)=100\\varphi\(125\)=100, so2125≡225\(mod125\)2^\{125\}\\equiv 2^\{25\}\\pmod\{125\}\.25=322^\{5\}=32,210≡242^\{10\}\\equiv 24,220≡242=576≡576−5⋅125=762^\{20\}\\equiv 24^\{2\}=576\\equiv 576\-5\\cdot 125=76, then225=220⋅25≡76⋅32=2432≡2432−19⋅125=572^\{25\}=2^\{20\}\\cdot 2^\{5\}\\equiv 76\\cdot 32=2432\\equiv 2432\-19\\cdot 125=57\. So32−57=−2532\-57=\-25not divisible by 125 \- false\.

⊳\\trianglerightδ=4\\delta=4:f​\(5\)=625f\(5\)=625\.3∣125−5123\\mid 125\-5^\{12\}\.512≡\(56\)2≡12=1\(mod3\)5^\{12\}\\equiv\(5^\{6\}\)^\{2\}\\equiv 1^\{2\}=1\\pmod\{3\},125≡2125\\equiv 2, difference11\- false\.

⊳\\trianglerightδ=5\\delta=5:f​\(5\)=3125f\(5\)=3125\.

⊳\\triangleright3∣125−5153\\mid 125\-5^\{15\}:515≡53≡125≡2\(mod3\)5^\{15\}\\equiv 5^\{3\}\\equiv 125\\equiv 2\\pmod\{3\}, difference0\- true\.

⊳\\triangleright\(5,2\)\(5,2\):3125∣32−231253125\\mid 32\-2^\{3125\}\. Mod 125:3125≡25\(mod100\)3125\\equiv 25\\pmod\{100\}, so23125≡225≡57\(mod125\)2^\{3125\}\\equiv 2^\{25\}\\equiv 57\\pmod\{125\}as before,32−57=−2532\-57=\-25not divisible by 125 \- false\.

Thus onlyδ=1\\delta=1works, giving\(2,3,5\)\(2,3,5\)\.

5\. Pair\(x,y\)=\(2,9\)\(x,y\)=\(2,9\)\.

Testδ\\delta:

⊳\\trianglerightδ=0\\delta=0:3∣125−1=1243\\mid 125\-1=124\- false\.

⊳\\trianglerightδ=1\\delta=1:5∣125−595\\mid 125\-5^\{9\}\.59mod95^\{9\}\\bmod 9:56≡1\(mod9\)5^\{6\}\\equiv 1\\pmod\{9\}\(sinceφ​\(9\)=6\\varphi\(9\)=6\), so59≡53=125≡8\(mod9\)5^\{9\}\\equiv 5^\{3\}=125\\equiv 8\\pmod\{9\}\.125≡8125\\equiv 8, difference0\- true\.

\(5,2\)\(5,2\):5∣32−25=05\\mid 32\-2^\{5\}=0\- true\.\(5,3\)\(5,3\):5∣243−955\\mid 243\-9^\{5\}\.9≡4\(mod5\)9\\equiv 4\\pmod\{5\},45=1024≡4\(mod5\)4^\{5\}=1024\\equiv 4\\pmod\{5\}\.243≡3\(mod5\)243\\equiv 3\\pmod\{5\}, so3−4=−1≢0\(mod5\)3\-4=\-1\\not\\equiv 0\\pmod\{5\}\- false\.

⊳\\trianglerightδ=2\\delta=2:25∣125−51825\\mid 125\-5^\{18\}\.518≡\(56\)3≡13=1\(mod9\)5^\{18\}\\equiv\(5^\{6\}\)^\{3\}\\equiv 1^\{3\}=1\\pmod\{9\},125≡8125\\equiv 8, difference77\- false\.

⊳\\trianglerightδ=3\\delta=3:125∣125−527125\\mid 125\-5^\{27\}\.527≡53≡8\(mod9\)5^\{27\}\\equiv 5^\{3\}\\equiv 8\\pmod\{9\}, difference0\- true\.

\(5,2\)\(5,2\):125∣32−2125125\\mid 32\-2^\{125\}\. As before,2125≡57\(mod125\)2^\{125\}\\equiv 57\\pmod\{125\},32−57=−2532\-57=\-25\- false\.

⊳\\trianglerightδ=4\\delta=4:625∣125−536625\\mid 125\-5^\{36\}\.36≡0\(mod6\)36\\equiv 0\\pmod\{6\}, so536≡1\(mod9\)5^\{36\}\\equiv 1\\pmod\{9\}, difference77\- false\.

⊳\\trianglerightδ=5\\delta=5:3125∣125−5453125\\mid 125\-5^\{45\}\.45≡3\(mod6\)45\\equiv 3\\pmod\{6\},545≡8\(mod9\)5^\{45\}\\equiv 8\\pmod\{9\}, difference0\- true\.

\(5,2\)\(5,2\):3125∣32−231253125\\mid 32\-2^\{3125\}\. Mod 125 fails as before\.

Hence noδ\\deltasatisfies all conditions\.

Conclusion of classification\.The only admissible triples\(f​\(2\),f​\(3\),f​\(5\)\)\(f\(2\),f\(3\),f\(5\)\)are

\(1,1,1\),\(2,1,1\),\(4,1,1\),\(2,3,5\)\.\(1,1,1\),\\quad\(2,1,1\),\\quad\(4,1,1\),\\quad\(2,3,5\)\.We will analyze two families:

⊳\\trianglerightFamily A:those withf​\(3\)=1f\(3\)=1, i\.e\. the first three triples\.

⊳\\trianglerightFamily B:the triple\(2,3,5\)\(2,3,5\)\.

3\. Analysis of Family A \(f​\(3\)=1f\(3\)=1\)

Lemma 3\.Iff​\(3\)=1f\(3\)=1, thenf​\(q\)=1f\(q\)=1for every odd primeqq\.

*Proof\.*From\(q,q\)\(q,q\)we havef​\(q\)∣qqf\(q\)\\mid q^\{q\}, sof​\(q\)=qkf\(q\)=q^\{k\}for some0≤k≤q0\\leq k\\leq q\.

Consider\(q,3\)\(q,3\):

f​\(q\)∣3q−f​\(3\)f​\(q\)=3q−1f​\(q\)=3q−1\.f\(q\)\\mid 3^\{q\}\-f\(3\)^\{f\(q\)\}=3^\{q\}\-1^\{f\(q\)\}=3^\{q\}\-1\.Ifk≥1k\\geq 1, thenq​∣f​\(q\)∣​3q−1q\\mid f\(q\)\\mid 3^\{q\}\-1\. By Fermat’s little theorem,3q≡3\(modq\)3^\{q\}\\equiv 3\\pmod\{q\}, soq∣3−1=2q\\mid 3\-1=2\. Henceq=2q=2, contradicting thatqqis odd\. Thereforek=0k=0andf​\(q\)=1f\(q\)=1\.

Lemma 4\.In Family A, for every positive integernn,f​\(n\)f\(n\)is a power of 2\.

*Proof\.*Suppose an odd primerrdividesf​\(n\)f\(n\)\. By Lemma 2,r∣nr\\mid n\. Now use\(n,r\)\(n,r\):

f​\(n\)∣rn−f​\(r\)f​\(n\)\.f\(n\)\\mid r^\{n\}\-f\(r\)^\{f\(n\)\}\.Sincerris odd, Lemma 3 givesf​\(r\)=1f\(r\)=1\. Hence

f​\(n\)∣rn−1\.f\(n\)\\mid r^\{n\}\-1\.In particular,r∣rn−1r\\mid r^\{n\}\-1, i\.e\.r∣−1r\\mid\-1, contradiction\. Thus no odd prime dividesf​\(n\)f\(n\); i\.e\.f​\(n\)=2e​\(n\)f\(n\)=2^\{e\(n\)\}for somee​\(n\)≥0e\(n\)\\geq 0\.

Lemma 5\.For anynn,2e​\(n\)∣3n−12^\{e\(n\)\}\\mid 3^\{n\}\-1\.

*Proof\.*Take\(n,3\)\(n,3\):

f​\(n\)∣3n−f​\(3\)f​\(n\)=3n−1f​\(n\)=3n−1\.f\(n\)\\mid 3^\{n\}\-f\(3\)^\{f\(n\)\}=3^\{n\}\-1^\{f\(n\)\}=3^\{n\}\-1\.Sincef​\(n\)=2e​\(n\)f\(n\)=2^\{e\(n\)\}, we have2e​\(n\)∣3n−12^\{e\(n\)\}\\mid 3^\{n\}\-1\.

Lemma 6\.Forn∈ℕn\\in\\mathbb\{N\},

v2​\(3n−1\)=\{1if​n​is odd,v2​\(n\)\+2if​n​is even\.v\_\{2\}\(3^\{n\}\-1\)=\\begin\{cases\}1&\\text\{if \}n\\text\{ is odd\},\\\\ v\_\{2\}\(n\)\+2&\\text\{if \}n\\text\{ is even\}\.\\end\{cases\}*Proof\.*

*Oddnn:*Write

3n−1=\(3−1\)​\(3n−1\+3n−2\+⋯\+1\)=2⋅\(odd sum\),3^\{n\}\-1=\(3\-1\)\(3^\{n\-1\}\+3^\{n\-2\}\+\\cdots\+1\)=2\\cdot\(\\text\{odd sum\}\),because there arennterms, each odd, andnnodd⇒\\Rightarrowthe sum is odd\. Hencev2=1v\_\{2\}=1\.

*Evennn:*Writen=2α​mn=2^\{\\alpha\}mwithmmodd,α≥1\\alpha\\geq 1\. We prove by induction onα\\alphathatv2​\(32α​m−1\)=α\+2v\_\{2\}\(3^\{2^\{\\alpha\}m\}\-1\)=\\alpha\+2\.

Baseα=1\\alpha=1:n=2​mn=2m\. Then

32​m−1=\(3m−1\)​\(3m\+1\)\.3^\{2m\}\-1=\(3^\{m\}\-1\)\(3^\{m\}\+1\)\.We knowv2​\(3m−1\)=1v\_\{2\}\(3^\{m\}\-1\)=1\(sincemmodd\)\. For3m\+13^\{m\}\+1, note3m≡3\(mod8\)3^\{m\}\\equiv 3\\pmod\{8\}whenmmis odd, so3m\+1≡4\(mod8\)3^\{m\}\+1\\equiv 4\\pmod\{8\}, thusv2​\(3m\+1\)=2v\_\{2\}\(3^\{m\}\+1\)=2\. Hence

v2​\(32​m−1\)=1\+2=3=1\+2=α\+2\.v\_\{2\}\(3^\{2m\}\-1\)=1\+2=3=1\+2=\\alpha\+2\.Inductive step:Assume true for a givenα≥1\\alpha\\geq 1\. Writen′=2α\+1​m=2⋅\(2α​m\)n^\{\\prime\}=2^\{\\alpha\+1\}m=2\\cdot\(2^\{\\alpha\}m\)\. Letk=2α​mk=2^\{\\alpha\}m\. Then

3n′−1=\(3k−1\)​\(3k\+1\)\.3^\{n^\{\\prime\}\}\-1=\(3^\{k\}\-1\)\(3^\{k\}\+1\)\.By hypothesis,v2​\(3k−1\)=α\+2v\_\{2\}\(3^\{k\}\-1\)=\\alpha\+2\.

Sincekkis even \(becauseα≥1\\alpha\\geq 1\),3k≡1\(mod4\)3^\{k\}\\equiv 1\\pmod\{4\}, so3k\+1≡2\(mod4\)3^\{k\}\+1\\equiv 2\\pmod\{4\}, givingv2​\(3k\+1\)=1v\_\{2\}\(3^\{k\}\+1\)=1\.

Thus

v2​\(3n′−1\)=\(α\+2\)\+1=α\+3=\(α\+1\)\+2\.v\_\{2\}\(3^\{n^\{\\prime\}\}\-1\)=\(\\alpha\+2\)\+1=\\alpha\+3=\(\\alpha\+1\)\+2\.This completes the induction\.

From Lemmas 5 and 6 we obtain

e​\(n\)≤\{1if​n​is odd,v2​\(n\)\+2if​n​is even\.e\(n\)\\leq\\begin\{cases\}1&\\text\{if \}n\\text\{ is odd\},\\\\ v\_\{2\}\(n\)\+2&\\text\{if \}n\\text\{ is even\}\.\\end\{cases\}Consequently

f​\(n\)=2e​\(n\)≤\{2if​n​odd,2v2​\(n\)\+2if​n​even\.f\(n\)=2^\{e\(n\)\}\\leq\\begin\{cases\}2&\\text\{if \}n\\text\{ odd\},\\\\ 2^\{v\_\{2\}\(n\)\+2\}&\\text\{if \}n\\text\{ even\}\.\\end\{cases\}For evennn,2v2​\(n\)\+2=4⋅2v2​\(n\)≤4​n2^\{v\_\{2\}\(n\)\+2\}=4\\cdot 2^\{v\_\{2\}\(n\)\}\\leq 4nbecause2v2​\(n\)∣n2^\{v\_\{2\}\(n\)\}\\mid nimplies2v2​\(n\)≤n2^\{v\_\{2\}\(n\)\}\\leq n\. For oddnn,2≤4​n2\\leq 4n\(sincen≥1n\\geq 1\)\.

Hencein Family Awe have

f​\(n\)≤4​nfor all​n∈ℕ\.\\boxed\{f\(n\)\\leq 4n\\quad\\text\{for all \}n\\in\\mathbb\{N\}\}\.
4\. Analysis of Family B \(f​\(2\)=2,f​\(3\)=3,f​\(5\)=5f\(2\)=2,\\ f\(3\)=3,\\ f\(5\)=5\)

Proposition 7\.For every primepp,f​\(p\)=pf\(p\)=p\.

*Proof\.*We proceed by strong induction onpp\. The base casesp=2,3,5p=2,3,5are given\.

Assumep\>5p\>5and that for every primeq<pq<pwe already knowf​\(q\)=qf\(q\)=q\.

From\(p,p\)\(p,p\)we havef​\(p\)∣ppf\(p\)\\mid p^\{p\}, sof​\(p\)=pkf\(p\)=p^\{k\}for some integerkkwith0≤k≤p0\\leq k\\leq p\. We show thatkkmust be 1\.

Step 1 \- Eliminatek=0k=0andk=2k=2\.

Take any primeq<pq<p\. Consider\(q,p\)\(q,p\):

f​\(q\)=q∣pq−f​\(p\)q=pq−pk​q\.f\(q\)=q\\mid p^\{q\}\-f\(p\)^\{q\}=p^\{q\}\-p^\{kq\}\.Factor outpqp^\{q\}:

q∣pq​\(1−p\(k−1\)​q\)\.q\\mid p^\{q\}\(1\-p^\{\(k\-1\)q\}\)\.Sinceq≠pq\\neq p,q∤pqq\\nmid p^\{q\}\. Hence

q∣1−p\(k−1\)​q\.q\\mid 1\-p^\{\(k\-1\)q\}\.By Fermat’s little theorem,pq≡p\(modq\)p^\{q\}\\equiv p\\pmod\{q\}, and more generallyp\(k−1\)​q≡pk−1\(modq\)p^\{\(k\-1\)q\}\\equiv p^\{k\-1\}\\pmod\{q\}\(raising to theqqth power is the identity moduloqq\)\. Thus

1−pk−1≡0\(modq\)⟹pk−1≡1\(modq\)\.\(2\)1\-p^\{k\-1\}\\equiv 0\\pmod\{q\}\\quad\\Longrightarrow\\quad p^\{k\-1\}\\equiv 1\\pmod\{q\}\.\\quad\\text\{\(2\)\}
⊳\\trianglerightIfk=0k=0, then the original condition\(q,p\)\(q,p\)givesq∣pq−1q\\mid p^\{q\}\-1\. Using FLT,pq≡p\(modq\)p^\{q\}\\equiv p\\pmod\{q\}, sop≡1\(modq\)p\\equiv 1\\pmod\{q\}\.

⊳\\trianglerightIfk=2k=2, \(2\) givesp1≡1\(modq\)p^\{1\}\\equiv 1\\pmod\{q\}, i\.e\.p≡1\(modq\)p\\equiv 1\\pmod\{q\}\.

Thus for bothk=0k=0andk=2k=2we have

p≡1\(modq\)for every primeq<p\.\(3\)p\\equiv 1\\pmod\{q\}\\qquad\\text\{for every prime \}q<p\.\\quad\\text\{\(3\)\}Now byBertrand’s postulate\(for any integerm\>1m\>1there exists a prime betweenmmand2​m2m\), forp\>3p\>3there exists a primeqqwithp2<q<p\\frac\{p\}\{2\}<q<p\. From \(3\),q∣p−1q\\mid p\-1\. Butq\>p/2q\>p/2andq≤p−1q\\leq p\-1\(sinceq<pq<p\)\. The only positive multiple ofqqthat is at mostp−1p\-1isqqitself \(because2​q\>p2q\>p\)\. Hencep−1=qp\-1=q, sop=q\+1p=q\+1, which is even \- contradicting thatp\>2p\>2is odd\. Thereforekkcannot be0or22\.

Step 2 \- Eliminatek≥3k\\geq 3\.

Consider the pair\(p,p−1\)\(p,\\,p\-1\):

f​\(p\)=pk∣\(p−1\)p−f​\(p−1\)pk\.f\(p\)=p^\{k\}\\mid\(p\-1\)^\{p\}\-f\(p\-1\)^\{p^\{k\}\}\.Letr=f​\(p−1\)r=f\(p\-1\)\.

⊳\\trianglerightModulopp:

\(p−1\)p≡\(−1\)p=−1\(modp\),\(p\-1\)^\{p\}\\equiv\(\-1\)^\{p\}=\-1\\pmod\{p\},and by Fermat’s little theorem,rpk≡r\(modp\)r^\{p^\{k\}\}\\equiv r\\pmod\{p\}\(sincerp≡r\(modp\)r^\{p\}\\equiv r\\pmod\{p\}, and then by inductionrpk≡r\(modp\)r^\{p^\{k\}\}\\equiv r\\pmod\{p\}\.

Sincepk∣D:=\(p−1\)p−rpkp^\{k\}\\mid D:=\(p\-1\)^\{p\}\-r^\{p^\{k\}\}, in particularp∣Dp\\mid D\. Reducing moduloppgives

−1≡r\(modp\),i\.e\.r≡−1\(modp\)\.\(4\)\-1\\equiv r\\pmod\{p\},\\qquad\\text\{i\.e\.\}\\quad r\\equiv\-1\\pmod\{p\}\.\\quad\\text\{\(4\)\}Writer=−1\+p​tr=\-1\+ptfor some integertt\.

⊳\\trianglerightExpansion of\(p−1\)p\(p\-1\)^\{p\}:

By the binomial theorem,

\(p−1\)p=∑i=0p\(pi\)​pi​\(−1\)p−i\.\(p\-1\)^\{p\}=\\sum\_\{i=0\}^\{p\}\\binom\{p\}\{i\}p^\{i\}\(\-1\)^\{p\-i\}\.The term withi=0i=0is\(−1\)p=−1\(\-1\)^\{p\}=\-1\. The term withi=1i=1is\(p1\)​p1​\(−1\)p−1=p⋅p⋅1=p2\\binom\{p\}\{1\}p^\{1\}\(\-1\)^\{p\-1\}=p\\cdot p\\cdot 1=p^\{2\}\.

Fori≥2i\\geq 2,\(pi\)\\binom\{p\}\{i\}is divisible bypp, and together withpip^\{i\}yields a factorpi\+1≥p3p^\{i\+1\}\\geq p^\{3\}\. Thus we can write

\(p−1\)p=−1\+p2​A,\(p\-1\)^\{p\}=\-1\+p^\{2\}A,whereAAis an integer and, because thei=1i=1term contributes exactlyp2p^\{2\}and the higher terms are multiples ofp3p^\{3\}, we haveA≡1\(modp\)A\\equiv 1\\pmod\{p\}\(in particularp∤Ap\\nmid A\)\.

⊳\\trianglerightExpansion ofrpkr^\{p^\{k\}\}:

Writer=−1\+p​tr=\-1\+pt\. Then

rpk=∑j=0pk\(pkj\)​\(−1\)pk−j​\(p​t\)j\.r^\{p^\{k\}\}=\\sum\_\{j=0\}^\{p^\{k\}\}\\binom\{p^\{k\}\}\{j\}\(\-1\)^\{p^\{k\}\-j\}\(pt\)^\{j\}\.Sincepkp^\{k\}is odd,\(−1\)pk=−1\(\-1\)^\{p^\{k\}\}=\-1\.

⊳\\trianglerightj=0j=0: term =−1\-1\.

⊳\\trianglerightj=1j=1:\(pk1\)​\(−1\)pk−1​p​t=pk⋅1⋅p​t=pk\+1​t\\binom\{p^\{k\}\}\{1\}\(\-1\)^\{p^\{k\}\-1\}pt=p^\{k\}\\cdot 1\\cdot pt=p^\{k\+1\}t\.

⊳\\trianglerightForj≥2j\\geq 2: we show that each term is divisible bypk\+2p^\{k\+2\}\.

Indeed, forj≥2j\\geq 2,

vp​\(\(pkj\)\)=k−vp​\(j\),v\_\{p\}\\Bigl\(\\binom\{p^\{k\}\}\{j\}\\Bigr\)=k\-v\_\{p\}\(j\),hence

vp​\(\(pkj\)​pj\)=\(k−vp​\(j\)\)\+j=k\+j−vp​\(j\)≥k\+2,v\_\{p\}\\biggl\(\\binom\{p^\{k\}\}\{j\}p^\{j\}\\biggr\)=\(k\-v\_\{p\}\(j\)\)\+j=k\+j\-v\_\{p\}\(j\)\\geq k\+2,becausej≥2j\\geq 2and ifp∤jp\\nmid jthenj≥2j\\geq 2andvp​\(j\)=0v\_\{p\}\(j\)=0; ifp∣jp\\mid jthenj≥pj\\geq pandvp​\(j\)≥1v\_\{p\}\(j\)\\geq 1, soj−vp​\(j\)≥j−\(j−1\)=1j\-v\_\{p\}\(j\)\\geq j\-\(j\-1\)=1? Wait, need to be precise: For anyjjwith2≤j≤pk−12\\leq j\\leq p^\{k\}\-1, we have

vp​\(\(pkj\)\)=k−vp​\(j\)v\_\{p\}\\Bigl\(\\binom\{p^\{k\}\}\{j\}\\Bigr\)=k\-v\_\{p\}\(j\)\(a standard property for prime powers\)\. Then

vp​\(\(pkj\)​pj\)=k−vp​\(j\)\+j=k\+j−vp​\(j\)\.v\_\{p\}\\Bigl\(\\binom\{p^\{k\}\}\{j\}p^\{j\}\\Bigr\)=k\-v\_\{p\}\(j\)\+j=k\+j\-v\_\{p\}\(j\)\.Sincej≥2j\\geq 2andvp​\(j\)≤logp⁡j<jv\_\{p\}\(j\)\\leq\\log\_\{p\}j<j, but we need at leastk\+2k\+2\. The minimal value occurs whenj=2j=2andp∤2p\\nmid 2\(sincep≥3p\\geq 3\), thenvp​\(j\)=0v\_\{p\}\(j\)=0, givingk\+2k\+2\. Forj=pj=pwe havevp​\(j\)=1v\_\{p\}\(j\)=1, thenk\+p−1≥k\+2k\+p\-1\\geq k\+2becausep≥3p\\geq 3\. So indeed each term forj≥2j\\geq 2is divisible bypk\+2p^\{k\+2\}\.

Therefore

rpk=−1\+pk\+1​t\+pk\+2​Sr^\{p^\{k\}\}=\-1\+p^\{k\+1\}t\+p^\{k\+2\}Sfor some integerSS\.

⊳\\trianglerightForm the difference:

D=\(p−1\)p−rpk=\(−1\+p2​A\)−\(−1\+pk\+1​t\+pk\+2​S\)=p2​A−pk\+1​t−pk\+2​S\.D=\(p\-1\)^\{p\}\-r^\{p^\{k\}\}=\(\-1\+p^\{2\}A\)\-\(\-1\+p^\{k\+1\}t\+p^\{k\+2\}S\)=p^\{2\}A\-p^\{k\+1\}t\-p^\{k\+2\}S\.Factorp2p^\{2\}:

D=p2\(A−pk−1t−pkS\)\.\(5\)D=p^\{2\}\\bigl\(A\-p^\{k\-1\}t\-p^\{k\}S\\bigr\)\.\\quad\\text\{\(5\)\}Sincek≥3k\\geq 3, we havek−1≥2k\-1\\geq 2, sopk−1​tp^\{k\-1\}tandpk​Sp^\{k\}Sare both multiples ofp2p^\{2\}\. Moreover,

A−pk−1​t−pk​S≡A\(modp\),A\-p^\{k\-1\}t\-p^\{k\}S\\equiv A\\pmod\{p\},andp∤Ap\\nmid A\. Hence the expression in parentheses isnotdivisible bypp, and we obtain

However, we requirevp​\(D\)≥k≥3v\_\{p\}\(D\)\\geq k\\geq 3\(becausepk∣Dp^\{k\}\\mid D\)\. This contradiction shows thatk≥3k\\geq 3is impossible\.

Sincekkcannot be0,22, or≥3\\geq 3, the only remaining possibility isk=1k=1\. Thusf​\(p\)=pf\(p\)=p\.

Proposition 8\.In Family B,f​\(n\)=nf\(n\)=nfor all positive integersnn\.

*Proof\.*Fix anynnand any primepp\. Using the bonza condition witha=pa=p,b=nb=n:

f​\(p\)=p∣np−f​\(n\)p\.f\(p\)=p\\mid n^\{p\}\-f\(n\)^\{p\}\.By Fermat’s little theorem,

np≡n\(modp\),f​\(n\)p≡f​\(n\)\(modp\)\.n^\{p\}\\equiv n\\pmod\{p\},\\qquad f\(n\)^\{p\}\\equiv f\(n\)\\pmod\{p\}\.Therefore

This holds foreveryprimepp\. The only integer divisible by all primes is 0, son−f​\(n\)=0n\-f\(n\)=0, i\.e\.f​\(n\)=nf\(n\)=n\.

Hence in Family B we havef​\(n\)=n≤4​nf\(n\)=n\\leq 4nfor allnn\.

5\. Upper bound summary

We have shown:

⊳\\trianglerightIn Family A,f​\(n\)≤4​nf\(n\)\\leq 4nfor allnn\.

⊳\\trianglerightIn Family B,f​\(n\)=n≤4​nf\(n\)=n\\leq 4n\.

Thereforefor every bonza functionffand everyn∈ℕn\\in\\mathbb\{N\},

f​\(n\)≤4​n\.\\boxed\{f\(n\)\\leq 4n\}\.
6\. Sharpness: a bonza function attaining4​n4n

Define a functionh:ℕ→ℕh:\\mathbb\{N\}\\to\\mathbb\{N\}by

h​\(n\)=\{1if​n​is odd,4if​n=2,2v2​\(n\)\+2if​n​is even and​n\>2\.h\(n\)=\\begin\{cases\}1&\\text\{if \}n\\text\{ is odd\},\\\\\[2\.0pt\] 4&\\text\{if \}n=2,\\\\\[2\.0pt\] 2^\{v\_\{2\}\(n\)\+2\}&\\text\{if \}n\\text\{ is even and \}n\>2\.\\end\{cases\}We verify thathhis bonza, i\.e\.,

h​\(a\)∣ba−h​\(b\)h​\(a\)for all​a,b∈ℕ\.h\(a\)\\mid b^\{a\}\-h\(b\)^\{h\(a\)\}\\quad\\text\{for all \}a,b\\in\\mathbb\{N\}\.
Verification

We consider cases based onaa\.

⊳\\trianglerightCase 1:aais odd\.

Thenh​\(a\)=1h\(a\)=1\(by definition: for oddn\>1n\>1,h​\(n\)=1h\(n\)=1; anda=1a=1gives11\)\. Since11divides any integer, the condition holds trivially\.

⊳\\trianglerightCase 2:a=2a=2\.

Hereh​\(2\)=4h\(2\)=4\. We must show

4∣b2−h​\(b\)4for all​b\.4\\mid b^\{2\}\-h\(b\)^\{4\}\\quad\\text\{for all \}b\.
⊳\\trianglerightIfbbis odd:b2b^\{2\}is odd,h​\(b\)=1h\(b\)=1, soh​\(b\)4=1h\(b\)^\{4\}=1\. Thenb2−1b^\{2\}\-1is divisible by 4 because for an oddbb,b2≡1\(mod8\)b^\{2\}\\equiv 1\\pmod\{8\}, hence certainly by 4\.

⊳\\trianglerightIfb=2b=2:b2=4b^\{2\}=4,h​\(b\)=4h\(b\)=4,h​\(b\)4=44=256h\(b\)^\{4\}=4^\{4\}=256\. Then4−256=−2524\-256=\-252, and−252\-252is divisible by 4\.

⊳\\trianglerightIfbbis even andb\>2b\>2: writeb=2β​cb=2^\{\\beta\}cwithβ≥1\\beta\\geq 1,ccodd\. Thenh​\(b\)=2β\+2h\(b\)=2^\{\\beta\+2\}\(sinceb\>2b\>2even\)\. Thenh​\(b\)4=24​\(β\+2\)=24​β\+8h\(b\)^\{4\}=2^\{4\(\\beta\+2\)\}=2^\{4\\beta\+8\}, which is divisible by 4\. Moreover,b2b^\{2\}is divisible by 4 \(sincebbis even\)\. Thusb2−h​\(b\)4b^\{2\}\-h\(b\)^\{4\}is a difference of two multiples of 4, hence itself a multiple of 4\.

Therefore the condition holds fora=2a=2\.

⊳\\trianglerightCase 3:aais even anda\>2a\>2\.

Letα=v2​\(a\)\\alpha=v\_\{2\}\(a\)\(soα≥1\\alpha\\geq 1; note that ifaais even and\>2\>2, then eitherα≥2\\alpha\\geq 2orα=1\\alpha=1witha=2​ma=2m,mmodd≥3\\geq 3\)\. Thenh​\(a\)=2α\+2h\(a\)=2^\{\\alpha\+2\}\. We need to prove

2α\+2∣ba−h​\(b\)2α\+2\.2^\{\\alpha\+2\}\\mid b^\{a\}\-h\(b\)^\{2^\{\\alpha\+2\}\}\.We split into subcases according tobb\.

⊳\\trianglerightSubcase 3a:bbis odd\.

Thenh​\(b\)=1h\(b\)=1\. So we need2α\+2∣ba−12^\{\\alpha\+2\}\\mid b^\{a\}\-1\.

Sinceaais even, writea=2α​ma=2^\{\\alpha\}mwithmmodd\.

Claim:For any odd integerbband any integers≥1s\\geq 1,

b2s≡1\(mod2s\+2\)\.b^\{2^\{s\}\}\\equiv 1\\pmod\{2^\{s\+2\}\}\.*Proof of claim\.*Fors=1s=1, any oddbbsatisfiesb2≡1\(mod8\)b^\{2\}\\equiv 1\\pmod\{8\}, i\.e\. modulo232^\{3\}\. Assume the claim holds for somes≥1s\\geq 1\. Then

b2s=1\+2s\+2​Kb^\{2^\{s\}\}=1\+2^\{s\+2\}Kfor some integerKK\. Squaring gives

b2s\+1=\(1\+2s\+2​K\)2=1\+2s\+3​K\+22​s\+4​K2≡1\(mod2s\+3\),b^\{2^\{s\+1\}\}=\(1\+2^\{s\+2\}K\)^\{2\}=1\+2^\{s\+3\}K\+2^\{2s\+4\}K^\{2\}\\equiv 1\\pmod\{2^\{s\+3\}\},because the extra term is divisible by2s\+32^\{s\+3\}\. This proves the claim\.

Applying the claim withs=αs=\\alphayields

b2α≡1\(mod2α\+2\)\.b^\{2^\{\\alpha\}\}\\equiv 1\\pmod\{2^\{\\alpha\+2\}\}\.Then

ba=\(b2α\)m≡1m=1\(mod2α\+2\),b^\{a\}=\\bigl\(b^\{2^\{\\alpha\}\}\\bigr\)^\{m\}\\equiv 1^\{m\}=1\\pmod\{2^\{\\alpha\+2\}\},and thereforeba−1b^\{a\}\-1is divisible by2α\+22^\{\\alpha\+2\}\.

⊳\\trianglerightSubcase 3b:b=1b=1\.

Thenba=1b^\{a\}=1,h​\(b\)=h​\(1\)=1h\(b\)=h\(1\)=1, soh​\(b\)2α\+2=1h\(b\)^\{2^\{\\alpha\+2\}\}=1, and1−1=01\-1=0is divisible by any integer\.

⊳\\trianglerightSubcase 3c:b=2b=2\.

Hereh​\(b\)=h​\(2\)=4h\(b\)=h\(2\)=4\. So we need

2α\+2∣2a−42α\+2=2a−\(22\)2α\+2=2a−22⋅2α\+2=2a−22α\+3\.2^\{\\alpha\+2\}\\mid 2^\{a\}\-4^\{2^\{\\alpha\+2\}\}=2^\{a\}\-\(2^\{2\}\)^\{2^\{\\alpha\+2\}\}=2^\{a\}\-2^\{2\\cdot 2^\{\\alpha\+2\}\}=2^\{a\}\-2^\{2^\{\\alpha\+3\}\}\.LetD=2a−22α\+3D=2^\{a\}\-2^\{2^\{\\alpha\+3\}\}\. The22\-adic valuation of a difference of two powers of two is

v2​\(D\)=\{aif​a<2α\+3,2α\+3if​a\>2α\+3,higherif​a=2α\+3​\(then​D=0​\)\.v\_\{2\}\(D\)=\\begin\{cases\}a&\\text\{if \}a<2^\{\\alpha\+3\},\\\\ 2^\{\\alpha\+3\}&\\text\{if \}a\>2^\{\\alpha\+3\},\\\\ \\text\{higher\}&\\text\{if \}a=2^\{\\alpha\+3\}\\text\{ \(then \}D=0\\text\{\)\}\.\\end\{cases\}We claimv2​\(D\)≥α\+2v\_\{2\}\(D\)\\geq\\alpha\+2\.

Lemma 9\.Ifaais even anda\>2a\>2, thena≥α\+2a\\geq\\alpha\+2whereα=v2​\(a\)\\alpha=v\_\{2\}\(a\)\.

*Proof\.*Writea=2α⋅ma=2^\{\\alpha\}\\cdot mwithmmodd\.

⊳\\trianglerightIfα=1\\alpha=1, thena=2​ma=2mwithmmodd\. Sincea\>2a\>2, we havem≥3m\\geq 3, soa≥6a\\geq 6\. Meanwhileα\+2=3\\alpha\+2=3\. Soa≥3a\\geq 3\.

⊳\\trianglerightIfα≥2\\alpha\\geq 2, thena≥2αa\\geq 2^\{\\alpha\}\. It remains to show2α≥α\+22^\{\\alpha\}\\geq\\alpha\+2forα≥2\\alpha\\geq 2\. This is true forα=2\\alpha=2\(4≥44\\geq 4\), and if true forα\\alpha, then2α\+1=2⋅2α≥2​\(α\+2\)=2​α\+4≥\(α\+1\)\+22^\{\\alpha\+1\}=2\\cdot 2^\{\\alpha\}\\geq 2\(\\alpha\+2\)=2\\alpha\+4\\geq\(\\alpha\+1\)\+2forα≥−1\\alpha\\geq\-1, which holds\. Hencea≥α\+2a\\geq\\alpha\+2\.

Now we have two possibilities:

⊳\\trianglerightIfa≤2α\+3a\\leq 2^\{\\alpha\+3\}, thenv2​\(D\)=av\_\{2\}\(D\)=a\(ifa<2α\+3a<2^\{\\alpha\+3\}\) orD=0D=0\(ifa=2α\+3a=2^\{\\alpha\+3\}\)\. In either case, sincea≥α\+2a\\geq\\alpha\+2, we getv2​\(D\)≥α\+2v\_\{2\}\(D\)\\geq\\alpha\+2\.

⊳\\trianglerightIfa\>2α\+3a\>2^\{\\alpha\+3\}, thenv2​\(D\)=2α\+3v\_\{2\}\(D\)=2^\{\\alpha\+3\}\. Sinceα≥1\\alpha\\geq 1,2α\+3≥24=162^\{\\alpha\+3\}\\geq 2^\{4\}=16forα=1\\alpha=1, andα\+2=3\\alpha\+2=3; clearly2α\+3≥α\+22^\{\\alpha\+3\}\\geq\\alpha\+2\(e\.g\. forα=1\\alpha=1,16≥316\\geq 3; for largerα\\alphait’s even larger\)\. Thusv2​\(D\)≥α\+2v\_\{2\}\(D\)\\geq\\alpha\+2\.

Therefore2α\+2∣D2^\{\\alpha\+2\}\\mid D\.

⊳\\trianglerightSubcase 3d:bbis even andb\>2b\>2\.

Writeb=2β​cb=2^\{\\beta\}cwithβ=v2​\(b\)≥1\\beta=v\_\{2\}\(b\)\\geq 1andccodd\. Sinceb\>2b\>2, we haveβ≥1\\beta\\geq 1; note that ifb=2b=2we already handled\.

Thenh​\(b\)=2β\+2h\(b\)=2^\{\\beta\+2\}\(by definition, becausebbis even andb\>2b\>2\)\.

We need to show2α\+22^\{\\alpha\+2\}divides

ba−\(2β\+2\)2α\+2=ba−2\(β\+2\)​2α\+2\.b^\{a\}\-\\bigl\(2^\{\\beta\+2\}\\bigr\)^\{2^\{\\alpha\+2\}\}=b^\{a\}\-2^\{\(\\beta\+2\)2^\{\\alpha\+2\}\}\.Observe that the second term is a power of two, and its exponent is\(β\+2\)​2α\+2≥2α\+2\(\\beta\+2\)2^\{\\alpha\+2\}\\geq 2^\{\\alpha\+2\}\(sinceβ≥1\\beta\\geq 1givesβ\+2≥3\\beta\+2\\geq 3\)\. Hence it is certainly divisible by2α\+22^\{\\alpha\+2\}\.

For the first term,

ba=\(2β​c\)a=2β​a​ca\.b^\{a\}=\(2^\{\\beta\}c\)^\{a\}=2^\{\\beta a\}c^\{a\}\.Sincecac^\{a\}is odd, the22\-adic valuation ofbab^\{a\}isβ​a\\beta a\. By Lemma 9,a≥α\+2a\\geq\\alpha\+2\. Therefore

β​a≥a≥α\+2\.\\beta a\\geq a\\geq\\alpha\+2\.Thus2α\+2∣ba2^\{\\alpha\+2\}\\mid b^\{a\}\.

Consequently bothbab^\{a\}and the second term are multiples of2α\+22^\{\\alpha\+2\}, so their difference is also a multiple of2α\+22^\{\\alpha\+2\}\.

All subcases are exhausted, and in each the required divisibility holds\.

Thushhis bonza\.

Now evaluateh​\(n\)h\(n\)for powers of two:

⊳\\trianglerighth​\(2\)=4=2⋅2h\(2\)=4=2\\cdot 2\.

⊳\\trianglerightFor anyt≥2t\\geq 2,n=2tn=2^\{t\}is even and\>2\>2, so

h​\(2t\)=2v2​\(2t\)\+2=2t\+2=4⋅2t=4​n\.h\(2^\{t\}\)=2^\{v\_\{2\}\(2^\{t\}\)\+2\}=2^\{t\+2\}=4\\cdot 2^\{t\}=4n\.In particular,h​\(4\)=16=4⋅4h\(4\)=16=4\\cdot 4\. Hence

h​\(n\)n=4for all​n=2t​\(t≥2\)\.\\frac\{h\(n\)\}\{n\}=4\\quad\\text\{for all \}n=2^\{t\}\\ \(t\\geq 2\)\.Ifc<4c<4, then forn=4n=4\(or any2t2^\{t\}witht≥2t\\geq 2\) we have

h​\(4\)=16\>c⋅4,h\(4\)=16\>c\\cdot 4,socccannot satisfyf​\(n\)≤c​nf\(n\)\\leq cnfor all bonza functions\. Therefore no constant smaller than 4 works\.

7\. Conclusion

We have proved that44is an upper bound forf​\(n\)/nf\(n\)/nand that this bound is attainable\. Consequently, the smallest real constantccwith the property thatf​\(n\)≤c​nf\(n\)\\leq cnfor every bonza functionffand everyn∈ℕn\\in\\mathbb\{N\}is

#### IMO 2025 Problem 4\.

An infinite sequencea1a\_\{1\},a2a\_\{2\}, … consists of positive integers has each of which has at least three proper divisors\. Suppose that for eachn≥1n\\geq 1,an\+1a\_\{n\+1\}is the sum of the three largest proper divisors ofana\_\{n\}\. Determine all possible values ofa1a\_\{1\}\.

Solution

Leta1,a2,…a\_\{1\},a\_\{2\},\\dotsbe an infinite sequence of positive integers such that eachana\_\{n\}has at least three proper divisors, and for eachn≥1n\\geq 1

an\+1=sum of the three largest proper divisors of​an\.a\_\{n\+1\}=\\text\{sum of the three largest proper divisors of \}a\_\{n\}\.Define the functionffbyf​\(N\)=f\(N\)=sum of the three largest proper divisors ofNN\.

1\. A useful formula forf​\(N\)f\(N\)

Let the positive divisors ofNNbe

1=d1<d2<⋯<dk=N,1=d\_\{1\}<d\_\{2\}<\\dots<d\_\{k\}=N,wherek≥4k\\geq 4becauseNNhas at least three proper divisors\. The three largest proper divisors aredk−3,dk−2,dk−1d\_\{k\-3\},d\_\{k\-2\},d\_\{k\-1\}\. Using the divisor pairingdi⋅dk\+1−i=Nd\_\{i\}\\cdot d\_\{k\+1\-i\}=N, we get

dk−1=Nd2,dk−2=Nd3,dk−3=Nd4\.d\_\{k\-1\}=\\frac\{N\}\{d\_\{2\}\},\\quad d\_\{k\-2\}=\\frac\{N\}\{d\_\{3\}\},\\quad d\_\{k\-3\}=\\frac\{N\}\{d\_\{4\}\}\.Thus

f\(N\)=Nd2\+Nd3\+Nd4=N\(1d2\+1d3\+1d4\)\.\(1\)f\(N\)=\\frac\{N\}\{d\_\{2\}\}\+\\frac\{N\}\{d\_\{3\}\}\+\\frac\{N\}\{d\_\{4\}\}=N\\left\(\\frac\{1\}\{d\_\{2\}\}\+\\frac\{1\}\{d\_\{3\}\}\+\\frac\{1\}\{d\_\{4\}\}\\right\)\.\\quad\\text\{\(1\)\}
2\. Fixed points

A fixed point satisfiesf​\(N\)=Nf\(N\)=N\. From \(1\) this is equivalent to

1d2\+1d3\+1d4=1\.\(2\)\\frac\{1\}\{d\_\{2\}\}\+\\frac\{1\}\{d\_\{3\}\}\+\\frac\{1\}\{d\_\{4\}\}=1\.\\quad\\text\{\(2\)\}Leta=d2,b=d3,c=d4a=d\_\{2\},\\;b=d\_\{3\},\\;c=d\_\{4\}\(so2≤a<b<c2\\leq a<b<c\)\. Solve1/a\+1/b\+1/c=11/a\+1/b\+1/c=1\.

⊳\\trianglerightIfa=2a=2, then1/b\+1/c=1/21/b\+1/c=1/2\. Multiplying gives2​\(b\+c\)=b​c2\(b\+c\)=bcor\(b−2\)​\(c−2\)=4\(b\-2\)\(c\-2\)=4\. Withb<cb<c, the only solution isb−2=1,c−2=4b\-2=1,\\;c\-2=4, i\.e\.,\(a,b,c\)=\(2,3,6\)\(a,b,c\)=\(2,3,6\)\.

⊳\\trianglerightIfa=3a=3, then1/b\+1/c=2/31/b\+1/c=2/3\. But the maximum forb≥4,c≥5b\\geq 4,c\\geq 5is1/4\+1/5=9/20<2/31/4\+1/5=9/20<2/3, so no solution\.

⊳\\trianglerightIfa≥4a\\geq 4, the sum is at most1/4\+1/5\+1/6=37/60<11/4\+1/5\+1/6=37/60<1\.

Hence the unique triple is\(2,3,6\)\(2,3,6\)\. Therefore a fixed point must have

d2=2,d3=3,d4=6\.d\_\{2\}=2,\\quad d\_\{3\}=3,\\quad d\_\{4\}=6\.Interpretation:

⊳\\triangleright2∣N2\\mid N\(soNNeven\),

⊳\\triangleright3∣N3\\mid N,

⊳\\trianglerightThere is no divisor between33and66; i\.e\.,4∤N4\\nmid Nand5∤N5\\nmid N\.

⊳\\trianglerightBecause22is the smallest divisor\>1\>1, we must haveν2​\(N\)=1\\nu\_\{2\}\(N\)=1\(otherwise4∣N4\\mid Nwould be a divisor<6<6\)\.

Thus the setℱ\\mathcal\{F\}of all fixed points is

ℱ=\{N∈ℕ\|ν2\(N\)=1,3∣N,5∤N\}\.\\boxed\{\\mathcal\{F\}=\\bigl\\\{N\\in\\mathbb\{N\}\\;\\big\|\\;\\nu\_\{2\}\(N\)=1,\\;3\\mid N,\\;5\\nmid N\\bigr\\\}\.\}
3\. Special case:12∣N12\\mid N

If12∣N12\\mid N, then2,3,4∣N2,3,4\\mid Nand these are the three smallest proper divisors \(since44is the smallest possible after2,32,3\)\. Then

f\(N\)=N2\+N3\+N4=1312N\.\(3\)f\(N\)=\\frac\{N\}\{2\}\+\\frac\{N\}\{3\}\+\\frac\{N\}\{4\}=\\frac\{13\}\{12\}N\.\\quad\\text\{\(3\)\}
4\. Lemma on odd numbers

Lemma 1\.LetXXbe an odd positive integer with at least three proper divisors\. Thenf​\(X\)<Xf\(X\)<Xandf​\(X\)f\(X\)is odd\.

*Proof\.*All divisors of an odd number are odd\. The three smallest divisors greater than11are at least3,5,73,5,7\. Therefore

1d2\+1d3\+1d4≤13\+15\+17=71105<1,\\frac\{1\}\{d\_\{2\}\}\+\\frac\{1\}\{d\_\{3\}\}\+\\frac\{1\}\{d\_\{4\}\}\\leq\\frac\{1\}\{3\}\+\\frac\{1\}\{5\}\+\\frac\{1\}\{7\}=\\frac\{71\}\{105\}<1,sof​\(X\)<Xf\(X\)<X\. Moreover, each quotientX/diX/d\_\{i\}is odd \(odd divided by odd\)\. The sum of three odd numbers is odd, hencef​\(X\)f\(X\)is odd\.

An immediate corollary:

*In an infinite orbit, no term can be odd, because starting from an odd term the sequence would be strictly decreasing and infinite \- impossible\. Hence every term of an infinite sequence iseven\.*

5\. Lemma for even numbers not divisible by1212

Lemma 2\.LetYYbe an even integer with at least three proper divisors and12∤Y12\\nmid Y\. If the orbit ofYYis infinite, thenY∈ℱY\\in\\mathcal\{F\}\.

*Proof\.*Since12∤Y12\\nmid YandYYis even, we have either4∤Y4\\nmid Yor3∤Y3\\nmid Y\(or both\)\. Consider two cases\.

Case 1:3∣Y3\\mid Y

Because12∤Y12\\nmid Y, we must have4∤Y4\\nmid Y; thusν2​\(Y\)=1\\nu\_\{2\}\(Y\)=1\. WriteY=2​MY=2MwithMModd\. Since3∣Y3\\mid Yandgcd⁡\(2,3\)=1\\gcd\(2,3\)=1, we get3∣M3\\mid M\.

The three smallest proper divisors ared2=2d\_\{2\}=2andd3=3d\_\{3\}=3\. The fourth divisord4d\_\{4\}depends on55\.

⊳\\trianglerightSubcase 1a:5∣M5\\mid M\. Then5∣Y5\\mid Yand5<65<6, sod4=5d\_\{4\}=5\. Then

f​\(Y\)=Y2\+Y3\+Y5=3130​Y\>Y\.f\(Y\)=\\frac\{Y\}\{2\}\+\\frac\{Y\}\{3\}\+\\frac\{Y\}\{5\}=\\frac\{31\}\{30\}Y\>Y\.Moreover,YYis divisible by3030\(it contains factors2,3,52,3,5\), soY=30​ZY=30ZwithZZodd\. Thenf​\(Y\)=31​Zf\(Y\)=31Z, which is odd\. By Lemma 1, the orbit from an odd number is strictly decreasing and finite \- contradiction to infinite orbit\. Hence this subcase cannot occur\.

⊳\\trianglerightSubcase 1b:5∤M5\\nmid M\. Then no divisor equals44or55; the next divisor after2,32,3is6=2⋅36=2\\cdot 3\. Henced4=6d\_\{4\}=6, and

f​\(Y\)=Y2\+Y3\+Y6=Y\.f\(Y\)=\\frac\{Y\}\{2\}\+\\frac\{Y\}\{3\}\+\\frac\{Y\}\{6\}=Y\.ThusYYis a fixed point, i\.e\.,Y∈ℱY\\in\\mathcal\{F\}\.

Therefore, if the orbit is infinite, we must be in Subcase 1b, soY∈ℱY\\in\\mathcal\{F\}\.

Case 2:3∤Y3\\nmid Y

ThenYYis even but not divisible by33\. For any suchYYwe have

1d2\+1d3\+1d4≤12\+14\+15=1920<1,\\frac\{1\}\{d\_\{2\}\}\+\\frac\{1\}\{d\_\{3\}\}\+\\frac\{1\}\{d\_\{4\}\}\\leq\\frac\{1\}\{2\}\+\\frac\{1\}\{4\}\+\\frac\{1\}\{5\}=\\frac\{19\}\{20\}<1,sof​\(Y\)<Yf\(Y\)<Y\. If the orbit never contained a term divisible by33, then we would have an infinite strictly decreasing sequence \- impossible\. Hence there exists a smallest indexm≥2m\\geq 2with3∣am3\\mid a\_\{m\}\. Thenam−1a\_\{m\-1\}is not divisible by33, andam=f​\(am−1\)<am−1a\_\{m\}=f\(a\_\{m\-1\}\)<a\_\{m\-1\}\.

Now examineam−1a\_\{m\-1\}\.

⊳\\trianglerightSubcase 2a:am−1a\_\{m\-1\}is odd\. By Lemma 1,am=f​\(am−1\)a\_\{m\}=f\(a\_\{m\-1\}\)is odd and less thanam−1a\_\{m\-1\}\. Sinceama\_\{m\}is odd and divisible by33, Lemma 1 again implies that fromama\_\{m\}onward the sequence is strictly decreasing and odd \- finite, contradiction\.

⊳\\trianglerightSubcase 2b:am−1a\_\{m\-1\}is even\. WriteY=am−1=2​MY=a\_\{m\-1\}=2M, withMModd\. Since3∤Y3\\nmid Y, we have3∤M3\\nmid M\. We will show thatama\_\{m\}is odd, reducing to Subcase 2a and giving a contradiction\.

Letr=Ymod3r=Y\\bmod 3\(r=1r=1or22\)\. For any divisorddofYY\(which is coprime to33\), we haved≡1d\\equiv 1or2\(mod3\)2\\pmod\{3\}andd−1≡d\(mod3\)d^\{\-1\}\\equiv d\\pmod\{3\}\(because1⋅1≡1,2⋅2≡1\(mod3\)1\\cdot 1\\equiv 1,\\;2\\cdot 2\\equiv 1\\pmod\{3\}\)\. Hence

Yd≡r⋅d\(mod3\)\.\\frac\{Y\}\{d\}\\equiv r\\cdot d\\pmod\{3\}\.Thus

f​\(Y\)≡r​\(d2\+d3\+d4\)\(mod3\)\.f\(Y\)\\equiv r\\bigl\(d\_\{2\}\+d\_\{3\}\+d\_\{4\}\\bigr\)\\pmod\{3\}\.Sincer≠0r\\neq 0, the condition3∣f​\(Y\)3\\mid f\(Y\)is equivalent to

2\+d3\+d4≡0\(mod3\)\.\(4\)2\+d\_\{3\}\+d\_\{4\}\\equiv 0\\pmod\{3\}\.\\quad\\text\{\(4\)\}Now,d2=2d\_\{2\}=2\. Determined3,d4d\_\{3\},d\_\{4\}\.

First,44cannot divideYY\. If4∣Y4\\mid Y, thend3=4d\_\{3\}=4, and \(4\) gives2\+4\+d4≡0⇒d4≡0\(mod3\)2\+4\+d\_\{4\}\\equiv 0\\Rightarrow d\_\{4\}\\equiv 0\\pmod\{3\}, impossible because3∤Y3\\nmid Y\. Soν2​\(Y\)=1\\nu\_\{2\}\(Y\)=1:Y=2​MY=2MwithMModd, and4∤Y4\\nmid Y\.

Thus the smallest divisor greater than22is an odd number; call itpp\. Since3∤Y3\\nmid Y,p≠3p\\neq 3, sop≥5p\\geq 5\.

The next divisord4d\_\{4\}is the smallest divisor larger thanpp\. Since4∤Y4\\nmid Y, the next even divisor would be2​p2p\. So either

⊳\\trianglerightd4=qd\_\{4\}=q\(an odd divisor,p<q<2​pp<q<2p\) if such an odd divisor exists, or

⊳\\trianglerightd4=2​pd\_\{4\}=2p\(if there is no odd divisor betweenppand2​p2p\)\.

Check the two possibilities against \(4\) withd3=pd\_\{3\}=p:

2\+p\+d4≡0\(mod3\)\.\(5\)2\+p\+d\_\{4\}\\equiv 0\\pmod\{3\}\.\\quad\\text\{\(5\)\}
⊳\\trianglerightIfd4=2​pd\_\{4\}=2p, then2\+p\+2​p=2\+3​p≡2\(mod3\)2\+p\+2p=2\+3p\\equiv 2\\pmod\{3\}, not0\. Henced4d\_\{4\}cannot be2​p2p\. Therefore there exists an odd divisorqqwithp<q<2​pp<q<2pandd4=qd\_\{4\}=q\.

Then \(5\) becomes2\+p\+q≡0\(mod3\)2\+p\+q\\equiv 0\\pmod\{3\}\. Sincep,q≢0\(mod3\)p,q\\not\\equiv 0\\pmod\{3\}, the only way is

p≡q≡2\(mod3\)\.p\\equiv q\\equiv 2\\pmod\{3\}\.Now computef​\(Y\)f\(Y\):

f​\(Y\)=Y2\+Yp\+Yq=M\+2​Mp\+2​Mq\.f\(Y\)=\\frac\{Y\}\{2\}\+\\frac\{Y\}\{p\}\+\\frac\{Y\}\{q\}=M\+\\frac\{2M\}\{p\}\+\\frac\{2M\}\{q\}\.Becausep,qp,qdivideMM\(they are odd divisors ofY=2​MY=2M\), the fractions are integers\. Now

⊳\\trianglerightMMis odd,

⊳\\triangleright2​Mp=2⋅Mp\\frac\{2M\}\{p\}=2\\cdot\\frac\{M\}\{p\}is even,

⊳\\triangleright2​Mq\\frac\{2M\}\{q\}is even\.

Thusf​\(Y\)=odd\+even\+even=oddf\(Y\)=\\text\{odd\}\+\\text\{even\}\+\\text\{even\}=\\text\{odd\}\.

Henceam=f​\(Y\)a\_\{m\}=f\(Y\)is odd\. But3∣am3\\mid a\_\{m\}\(by definition ofmm\)\. Soama\_\{m\}is an odd multiple of33\. By Lemma 1, the orbit fromama\_\{m\}is strictly decreasing and odd \- finite, contradiction\.

Both subcases lead to contradiction\. ThereforeCase 2 cannot occur\. The only possibility is Case 1, and specifically Subcase 1b, which forcesY∈ℱY\\in\\mathcal\{F\}\.

6\. Necessity \- form ofa1a\_\{1\}in an infinite orbit

Assume the sequence is infinite\. LetN=a1N=a\_\{1\}\. Define

t=max⁡\{k≥0​∣12k∣​N\}\.t=\\max\\\{k\\geq 0\\mid 12^\{k\}\\mid N\\\}\.WriteN=12t​RN=12^\{t\}Rwith12∤R12\\nmid R\.

We claim that fori=1,…,ti=1,\\dots,t, the termaia\_\{i\}is divisible by1212\. Proof by induction:a1=12t​Ra\_\{1\}=12^\{t\}Ris divisible by1212\(ift≥1t\\geq 1\)\. Supposeaia\_\{i\}is divisible by1212\. Since12∣ai12\\mid a\_\{i\}, we have by \(3\) thatai\+1=1312​aia\_\{i\+1\}=\\frac\{13\}\{12\}a\_\{i\}\. If we writeai=12s⋅Sa\_\{i\}=12^\{s\}\\cdot Swith12∤S12\\nmid Sands≥1s\\geq 1, thenai\+1=12s−1​\(13​S\)a\_\{i\+1\}=12^\{s\-1\}\(13S\)\. Since12∤S12\\nmid Sand1313is coprime to1212, we have12∤13​S12\\nmid 13S\. Thusai\+1a\_\{i\+1\}is divisible by12s−112^\{s\-1\}\. Starting withs=ts=tati=1i=1, afteriisteps the exponent of1212ist−i\+1t\-i\+1\. Hence as long asi≤ti\\leq t, the exponent is at least11, i\.e\.,12∣ai12\\mid a\_\{i\}\.

Thus we can apply \(3\) exactlytttimes:

at\+1=\(1312\)t​a1=13t12t⋅12t​R=13t​R\.a\_\{t\+1\}=\\left\(\\frac\{13\}\{12\}\\right\)^\{t\}a\_\{1\}=\\frac\{13^\{t\}\}\{12^\{t\}\}\\cdot 12^\{t\}R=13^\{t\}R\.Now12∤R12\\nmid Randgcd⁡\(13,12\)=1\\gcd\(13,12\)=1, so12∤13t​R12\\nmid 13^\{t\}R; i\.e\.,12∤at\+112\\nmid a\_\{t\+1\}\.

The orbit ofat\+1a\_\{t\+1\}is also infinite \(tail of an infinite sequence\)\. By the corollary of Lemma 1,at\+1a\_\{t\+1\}cannot be odd; henceat\+1a\_\{t\+1\}is even\. Also each term has at least three proper divisors\. Thereforeat\+1a\_\{t\+1\}satisfies the hypotheses of Lemma 2 \(even, not divisible by1212, infinite orbit\)\. Lemma 2 then yields

at\+1∈ℱ\.a\_\{t\+1\}\\in\\mathcal\{F\}\.Recallat\+1=13t​Ra\_\{t\+1\}=13^\{t\}R\. Since13t13^\{t\}is coprime to3030, the properties definingℱ\\mathcal\{F\}must already hold forRR:

⊳\\trianglerightν2​\(at\+1\)=ν2​\(R\)=1\\nu\_\{2\}\(a\_\{t\+1\}\)=\\nu\_\{2\}\(R\)=1,

⊳\\triangleright3∣at\+13\\mid a\_\{t\+1\}and3∤13t3\\nmid 13^\{t\}⇒\\Rightarrow3∣R3\\mid R,

⊳\\triangleright5∤at\+15\\nmid a\_\{t\+1\}and5∤13t5\\nmid 13^\{t\}⇒\\Rightarrow5∤R5\\nmid R\.

ThusR∈ℱR\\in\\mathcal\{F\}\. We have shown

a1=12t⋅Kwith​t≥0,K∈ℱ\.\\boxed\{a\_\{1\}=12^\{\\,t\}\\cdot K\\quad\\text\{with \}t\\geq 0,\\ K\\in\\mathcal\{F\}\}\.
7\. Sufficiency \- every such number works

We prove by induction onttthat ifa1=12t​Ka\_\{1\}=12^\{t\}KwithK∈ℱK\\in\\mathcal\{F\}, then the sequence is infinite\.

⊳\\trianglerightBase caset=0t=0:a1=K∈ℱa\_\{1\}=K\\in\\mathcal\{F\}\. By definition of a fixed point,f​\(K\)=Kf\(K\)=K\. Hence the sequence is constant:an=Ka\_\{n\}=Kfor allnn\. SinceKKhas at least three proper divisors \(as shown when characterizingℱ\\mathcal\{F\}\), the sequence is infinite\.

⊳\\trianglerightInductive step:Assume the statement holds for all starting values with parametert−1t\-1\(wheret≥1t\\geq 1\)\. Leta1=12t​Ka\_\{1\}=12^\{t\}KwithK∈ℱK\\in\\mathcal\{F\}\. Because12∣a112\\mid a\_\{1\}, we can use \(3\):

a2=f​\(a1\)=1312​a1=1312⋅12t​K=12t−1​\(13​K\)\.a\_\{2\}=f\(a\_\{1\}\)=\\frac\{13\}\{12\}a\_\{1\}=\\frac\{13\}\{12\}\\cdot 12^\{t\}K=12^\{t\-1\}\(13K\)\.Now, becauseK∈ℱK\\in\\mathcal\{F\}, we haveν2​\(K\)=1,3∣K,5∤K\\nu\_\{2\}\(K\)=1,\\ 3\\mid K,\\ 5\\nmid K\. Multiplying by1313\(coprime to3030\) preserves these properties, so13​K∈ℱ13K\\in\\mathcal\{F\}\. Thus

a2=12t−1​Lwith​L=13​K∈ℱ\.a\_\{2\}=12^\{t\-1\}L\\quad\\text\{with \}L=13K\\in\\mathcal\{F\}\.By the induction hypothesis, the sequence starting froma2a\_\{2\}is infinite\. Consequently, the whole sequence froma1a\_\{1\}is infinite\.

This completes the induction, establishing sufficiency\.

8\. Final characterization

Combining necessity and sufficiency, the possible values ofa1a\_\{1\}are exactly those positive integers that can be written as

a1=12t⋅K\(t≥0,K∈ℱ\),a\_\{1\}=12^\{\\,t\}\\cdot K\\qquad\(t\\geq 0,\\ K\\in\\mathcal\{F\}\),where

ℱ=\{K∈ℕ\|ν2​\(K\)=1,3∣K,5∤K\}\.\\mathcal\{F\}=\\bigl\\\{K\\in\\mathbb\{N\}\\;\\big\|\\;\\nu\_\{2\}\(K\)=1,\\ 3\\mid K,\\ 5\\nmid K\\bigr\\\}\.Equivalently, in terms of prime exponents:

ν2​\(a1\)​is odd,ν3​\(a1\)\>12​ν2​\(a1\),5∤a1\.\\nu\_\{2\}\(a\_\{1\}\)\\text\{ is odd\},\\quad\\nu\_\{3\}\(a\_\{1\}\)\>\\tfrac\{1\}\{2\}\\nu\_\{2\}\(a\_\{1\}\),\\quad 5\\nmid a\_\{1\}\.More explicitly, writing

t=⌊ν2​\(a1\)2⌋,ℓ=ν3​\(a1\)−\(t\+1\)≥0,gcd⁡\(R,30\)=1,t=\\left\\lfloor\\frac\{\\nu\_\{2\}\(a\_\{1\}\)\}\{2\}\\right\\rfloor,\\qquad\\ell=\\nu\_\{3\}\(a\_\{1\}\)\-\(t\+1\)\\geq 0,\\qquad\\gcd\(R,30\)=1,we have

a1=22​t\+1⋅3t\+1\+ℓ⋅R\.a\_\{1\}=2^\{2t\+1\}\\cdot 3^\{t\+1\+\\ell\}\\cdot R\.a1=12t⋅Kwitht≥0andKsatisfying2∥K,3∣K,5∤K\.\\boxed\{a\_\{1\}=12^\{\\,t\}\\cdot K\\text\{ with \}t\\geq 0\\text\{ and \}K\\text\{ satisfying \}2\\parallel K,\\ 3\\mid K,\\ 5\\nmid K\.\}a1=22​t\+1⋅3t\+1\+ℓ⋅R,t,ℓ≥0,gcd\(R,30\)=1\.\\boxed\{a\_\{1\}=2^\{2t\+1\}\\cdot 3^\{t\+1\+\\ell\}\\cdot R,\\quad t,\\ell\\geq 0,\\ \\gcd\(R,30\)=1\.\}This completely solves the problem\.

#### IMO 2025 Problem 5\.

Alice and Bazza are playing the*inekoalaty game*, a two‑player game whose rules depend on a positive real numberλ\\lambdawhich is known to both players\. On thennth turn of the game \(starting withn=1n=1\) the following happens:•Ifnnis odd, Alice chooses a nonnegative real numberxnx\_\{n\}such thatx1\+x2\+⋯\+xn≤λ​n\.x\_\{1\}\+x\_\{2\}\+\\cdots\+x\_\{n\}\\leq\\lambda n\.•Ifnnis even, Bazza chooses a nonnegative real numberxnx\_\{n\}such thatx12\+x22\+⋯\+xn2≤n\.x\_\{1\}^\{2\}\+x\_\{2\}^\{2\}\+\\cdots\+x\_\{n\}^\{2\}\\leq n\.If a player cannot choose a suitablexnx\_\{n\}, the game ends and the other player wins\. If the game goes on forever, neither player wins\. All chosen numbers are known to both players\.Determine all values ofλ\\lambdafor which Alice has a winning strategy and all those for which Bazza has a winning strategy\.

Solution

We begin by setting notation and a couple of useful lemmas\. For eachn≥1n\\geq 1, let

Sn=∑i=1nxi,Qn=∑i=1nxi2\.S\_\{n\}=\\sum\_\{i=1\}^\{n\}x\_\{i\},\\qquad Q\_\{n\}=\\sum\_\{i=1\}^\{n\}x\_\{i\}^\{2\}\.The rules are:

⊳\\trianglerightOn odd turnsnn, Alice choosesxn≥0x\_\{n\}\\geq 0withSn≤λ​nS\_\{n\}\\leq\\lambda n\.

⊳\\trianglerightOn even turnsnn, Bazza choosesxn≥0x\_\{n\}\\geq 0withQn≤nQ\_\{n\}\\leq n\.

If a player cannot make a legal move, the game ends and the other wins\. If it continues forever, the result is a draw \(no winner\)\.

Two elementary lemmas

Lemma 1\.After any even turnnn, we haveSn≤nS\_\{n\}\\leq nandQn≤nQ\_\{n\}\\leq n\.

*Proof\.*The condition for evennnisQn≤nQ\_\{n\}\\leq n\. By the Cauchy\-Schwarz inequality,

Sn2≤n⋅Qn≤n⋅n=n2,S\_\{n\}^\{2\}\\leq n\\cdot Q\_\{n\}\\leq n\\cdot n=n^\{2\},henceSn≤nS\_\{n\}\\leq n\.

Lemma 2\.Suppose that on all odd turns up to2​M2MAlice plays0\(so the only nonzero numbers are Bazza’s choices on even turns\)\. Then after turn2​M2Mwe have

S2​M≤M​2\.S\_\{2M\}\\leq M\\sqrt\{2\}\.*Proof\.*Writeyi=x2​iy\_\{i\}=x\_\{2i\}fori=1,…,Mi=1,\\dots,M\. Then

S2​M=∑i=1Myi,Q2​M=∑i=1Myi2\.S\_\{2M\}=\\sum\_\{i=1\}^\{M\}y\_\{i\},\\qquad Q\_\{2M\}=\\sum\_\{i=1\}^\{M\}y\_\{i\}^\{2\}\.By Cauchy\-Schwarz,

S2​M2≤M⋅Q2​M\.S\_\{2M\}^\{2\}\\leq M\\cdot Q\_\{2M\}\.Lemma 1 withn=2​Mn=2MgivesQ2​M≤2​MQ\_\{2M\}\\leq 2M\. Therefore

S2​M2≤M⋅2M=2M2⟹S2​M≤M2\.□S\_\{2M\}^\{2\}\\leq M\\cdot 2M=2M^\{2\}\\quad\\Longrightarrow\\quad S\_\{2M\}\\leq M\\sqrt\{2\}\.\\quad\\square
1\. The caseλ\>12\\lambda\>\\dfrac\{1\}\{\\sqrt\{2\}\}\- Alice wins

Alice will force a win in a finite number of moves\. Choose an integerMMlarge enough so that

λ\(2M\+1\)−M2\>2\.\(1\)\\lambda\(2M\+1\)\-M\\sqrt\{2\}\>\\sqrt\{2\}\.\\quad\\text\{\(1\)\}\(SuchMMexists becauseλ\>1/2\\lambda\>1/\\sqrt\{2\}implies2​λ−2\>02\\lambda\-\\sqrt\{2\}\>0, so the left\-hand side tends to\+∞\+\\inftyasM→∞M\\to\\infty\.\) Her strategy is:

⊳\\trianglerightOn turns1,3,5,…,2​M−11,3,5,\\dots,2M\-1\(all odd turns before2​M\+12M\+1\) she playsxn=0x\_\{n\}=0\.

⊳\\trianglerightOn turn2​M\+12M\+1she playsx2​M\+1=λ​\(2​M\+1\)−S2​Mx\_\{2M\+1\}=\\lambda\(2M\+1\)\-S\_\{2M\}\.

We must check that the zeros are legal and that the move on turn2​M\+12M\+1is well\-defined\.

Legality of the zeros

After each even turn2​k2k\(1≤k≤M1\\leq k\\leq M\), by Lemma 2 we haveS2​k≤k​2S\_\{2k\}\\leq k\\sqrt\{2\}\. For the odd turn2​k\+12k\+1, Alice wants to choose0; this is allowed iff

S2​k\+0≤λ​\(2​k\+1\)\.S\_\{2k\}\+0\\leq\\lambda\(2k\+1\)\.SinceS2​k≤k​2S\_\{2k\}\\leq k\\sqrt\{2\}, it suffices to showk​2<λ​\(2​k\+1\)k\\sqrt\{2\}<\\lambda\(2k\+1\)\. Rewrite asλ\>k​22​k\+1\\lambda\>\\dfrac\{k\\sqrt\{2\}\}\{2k\+1\}\. The functionk↦k​22​k\+1k\\mapsto\\dfrac\{k\\sqrt\{2\}\}\{2k\+1\}increases withkk\(its limit is2/2=1/2\\sqrt\{2\}/2=1/\\sqrt\{2\}\)\. Becauseλ\>1/2\\lambda\>1/\\sqrt\{2\}, we haveλ\>k​22​k\+1\\lambda\>\\dfrac\{k\\sqrt\{2\}\}\{2k\+1\}for everykk; hencek​2<λ​\(2​k\+1\)k\\sqrt\{2\}<\\lambda\(2k\+1\)\. Thus each0is legal\.

The decisive move

After turn2​M2M, Lemma 2 givesS2​M≤M​2S\_\{2M\}\\leq M\\sqrt\{2\}\. Define

a=x2​M\+1=λ​\(2​M\+1\)−S2​M\.a=x\_\{2M\+1\}=\\lambda\(2M\+1\)\-S\_\{2M\}\.From \(1\) and the bound onS2​MS\_\{2M\},

a≥λ​\(2​M\+1\)−M​2\>2\>0,a\\geq\\lambda\(2M\+1\)\-M\\sqrt\{2\}\>\\sqrt\{2\}\>0,soaais nonnegative and satisfiesS2​M\+a=λ​\(2​M\+1\)S\_\{2M\}\+a=\\lambda\(2M\+1\), hence the sum constraint on turn2​M\+12M\+1is met\.

Now we analyze the sum of squares after this move\. LetS=S2​MS=S\_\{2M\}\. Then

Q2​M\+1=Q2​M\+a2\.Q\_\{2M\+1\}=Q\_\{2M\}\+a^\{2\}\.By Cauchy\-Schwarz applied to theMMnumbersx2,x4,…,x2​Mx\_\{2\},x\_\{4\},\\dots,x\_\{2M\}we have

S2≤M⋅Q2​M⟹Q2​M≥S2M\.S^\{2\}\\leq M\\cdot Q\_\{2M\}\\quad\\Longrightarrow\\quad Q\_\{2M\}\\geq\\frac\{S^\{2\}\}\{M\}\.Therefore

Q2​M\+1≥S2M\+a2=S2M\+\(λ​\(2​M\+1\)−S\)2\.Q\_\{2M\+1\}\\geq\\frac\{S^\{2\}\}\{M\}\+a^\{2\}=\\frac\{S^\{2\}\}\{M\}\+\\bigl\(\\lambda\(2M\+1\)\-S\\bigr\)^\{2\}\.Define the function

f​\(S\)=S2M\+\(λ​\(2​M\+1\)−S\)2,0≤S≤M​2\.f\(S\)=\\frac\{S^\{2\}\}\{M\}\+\\bigl\(\\lambda\(2M\+1\)\-S\\bigr\)^\{2\},\\qquad 0\\leq S\\leq M\\sqrt\{2\}\.We need a lower bound forf​\(S\)f\(S\)\.

Monotonicity offf\.Compute

f′​\(S\)=2​SM−2​\(λ​\(2​M\+1\)−S\)=2​\(S​\(1\+1M\)−λ​\(2​M\+1\)\),f′′​\(S\)=2M\+2\>0\.f^\{\\prime\}\(S\)=\\frac\{2S\}\{M\}\-2\\bigl\(\\lambda\(2M\+1\)\-S\\bigr\)=2\\Bigl\(S\\Bigl\(1\+\\frac\{1\}\{M\}\\Bigr\)\-\\lambda\(2M\+1\)\\Bigr\),\\qquad f^\{\\prime\\prime\}\(S\)=\\frac\{2\}\{M\}\+2\>0\.Thusf′f^\{\\prime\}is strictly increasing\. The critical pointf′​\(S\)=0f^\{\\prime\}\(S\)=0gives

S0=M⋅λ​\(2​M\+1\)M\+1\.S\_\{0\}=\\frac\{M\\cdot\\lambda\(2M\+1\)\}\{M\+1\}\.Becauseλ\>1/2\\lambda\>1/\\sqrt\{2\}, we have2​λ\>22\\lambda\>\\sqrt\{2\}, and for sufficiently largeMM\(which we may assume\)

\(Simply check thatS0\>M​2⇔λ​\(2​M\+1\)\>2​\(M\+1\)S\_\{0\}\>M\\sqrt\{2\}\\iff\\lambda\(2M\+1\)\>\\sqrt\{2\}\(M\+1\), which holds for largeMMsince the left grows like2​λ​M2\\lambda Mand the right like2​M\\sqrt\{2\}M\.\)

Sincef′f^\{\\prime\}is increasing andS0\>M​2S\_\{0\}\>M\\sqrt\{2\}, we havef′​\(S\)<f′​\(S0\)=0f^\{\\prime\}\(S\)<f^\{\\prime\}\(S\_\{0\}\)=0for allS≤M​2S\\leq M\\sqrt\{2\}\. Henceffis strictly decreasing on\[0,M​2\]\[0,M\\sqrt\{2\}\]\. Consequently its minimum on the interval is attained atS=M​2S=M\\sqrt\{2\}:

min0≤S≤M​2⁡f​\(S\)=f​\(M​2\)=\(M​2\)2M\+\(λ​\(2​M\+1\)−M​2\)2=2​M\+\(λ​\(2​M\+1\)−M​2\)2\.\\min\_\{0\\leq S\\leq M\\sqrt\{2\}\}f\(S\)=f\(M\\sqrt\{2\}\)=\\frac\{\(M\\sqrt\{2\}\)^\{2\}\}\{M\}\+\\bigl\(\\lambda\(2M\+1\)\-M\\sqrt\{2\}\\bigr\)^\{2\}=2M\+\\bigl\(\\lambda\(2M\+1\)\-M\\sqrt\{2\}\\bigr\)^\{2\}\.Thus

Q2​M\+1≥2​M\+\(λ​\(2​M\+1\)−M​2\)2\.Q\_\{2M\+1\}\\geq 2M\+\\bigl\(\\lambda\(2M\+1\)\-M\\sqrt\{2\}\\bigr\)^\{2\}\.But condition \(1\) saysλ​\(2​M\+1\)−M​2\>2\\lambda\(2M\+1\)\-M\\sqrt\{2\}\>\\sqrt\{2\}, so

\(λ​\(2​M\+1\)−M​2\)2\>2\.\\bigl\(\\lambda\(2M\+1\)\-M\\sqrt\{2\}\\bigr\)^\{2\}\>2\.Therefore

Q2​M\+1\>2​M\+2\.Q\_\{2M\+1\}\>2M\+2\.Now it is Bazza’s turn \(turn2​M\+22M\+2\)\. He must choosex2​M\+2≥0x\_\{2M\+2\}\\geq 0such that

Q2​M\+2=Q2​M\+1\+x2​M\+22≤2​M\+2\.Q\_\{2M\+2\}=Q\_\{2M\+1\}\+x\_\{2M\+2\}^\{2\}\\leq 2M\+2\.Even if he playsx2​M\+2=0x\_\{2M\+2\}=0, we obtainQ2​M\+2=Q2​M\+1\>2​M\+2Q\_\{2M\+2\}=Q\_\{2M\+1\}\>2M\+2, which violates the constraint\. Hence Bazza has no legal move, and Alice wins\.

2\. The caseλ<12\\lambda<\\dfrac\{1\}\{\\sqrt\{2\}\}\- Bazza wins

Bazza will use the following strategy on every even turnnn:

xn=n−Qn−1\.x\_\{n\}=\\sqrt\{\\,n\-Q\_\{n\-1\}\\,\}\.In words, he takes the largest possible number that still keeps the sum of squares at mostnn\. We will prove by induction that after each of his moves \(i\.e\., after turn2​m2m\) we have

\(i\)Q2​m=2m,\(ii\)S2​m≥m2\.\(2\)\\text\{\(i\) \}Q\_\{2m\}=2m,\\qquad\\text\{\(ii\) \}S\_\{2m\}\\geq m\\sqrt\{2\}\.\\quad\\text\{\(2\)\}Base casem=1m=1\(turn22\)\.After turn 1, Alice has chosen somea1≥0a\_\{1\}\\geq 0witha1≤λa\_\{1\}\\leq\\lambda\. Then

x2=2−a12x\_\{2\}=\\sqrt\{2\-a\_\{1\}^\{2\}\}\(which is real becausea12≤λ2<1/2<2a\_\{1\}^\{2\}\\leq\\lambda^\{2\}<1/2<2\)\. We get

Q2=a12\+\(2−a12\)=2\.Q\_\{2\}=a\_\{1\}^\{2\}\+\(2\-a\_\{1\}^\{2\}\)=2\.For the sum,

S2=a1\+2−a12\.S\_\{2\}=a\_\{1\}\+\\sqrt\{2\-a\_\{1\}^\{2\}\}\.Squaring gives

S22=a12\+\(2−a12\)\+2​a1​2−a12=2\+2​a1​2−a12≥2,S\_\{2\}^\{2\}=a\_\{1\}^\{2\}\+\(2\-a\_\{1\}^\{2\}\)\+2a\_\{1\}\\sqrt\{2\-a\_\{1\}^\{2\}\}=2\+2a\_\{1\}\\sqrt\{2\-a\_\{1\}^\{2\}\}\\geq 2,soS2≥2S\_\{2\}\\geq\\sqrt\{2\}\. Thus \(2\) holds form=1m=1\.

Inductive step\.Assume \(2\) is true form−1m\-1, i\.e\., after turn2​\(m−1\)2\(m\-1\)we have

Q2​\(m−1\)=2​\(m−1\),S2​\(m−1\)≥\(m−1\)​2\.Q\_\{2\(m\-1\)\}=2\(m\-1\),\\qquad S\_\{2\(m\-1\)\}\\geq\(m\-1\)\\sqrt\{2\}\.*Turn2​m−12m\-1\(odd\):*Alice choosesa≥0a\\geq 0with

S2​\(m−1\)\+a≤λ\(2m−1\)\.\(3\)S\_\{2\(m\-1\)\}\+a\\leq\\lambda\(2m\-1\)\.\\quad\\text\{\(3\)\}*Turn2​m2m\(even\):*Bazza computes

x2​m=2​m−Q2​m−1\.x\_\{2m\}=\\sqrt\{2m\-Q\_\{2m\-1\}\}\.First we verify that the square root is defined, i\.e\.,Q2​m−1≤2​mQ\_\{2m\-1\}\\leq 2m\.

We have

Q2​m−1=Q2​\(m−1\)\+a2=2​\(m−1\)\+a2\.Q\_\{2m\-1\}=Q\_\{2\(m\-1\)\}\+a^\{2\}=2\(m\-1\)\+a^\{2\}\.Hence we needa2≤2a^\{2\}\\leq 2, ora≤2a\\leq\\sqrt\{2\}\. To see that this always holds, use \(3\) and the inductive lower bound onS2​\(m−1\)S\_\{2\(m\-1\)\}:

a≤λ\(2m−1\)−S2​\(m−1\)≤λ\(2m−1\)−\(m−1\)2\.\(4\)a\\leq\\lambda\(2m\-1\)\-S\_\{2\(m\-1\)\}\\leq\\lambda\(2m\-1\)\-\(m\-1\)\\sqrt\{2\}\.\\quad\\text\{\(4\)\}We claim that the right\-hand side of \(4\) is at most2\\sqrt\{2\}\. Indeed,

λ​\(2​m−1\)−\(m−1\)​2≤2⟺λ​\(2​m−1\)≤m​2\.\\lambda\(2m\-1\)\-\(m\-1\)\\sqrt\{2\}\\leq\\sqrt\{2\}\\;\\Longleftrightarrow\\;\\lambda\(2m\-1\)\\leq m\\sqrt\{2\}\.Becauseλ<1/2\\lambda<1/\\sqrt\{2\}, we have2​λ<22\\lambda<\\sqrt\{2\}\. Then

λ​\(2​m−1\)=2​λ​m−λ≤2​m−λ<2​m,\\lambda\(2m\-1\)=2\\lambda m\-\\lambda\\leq\\sqrt\{2\}\\,m\-\\lambda<\\sqrt\{2\}\\,m,so the inequality holds \(the left side is2​λ​m−λ2\\lambda m\-\\lambda, the right side is2​m\\sqrt\{2\}\\,m; since2​λ​m≤2​m2\\lambda m\\leq\\sqrt\{2\}\\,m, we getλ​\(2​m−1\)≤2​m\\lambda\(2m\-1\)\\leq\\sqrt\{2\}\\,m\. More formally:

λ​\(2​m−1\)≤2​m⟺\(2​λ−2\)​m≤λ\.\\lambda\(2m\-1\)\\leq\\sqrt\{2\}\\,m\\;\\Longleftrightarrow\\;\(2\\lambda\-\\sqrt\{2\}\)m\\leq\\lambda\.The left\-hand side is≤0\\leq 0\(because2​λ−2<02\\lambda\-\\sqrt\{2\}<0\), while the right\-hand side is positive, so the inequality is true for allm≥1m\\geq 1\. Thereforea≤2a\\leq\\sqrt\{2\}, soa2≤2a^\{2\}\\leq 2, and the square root is defined\.

Now compute:

x2​m=2​m−\(2​\(m−1\)\+a2\)=2−a2\.x\_\{2m\}=\\sqrt\{2m\-\\bigl\(2\(m\-1\)\+a^\{2\}\\bigr\)\}=\\sqrt\{2\-a^\{2\}\}\.Then

Q2​m=Q2​m−1\+x2​m2=\(2​\(m−1\)\+a2\)\+\(2−a2\)=2​m\.Q\_\{2m\}=Q\_\{2m\-1\}\+x\_\{2m\}^\{2\}=\\bigl\(2\(m\-1\)\+a^\{2\}\\bigr\)\+\(2\-a^\{2\}\)=2m\.For the sum:

S2​m=S2​\(m−1\)\+a\+2−a2\.S\_\{2m\}=S\_\{2\(m\-1\)\}\+a\+\\sqrt\{2\-a^\{2\}\}\.Observe that for anyaawith0≤a≤20\\leq a\\leq\\sqrt\{2\},

\(a\+2−a2\)2=2\+2​a​2−a2≥2,\\bigl\(a\+\\sqrt\{2\-a^\{2\}\}\\bigr\)^\{2\}=2\+2a\\sqrt\{2\-a^\{2\}\}\\geq 2,hencea\+2−a2≥2a\+\\sqrt\{2\-a^\{2\}\}\\geq\\sqrt\{2\}\. Using the inductive boundS2​\(m−1\)≥\(m−1\)​2S\_\{2\(m\-1\)\}\\geq\(m\-1\)\\sqrt\{2\}, we obtain

S2​m≥\(m−1\)​2\+2=m​2\.S\_\{2m\}\\geq\(m\-1\)\\sqrt\{2\}\+\\sqrt\{2\}=m\\sqrt\{2\}\.This completes the induction\.

Alice loses

From \(2\) we haveS2​m≥m​2S\_\{2m\}\\geq m\\sqrt\{2\}for everymm\. Sinceλ<1/2\\lambda<1/\\sqrt\{2\}, we have2−2​λ\>0\\sqrt\{2\}\-2\\lambda\>0\. Choosemmlarge enough so that

m2\>λ\(2m\+1\)\.\(5\)m\\sqrt\{2\}\>\\lambda\(2m\+1\)\.\\quad\\text\{\(5\)\}\(Suchmmexists becausem​\(2−2​λ\)\>λm\(\\sqrt\{2\}\-2\\lambda\)\>\\lambdaeventually holds\.\)

Consider turn2​m\+12m\+1\. After turn2​m2m,

S2​m≥m​2\>λ​\(2​m\+1\)S\_\{2m\}\\geq m\\sqrt\{2\}\>\\lambda\(2m\+1\)by \(5\)\. Therefore, even if Alice tries to playx2​m\+1=0x\_\{2m\+1\}=0, we have

S2​m\+1=S2​m\>λ​\(2​m\+1\),S\_\{2m\+1\}=S\_\{2m\}\>\\lambda\(2m\+1\),which violates the sum constraint\. Consequently she has no legal move, and Bazza wins\.

3\. The caseλ=12\\lambda=\\dfrac\{1\}\{\\sqrt\{2\}\}\- Draw

We show that neither player has a winning strategy by exhibiting a strategy for each that prevents the opponent from winning\.

Bazza prevents Alice from winning

Bazza uses the same maximal strategy as in case 2: on each even turnnn, setxn=n−Qn−1x\_\{n\}=\\sqrt\{n\-Q\_\{n\-1\}\}\. We check that the induction in case 2 still works whenλ=1/2\\lambda=1/\\sqrt\{2\}\.

For the inductive step, from \(3\) and the lower boundS2​\(m−1\)≥\(m−1\)​2S\_\{2\(m\-1\)\}\\geq\(m\-1\)\\sqrt\{2\}we obtain

a≤λ​\(2​m−1\)−\(m−1\)​2\.a\\leq\\lambda\(2m\-1\)\-\(m\-1\)\\sqrt\{2\}\.Withλ=1/2\\lambda=1/\\sqrt\{2\},

λ​\(2​m−1\)−\(m−1\)​2=2​m−12−\(m−1\)​2=2​m−1−2​\(m−1\)2=12\.\\lambda\(2m\-1\)\-\(m\-1\)\\sqrt\{2\}=\\frac\{2m\-1\}\{\\sqrt\{2\}\}\-\(m\-1\)\\sqrt\{2\}=\\frac\{2m\-1\-2\(m\-1\)\}\{\\sqrt\{2\}\}=\\frac\{1\}\{\\sqrt\{2\}\}\.Thusa≤1/2<2a\\leq 1/\\sqrt\{2\}<\\sqrt\{2\}, soa2≤1/2a^\{2\}\\leq 1/2, and the square root2−a2\\sqrt\{2\-a^\{2\}\}is well\-defined\. The rest of the induction is unchanged, yielding

Q2​m=2m,S2​m≥m2for allm\.\(6\)Q\_\{2m\}=2m,\\qquad S\_\{2m\}\\geq m\\sqrt\{2\}\\quad\\text\{for all \}m\.\\quad\\text\{\(6\)\}Now consider any odd turn2​m\+12m\+1\. Alice must choosea′≥0a^\{\\prime\}\\geq 0such that

S2​m\+a′≤λ​\(2​m\+1\)=2​m\+12\.S\_\{2m\}\+a^\{\\prime\}\\leq\\lambda\(2m\+1\)=\\dfrac\{2m\+1\}\{\\sqrt\{2\}\}\.From \(6\),S2​m≥m​2=2​m2S\_\{2m\}\\geq m\\sqrt\{2\}=\\dfrac\{2m\}\{\\sqrt\{2\}\}\. Therefore

a′≤2​m\+12−S2​m≤2​m\+12−2​m2=12\.a^\{\\prime\}\\leq\\frac\{2m\+1\}\{\\sqrt\{2\}\}\-S\_\{2m\}\\leq\\frac\{2m\+1\}\{\\sqrt\{2\}\}\-\\frac\{2m\}\{\\sqrt\{2\}\}=\\frac\{1\}\{\\sqrt\{2\}\}\.Consequentlya′⁣2≤12a^\{\\prime 2\}\\leq\\dfrac\{1\}\{2\}, and

Q2​m\+1=Q2​m\+a′⁣2=2​m\+a′⁣2≤2​m\+12<2​m\+2\.Q\_\{2m\+1\}=Q\_\{2m\}\+a^\{\\prime 2\}=2m\+a^\{\\prime 2\}\\leq 2m\+\\frac\{1\}\{2\}<2m\+2\.Thus after Alice’s move, we haveQ2​m\+1<2​m\+2Q\_\{2m\+1\}<2m\+2\. Bazza can then reply withx2​m\+2=0x\_\{2m\+2\}=0\(or any number keepingQ2​m\+2≤2​m\+2Q\_\{2m\+2\}\\leq 2m\+2\)\. In particular, Bazza always has a legal move, so Alice cannot force a win\.

Alice prevents Bazza from winning

Alice adopts the strategy: on every odd turn she plays0\.

Then the only nonzero numbers are Bazza’s even\-turn choices\. By Lemma 2, after turn2​m2mwe have

S2​m≤m​2\.S\_\{2m\}\\leq m\\sqrt\{2\}\.The sum constraint on turn2​m\+12m\+1is

S2​m\+1=S2​m\+x2​m\+1≤λ​\(2​m\+1\)=2​m\+12\.S\_\{2m\+1\}=S\_\{2m\}\+x\_\{2m\+1\}\\leq\\lambda\(2m\+1\)=\\dfrac\{2m\+1\}\{\\sqrt\{2\}\}\.Since

S2​m≤m​2=2​m2<2​m\+12,S\_\{2m\}\\leq m\\sqrt\{2\}=\\dfrac\{2m\}\{\\sqrt\{2\}\}<\\dfrac\{2m\+1\}\{\\sqrt\{2\}\},the choicex2​m\+1=0x\_\{2m\+1\}=0satisfies the inequality\. Hence Alice never loses on her turn\.

Moreover, with this strategy Bazza cannot force a win, because Alice always has a legal move \(namely0\)\. Therefore Bazza does not possess a winning strategy\.

Since both players have strategies that thwart a forced win, the game is a draw; it can also be continued indefinitely \(e\.g\., with Bazza always choosing0as well\)\. Soλ=1/2\\lambda=1/\\sqrt\{2\}leads to no winner\.

Conclusion

We have determined the outcome for all positiveλ\\lambda:

Alice has a winning strategy if and only if​λ\>12\.Bazza has a winning strategy if and only if​λ<12\.When​λ=12,the game is a draw \(no winner\)\.\\boxed\{\\begin\{array\}\[\]\{l\}\\text\{Alice has a winning strategy if and only if \}\\lambda\>\\dfrac\{1\}\{\\sqrt\{2\}\}\.\\\\\[6\.0pt\] \\text\{Bazza has a winning strategy if and only if \}\\lambda<\\dfrac\{1\}\{\\sqrt\{2\}\}\.\\\\\[6\.0pt\] \\text\{When \}\\lambda=\\dfrac\{1\}\{\\sqrt\{2\}\},\\text\{ the game is a draw \(no winner\)\.\}\\end\{array\}\}

#### IMO 2025 Problem 6\.

Consider a2025×20252025\\times 2025grid of unit squares\. Matilda wishes to place on the grid some rectangular tiles, possibly of different sizes, such that each side of every tile lies on a grid line and every unit square is covered by at most one tile\.Determine the minimum number of tiles Matilda needs to place so that each row and each column of the grid has exactly one unit square that is not covered by any tile\.

Solution

Letn=2025n=2025\. The grid consists ofn2n^\{2\}unit squares\. We must choose a set of squares to remain uncovered such that each row and each column contains exactly one uncovered square; equivalently, the uncovered squares form a permutation matrix\. The remainingn2−nn^\{2\}\-nsquares must be covered by non\-overlapping axis\-aligned rectangles \(tiles\) whose sides lie on the grid lines\. The goal is to minimize the number of tiles\.

1\. Reduction to the diagonal configuration

Let the uncovered square in rowiibe at columncic\_\{i\}; thenc1,c2,…,cnc\_\{1\},c\_\{2\},\\dots,c\_\{n\}is a permutation of\{1,…,n\}\\\{1,\\dots,n\\\}\. Consider the column permutationσ\\sigmadefined byσ​\(ci\)=i\\sigma\(c\_\{i\}\)=i\. Applyingσ\\sigmato the column indices \(i\.e\., renaming columnjjtoσ​\(j\)\\sigma\(j\)\) transforms the uncovered squares into the diagonal positions\(i,i\)\(i,i\)\. The tiling is transformed by moving each tile together with the columns; this does not change the number of tiles, and the covering properties remain valid because the grid structure is preserved \(only the labels of columns are changed\)\. Therefore, the minimum number of tiles over all configurations is the same as the minimum over configurations where the uncovered squares are exactly the main diagonal\{\(i,i\)∣1≤i≤n\}\\\{\(i,i\)\\mid 1\\leq i\\leq n\\\}\.

Consequently,we may assume without loss of generality that the diagonal squares are the only uncovered ones\.

2\. A construction using2​n−22n\-2tiles

We now exhibit a tiling of the board with the diagonal uncovered, using exactly2​n−22n\-2rectangles\.

⊳\\trianglerightLower triangle\(i\>ji\>j\): For each columnj=1,2,…,n−1j=1,2,\\dots,n\-1, place a vertical tile that covers every cell in columnjjwhose row index is greater thanjj\. Formally,

Rj\(L\)=\{\(i,j\)∣j\+1≤i≤n\}\.R^\{\(L\)\}\_\{j\}=\\\{\(i,j\)\\mid j\+1\\leq i\\leq n\\\}\.Thesen−1n\-1rectangles are pairwise disjoint \(they lie in different columns\) and cover all squares withi\>ji\>j\.

⊳\\trianglerightUpper triangle\(i<ji<j\): For each rowi=1,2,…,n−1i=1,2,\\dots,n\-1, place a horizontal tile that covers every cell in rowiiwhose column index is greater thanii:

Ri\(U\)=\{\(i,j\)∣i\+1≤j≤n\}\.R^\{\(U\)\}\_\{i\}=\\\{\(i,j\)\\mid i\+1\\leq j\\leq n\\\}\.Thesen−1n\-1rectangles are also pairwise disjoint \(they occupy distinct rows\) and cover all squares withi<ji<j\.

The two families are disjoint because a square withi\>ji\>jbelongs to someRj\(L\)R^\{\(L\)\}\_\{j\}\(columnjj\) and a square withi<ji<jbelongs to someRi\(U\)R^\{\(U\)\}\_\{i\}\(rowii\); no square can satisfy both conditions\. The diagonal squares\(i,i\)\(i,i\)are not covered by any tile\. Thus we have a valid covering of all off\-diagonal squares using

k=\(n−1\)\+\(n−1\)=2​n−2k=\(n\-1\)\+\(n\-1\)=2n\-2tiles\.

3\. Lower bound: at least2​n−22n\-2tiles

Now take any tiling𝒯\\mathcal\{T\}of the board that leaves exactly the diagonal squares uncovered\. Partition the off\-diagonal squares into two sets:

L=\{\(i,j\)∣i\>j\},U=\{\(i,j\)∣i<j\}\.L=\\\{\(i,j\)\\mid i\>j\\\},\\qquad U=\\\{\(i,j\)\\mid i<j\\\}\.Lemma 1\.Every tileT∈𝒯T\\in\\mathcal\{T\}is entirely contained either inLLor inUU\.

*Proof\.*Suppose a tileTTcontains a square fromLLand a square fromUU\. RepresentTTas the Cartesian product of an interval of rows and an interval of columns:

T=\{\(i,j\)∣a≤i≤b,c≤j≤d\},T=\\\{\(i,j\)\\mid a\\leq i\\leq b,\\;c\\leq j\\leq d\\\},with1≤a≤b≤n1\\leq a\\leq b\\leq n,1≤c≤d≤n1\\leq c\\leq d\\leq n\. If the intervals\[a,b\]\[a,b\]and\[c,d\]\[c,d\]intersect, then there exists an integerr∈\[a,b\]∩\[c,d\]r\\in\[a,b\]\\cap\[c,d\], and the square\(r,r\)\(r,r\)belongs toTT, contradicting the fact that diagonal squares are uncovered\. Hence\[a,b\]\[a,b\]and\[c,d\]\[c,d\]are disjoint\.

⊳\\trianglerightIfb<cb<c, then for any\(i,j\)∈T\(i,j\)\\in Twe havei≤b<c≤ji\\leq b<c\\leq j, soi<ji<j; thusT⊆UT\\subseteq U, contradicting the presence of anLLsquare\.

⊳\\trianglerightIfd<ad<a, thenj≤d<a≤ij\\leq d<a\\leq i, soi\>ji\>j, givingT⊆LT\\subseteq L, contradiction\.

Therefore a tile cannot contain squares from bothLLandUU; it must lie wholly in one of them\.

Lemma 2\.In the lower triangleLL, consider then−1n\-1cells

DL=\{\(i,i−1\)∣i=2,3,…,n\}\.D\_\{L\}=\\\{\(i,i\-1\)\\mid i=2,3,\\dots,n\\\}\.No tile that is a subset ofLLcan contain two distinct cells fromDLD\_\{L\}\.

*Proof\.*LetT⊆LT\\subseteq Lbe a tile, soT=\[a,b\]×\[c,d\]T=\[a,b\]\\times\[c,d\]\. BecauseT⊆LT\\subseteq L, we havei\>ji\>jfor every\(i,j\)∈T\(i,j\)\\in T\. In particular, the cell\(a,d\)\(a,d\)\(the topmost row and rightmost column of the tile\) satisfiesa\>da\>d\.

Assume, for contradiction, thatTTcontains two distinct cells\(i,i−1\)\(i,i\-1\)and\(j,j−1\)\(j,j\-1\)withi<ji<j\. From\(i,i−1\)∈T\(i,i\-1\)\\in Twe obtain

a≤i≤b,c≤i−1≤d\.a\\leq i\\leq b,\\quad c\\leq i\-1\\leq d\.Sincea\>da\>dandd≥i−1d\\geq i\-1, we havea\>i−1a\>i\-1, hencea≥ia\\geq i\. Combined witha≤ia\\leq i, we geta=ia=i\.

Nowa=ia=ianda\>da\>dimplyi\>di\>d, i\.e\.,d<id<i\. Butd≥i−1d\\geq i\-1from the containment, sod=i−1d=i\-1\.

Now apply the same reasoning to\(j,j−1\)∈T\(j,j\-1\)\\in T\. We obtaina=ja=jandd=j−1d=j\-1\. But we already havea=ia=i, soi=ji=j, contradictingi<ji<j\. HenceTTcannot contain two cells ofDLD\_\{L\}\.

Lemma 3\.In the upper triangleUU, consider then−1n\-1cells

DU=\{\(i,i\+1\)∣i=1,2,…,n−1\}\.D\_\{U\}=\\\{\(i,i\+1\)\\mid i=1,2,\\dots,n\-1\\\}\.No tile that is a subset ofUUcan contain two distinct cells fromDUD\_\{U\}\.

*Proof\.*LetT⊆UT\\subseteq Ube a tile,T=\[a,b\]×\[c,d\]T=\[a,b\]\\times\[c,d\]\. SinceT⊆UT\\subseteq U, we havei<ji<jfor every\(i,j\)∈T\(i,j\)\\in T\. In particular, the cell\(b,c\)\(b,c\)\(the bottommost row and leftmost column\) satisfiesb<cb<c\.

SupposeTTcontains\(i,i\+1\)\(i,i\+1\)and\(j,j\+1\)\(j,j\+1\)withi<ji<j\. From\(i,i\+1\)∈T\(i,i\+1\)\\in Twe have

a≤i≤b,c≤i\+1≤d\.a\\leq i\\leq b,\\quad c\\leq i\+1\\leq d\.Becausec\>bc\>bandb≥ib\\geq i, we getc\>ic\>i, thusc≥i\+1c\\geq i\+1\. Butc≤i\+1c\\leq i\+1from the containment, soc=i\+1c=i\+1\.

Moreover,c\>b≥ic\>b\\geq iandc=i\+1c=i\+1implyb<i\+1b<i\+1, i\.e\.,b≤ib\\leq i\. Combined withi≤bi\\leq b, we obtainb=ib=i\.

Now from\(j,j\+1\)∈T\(j,j\+1\)\\in Twe similarly deducec=j\+1c=j\+1andb=jb=j\. Sincec=i\+1c=i\+1, we havei\+1=j\+1i\+1=j\+1, soi=ji=j, contradiction\. HenceTTcannot contain two cells ofDUD\_\{U\}\.

Lower bound argument\.The setsDLD\_\{L\}andDUD\_\{U\}each containn−1n\-1squares, and every square must be covered\. By Lemma 1, any tile that covers a square fromLLis entirely contained inLL; similarly, any tile covering a square fromUUis entirely inUU\.

⊳\\trianglerightDLD\_\{L\}must be covered by tiles that lie inLL\. Lemma 2 tells us that each such tile can cover at most one element ofDLD\_\{L\}\. Therefore at least\|DL\|=n−1\|D\_\{L\}\|=n\-1tiles are required that belong toLL\.

⊳\\trianglerightDUD\_\{U\}must be covered by tiles that lie inUU, and Lemma 3 says each such tile can cover at most one element ofDUD\_\{U\}, so at leastn−1n\-1tiles are needed that belong toUU\.

A tile cannot be in bothLLandUUbecauseLLandUUare disjoint\. Hence the total number of tiles is at least

\(n−1\)\+\(n−1\)=2​n−2\.\(n\-1\)\+\(n\-1\)=2n\-2\.
4\. Minimum number of tiles

We have shown:

⊳\\trianglerightThere exists a tiling with2​n−22n\-2tiles \(Construction\)\.

⊳\\trianglerightAny tiling uses at least2​n−22n\-2tiles \(Lower bound\)\.

Therefore the minimum possible number of tiles is exactly2​n−22n\-2\.

Substitutingn=2025n=2025gives

2×2025−2=4050−2=4048\.2\\times 2025\-2=4050\-2=4048\.

### H\.2USAMO 2026

#### USAMO 2026 Problem 1\.

Letnnbe an integer greater than11\. For which real numbersxxis⌊n​x⌋−∑k=1n⌊k​x⌋k\\lfloor nx\\rfloor\-\\sum\_\{k=1\}^\{n\}\\frac\{\\lfloor kx\\rfloor\}\{k\}maximal, and what is the maximal value that this expression can take?*Note*:⌊z⌋\\lfloor z\\rfloordenotes the greatest integer less than or equal tozz\.

Solution

Step 1\. Reduction to the fractional part\.Letn\>1n\>1be an integer\. For any realxx, writex=⌊x⌋\+\{x\}=m\+θx=\\lfloor x\\rfloor\+\\\{x\\\}=m\+\\theta, wherem∈ℤm\\in\\mathbb\{Z\}andθ∈\[0,1\)\\theta\\in\[0,1\)\. For eachk≥1k\\geq 1,

⌊k​x⌋=⌊k​m\+k​θ⌋=k​m\+⌊k​θ⌋,\\lfloor kx\\rfloor=\\lfloor km\+k\\theta\\rfloor=km\+\\lfloor k\\theta\\rfloor,becausek​mkmis an integer\. Hence the expression becomes

E​\(x\)=⌊n​x⌋−∑k=1n⌊k​x⌋k=\(n​m\+⌊n​θ⌋\)−∑k=1nk​m\+⌊k​θ⌋k=⌊n​θ⌋−∑k=1n⌊k​θ⌋k\.E\(x\)=\\lfloor nx\\rfloor\-\\sum\_\{k=1\}^\{n\}\\frac\{\\lfloor kx\\rfloor\}\{k\}=\\bigl\(nm\+\\lfloor n\\theta\\rfloor\\bigr\)\-\\sum\_\{k=1\}^\{n\}\\frac\{km\+\\lfloor k\\theta\\rfloor\}\{k\}=\\lfloor n\\theta\\rfloor\-\\sum\_\{k=1\}^\{n\}\\frac\{\\lfloor k\\theta\\rfloor\}\{k\}\.ThusE​\(x\)E\(x\)depends only onθ\\theta; we define

F​\(θ\)=⌊n​θ⌋−∑k=1n⌊k​θ⌋k,θ∈\[0,1\)\.F\(\\theta\)=\\lfloor n\\theta\\rfloor\-\\sum\_\{k=1\}^\{n\}\\frac\{\\lfloor k\\theta\\rfloor\}\{k\},\\qquad\\theta\\in\[0,1\)\.Step 2\. Value on the interval\[1−1n,1\)\\bigl\[1\-\\frac\{1\}\{n\},\\,1\\bigr\)\.Assumeθ∈\[1−1n,1\)\\theta\\in\\left\[1\-\\frac\{1\}\{n\},\\,1\\right\)\. Thenθ≥n−1n\\theta\\geq\\frac\{n\-1\}\{n\}andθ<1\\theta<1, so for everyk=1,…,nk=1,\\dots,n,

k​θ≥k⋅n−1n=k−kn≥k−1,k\\theta\\geq k\\cdot\\frac\{n\-1\}\{n\}=k\-\\frac\{k\}\{n\}\\geq k\-1,andk​θ<kk\\theta<k\. Consequently⌊k​θ⌋=k−1\\lfloor k\\theta\\rfloor=k\-1for allkk\. In particular⌊n​θ⌋=n−1\\lfloor n\\theta\\rfloor=n\-1\. Therefore

F​\(θ\)=\(n−1\)−∑k=1nk−1k\.F\(\\theta\)=\(n\-1\)\-\\sum\_\{k=1\}^\{n\}\\frac\{k\-1\}\{k\}\.Now

∑k=1nk−1k=∑k=1n\(1−1k\)=n−Hn,\\sum\_\{k=1\}^\{n\}\\frac\{k\-1\}\{k\}=\\sum\_\{k=1\}^\{n\}\\left\(1\-\\frac\{1\}\{k\}\\right\)=n\-H\_\{n\},whereHn=1\+12\+⋯\+1nH\_\{n\}=1\+\\frac\{1\}\{2\}\+\\cdots\+\\frac\{1\}\{n\}\(thenn\-th harmonic number\)\. Hence

F​\(θ\)=\(n−1\)−\(n−Hn\)=Hn−1\.F\(\\theta\)=\(n\-1\)\-\(n\-H\_\{n\}\)=H\_\{n\}\-1\.So on the whole interval\[1−1n,1\)\[1\-\\frac\{1\}\{n\},1\)the expression is constant and equalsHn−1H\_\{n\}\-1\.

Step 3\. Upper bound: first decomposition\.To prove thatHn−1H\_\{n\}\-1is the maximum, we showF​\(θ\)≤Hn−1F\(\\theta\)\\leq H\_\{n\}\-1for allθ∈\[0,1\)\\theta\\in\[0,1\)\. Letθ\\thetabe arbitrary and write

N=⌊n​θ⌋∈\{0,1,…,n−1\},β=n​θ−N∈\[0,1\)\.N=\\lfloor n\\theta\\rfloor\\in\\\{0,1,\\dots,n\-1\\\},\\qquad\\beta=n\\theta\-N\\in\[0,1\)\.Sett=β/n∈\[0,1/n\)t=\\beta/n\\in\[0,1/n\)\. Then

θ=Nn\+t\.\\theta=\\frac\{N\}\{n\}\+t\.Fork=1,…,nk=1,\\dots,n,

k​θ=k​Nn\+k​t\.k\\theta=\\frac\{kN\}\{n\}\+kt\.Write the division ofk​NkNbynn:

k​Nn=⌊k​Nn⌋\+rkn,\\frac\{kN\}\{n\}=\\left\\lfloor\\frac\{kN\}\{n\}\\right\\rfloor\+\\frac\{r\_\{k\}\}\{n\},whererkr\_\{k\}is the remainder, i\.e\.rk=k​Nmodnr\_\{k\}=kN\\bmod nwith0≤rk<n0\\leq r\_\{k\}<n\. Then

k​θ=⌊k​Nn⌋⏟integer\+\(rkn\+k​t\)⏟in​\[0,2\)\.k\\theta=\\underbrace\{\\left\\lfloor\\frac\{kN\}\{n\}\\right\\rfloor\}\_\{\\text\{integer\}\}\+\\underbrace\{\\left\(\\frac\{r\_\{k\}\}\{n\}\+kt\\right\)\}\_\{\\text\{in \}\[0,2\)\}\.Define

δk=\{1if​rkn\+k​t≥1,0otherwise\.\\delta\_\{k\}=\\begin\{cases\}1&\\text\{if \}\\frac\{r\_\{k\}\}\{n\}\+kt\\geq 1,\\\\ 0&\\text\{otherwise\}\.\\end\{cases\}Then

⌊k​θ⌋=⌊k​Nn⌋\+δk\.\\lfloor k\\theta\\rfloor=\\left\\lfloor\\frac\{kN\}\{n\}\\right\\rfloor\+\\delta\_\{k\}\.Substituting intoF​\(θ\)F\(\\theta\)gives

F​\(θ\)\\displaystyle F\(\\theta\)=N−∑k=1n⌊k​Nn⌋\+δkk\\displaystyle=N\-\\sum\_\{k=1\}^\{n\}\\frac\{\\left\\lfloor\\frac\{kN\}\{n\}\\right\\rfloor\+\\delta\_\{k\}\}\{k\}=N−∑k=1n⌊k​Nn⌋k⏟A​\(N\)−∑k=1nδkk\.\\displaystyle=\\underbrace\{N\-\\sum\_\{k=1\}^\{n\}\\frac\{\\left\\lfloor\\frac\{kN\}\{n\}\\right\\rfloor\}\{k\}\}\_\{A\(N\)\}\\;\-\\;\\sum\_\{k=1\}^\{n\}\\frac\{\\delta\_\{k\}\}\{k\}\.Since eachδk/k≥0\\delta\_\{k\}/k\\geq 0, we obtain the important inequality

F\(θ\)≤A\(N\),whereA\(N\)=N−∑k=1n⌊k​Nn⌋k\.\(1\)F\(\\theta\)\\leq A\(N\),\\qquad\\text\{where\}\\qquad A\(N\)=N\-\\sum\_\{k=1\}^\{n\}\\frac\{\\left\\lfloor\\frac\{kN\}\{n\}\\right\\rfloor\}\{k\}\.\\quad\\text\{\(1\)\}Thus it suffices to proveA​\(N\)≤Hn−1A\(N\)\\leq H\_\{n\}\-1for allN=0,1,…,n−1N=0,1,\\dots,n\-1\.

Step 4\. SimplifyingA​\(N\)A\(N\)\.Using⌊k​Nn⌋=k​N−rkn\\left\\lfloor\\frac\{kN\}\{n\}\\right\\rfloor=\\frac\{kN\-r\_\{k\}\}\{n\}, we compute

A​\(N\)\\displaystyle A\(N\)=N−∑k=1nk​N−rkn​k\\displaystyle=N\-\\sum\_\{k=1\}^\{n\}\\frac\{kN\-r\_\{k\}\}\{nk\}=N−∑k=1n\(Nn−rkn​k\)\\displaystyle=N\-\\sum\_\{k=1\}^\{n\}\\left\(\\frac\{N\}\{n\}\-\\frac\{r\_\{k\}\}\{nk\}\\right\)=N−\(Nn​∑k=1n1\)\+1n​∑k=1nrkk\\displaystyle=N\-\\left\(\\frac\{N\}\{n\}\\sum\_\{k=1\}^\{n\}1\\right\)\+\\frac\{1\}\{n\}\\sum\_\{k=1\}^\{n\}\\frac\{r\_\{k\}\}\{k\}=N−N\+1n∑k=1nrkk=1n∑k=1nrkk\.\(2\)\\displaystyle=N\-N\+\\frac\{1\}\{n\}\\sum\_\{k=1\}^\{n\}\\frac\{r\_\{k\}\}\{k\}=\\frac\{1\}\{n\}\\sum\_\{k=1\}^\{n\}\\frac\{r\_\{k\}\}\{k\}\.\\quad\\text\{\(2\)\}Step 5\. Introducing the greatest common divisor\.Letd=gcd⁡\(N,n\)d=\\gcd\(N,n\)\. Write

n=d​n1,N=d​N1,n=d\\,n\_\{1\},\\qquad N=d\\,N\_\{1\},withgcd⁡\(N1,n1\)=1\\gcd\(N\_\{1\},n\_\{1\}\)=1\.

Becauserkr\_\{k\}is the remainder ofk​NkNmodulonn, we haverk=d⋅skr\_\{k\}=d\\cdot s\_\{k\}, where

sk=\(k​N1\)modn1,0≤sk<n1\.s\_\{k\}=\(kN\_\{1\}\)\\bmod n\_\{1\},\\quad 0\\leq s\_\{k\}<n\_\{1\}\.Moreover, askkruns from11tonn, the valuessks\_\{k\}take each integer from0ton1−1n\_\{1\}\-1exactlyddtimes \(this follows from the fact thatk↦k​N1\(modn1\)k\\mapsto kN\_\{1\}\\pmod\{n\_\{1\}\}is a bijection over each block of lengthn1n\_\{1\}\)\. Therefore

1n∑k=1nrkk=1n∑k=1nd​skk=dn∑k=1nskk=1n1∑k=1nskk\.\(3\)\\frac\{1\}\{n\}\\sum\_\{k=1\}^\{n\}\\frac\{r\_\{k\}\}\{k\}=\\frac\{1\}\{n\}\\sum\_\{k=1\}^\{n\}\\frac\{d\\,s\_\{k\}\}\{k\}=\\frac\{d\}\{n\}\\sum\_\{k=1\}^\{n\}\\frac\{s\_\{k\}\}\{k\}=\\frac\{1\}\{n\_\{1\}\}\\sum\_\{k=1\}^\{n\}\\frac\{s\_\{k\}\}\{k\}\.\\quad\\text\{\(3\)\}Now decomposekkask=i​n1\+jk=in\_\{1\}\+jwithi=0,1,…,d−1i=0,1,\\dots,d\-1andj=1,2,…,n1j=1,2,\\dots,n\_\{1\}\. Thensk=sjs\_\{k\}=s\_\{j\}becausesi​n1\+j=\(\(i​n1\+j\)​N1\)modn1=\(j​N1\)modn1=sjs\_\{in\_\{1\}\+j\}=\(\(in\_\{1\}\+j\)N\_\{1\}\)\\bmod n\_\{1\}=\(jN\_\{1\}\)\\bmod n\_\{1\}=s\_\{j\}\. Hence

∑k=1nskk=∑j=1n1sj​∑i=0d−11i​n1\+j\.\\sum\_\{k=1\}^\{n\}\\frac\{s\_\{k\}\}\{k\}=\\sum\_\{j=1\}^\{n\_\{1\}\}s\_\{j\}\\sum\_\{i=0\}^\{d\-1\}\\frac\{1\}\{in\_\{1\}\+j\}\.Define the weights

wj=∑i=0d−11i​n1\+j,j=1,…,n1\.w\_\{j\}=\\sum\_\{i=0\}^\{d\-1\}\\frac\{1\}\{in\_\{1\}\+j\},\\qquad j=1,\\dots,n\_\{1\}\.Observe thatwjw\_\{j\}is strictly decreasing injj\(since largerjjmakes denominators larger\)\. Then

A\(N\)=1n1∑j=1n1sjwj\.\(4\)A\(N\)=\\frac\{1\}\{n\_\{1\}\}\\sum\_\{j=1\}^\{n\_\{1\}\}s\_\{j\}w\_\{j\}\.\\quad\\text\{\(4\)\}Step 6\. Applying the rearrangement inequality\.The set\{s1,…,sn1\}\\\{s\_\{1\},\\dots,s\_\{n\_\{1\}\}\\\}is a permutation of\{0,1,…,n1−1\}\\\{0,1,\\dots,n\_\{1\}\-1\\\}\(this is true for the indicesj=1,…,n1j=1,\\dots,n\_\{1\}, becausej​N1modn1jN\_\{1\}\\bmod n\_\{1\}runs through all residues exactly once\)\. The weightswjw\_\{j\}are decreasing\. By the rearrangement inequality, the sum∑sj​wj\\sum s\_\{j\}w\_\{j\}is maximized when thesjs\_\{j\}are also arranged in decreasing order\. The decreasing order of the numbers0,1,…,n1−10,1,\\dots,n\_\{1\}\-1iss1=n1−1s\_\{1\}=n\_\{1\}\-1,s2=n1−2s\_\{2\}=n\_\{1\}\-2, …,sn1=0s\_\{n\_\{1\}\}=0\. Hence for any permutation we have

∑j=1n1sjwj≤∑j=1n1\(n1−j\)wj\.\(5\)\\sum\_\{j=1\}^\{n\_\{1\}\}s\_\{j\}w\_\{j\}\\leq\\sum\_\{j=1\}^\{n\_\{1\}\}\(n\_\{1\}\-j\)w\_\{j\}\.\\quad\\text\{\(5\)\}Consequently,

A\(N\)≤1n1∑j=1n1\(n1−j\)wj\.\(6\)A\(N\)\\leq\\frac\{1\}\{n\_\{1\}\}\\sum\_\{j=1\}^\{n\_\{1\}\}\(n\_\{1\}\-j\)w\_\{j\}\.\\quad\\text\{\(6\)\}Step 7\. Relating to the harmonic number\.Compute the right\-hand side:

∑j=1n1\(n1−j\)wj=∑j=1n1\(n1−j\)∑i=0d−11i​n1\+j=∑i=0d−1∑j=1n1n1−ji​n1\+j=:S\.\\sum\_\{j=1\}^\{n\_\{1\}\}\(n\_\{1\}\-j\)w\_\{j\}=\\sum\_\{j=1\}^\{n\_\{1\}\}\(n\_\{1\}\-j\)\\sum\_\{i=0\}^\{d\-1\}\\frac\{1\}\{in\_\{1\}\+j\}=\\sum\_\{i=0\}^\{d\-1\}\\sum\_\{j=1\}^\{n\_\{1\}\}\\frac\{n\_\{1\}\-j\}\{in\_\{1\}\+j\}=:S\.Now the harmonic numberHnH\_\{n\}can be written as

Hn=∑i=0d−1∑j=1n11i​n1\+j\.H\_\{n\}=\\sum\_\{i=0\}^\{d\-1\}\\sum\_\{j=1\}^\{n\_\{1\}\}\\frac\{1\}\{in\_\{1\}\+j\}\.Thus

n1​Hn=∑i=0d−1∑j=1n1n1i​n1\+j\.n\_\{1\}H\_\{n\}=\\sum\_\{i=0\}^\{d\-1\}\\sum\_\{j=1\}^\{n\_\{1\}\}\\frac\{n\_\{1\}\}\{in\_\{1\}\+j\}\.Consider the difference

n1​Hn−n1−S=∑i=0d−1∑j=1n1\(n1i​n1\+j−n1−ji​n1\+j\)−n1=∑i=0d−1∑j=1n1ji​n1\+j−n1\.n\_\{1\}H\_\{n\}\-n\_\{1\}\-S=\\sum\_\{i=0\}^\{d\-1\}\\sum\_\{j=1\}^\{n\_\{1\}\}\\left\(\\frac\{n\_\{1\}\}\{in\_\{1\}\+j\}\-\\frac\{n\_\{1\}\-j\}\{in\_\{1\}\+j\}\\right\)\-n\_\{1\}=\\sum\_\{i=0\}^\{d\-1\}\\sum\_\{j=1\}^\{n\_\{1\}\}\\frac\{j\}\{in\_\{1\}\+j\}\-n\_\{1\}\.Define

T=∑i=0d−1∑j=1n1ji​n1\+j\.\(7\)T=\\sum\_\{i=0\}^\{d\-1\}\\sum\_\{j=1\}^\{n\_\{1\}\}\\frac\{j\}\{in\_\{1\}\+j\}\.\\quad\\text\{\(7\)\}ThenS≤n1​Hn−n1S\\leq n\_\{1\}H\_\{n\}\-n\_\{1\}is equivalent toT≥n1T\\geq n\_\{1\}\.

Step 8\. Proof thatT≥n1T\\geq n\_\{1\}\.Rewrite each term:

ji​n1\+j=1−i​n1i​n1\+j\.\\frac\{j\}\{in\_\{1\}\+j\}=1\-\\frac\{in\_\{1\}\}\{in\_\{1\}\+j\}\.Hence

T\\displaystyle T=∑i=0d−1∑j=1n1\(1−i​n1i​n1\+j\)\\displaystyle=\\sum\_\{i=0\}^\{d\-1\}\\sum\_\{j=1\}^\{n\_\{1\}\}\\left\(1\-\\frac\{in\_\{1\}\}\{in\_\{1\}\+j\}\\right\)=∑i=0d−1∑j=1n11−∑i=0d−1∑j=1n1i​n1i​n1\+j\\displaystyle=\\sum\_\{i=0\}^\{d\-1\}\\sum\_\{j=1\}^\{n\_\{1\}\}1\\;\-\\;\\sum\_\{i=0\}^\{d\-1\}\\sum\_\{j=1\}^\{n\_\{1\}\}\\frac\{in\_\{1\}\}\{in\_\{1\}\+j\}=d​n1−n1​∑i=0d−1i​∑j=1n11i​n1\+j\.\\displaystyle=dn\_\{1\}\\;\-\\;n\_\{1\}\\sum\_\{i=0\}^\{d\-1\}i\\sum\_\{j=1\}^\{n\_\{1\}\}\\frac\{1\}\{in\_\{1\}\+j\}\.Let

Q=∑i=0d−1i​∑j=1n11i​n1\+j\.Q=\\sum\_\{i=0\}^\{d\-1\}i\\sum\_\{j=1\}^\{n\_\{1\}\}\\frac\{1\}\{in\_\{1\}\+j\}\.The term withi=0i=0is0\. Fori≥1i\\geq 1, note thati​n1\+j\>i​n1in\_\{1\}\+j\>in\_\{1\}for allj≥1j\\geq 1, so

∑j=1n11i​n1\+j<∑j=1n11i​n1=n1i​n1=1i\.\\sum\_\{j=1\}^\{n\_\{1\}\}\\frac\{1\}\{in\_\{1\}\+j\}<\\sum\_\{j=1\}^\{n\_\{1\}\}\\frac\{1\}\{in\_\{1\}\}=\\frac\{n\_\{1\}\}\{in\_\{1\}\}=\\frac\{1\}\{i\}\.Multiplying byii\(positive\) gives

i​∑j=1n11i​n1\+j<1\.i\\sum\_\{j=1\}^\{n\_\{1\}\}\\frac\{1\}\{in\_\{1\}\+j\}<1\.Summing these inequalities fori=1,2,…,d−1i=1,2,\\dots,d\-1yields

Q<∑i=1d−11=d−1\.\(8\)Q<\\sum\_\{i=1\}^\{d\-1\}1=d\-1\.\\quad\\text\{\(8\)\}Ifd=1d=1, thenQ=0Q=0\(the sum overiiis empty\)\. Then

T=d​n1−n1​Q=n1\(since​d=1\)\.T=dn\_\{1\}\-n\_\{1\}Q=n\_\{1\}\\qquad\(\\text\{since \}d=1\)\.Ifd≥2d\\geq 2, we haveQ<d−1Q<d\-1, hence

T=d​n1−n1​Q\>d​n1−n1​\(d−1\)=n1\.T=dn\_\{1\}\-n\_\{1\}Q\>dn\_\{1\}\-n\_\{1\}\(d\-1\)=n\_\{1\}\.Thus in all casesT≥n1T\\geq n\_\{1\}, with equality exactly whend=1d=1\.

Step 9\. Final bound forA​\(N\)A\(N\)\.From the equivalence established in Step 7,T≥n1T\\geq n\_\{1\}impliesS≤n1​Hn−n1S\\leq n\_\{1\}H\_\{n\}\-n\_\{1\}\. Using \(6\) we obtain

A​\(N\)≤1n1​S≤1n1​\(n1​Hn−n1\)=Hn−1\.A\(N\)\\leq\\frac\{1\}\{n\_\{1\}\}S\\leq\\frac\{1\}\{n\_\{1\}\}\(n\_\{1\}H\_\{n\}\-n\_\{1\}\)=H\_\{n\}\-1\.HenceA​\(N\)≤Hn−1A\(N\)\\leq H\_\{n\}\-1for everyNN\.

Step 10\. Equality conditions\.We have the chain

F​\(θ\)≤A​\(N\)≤Hn−1\.F\(\\theta\)\\leq A\(N\)\\leq H\_\{n\}\-1\.To haveF​\(θ\)=Hn−1F\(\\theta\)=H\_\{n\}\-1, both inequalities must be equalities\.

⊳\\trianglerightEquality inF​\(θ\)≤A​\(N\)F\(\\theta\)\\leq A\(N\)requires∑k=1nδk/k=0\\sum\_\{k=1\}^\{n\}\\delta\_\{k\}/k=0, i\.e\.,δk=0\\delta\_\{k\}=0for allkk\. This meansrkn\+k​t<1\\frac\{r\_\{k\}\}\{n\}\+kt<1for everykk\.

⊳\\trianglerightEquality inA​\(N\)≤Hn−1A\(N\)\\leq H\_\{n\}\-1requires that the upper bound we derived be attained\. From the derivation, this happens exactly whend=1d=1\(i\.e\.,gcd⁡\(N,n\)=1\\gcd\(N,n\)=1\) and the rearrangement inequality \(5\) is sharp\. Sincewjw\_\{j\}is strictly decreasing, the maximum of∑sj​wj\\sum s\_\{j\}w\_\{j\}is achieved when thesjs\_\{j\}are also decreasing\. Ford=1d=1, the set\{s1,…,sn\}\\\{s\_\{1\},\\dots,s\_\{n\}\\\}\(sincen1=nn\_\{1\}=n\) is a permutation of\{0,1,…,n−1\}\\\{0,1,\\dots,n\-1\\\}\. The decreasing order corresponds tosj=n−js\_\{j\}=n\-jforj=1,…,nj=1,\\dots,n\(withsn=0s\_\{n\}=0\)\. This is equivalent to\(j​N\)modn=n−j\(jN\)\\bmod n=n\-jfor allj=1,…,nj=1,\\dots,n\. This congruence simplifies toj​N≡−j\(modn\)jN\\equiv\-j\\pmod\{n\}, i\.e\.,N≡−1\(modn\)N\\equiv\-1\\pmod\{n\}\. Because0≤N≤n−10\\leq N\\leq n\-1, this forcesN=n−1N=n\-1\.

Now check that forN=n−1N=n\-1the conditionδk=0\\delta\_\{k\}=0indeed holds\. WriteN=n−1N=n\-1, thend=gcd⁡\(n−1,n\)=1d=\\gcd\(n\-1,n\)=1\. Fork=1,…,nk=1,\\dots,n,rk=\(k​\(n−1\)\)modn=n−kr\_\{k\}=\(k\(n\-1\)\)\\bmod n=n\-k\(fork<nk<n\) andrn=0r\_\{n\}=0\. Then

rkn\+k​t=n−kn\+k​t\(or​0\+n​t​for​k=n\)\.\\frac\{r\_\{k\}\}\{n\}\+kt=\\frac\{n\-k\}\{n\}\+kt\\quad\(\\text\{or \}0\+nt\\text\{ for \}k=n\)\.Sincet<1/nt<1/n, we haven−kn\+k​t<n−kn\+k⋅1n=1,\\frac\{n\-k\}\{n\}\+kt<\\frac\{n\-k\}\{n\}\+k\\cdot\\frac\{1\}\{n\}=1,and fork=nk=n,n​t<1nt<1\. Henceδk=0\\delta\_\{k\}=0for allkk\. ThereforeF​\(θ\)=A​\(N\)=Hn−1F\(\\theta\)=A\(N\)=H\_\{n\}\-1\.

Conversely, ifθ∈\[1−1n,1\)\\theta\\in\\left\[1\-\\frac\{1\}\{n\},\\,1\\right\), thenN=n−1N=n\-1andt=θ−\(n−1\)/n∈\[0,1/n\)t=\\theta\-\(n\-1\)/n\\in\[0,1/n\)\. As just verified, this yieldsF​\(θ\)=Hn−1F\(\\theta\)=H\_\{n\}\-1\.

Thus the maximum valueHn−1H\_\{n\}\-1is attained precisely when the fractional part ofxxlies in the interval\[1−1/n,1\)\[1\-1/n,\\,1\)\.

Step 11\. Conclusion\.The maximal value of the expression isHn−1H\_\{n\}\-1, whereHn=1\+12\+⋯\+1nH\_\{n\}=1\+\\frac\{1\}\{2\}\+\\cdots\+\\frac\{1\}\{n\}, and this maximum occurs exactly for those real numbersxxwhose fractional part belongs to\[1−1/n,1\)\[1\-1/n,\\,1\)\.

Hn−1,where​Hn=1\+12\+⋯\+1n,and the maximum is attained iff​\{x\}∈\[1−1n,1\)\\boxed\{H\_\{n\}\-1,\\quad\\text\{where \}H\_\{n\}=1\+\\frac\{1\}\{2\}\+\\cdots\+\\frac\{1\}\{n\},\\ \\text\{and the maximum is attained iff \}\\\{x\\\}\\in\\left\[1\-\\frac\{1\}\{n\},\\,1\\right\)\}

#### USAMO 2026 Problem 2\.

Annie is playing a game where she starts with a row of positive integers, written on a blackboard, each of which is a power of22\. On each turn, she can erase two adjacent numbers and replace them with a power of22that is greater than either of the erased numbers\. This shortens the row of numbers, and she continues to take turns until only one number remains\. Annie wins the game if the final remaining number is less than44times the sum of the original numbers\. Is it always possible for Annie to win, regardless of the starting row of numbers?

Excerpted solution: the original response is very long, so only the beginning and ending portions are shown\.

Solution

We will prove that Annie can always win, i\.e\., for any initial row of powers of two, there exists a sequence of moves that ends with a single numberFFsatisfyingF<4​SF<4S, whereSSis the sum of the original numbers\.

LetSSbe the sum of the numbers, and letk=⌊log2⁡S⌋k=\\lfloor\\log\_\{2\}S\\rfloor, so that

2k≤S<2k\+1\.2^\{k\}\\leq S<2^\{k\+1\}\.The proof proceeds by strong induction on the number of termsnn\. We will show the following statement:

*Claim\.For any list ofnnpowers of two, there exists a sequence of moves such that the final numberFFsatisfiesF≤4⋅2kF\\leq 4\\cdot 2^\{k\}\.*

Once the Claim is established, we obtain Annie’s victory:

⊳\\trianglerightIfSSisnota power of two, then2k<S2^\{k\}<S, hence4⋅2k<4​S4\\cdot 2^\{k\}<4Sand thereforeF<4​SF<4S\.

⊳\\trianglerightIfSSis a power of two, the same inductive proof \(when analysed more carefully\) actually yields the sharper boundF≤2​SF\\leq 2S, which is certainly<4​S<4SforS\>0S\>0\. \(We will comment on this improvement at the end\.\)

Thus the main task is to prove the Claim by induction onnn\.

Base casen=1n=1

There is only the number itself, soF=SF=S\. SinceS≤2k\+1≤4⋅2kS\\leq 2^\{k\+1\}\\leq 4\\cdot 2^\{k\}, the Claim holds\.

Inductive step

Assume the Claim holds for every list of size less thannn\. Consider a list of sizenn\. Let its sum beSSand letk=⌊log2⁡S⌋k=\\lfloor\\log\_\{2\}S\\rfloor\.

Case 1: There exists an adjacent equal pair

Choose such a pair, sayai=ai\+1=aa\_\{i\}=a\_\{i\+1\}=a\. Replace them by2​a2a\(the smallest allowed power of two greater thanaa\)\. The sum does not change, sokkremains the same\. The new list has sizen−1n\-1, so by the induction hypothesis it can be reduced to a final number≤4⋅2k\\leq 4\\cdot 2^\{k\}\. Hence the same holds for the original list\.

Case 2: No two adjacent numbers are equal

In this case we choose an adjacent pair\(ai,ai\+1\)\(a\_\{i\},a\_\{i\+1\}\)for whichmax⁡\(ai,ai\+1\)\\max\(a\_\{i\},a\_\{i\+1\}\)is as small as possible; denote the two numbers byxxandyywithx<yx<y\. Replace them by2​y2y\(again the smallest possible power of two greater thanyy\)\. Let the new sum be

S′=S\+\(y−x\)\.S^\{\\prime\}=S\+\(y\-x\)\.Setk′=⌊log2⁡S′⌋k^\{\\prime\}=\\lfloor\\log\_\{2\}S^\{\\prime\}\\rfloor\.

We now consider two subcases\.

Subcase 2a:k′=kk^\{\\prime\}=k

ThenS′<2k\+1S^\{\\prime\}<2^\{k\+1\}, so the floor has not increased\. The new list has sizen−1n\-1; by the induction hypothesis it can be reduced to a final number≤4⋅2k′=4⋅2k\\leq 4\\cdot 2^\{k^\{\\prime\}\}=4\\cdot 2^\{k\}\. So the original list also admits such a strategy\.

Subcase 2b:k′=k\+1k^\{\\prime\}=k\+1

In this situation2k\+1≤S′<2k\+22^\{k\+1\}\\leq S^\{\\prime\}<2^\{k\+2\}\. First note thaty≤2ky\\leq 2^\{k\}; otherwisey≥2k\+1y\\geq 2^\{k\+1\}would forceS≥y≥2k\+1S\\geq y\\geq 2^\{k\+1\}, contradictingS<2k\+1S<2^\{k\+1\}\. Consequently2​y≤2k\+12y\\leq 2^\{k\+1\}\.

Now examine the list after the merge; it has sizen−1n\-1and sumS′S^\{\\prime\}\. Although its floor isk\+1k\+1, we can still use the induction hypothesis \- not directly on this list \(which would give a bound of4⋅2k\+1=8⋅2k4\\cdot 2^\{k\+1\}=8\\cdot 2^\{k\}\), but we can instead apply the induction hypothesis to therestof the list \- i\.e\., the numbers that were not part of the merged pair\. However, after the merge the list is contiguous, so the ”rest” is not isolated\. Instead we perform the following two step reduction:

1\.Reduce the part of the list that does not contain the new number2​y2y\(i\.e\., all other numbers\) to a single numberTT\. Because the rest consists of several \(possibly one\) contiguous blocks, we can reduce each block independently to a single number using the induction hypothesis \(the blocks have size<n<n\)\. By doing so we obtain a list consisting of2​y2yand the numbersT1,T2,…T\_\{1\},T\_\{2\},\\dots\(one per block\)\. EachTjT\_\{j\}satisfiesTj≤4⋅2kT\_\{j\}\\leq 4\\cdot 2^\{k\}because the sum of the corresponding block is<2k\+1<2^\{k\+1\}\(since the total sumS<2k\+1S<2^\{k\+1\}and we have removed at least the two numbersx,yx,ywithy≥2y\\geq 2, so the block sum is at most2k\+1−32^\{k\+1\}\-3? Actually a precise bound is not needed; we only need that every suchTjT\_\{j\}is≤4⋅2k\\leq 4\\cdot 2^\{k\}\. Indeed, for any block, its sum is less than2k\+12^\{k\+1\}, so by the induction hypothesis \(applied to that block\) we obtain a number≤4⋅2k\\leq 4\\cdot 2^\{k\}\(its floor is at mostkk, so the bound is4⋅2k4\\cdot 2^\{k\}\)\.

After this reduction, the list contains2​y2yand several numbersTjT\_\{j\}, each≤4⋅2k\\leq 4\\cdot 2^\{k\}\. Now we can merge these numbers one by one\. A crucial observation is that2​y≤2k\+1≤4⋅2k2y\\leq 2^\{k\+1\}\\leq 4\\cdot 2^\{k\}fork≥0k\\geq 0\. Hence all numbers present are at most4⋅2k4\\cdot 2^\{k\}\. When we merge two numbers each at most4⋅2k4\\cdot 2^\{k\}, the result is at most2⋅4⋅2k=8⋅2k2\\cdot 4\\cdot 2^\{k\}=8\\cdot 2^\{k\}, which is still too large\. To keep the bound at4⋅2k4\\cdot 2^\{k\}, we must merge in a way that the maximum never exceeds2k\+12^\{k\+1\}\. Notice that2​y≤2k\+12y\\leq 2^\{k\+1\}\. If we can ensure that everyTjT\_\{j\}is also≤2k\+1\\leq 2^\{k\+1\}, then merging any two of them yields at most2k\+2=4⋅2k2^\{k\+2\}=4\\cdot 2^\{k\}\. Therefore the problem reduces to showing that each block can be reduced to a number≤2k\+1\\leq 2^\{k\+1\}rather than the coarser4⋅2k4\\cdot 2^\{k\}\.

But can we guarantee≤2k\+1\\leq 2^\{k\+1\}? Since the sum of a block is<2k\+1<2^\{k\+1\}, a direct application of the induction hypothesis would only give≤4⋅2k\\leq 4\\cdot 2^\{k\}\. However, we can prove a slightly stronger lemma that will serve our purpose:

Lemma\.If a list has sumTTwith2L<T<2L\+12^\{L\}<T<2^\{L\+1\}, then there exists a strategy with final number≤2L\+1\\leq 2^\{L\+1\}providedthatT≤3⋅2L−1−1T\\leq 3\\cdot 2^\{L\-1\}\-1\. \(This lemma will be proved by induction onnnin a moment\.\)

In our situation, each block sum is at mostS−x−y<2k\+1S\-x\-y<2^\{k\+1\}\. Moreover, because we are in Subcase 2b, the original list has no adjacent equal and we chose the pair with smallest maximum\. A detailed analysis \(which we will carry out for the lemma\) shows that the sum of each block is actually≤3⋅2k−1−1\\leq 3\\cdot 2^\{k\-1\}\-1\. Hence the Lemma applies, givingTj≤2k\+1T\_\{j\}\\leq 2^\{k\+1\}\.

2\.With all numbers≤2k\+1\\leq 2^\{k\+1\}, merging any two yields at most2k\+2=4⋅2k2^\{k\+2\}=4\\cdot 2^\{k\}\. By continuing this process we eventually obtain a single number≤4⋅2k\\leq 4\\cdot 2^\{k\}\.

Thus, if the Lemma holds, the Claim is established in Subcase 2b as well\.

Proof of the Lemma

We now prove the Lemma stated above\.

Lemma\.LetL≥2L\\geq 2be an integer\. For any list of powers of two whose sumTTsatisfies

2L<T≤3⋅2L−1−1,2^\{L\}<T\\leq 3\\cdot 2^\{L\-1\}\-1,there exists a sequence of moves that ends with a number≤2L\+1\\leq 2^\{L\+1\}\.

*Proof\.*We use strong induction on the number of termsnn\.

⊳\\trianglerightBasen=2n=2\.The two numbers are powers of two, saya≤ba\\leq b\. SinceT=a\+b\>2LT=a\+b\>2^\{L\}, we must haveb≥2Lb\\geq 2^\{L\}\. Ifb\>2Lb\>2^\{L\}, thenb≥2L\+1b\\geq 2^\{L\+1\}, implyingT≥2L\+1T\\geq 2^\{L\+1\}, which contradictsT≤3⋅2L−1−1<2L\+1T\\leq 3\\cdot 2^\{L\-1\}\-1<2^\{L\+1\}\(forL≥2L\\geq 2\)\. Henceb=2Lb=2^\{L\}\. Thena=T−2L≤2L−1−1a=T\-2^\{L\}\\leq 2^\{L\-1\}\-1\. The only legal move is to replace\(a,2L\)\(a,2^\{L\}\)with a power of two greater than2L2^\{L\}; the smallest such is2L\+12^\{L\+1\}\. By choosing that, we obtain a final number exactly2L\+12^\{L\+1\}, which certainly is≤2L\+1\\leq 2^\{L\+1\}\.

⊳\\trianglerightInductive step\.Assume the Lemma holds for all lists of size smaller thannn\. Consider a list of sizennwith sumTTin the prescribed interval\.

*If there is an adjacent equal pair*, merge it\. The sum stays the same, soTTremains in the interval, and the size becomesn−1n\-1\. By the induction hypothesis we get a final number≤2L\+1\\leq 2^\{L\+1\}\.

*If there is no adjacent equal pair*, let\(x,y\)\(x,y\)be the adjacent pair with the smallest maximum;x<yx<y\. Merge them to2​y2yand obtain a new list of sizen−1n\-1with sumT′=T\+\(y−x\)T^\{\\prime\}=T\+\(y\-x\)\.

We need to show thatT′T^\{\\prime\}still lies in an interval that allows us to apply the induction hypothesis \(possibly with a differentLL\) and still obtain the bound2L\+12^\{L\+1\}\.

First, boundyy\.Because the list has no adjacent equal numbers and its sum is less than2L\+12^\{L\+1\}, the maximum element cannot exceed2L−12^\{L\-1\}if we want to stay within the interval after the merge? Actually we can havey=2Ly=2^\{L\}, but then the structure of the list is very restricted\. We analyse the two possibilities:

⊳\\trianglerighty≤2L−1y\\leq 2^\{L\-1\}\.Then2​y≤2L2y\\leq 2^\{L\}\. Also,y−x≤2L−1−1y\-x\\leq 2^\{L\-1\}\-1\. Consequently

T′≤T\+\(2L−1−1\)≤\(3⋅2L−1−1\)\+\(2L−1−1\)=4⋅2L−1−2=2L\+1−2\.T^\{\\prime\}\\leq T\+\(2^\{L\-1\}\-1\)\\leq\(3\\cdot 2^\{L\-1\}\-1\)\+\(2^\{L\-1\}\-1\)=4\\cdot 2^\{L\-1\}\-2=2^\{L\+1\}\-2\.Moreover,T′\>T\>2LT^\{\\prime\}\>T\>2^\{L\}\. HenceT′T^\{\\prime\}lies in\(2L,2L\+1\)\(2^\{L\},2^\{L\+1\}\)\. IfT′≤3⋅2L−1−1T^\{\\prime\}\\leq 3\\cdot 2^\{L\-1\}\-1, thenT′T^\{\\prime\}is still in the original interval forLL, and the induction hypothesis \(applied to the list of sizen−1n\-1\) givesF≤2L\+1F\\leq 2^\{L\+1\}\. IfT′\>3⋅2L−1−1T^\{\\prime\}\>3\\cdot 2^\{L\-1\}\-1, thenT′T^\{\\prime\}is in\(3⋅2L−1,2L\+1\)\(3\\cdot 2^\{L\-1\},2^\{L\+1\}\)\. In this range, we claim thatT′T^\{\\prime\}actually belongs to the interval forL\+1L\+1, i\.e\.,2L\+1<T′≤3⋅2L−12^\{L\+1\}<T^\{\\prime\}\\leq 3\\cdot 2^\{L\}\-1? ButT′<2L\+1T^\{\\prime\}<2^\{L\+1\}, so it cannot be in\(2L\+1,…\]\(2^\{L\+1\},\\dots\]\. However, observe thatT′≥3⋅2L−1\+1T^\{\\prime\}\\geq 3\\cdot 2^\{L\-1\}\+1\(since it exceeds the upper bound\)\. But thenT′≥3⋅2L−1\+1T^\{\\prime\}\\geq 3\\cdot 2^\{L\-1\}\+1\. ForL≥2L\\geq 2,3⋅2L−1\+13\\cdot 2^\{L\-1\}\+1may be less than2L\+12^\{L\+1\}, but we need a bound of2L\+12^\{L\+1\}\. Since we are only required to obtain some final number≤2L\+1\\leq 2^\{L\+1\}, we can simply note thatT′<2L\+1T^\{\\prime\}<2^\{L\+1\}and therefore the floor ofT′T^\{\\prime\}is at mostLL\. If the floor isLL\(i\.e\.,T′<2L\+1T^\{\\prime\}<2^\{L\+1\}\), then we can apply the induction hypothesis for the sameLL? The induction hypothesis requires the sum to be\>2L\>2^\{L\}\. It is, but we also need the upper bound3⋅2L−1−13\\cdot 2^\{L\-1\}\-1to use the Lemma? Actually the Lemma we are proving is exactly: for any list with sum in\(2L,3⋅2L−1−1\]\(2^\{L\},3\\cdot 2^\{L\-1\}\-1\]we can achieve≤2L\+1\\leq 2^\{L\+1\}\. IfT′T^\{\\prime\}falls outside this interval \(i\.e\.,T′\>3⋅2L−1−1T^\{\\prime\}\>3\\cdot 2^\{L\-1\}\-1\), the Lemma does not directly apply\. However, we can then apply the induction hypothesis for adifferentstatement: for any list with sum<2L\+1<2^\{L\+1\}we can achieve≤2L\+1\\leq 2^\{L\+1\}\. But is that universally true? No, as the example\[8,4,2\]\[8,4,2\]withL=3L=3\(sum 14\) shows that the minimum final is 32, which is\>16=2L\+1\>16=2^\{L\+1\}\. So we cannot claim that for every list with sum<2L\+1<2^\{L\+1\}we can get≤2L\+1\\leq 2^\{L\+1\}\. Hence we must be more careful\.

⊳\\trianglerighty=2Ly=2^\{L\}\.Then, because there are no adjacent equal numbers and the sum is at most3⋅2L−1−1<2L\+13\\cdot 2^\{L\-1\}\-1<2^\{L\+1\}, the list can contain at most one copy of2L2^\{L\}\(two would sum to at least2L\+12^\{L\+1\}\)\. Moreover, to avoid any adjacent pair with maximum<2L<2^\{L\}, every other number must be adjacent only to2L2^\{L\}\(otherwise a pair of smaller numbers would have maximum<2L<2^\{L\}\)\. This forces the list to be of the form

\[a1,a2,…,ap,2L,b1,b2,…,bq\]\[\\,a\_\{1\},a\_\{2\},\\dots,a\_\{p\},2^\{L\},b\_\{1\},b\_\{2\},\\dots,b\_\{q\}\\,\]where allaia\_\{i\}andbjb\_\{j\}are powers of two less than2L2^\{L\}, and no two of theaia\_\{i\}\(or two of thebjb\_\{j\}\) are adjacent; in particular,p≤1p\\leq 1andq≤1q\\leq 1\. Thus the list has at most three elements\. For size 2, the list is\[2L,c\]\[2^\{L\},c\]or\[c,2L\]\[c,2^\{L\}\]withc<2Lc<2^\{L\}\. That case is already covered by the base \(size 2\)\. For size 3, the list is\[a,2L,b\]\[a,2^\{L\},b\]witha,ba,bpowers of two,a,b<2La,b<2^\{L\}, anda\+b≤2L−1−1a\+b\\leq 2^\{L\-1\}\-1\(since the total sum is at most3⋅2L−1−13\\cdot 2^\{L\-1\}\-1\)\. Now, the pair with smallest maximum is either\(a,2L\)\(a,2^\{L\}\)or\(2L,b\)\(2^\{L\},b\), both have maximum2L2^\{L\}\. Choose one, say\(a,2L\)\(a,2^\{L\}\)\. Merging gives2L\+12^\{L\+1\}and the list becomes\[2L\+1,b\]\[2^\{L\+1\},b\]\. Its sum is2L\+1\+b2^\{L\+1\}\+b\. Sinceb≤2L−1−1b\\leq 2^\{L\-1\}\-1, we have

2L\+1<2L\+1\+b≤2L\+1\+2L−1−1=5⋅2L−1−1\.2^\{L\+1\}<2^\{L\+1\}\+b\\leq 2^\{L\+1\}\+2^\{L\-1\}\-1=5\\cdot 2^\{L\-1\}\-1\.ForL≥2L\\geq 2,5⋅2L−1−15\\cdot 2^\{L\-1\}\-1is still less than6⋅2L−1−1=3⋅2L−16\\cdot 2^\{L\-1\}\-1=3\\cdot 2^\{L\}\-1, so the sum lies in\(2L\+1,3⋅2L−1\]\(2^\{L\+1\},3\\cdot 2^\{L\}\-1\]\. Now, by the induction hypothesis \(the Lemma for the larger indexL\+1L\+1\), we can reduce this list to a number≤2\(L\+1\)\+1=2L\+2\\leq 2^\{\(L\+1\)\+1\}=2^\{L\+2\}\. But we need≤2L\+1\\leq 2^\{L\+1\}\. However, we are not forced to apply the Lemma forL\+1L\+1; we can directly finish the game: from\[2L\+1,b\]\[2^\{L\+1\},b\], the only move is to merge the two numbers \(they are different becauseb<2L<2L\+1b<2^\{L\}<2^\{L\+1\}\)\. The smallest possible result is2L\+22^\{L\+2\}\(since we must replace them with a power of two greater thanmax⁡\(2L\+1,b\)=2L\+1\\max\(2^\{L\+1\},b\)=2^\{L\+1\}, so the smallest is2L\+22^\{L\+2\}\. That would giveF=2L\+2F=2^\{L\+2\}, which is larger than2L\+12^\{L\+1\}\. But we are free to choose a larger number if we wish; however, to achieve the bound≤2L\+1\\leq 2^\{L\+1\}we would need to obtain exactly2L\+12^\{L\+1\}, which is impossible because the numbers are different and the rule requires a numbergreater thanboth, so the result must be at least2L\+22^\{L\+2\}\. This suggests that the Lemma might be false for lists of size 3 of this form? Let’s test with concrete numbers:L=3L=3, so2L=82^\{L\}=8\. Choosea=2a=2,b=2b=2? Buta\+b≤2L−1−1=3a\+b\\leq 2^\{L\-1\}\-1=3, soa=2,b=1a=2,b=1gives sum8\+2\+1=118\+2\+1=11, which is within the interval\(8,11\]\(8,11\]\. According to the Lemma, there should be a strategy yieldingF≤2L\+1=16F\\leq 2^\{L\+1\}=16\. But earlier we saw that from\[2,8,1\]\[2,8,1\]the minimum final is 32, not 16\. Let’s verify carefully:

List:\[2,8,1\]\[2,8,1\]\. No adjacent equal\. The pair with smallest maximum: pairs are\(2,8\)\(2,8\)max=8,\(8,1\)\(8,1\)max=8\. Suppose we merge\(2,8\)\(2,8\)to1616\(smallest possible\)\. Resulting list:\[16,1\]\[16,1\]\. Now merge\(16,1\)\(16,1\): the smallest power of two greater than 16 is 32\. So final 32\. If we merge\(8,1\)\(8,1\)first, we get\[2,16\]\[2,16\]then merge to 32\. So indeed the minimum is 32\. Therefore the Lemma, as stated, isfalseforL=3L=3with the list\[2,8,1\]\[2,8,1\]\. Indeed,2L=82^\{L\}=8,3⋅2L−1−1=113\\cdot 2^\{L\-1\}\-1=11, andT=11T=11is allowed\. But we cannot achieve≤16\\leq 16; the best is 32\.

Remaining solution omitted because the original response is substantially longer than the other appendix examples\.

#### USAMO 2026 Problem 3\.

LetA​B​CABCbe an acute scalene triangle with no angle equal to60∘60^\{\\circ\}\. Letω\\omegabe the circumcircle ofA​B​CABC\. LetΔB\\Delta\_\{B\}be the equilateral triangle with three vertices onω\\omega, one of which isBB\. LetℓB\\ell\_\{B\}be the line through the two vertices ofΔB\\Delta\_\{B\}other thanBB\. LetΔC\\Delta\_\{C\}andℓC\\ell\_\{C\}be defined analogously\. LetYYbe the intersection ofA​CACandℓB\\ell\_\{B\}, and letZZbe the intersection ofA​BABandℓC\\ell\_\{C\}\.LetNNbe the midpoint of minor arcB​CBConω\\omega\. Letℛ\\mathcal\{R\}be the triangle formed byℓB\\ell\_\{B\},ℓC\\ell\_\{C\}, and the tangent toω\\omegathroughNN\. Prove that the circumcircle ofA​Y​ZAYZand the incircle ofℛ\\mathcal\{R\}are tangent\.

Solution

We are given an acute scalene triangleA​B​CABCwith no angle equal to60∘60^\{\\circ\}\. Its circumcircle isω\\omega\. For vertexBBwe construct the equilateral triangleΔB\\Delta\_\{B\}inscribed inω\\omega\(one vertex isBB\); letℓB\\ell\_\{B\}be the line through the two other vertices\. Similarly,ΔC\\Delta\_\{C\}andℓC\\ell\_\{C\}\. Define

Y=A​C∩ℓB,Z=A​B∩ℓC\.Y=AC\\cap\\ell\_\{B\},\\qquad Z=AB\\cap\\ell\_\{C\}\.LetNNbe the midpoint of the minor arcB​CBC\(the arc not containingAA\)\. Letℛ\\mathcal\{R\}be the triangle bounded byℓB\\ell\_\{B\},ℓC\\ell\_\{C\}and the tangent toω\\omegaatNN\. We must prove that the circumcircle of△​A​Y​Z\\triangle AYZand the incircle of△​ℛ\\triangle\\mathcal\{R\}are tangent\.

1\. Complex numbers on the unit circle

Place the circumcircleω\\omegaas the unit circle in the complex plane\. Denote the points by

A=a,B=b,C=c,\|a\|=\|b\|=\|c\|=1\.A=a,\\quad B=b,\\quad C=c,\\qquad\|a\|=\|b\|=\|c\|=1\.The triangle is acute, scalene and none of its angles is60∘60^\{\\circ\}\.

Lemma 1 \(Line through two points on the unit circle\)\.For distinctu,vu,von the unit circle, the lineu​vuvis given by

z\+u​v​z¯=u\+v\.z\+uv\\,\\bar\{z\}=u\+v\.*Proof\.*A pointzzis collinear withuuandvviffz−uv−u\\frac\{z\-u\}\{v\-u\}is real\. Taking conjugates and usingu¯=1/u\\bar\{u\}=1/u,v¯=1/v\\bar\{v\}=1/vgives the stated equation\.□\\square

2\. Equations of the relevant lines

Letζ=e2​π​i/3\\zeta=e^\{2\\pi i/3\}\. The equilateral triangleΔB\\Delta\_\{B\}inscribed inω\\omegawith vertexBBhas the other two verticesB​ζB\\zetaandB​ζ2B\\zeta^\{2\}\. The side oppositeBBisℓB\\ell\_\{B\}, the line throughB​ζB\\zetaandB​ζ2B\\zeta^\{2\}\. Applying Lemma 1 withu=b​ζu=b\\zeta,v=b​ζ2v=b\\zeta^\{2\}yieldsu​v=b2uv=b^\{2\}andu\+v=b​\(ζ\+ζ2\)=−bu\+v=b\(\\zeta\+\\zeta^\{2\}\)=\-b\. Hence

ℓB:z\+b2z¯=−b\.\(1\)\\ell\_\{B\}:\\;z\+b^\{2\}\\bar\{z\}=\-b\.\\quad\\text\{\(1\)\}Analogously,

ℓC:z\+c2z¯=−c\.\(2\)\\ell\_\{C\}:\\;z\+c^\{2\}\\bar\{z\}=\-c\.\\quad\\text\{\(2\)\}The pointNNis the midpoint of the minor arcB​CBC\(not containingAA\)\. On the unit circle the midpoint of an arc has the property that its square equals the product of the endpoints:

N2=bc\.\(3\)N^\{2\}=bc\.\\quad\\text\{\(3\)\}The tangent toω\\omegaatNNis given by

z\+N2​z¯=2​N\.z\+N^\{2\}\\bar\{z\}=2N\.Using \(3\) we obtain

tangent atN:z\+bcz¯=2N\.\(4\)\\text\{tangent at \}N:\\;z\+bc\\,\\bar\{z\}=2N\.\\quad\\text\{\(4\)\}
3\. Intersection pointsYYandZZ

The lineA​CAChas equation \(by Lemma 1\)

AC:z\+acz¯=a\+c\.\(5\)AC:\\;z\+ac\\,\\bar\{z\}=a\+c\.\\quad\\text\{\(5\)\}IntersectA​CACwithℓB\\ell\_\{B\}\. Subtract \(5\) from \(1\):

\(b2−a​c\)​z¯=−b−\(a\+c\)=−\(a\+b\+c\)\.\(b^\{2\}\-ac\)\\,\\bar\{z\}=\-b\-\(a\+c\)=\-\(a\+b\+c\)\.Set

S=a\+b\+c,D1=a​c−b2\.S=a\+b\+c,\\qquad D\_\{1\}=ac\-b^\{2\}\.Thenz¯=S/D1\\bar\{z\}=S/D\_\{1\}\. Substitute into \(1\) to find the coordinateyy:

y=−b−b2​y¯=−b−b2​SD1=−b​D1\+b2​SD1\.y=\-b\-b^\{2\}\\bar\{y\}=\-b\-\\frac\{b^\{2\}S\}\{D\_\{1\}\}=\-\\frac\{bD\_\{1\}\+b^\{2\}S\}\{D\_\{1\}\}\.Compute the numerator:

b​D1\+b2​S=b​\(a​c−b2\)\+b2​\(a\+b\+c\)=a​b​c−b3\+a​b2\+b3\+b2​c=b​\(a​c\+a​b\+b​c\)=b​T,bD\_\{1\}\+b^\{2\}S=b\(ac\-b^\{2\}\)\+b^\{2\}\(a\+b\+c\)=abc\-b^\{3\}\+ab^\{2\}\+b^\{3\}\+b^\{2\}c=b\(ac\+ab\+bc\)=bT,where

T=a​b\+b​c\+c​a\.T=ab\+bc\+ca\.Thus

y=−b​TD1,y¯=SD1\.\(6\)y=\-\\frac\{bT\}\{D\_\{1\}\},\\qquad\\bar\{y\}=\\frac\{S\}\{D\_\{1\}\}\.\\quad\\text\{\(6\)\}By symmetry, intersectingA​BAB\(equationz\+a​b​z¯=a\+bz\+ab\\,\\bar\{z\}=a\+b\) withℓC\\ell\_\{C\}gives

z=−c​TD2,z¯=SD2,D2=ab−c2\.\(7\)z=\-\\frac\{cT\}\{D\_\{2\}\},\\qquad\\bar\{z\}=\\frac\{S\}\{D\_\{2\}\},\\qquad D\_\{2\}=ab\-c^\{2\}\.\\quad\\text\{\(7\)\}
4\. A convenient rotation

The configuration is invariant under rotations of the circle\. We choose the rotation so thatN=1N=1\. Then \(3\) yieldsb​c=1bc=1, hencec=b¯c=\\bar\{b\}\. Write

b=ei​α,c=e−i​α,b=e^\{i\\alpha\},\\qquad c=e^\{\-i\\alpha\},whereα=∠​A\\alpha=\\angle A\(by the inscribed angle theorem, the central angle subtended byB​CBCis2​α2\\alpha\)\. Since the triangle is acute,α∈\(0∘,90∘\)\\alpha\\in\(0^\{\\circ\},90^\{\\circ\}\)and, by hypothesis,α≠60∘\\alpha\\neq 60^\{\\circ\}\. Let

The remaining vertex isA=a=ei​θA=a=e^\{i\\theta\}\. BecauseN=1N=1lies on the minor arcB​CBCnot containingAA, the argumentθ\\thetasatisfiesθ∈\(α,2​π−α\)\\theta\\in\(\\alpha,\\,2\\pi\-\\alpha\)\(or its symmetric equivalent\)\.

Now compute the quantities that appear in \(6\), \(7\) in this coordinate system:

S\\displaystyle S=a\+b\+c=a\+ei​α\+e−i​α=a\+2​k,\\displaystyle=a\+b\+c=a\+e^\{i\\alpha\}\+e^\{\-i\\alpha\}=a\+2k,T\\displaystyle T=a​b\+b​c\+c​a=a​ei​α\+1\+a​e−i​α=1\+2​k​a,\\displaystyle=ab\+bc\+ca=ae^\{i\\alpha\}\+1\+ae^\{\-i\\alpha\}=1\+2ka,Q\\displaystyle Q=b2\+b​c\+c2=ei​2​α\+1\+e−i​2​α=2​cos⁡2​α\+1=4​k2−1,\\displaystyle=b^\{2\}\+bc\+c^\{2\}=e^\{i2\\alpha\}\+1\+e^\{\-i2\\alpha\}=2\\cos 2\\alpha\+1=4k^\{2\}\-1,D1\\displaystyle D\_\{1\}=a​c−b2=a​e−i​α−ei​2​α,\\displaystyle=ac\-b^\{2\}=ae^\{\-i\\alpha\}\-e^\{i2\\alpha\},D2\\displaystyle D\_\{2\}=a​b−c2=a​ei​α−e−i​2​α,\\displaystyle=ab\-c^\{2\}=ae^\{i\\alpha\}\-e^\{\-i2\\alpha\},Δ\\displaystyle\\Delta=D1​D2=\(a​e−i​α−ei​2​α\)​\(a​ei​α−e−i​2​α\)=a2−2​a​cos⁡3​α\+1\.\\displaystyle=D\_\{1\}D\_\{2\}=\(ae^\{\-i\\alpha\}\-e^\{i2\\alpha\}\)\(ae^\{i\\alpha\}\-e^\{\-i2\\alpha\}\)=a^\{2\}\-2a\\cos 3\\alpha\+1\.
5\. A key algebraic identity

Lemma 2\.With the above notation,

\(a\+2k\)2=QT\+Δ\.\(8\)\(a\+2k\)^\{2\}=Q\\,T\+\\Delta\.\\quad\\text\{\(8\)\}*Proof\.*ExpandQ​T\+ΔQ\\,T\+\\Delta:

Q​T\\displaystyle Q\\,T=\(4​k2−1\)​\(1\+2​k​a\)=\(4​k2−1\)\+2​k​\(4​k2−1\)​a,\\displaystyle=\(4k^\{2\}\-1\)\(1\+2ka\)=\(4k^\{2\}\-1\)\+2k\(4k^\{2\}\-1\)a,Δ\\displaystyle\\Delta=a2−2​a​cos⁡3​α\+1\.\\displaystyle=a^\{2\}\-2a\\cos 3\\alpha\+1\.Recallcos⁡3​α=4​k3−3​k\\cos 3\\alpha=4k^\{3\}\-3k\. Then

Q​T\+Δ\\displaystyle Q\\,T\+\\Delta=a2\+\[2​k​\(4​k2−1\)−2​cos⁡3​α\]​a\+\[\(4​k2−1\)\+1\]\\displaystyle=a^\{2\}\+\\bigl\[2k\(4k^\{2\}\-1\)\-2\\cos 3\\alpha\\bigr\]a\+\\bigl\[\(4k^\{2\}\-1\)\+1\\bigr\]=a2\+\[8​k3−2​k−\(8​k3−6​k\)\]​a\+4​k2\\displaystyle=a^\{2\}\+\\bigl\[8k^\{3\}\-2k\-\(8k^\{3\}\-6k\)\\bigr\]a\+4k^\{2\}=a2\+4ka\+4k2=\(a\+2k\)2\.□\\displaystyle=a^\{2\}\+4ka\+4k^\{2\}=\(a\+2k\)^\{2\}\.\\quad\\square
6\. The circumcenterOOof△​A​Y​Z\\triangle AYZ

We claim that the point

O=−a​Q​TΔ\(9\)O=\-\\frac\{aQT\}\{\\Delta\}\\quad\\text\{\(9\)\}is the circumcenter of△​A​Y​Z\\triangle AYZ\. Let us verify that it is equidistant fromAA,YYandZZ\.

Distance toAA

O−a=−a​Q​TΔ−a=−a​\(Q​TΔ\+1\)=−a​\(Q​T\+Δ\)Δ\.O\-a=\-\\frac\{aQT\}\{\\Delta\}\-a=\-a\\left\(\\frac\{QT\}\{\\Delta\}\+1\\right\)=\-\\frac\{a\(QT\+\\Delta\)\}\{\\Delta\}\.By Lemma 2,Q​T\+Δ=\(a\+2​k\)2QT\+\\Delta=\(a\+2k\)^\{2\}, so

O−a=−a​\(a\+2​k\)2Δ\.\(10\)O\-a=\-\\frac\{a\(a\+2k\)^\{2\}\}\{\\Delta\}\.\\quad\\text\{\(10\)\}Hence

\|O−a\|=\|a\+2​k\|2\|Δ\|\.\(11\)\|O\-a\|=\\frac\{\|a\+2k\|^\{2\}\}\{\|\\Delta\|\}\.\\quad\\text\{\(11\)\}
Distance toYY

Using \(6\),y¯=S/D1\\bar\{y\}=S/D\_\{1\}andy=−b​T/D1y=\-bT/D\_\{1\}\. Compute

O−y=−a​Q​TΔ\+b​TD1=T​\(−a​QΔ\+bD1\)\.O\-y=\-\\frac\{aQT\}\{\\Delta\}\+\\frac\{bT\}\{D\_\{1\}\}=T\\left\(\-\\frac\{aQ\}\{\\Delta\}\+\\frac\{b\}\{D\_\{1\}\}\\right\)\.SinceΔ=D1​D2\\Delta=D\_\{1\}D\_\{2\},

−a​QΔ\+bD1=−a​QD1​D2\+b​D2D1​D2=−a​Q\+b​D2Δ\.\-\\frac\{aQ\}\{\\Delta\}\+\\frac\{b\}\{D\_\{1\}\}=\\frac\{\-aQ\}\{D\_\{1\}D\_\{2\}\}\+\\frac\{bD\_\{2\}\}\{D\_\{1\}D\_\{2\}\}=\\frac\{\-aQ\+bD\_\{2\}\}\{\\Delta\}\.Thus

O−y=T−a​Q\+b​D2Δ\.\(12\)O\-y=T\\,\\frac\{\-aQ\+bD\_\{2\}\}\{\\Delta\}\.\\quad\\text\{\(12\)\}Now computeb​D2−a​QbD\_\{2\}\-aQ:

b​D2\\displaystyle bD\_\{2\}=ei​α​\(a​ei​α−e−i​2​α\)=a​ei​2​α−e−i​α,\\displaystyle=e^\{i\\alpha\}\(ae^\{i\\alpha\}\-e^\{\-i2\\alpha\}\)=ae^\{i2\\alpha\}\-e^\{\-i\\alpha\},a​Q\\displaystyle aQ=a​\(ei​2​α\+1\+e−i​2​α\)=a​ei​2​α\+a\+a​e−i​2​α,\\displaystyle=a\(e^\{i2\\alpha\}\+1\+e^\{\-i2\\alpha\}\)=ae^\{i2\\alpha\}\+a\+ae^\{\-i2\\alpha\},∴b​D2−a​Q\\displaystyle\\therefore\\;bD\_\{2\}\-aQ=−e−i​α−a−a​e−i​2​α\.\\displaystyle=\-e^\{\-i\\alpha\}\-a\-ae^\{\-i2\\alpha\}\.On the other hand,

−e−i​α​T=−e−i​α​\(1\+2​k​a\)=−e−i​α−2​k​a​e−i​α=−e−i​α−a​\(ei​α\+e−i​α\)​e−i​α=−e−i​α−a​\(1\+e−i​2​α\)=−e−i​α−a−a​e−i​2​α\.\-e^\{\-i\\alpha\}T=\-e^\{\-i\\alpha\}\(1\+2ka\)=\-e^\{\-i\\alpha\}\-2kae^\{\-i\\alpha\}=\-e^\{\-i\\alpha\}\-a\(e^\{i\\alpha\}\+e^\{\-i\\alpha\}\)e^\{\-i\\alpha\}=\-e^\{\-i\\alpha\}\-a\(1\+e^\{\-i2\\alpha\}\)=\-e^\{\-i\\alpha\}\-a\-ae^\{\-i2\\alpha\}\.Therefore

b​D2−a​Q=−e−i​α​T,bD\_\{2\}\-aQ=\-e^\{\-i\\alpha\}T,and consequently

\|b​D2−a​Q\|=\|T\|\.\|bD\_\{2\}\-aQ\|=\|T\|\.From \(12\) we obtain

\|O−y\|=\|T\|2\|Δ\|\.\(13\)\|O\-y\|=\\frac\{\|T\|^\{2\}\}\{\|\\Delta\|\}\.\\quad\\text\{\(13\)\}
Distance toZZ

A completely symmetric computation \(interchangingbbwithcc,D1D\_\{1\}withD2D\_\{2\}\) gives

\|O−z\|=\|T\|2\|Δ\|\.\(14\)\|O\-z\|=\\frac\{\|T\|^\{2\}\}\{\|\\Delta\|\}\.\\quad\\text\{\(14\)\}
Equality of the distances

Now note that

\|T\|2=\(1\+2​k​a\)​\(1\+2​k​a¯\)=1\+2​k​\(a\+a¯\)\+4​k2=1\+2​k​t\+4​k2,t=a\+a¯=2​cos⁡θ\.\|T\|^\{2\}=\(1\+2ka\)\(1\+2k\\bar\{a\}\)=1\+2k\(a\+\\bar\{a\}\)\+4k^\{2\}=1\+2kt\+4k^\{2\},\\qquad t=a\+\\bar\{a\}=2\\cos\\theta\.But also

\|a\+2​k\|2=\(a\+2​k\)​\(a¯\+2​k\)=1\+2​k​\(a\+a¯\)\+4​k2=\|T\|2\.\|a\+2k\|^\{2\}=\(a\+2k\)\(\\bar\{a\}\+2k\)=1\+2k\(a\+\\bar\{a\}\)\+4k^\{2\}=\|T\|^\{2\}\.Thus

\|O−a\|=\|a\+2​k\|2\|Δ\|,\|O−y\|=\|T\|2\|Δ\|=\|a\+2​k\|2\|Δ\|,\|O−z\|=\|a\+2​k\|2\|Δ\|\.\|O\-a\|=\\frac\{\|a\+2k\|^\{2\}\}\{\|\\Delta\|\},\\qquad\|O\-y\|=\\frac\{\|T\|^\{2\}\}\{\|\\Delta\|\}=\\frac\{\|a\+2k\|^\{2\}\}\{\|\\Delta\|\},\\qquad\|O\-z\|=\\frac\{\|a\+2k\|^\{2\}\}\{\|\\Delta\|\}\.HenceOOis indeed the circumcenter of△​A​Y​Z\\triangle AYZ, and the circumradius is

R=\|a\+2​k\|2\|Δ\|\.\(15\)R=\\frac\{\|a\+2k\|^\{2\}\}\{\|\\Delta\|\}\.\\quad\\text\{\(15\)\}
7\. Cartesian description of triangleℛ\\mathcal\{R\}

Now we work in the rotated coordinate system whereN=1N=1,b=ei​αb=e^\{i\\alpha\},c=e−i​αc=e^\{\-i\\alpha\}\. Write a complex numberz=x\+i​yz=x\+iy\.

Equation ofℓB\\ell\_\{B\}

From \(1\):z\+b2​z¯=−bz\+b^\{2\}\\bar\{z\}=\-b\. Substitutingb=cos⁡α\+i​sin⁡αb=\\cos\\alpha\+i\\sin\\alpha,b2=cos⁡2​α\+i​sin⁡2​αb^\{2\}=\\cos 2\\alpha\+i\\sin 2\\alpha, we separate real and imaginary parts:

\{x​\(1\+cos⁡2​α\)\+y​sin⁡2​α=−cos⁡α,x​sin⁡2​α\+y​\(1−cos⁡2​α\)=−sin⁡α\.\\begin\{cases\}x\(1\+\\cos 2\\alpha\)\+y\\sin 2\\alpha=\-\\cos\\alpha,\\\\\[5\.69054pt\] x\\sin 2\\alpha\+y\(1\-\\cos 2\\alpha\)=\-\\sin\\alpha\.\\end\{cases\}Using the identities

1\+cos⁡2​α=2​cos2⁡α,sin⁡2​α=2​sin⁡α​cos⁡α,1−cos⁡2​α=2​sin2⁡α,1\+\\cos 2\\alpha=2\\cos^\{2\}\\alpha,\\quad\\sin 2\\alpha=2\\sin\\alpha\\cos\\alpha,\\quad 1\-\\cos 2\\alpha=2\\sin^\{2\}\\alpha,and dividing the first equation by2​cos⁡α2\\cos\\alpha\(sincecos⁡α\>0\\cos\\alpha\>0\) and the second by2​sin⁡α2\\sin\\alpha\(sincesin⁡α\>0\\sin\\alpha\>0\), both reduce to

xcosα\+ysinα=−12\.\(16\)x\\cos\\alpha\+y\\sin\\alpha=\-\\frac\{1\}\{2\}\.\\quad\\text\{\(16\)\}
Equation ofℓC\\ell\_\{C\}

Similarly, forℓC\\ell\_\{C\}given byz\+c2​z¯=−cz\+c^\{2\}\\bar\{z\}=\-cwithc=cos⁡α−i​sin⁡αc=\\cos\\alpha\-i\\sin\\alpha,c2=cos⁡2​α−i​sin⁡2​αc^\{2\}=\\cos 2\\alpha\-i\\sin 2\\alpha\. Separating real and imaginary parts yields

\{x​\(1\+cos⁡2​α\)−y​sin⁡2​α=−cos⁡α,−x​sin⁡2​α\+y​\(1−cos⁡2​α\)=sin⁡α\.\\begin\{cases\}x\(1\+\\cos 2\\alpha\)\-y\\sin 2\\alpha=\-\\cos\\alpha,\\\\\[5\.69054pt\] \-x\\sin 2\\alpha\+y\(1\-\\cos 2\\alpha\)=\\sin\\alpha\.\\end\{cases\}Again, dividing appropriately we obtain

xcosα−ysinα=−12\.\(17\)x\\cos\\alpha\-y\\sin\\alpha=\-\\frac\{1\}\{2\}\.\\quad\\text\{\(17\)\}
Equation of the tangent atN=1N=1

SinceN=1N=1lies on the unit circle, its tangent is the linez\+z¯=2z\+\\bar\{z\}=2, i\.e\.,

x=1\.\(18\)x=1\.\\quad\\text\{\(18\)\}Thusℛ\\mathcal\{R\}is the triangle bounded by the three lines \(16\), \(17\) and \(18\)\.

8\. Vertices and interior ofℛ\\mathcal\{R\}

The three vertices are the pairwise intersections of the lines\.

⊳\\trianglerightP=ℓB∩ℓCP=\\ell\_\{B\}\\cap\\ell\_\{C\}: solving \(16\) and \(17\) gives

2​x​cos⁡α=−1⟹x=−12​cos⁡α,y=0\.2x\\cos\\alpha=\-1\\;\\Longrightarrow\\;x=\-\\frac\{1\}\{2\\cos\\alpha\},\\qquad y=0\.
⊳\\trianglerightQ′=ℓC∩tangent​x=1Q^\{\\prime\}=\\ell\_\{C\}\\cap\\text\{tangent \}x=1: plugx=1x=1into \(17\):

cos⁡α−y​sin⁡α=−12⟹y=cos⁡α\+12sin⁡α\.\\cos\\alpha\-y\\sin\\alpha=\-\\frac\{1\}\{2\}\\;\\Longrightarrow\\;y=\\frac\{\\cos\\alpha\+\\frac\{1\}\{2\}\}\{\\sin\\alpha\}\.
⊳\\trianglerightR′=ℓB∩tangent​x=1R^\{\\prime\}=\\ell\_\{B\}\\cap\\text\{tangent \}x=1: plugx=1x=1into \(16\):

cos⁡α\+y​sin⁡α=−12⟹y=−cos⁡α\+12sin⁡α\.\\cos\\alpha\+y\\sin\\alpha=\-\\frac\{1\}\{2\}\\;\\Longrightarrow\\;y=\-\\frac\{\\cos\\alpha\+\\frac\{1\}\{2\}\}\{\\sin\\alpha\}\.Hence the triangleℛ\\mathcal\{R\}has vertices

P​\(−12​cos⁡α,0\),Q′​\(1,cos⁡α\+12sin⁡α\),R′​\(1,−cos⁡α\+12sin⁡α\)\.P\\left\(\-\\frac\{1\}\{2\\cos\\alpha\},0\\right\),\\quad Q^\{\\prime\}\\left\(1,\\frac\{\\cos\\alpha\+\\frac\{1\}\{2\}\}\{\\sin\\alpha\}\\right\),\\quad R^\{\\prime\}\\left\(1,\-\\frac\{\\cos\\alpha\+\\frac\{1\}\{2\}\}\{\\sin\\alpha\}\\right\)\.To determine which side of each line constitutes the interior, we test the vertex opposite that line\.

⊳\\trianglerightFor lineℓB\\ell\_\{B\}, the opposite vertex isQ′Q^\{\\prime\}\. Compute the left\-hand side of \(16\) atQ′Q^\{\\prime\}:

x​cos⁡α\+y​sin⁡α=1⋅cos⁡α\+\(cos⁡α\+12sin⁡α\)​sin⁡α=cos⁡α\+cos⁡α\+12=2​cos⁡α\+12\>−12\.x\\cos\\alpha\+y\\sin\\alpha=1\\cdot\\cos\\alpha\+\\left\(\\frac\{\\cos\\alpha\+\\frac\{1\}\{2\}\}\{\\sin\\alpha\}\\right\)\\sin\\alpha=\\cos\\alpha\+\\cos\\alpha\+\\frac\{1\}\{2\}=2\\cos\\alpha\+\\frac\{1\}\{2\}\>\-\\frac\{1\}\{2\}\.Therefore the interior ofℛ\\mathcal\{R\}satisfies

xcosα\+ysinα\>−12\.\(19a\)x\\cos\\alpha\+y\\sin\\alpha\>\-\\frac\{1\}\{2\}\.\\quad\\text\{\(19a\)\}
⊳\\trianglerightFor lineℓC\\ell\_\{C\}, opposite vertex isR′R^\{\\prime\}\. AtR′R^\{\\prime\}:

x​cos⁡α−y​sin⁡α=1⋅cos⁡α−\(−cos⁡α\+12sin⁡α\)​sin⁡α=cos⁡α\+cos⁡α\+12=2​cos⁡α\+12\>−12,x\\cos\\alpha\-y\\sin\\alpha=1\\cdot\\cos\\alpha\-\\left\(\-\\frac\{\\cos\\alpha\+\\frac\{1\}\{2\}\}\{\\sin\\alpha\}\\right\)\\sin\\alpha=\\cos\\alpha\+\\cos\\alpha\+\\frac\{1\}\{2\}=2\\cos\\alpha\+\\frac\{1\}\{2\}\>\-\\frac\{1\}\{2\},so the interior satisfies

xcosα−ysinα\>−12\.\(19b\)x\\cos\\alpha\-y\\sin\\alpha\>\-\\frac\{1\}\{2\}\.\\quad\\text\{\(19b\)\}
⊳\\trianglerightFor the tangentx=1x=1, opposite vertex isPPwithxP=−12​cos⁡α<1x\_\{P\}=\-\\frac\{1\}\{2\\cos\\alpha\}<1, thus the interior satisfies

x<1\.\(19c\)x<1\.\\quad\\text\{\(19c\)\}Thereforeℛ=\{\(x,y\):x​cos⁡α\+y​sin⁡α\>−12,x​cos⁡α−y​sin⁡α\>−12,x<1\}\\mathcal\{R\}=\\bigl\\\{\(x,y\):x\\cos\\alpha\+y\\sin\\alpha\>\-\\frac\{1\}\{2\},\\;x\\cos\\alpha\-y\\sin\\alpha\>\-\\frac\{1\}\{2\},\\;x<1\\bigr\\\}\.

9\. IncenterIIand inradiusrrofℛ\\mathcal\{R\}

The triangle is symmetric with respect to thexx\-axis \(the linesℓB\\ell\_\{B\}andℓC\\ell\_\{C\}are symmetric, the tangent is vertical\)\. Hence the incenter lies on thexx\-axis:I=\(p,0\)I=\(p,0\)\.

For a point\(x,0\)\(x,0\)insideℛ\\mathcal\{R\}\(so satisfying the inequalities\), the distances to the three lines are:

⊳\\trianglerightToℓB\\ell\_\{B\}: the line isx​cos⁡α\+y​sin⁡α\+12=0x\\cos\\alpha\+y\\sin\\alpha\+\\frac\{1\}\{2\}=0; distance=x​cos⁡α\+12cos2⁡α\+sin2⁡α=x​cos⁡α\+12=\\dfrac\{x\\cos\\alpha\+\\frac\{1\}\{2\}\}\{\\sqrt\{\\cos^\{2\}\\alpha\+\\sin^\{2\}\\alpha\}\}=x\\cos\\alpha\+\\frac\{1\}\{2\}\(since the interior givesx​cos⁡α\+12\>0x\\cos\\alpha\+\\frac\{1\}\{2\}\>0\)\.

⊳\\trianglerightToℓC\\ell\_\{C\}: similarly, distance=x​cos⁡α\+12=x\\cos\\alpha\+\\frac\{1\}\{2\}\(becausex​cos⁡α−0\+12=x​cos⁡α\+12\>0x\\cos\\alpha\-0\+\\frac\{1\}\{2\}=x\\cos\\alpha\+\\frac\{1\}\{2\}\>0\)\.

⊳\\trianglerightTo the tangentx=1x=1: distance=1−x=1\-x\(sincex<1x<1\)\.

Setting these equal gives

x​cos⁡α\+12=1−x⟹x​\(cos⁡α\+1\)=12⟹x=12​\(1\+cos⁡α\)\.x\\cos\\alpha\+\\frac\{1\}\{2\}=1\-x\\quad\\Longrightarrow\\quad x\(\\cos\\alpha\+1\)=\\frac\{1\}\{2\}\\quad\\Longrightarrow\\quad x=\\frac\{1\}\{2\(1\+\\cos\\alpha\)\}\.Thus the incenter is

I=\(12​\(1\+cos⁡α\),0\)=\(p,0\),p=12​\(1\+k\)\.\(20\)I=\\left\(\\frac\{1\}\{2\(1\+\\cos\\alpha\)\},\\,0\\right\)=\(p,\\,0\),\\qquad p=\\frac\{1\}\{2\(1\+k\)\}\.\\quad\\text\{\(20\)\}The common distance is the inradius:

r=1−p=1−12​\(1\+k\)=2​\(1\+k\)−12​\(1\+k\)=1\+2​k2​\(1\+k\)\.\(21\)r=1\-p=1\-\\frac\{1\}\{2\(1\+k\)\}=\\frac\{2\(1\+k\)\-1\}\{2\(1\+k\)\}=\\frac\{1\+2k\}\{2\(1\+k\)\}\.\\quad\\text\{\(21\)\}
10\. DistanceO​IOI

We have the circumcenter

O=−a​Q​TΔ,O=\-\\frac\{aQT\}\{\\Delta\},and the incenter

I=p=12​\(1\+cos⁡α\)=1D0,I=p=\\frac\{1\}\{2\(1\+\\cos\\alpha\)\}=\\frac\{1\}\{D\_\{0\}\},where we denote

D0=2​\(1\+cos⁡α\)=2​\(1\+k\)\.D\_\{0\}=2\(1\+\\cos\\alpha\)=2\(1\+k\)\.Then

O−I=−a​Q​TΔ−p=−a​Q​T\+p​ΔΔ\.O\-I=\-\\frac\{aQT\}\{\\Delta\}\-p=\-\\frac\{aQT\+p\\Delta\}\{\\Delta\}\.SetU=a​Q​T\+p​ΔU=aQT\+p\\Delta; thenO​I=\|U\|/\|Δ\|OI=\|U\|/\|\\Delta\|\.

Now substitute the expressions fora​Q​TaQT,ppandΔ\\Delta\. Write

a​Q​T=a​\(4​k2−1\)​\(1\+2​k​a\)=\(4​k2−1\)​a\+2​k​\(4​k2−1\)​a2\.aQT=a\(4k^\{2\}\-1\)\(1\+2ka\)=\(4k^\{2\}\-1\)a\+2k\(4k^\{2\}\-1\)a^\{2\}\.Alsop=1/D0p=1/D\_\{0\}andΔ=a2−2​a​cos⁡3​α\+1\\Delta=a^\{2\}\-2a\\cos 3\\alpha\+1\.

Thus

U=\[2k\(4k2−1\)\+p\]a2\+\[\(4k2−1\)−2pcos3α\]a\+p\.\(22\)U=\\bigl\[2k\(4k^\{2\}\-1\)\+p\\bigr\]a^\{2\}\+\\bigl\[\(4k^\{2\}\-1\)\-2p\\cos 3\\alpha\\bigr\]a\+p\.\\quad\\text\{\(22\)\}Compute the coefficients with denominatorD0=2​\(1\+k\)D\_\{0\}=2\(1\+k\)\.

⊳\\trianglerightCoefficient ofa2a^\{2\}:

A2=2​k​\(4​k2−1\)\+1D0\.A\_\{2\}=2k\(4k^\{2\}\-1\)\+\\frac\{1\}\{D\_\{0\}\}\.Write as a single fraction:

A2=\(2​k​\(4​k2−1\)\)​D0\+1D0\.A\_\{2\}=\\frac\{\(2k\(4k^\{2\}\-1\)\)D\_\{0\}\+1\}\{D\_\{0\}\}\.Compute\(2​k​\(4​k2−1\)\)​D0=\(8​k3−2​k\)⋅2​\(1\+k\)=16​k4\+16​k3−4​k2−4​k\.\(2k\(4k^\{2\}\-1\)\)D\_\{0\}=\(8k^\{3\}\-2k\)\\cdot 2\(1\+k\)=16k^\{4\}\+16k^\{3\}\-4k^\{2\}\-4k\.Adding 1 gives

16​k4\+16​k3−4​k2−4​k\+1\.16k^\{4\}\+16k^\{3\}\-4k^\{2\}\-4k\+1\.Notice that

X=4​k2\+2​k−1,X=4k^\{2\}\+2k\-1,then

X2=\(4​k2\+2​k−1\)2=16​k4\+16​k3−4​k2−4​k\+1\.X^\{2\}=\(4k^\{2\}\+2k\-1\)^\{2\}=16k^\{4\}\+16k^\{3\}\-4k^\{2\}\-4k\+1\.Hence

A2=X2D0\.\(23\)A\_\{2\}=\\frac\{X^\{2\}\}\{D\_\{0\}\}\.\\quad\\text\{\(23\)\}
⊳\\trianglerightCoefficient ofaa:

A1=\(4​k2−1\)−2​p​cos⁡3​α=\(4​k2−1\)−2​cos⁡3​αD0\.A\_\{1\}=\(4k^\{2\}\-1\)\-2p\\cos 3\\alpha=\(4k^\{2\}\-1\)\-\\frac\{2\\cos 3\\alpha\}\{D\_\{0\}\}\.Write as

A1=\(4​k2−1\)​D0−2​cos⁡3​αD0\.A\_\{1\}=\\frac\{\(4k^\{2\}\-1\)D\_\{0\}\-2\\cos 3\\alpha\}\{D\_\{0\}\}\.Compute\(4​k2−1\)​D0=\(4​k2−1\)⋅2​\(1\+k\)=8​k3\+8​k2−2​k−2\.\(4k^\{2\}\-1\)D\_\{0\}=\(4k^\{2\}\-1\)\\cdot 2\(1\+k\)=8k^\{3\}\+8k^\{2\}\-2k\-2\.And2​cos⁡3​α=2​\(4​k3−3​k\)=8​k3−6​k\.2\\cos 3\\alpha=2\(4k^\{3\}\-3k\)=8k^\{3\}\-6k\.Subtract:

\(8​k3\+8​k2−2​k−2\)−\(8​k3−6​k\)=8​k2\+4​k−2=2​\(4​k2\+2​k−1\)=2​X\.\(8k^\{3\}\+8k^\{2\}\-2k\-2\)\-\(8k^\{3\}\-6k\)=8k^\{2\}\+4k\-2=2\(4k^\{2\}\+2k\-1\)=2X\.Therefore

A1=2​XD0\.\(24\)A\_\{1\}=\\frac\{2X\}\{D\_\{0\}\}\.\\quad\\text\{\(24\)\}
⊳\\trianglerightConstant term:p=1D0p=\\dfrac\{1\}\{D\_\{0\}\}\.

Substituting \(23\) and \(24\) into \(22\) yields

U=X2D0a2\+2​XD0a\+1D0=\(X​a\+1\)2D0\.\(25\)U=\\frac\{X^\{2\}\}\{D\_\{0\}\}a^\{2\}\+\\frac\{2X\}\{D\_\{0\}\}a\+\\frac\{1\}\{D\_\{0\}\}=\\frac\{\(Xa\+1\)^\{2\}\}\{D\_\{0\}\}\.\\quad\\text\{\(25\)\}Consequently,

OI=\|U\|\|Δ\|=\|X​a\+1\|2D0​\|Δ\|\.\(26\)OI=\\frac\{\|U\|\}\{\|\\Delta\|\}=\\frac\{\|Xa\+1\|^\{2\}\}\{D\_\{0\}\\,\|\\Delta\|\}\.\\quad\\text\{\(26\)\}
11\. RelatingO​IOItoRRandrr

Recall the circumradius of△​A​Y​Z\\triangle AYZ:

R=\|a\+2​k\|2\|Δ\|\.\(27\)R=\\frac\{\|a\+2k\|^\{2\}\}\{\|\\Delta\|\}\.\\quad\\text\{\(27\)\}The inradius ofℛ\\mathcal\{R\}is

r=1\+2​kD0\.\(28\)r=\\frac\{1\+2k\}\{D\_\{0\}\}\.\\quad\\text\{\(28\)\}Now compute

\|X​a\+1\|2D0=\|a\+2​k\|2\+L,\\frac\{\|Xa\+1\|^\{2\}\}\{D\_\{0\}\}=\|a\+2k\|^\{2\}\+L,where

L=\|X​a\+1\|2D0−\|a\+2​k\|2\.L=\\frac\{\|Xa\+1\|^\{2\}\}\{D\_\{0\}\}\-\|a\+2k\|^\{2\}\.From \(26\),

OI=\|a\+2​k\|2\+L\|Δ\|=R\+L\|Δ\|\.\(29\)OI=\\frac\{\|a\+2k\|^\{2\}\+L\}\{\|\\Delta\|\}=R\+\\frac\{L\}\{\|\\Delta\|\}\.\\quad\\text\{\(29\)\}Thus if we can showL=±r​\|Δ\|L=\\pm r\\,\|\\Delta\|, we will haveO​I=R±rOI=R\\pm r, which implies tangency\.

Lett=a\+a¯=2​cos⁡θt=a\+\\bar\{a\}=2\\cos\\theta\. Then

\|X​a\+1\|2=\(X​a\+1\)​\(X​a¯\+1\)=X2\+X​t\+1\.\|Xa\+1\|^\{2\}=\(Xa\+1\)\(X\\bar\{a\}\+1\)=X^\{2\}\+Xt\+1\.Also

\|a\+2​k\|2=\(a\+2​k\)​\(a¯\+2​k\)=1\+2​k​t\+4​k2\.\|a\+2k\|^\{2\}=\(a\+2k\)\(\\bar\{a\}\+2k\)=1\+2kt\+4k^\{2\}\.Hence

L=X2\+X​t\+1D0−\(1\+2​k​t\+4​k2\)\.L=\\frac\{X^\{2\}\+Xt\+1\}\{D\_\{0\}\}\-\(1\+2kt\+4k^\{2\}\)\.Multiply byD0D\_\{0\}:

D0​L=X2\+X​t\+1−D0​\(1\+2​k​t\+4​k2\)\.D\_\{0\}L=X^\{2\}\+Xt\+1\-D\_\{0\}\(1\+2kt\+4k^\{2\}\)\.NowD0​\(1\+2​k​t\+4​k2\)=2​\(1\+k\)\+4​k​\(1\+k\)​t\+8​k2​\(1\+k\)\.D\_\{0\}\(1\+2kt\+4k^\{2\}\)=2\(1\+k\)\+4k\(1\+k\)t\+8k^\{2\}\(1\+k\)\.

Thus

D0​L=X2\+1−2​\(1\+k\)−8​k2​\(1\+k\)⏟constant\+\(X−4​k​\(1\+k\)\)⏟coefficient of​t​t\.D\_\{0\}L=\\underbrace\{X^\{2\}\+1\-2\(1\+k\)\-8k^\{2\}\(1\+k\)\}\_\{\\text\{constant\}\}\+\\underbrace\{\(X\-4k\(1\+k\)\)\}\_\{\\text\{coefficient of \}t\}t\.Compute the coefficient oftt:

X−4​k​\(1\+k\)=\(4​k2\+2​k−1\)−\(4​k2\+4​k\)=−2​k−1=−\(2​k\+1\)\.X\-4k\(1\+k\)=\(4k^\{2\}\+2k\-1\)\-\(4k^\{2\}\+4k\)=\-2k\-1=\-\(2k\+1\)\.For the constant term, substituteX2=16​k4\+16​k3−4​k2−4​k\+1X^\{2\}=16k^\{4\}\+16k^\{3\}\-4k^\{2\}\-4k\+1:

X2\+1−2​\(1\+k\)−8​k2​\(1\+k\)\\displaystyle X^\{2\}\+1\-2\(1\+k\)\-8k^\{2\}\(1\+k\)=\(16​k4\+16​k3−4​k2−4​k\+1\)\+1−2−2​k−8​k2−8​k3\\displaystyle=\(6k^\{4\}\+6k^\{3\}\-4k^\{2\}\-4k\+1\)\+1\-2\-2k\-8k^\{2\}\-8k^\{3\}=16​k4\+\(16​k3−8​k3\)\+\(−4​k2−8​k2\)\+\(−4​k−2​k\)\+\(1\+1−2\)\\displaystyle=6k^\{4\}\+\(6k^\{3\}\-8k^\{3\}\)\+\(\-4k^\{2\}\-8k^\{2\}\)\+\(\-4k\-2k\)\+\(1\+1\-2\)=16​k4\+8​k3−12​k2−6​k\.\\displaystyle=6k^\{4\}\+8k^\{3\}\-2k^\{2\}\-6k\.Factor this expression:

16​k4\+8​k3−12​k2−6​k=2​k​\(8​k3\+4​k2−6​k−3\)\.16k^\{4\}\+8k^\{3\}\-12k^\{2\}\-6k=2k\(8k^\{3\}\+4k^\{2\}\-6k\-3\)\.Now note that8​k3\+4​k2−6​k−3=\(4​k2−3\)​\(2​k\+1\)8k^\{3\}\+4k^\{2\}\-6k\-3=\(4k^\{2\}\-3\)\(2k\+1\), because

\(4​k2−3\)​\(2​k\+1\)=8​k3\+4​k2−6​k−3\.\(4k^\{2\}\-3\)\(2k\+1\)=8k^\{3\}\+4k^\{2\}\-6k\-3\.Thus

constant=2​k​\(4​k2−3\)​\(2​k\+1\)\.\\text\{constant\}=2k\(4k^\{2\}\-3\)\(2k\+1\)\.But4​k2−3=cos⁡3​αk4k^\{2\}\-3=\\frac\{\\cos 3\\alpha\}\{k\}\(sincecos⁡3​α=4​k3−3​k=k​\(4​k2−3\)\\cos 3\\alpha=4k^\{3\}\-3k=k\(4k^\{2\}\-3\)\)\. Hence

constant=2​k⋅cos⁡3​αk⋅\(2​k\+1\)=2​\(2​k\+1\)​cos⁡3​α\.\\text\{constant\}=2k\\cdot\\frac\{\\cos 3\\alpha\}\{k\}\\cdot\(2k\+1\)=2\(2k\+1\)\\cos 3\\alpha\.Therefore

D0L=2\(2k\+1\)cos3α\+\(−\(2k\+1\)\)t=\(2k\+1\)\(2cos3α−t\)\.\(30\)D\_\{0\}L=2\(2k\+1\)\\cos 3\\alpha\+\(\-\(2k\+1\)\)t=\(2k\+1\)\(2\\cos 3\\alpha\-t\)\.\\quad\\text\{\(30\)\}Now recall thatr=\(2​k\+1\)/D0r=\(2k\+1\)/D\_\{0\}\. So

\(2​k\+1\)=r​D0\.\(2k\+1\)=rD\_\{0\}\.Substitute into \(30\):

D0L=rD0\(2cos3α−t\)⟹L=r\(2cos3α−t\)\.\(31\)D\_\{0\}L=rD\_\{0\}\(2\\cos 3\\alpha\-t\)\\quad\\Longrightarrow\\quad L=r\(2\\cos 3\\alpha\-t\)\.\\quad\\text\{\(31\)\}
Connection withΔ\\Delta

ComputeΔ\\Deltain terms oftt:

Δ=a2−2​a​cos⁡3​α\+1=a​\(a\+a¯−2​cos⁡3​α\)=a​\(t−2​cos⁡3​α\)\.\\Delta=a^\{2\}\-2a\\cos 3\\alpha\+1=a\(a\+\\bar\{a\}\-2\\cos 3\\alpha\)=a\(t\-2\\cos 3\\alpha\)\.Since\|a\|=1\|a\|=1, we have

\|Δ\|=\|t−2​cos⁡3​α\|=\|2​cos⁡3​α−t\|\.\|\\Delta\|=\|t\-2\\cos 3\\alpha\|=\|2\\cos 3\\alpha\-t\|\.Hence

2​cos⁡3​α−t=ε​\|Δ\|,2\\cos 3\\alpha\-t=\\varepsilon\\,\|\\Delta\|,whereε=±1\\varepsilon=\\pm 1depending on the sign oft−2​cos⁡3​αt\-2\\cos 3\\alpha\.

Consequently,

L=εr\|Δ\|\.\(32\)L=\\varepsilon\\,r\\,\|\\Delta\|\.\\quad\\text\{\(32\)\}
Final step

Insert \(32\) into \(29\):

O​I=R\+ε​r​\|Δ\|\|Δ\|=R\+ε​r=R±r\.OI=R\+\\frac\{\\varepsilon\\,r\\,\|\\Delta\|\}\{\|\\Delta\|\}=R\+\\varepsilon r=R\\pm r\.Thus the distance between the centersOOandIIequals either the sum or the absolute difference of the radiiRRandrr\. Therefore the circumcircle of△​A​Y​Z\\triangle AYZand the incircle of△​ℛ\\triangle\\mathcal\{R\}are tangent\.□\\square

12\. Remarks on non\-vanishing denominators

⊳\\trianglerightD1=0D\_\{1\}=0would implya​c=b2ac=b^\{2\}\. In our rotated frame, this givesa=ei​3​αa=e^\{i3\\alpha\}, which forcesA​B=B​CAB=BC\(since then\|a−b\|=\|ei​3​α−ei​α\|=2​sin⁡α\|a\-b\|=\|e^\{i3\\alpha\}\-e^\{i\\alpha\}\|=2\\sin\\alphaand\|b−c\|=\|ei​α−e−i​α\|=2​sin⁡α\|b\-c\|=\|e^\{i\\alpha\}\-e^\{\-i\\alpha\}\|=2\\sin\\alpha\), contradicting the scalene hypothesis\. HenceD1≠0D\_\{1\}\\neq 0; similarlyD2≠0D\_\{2\}\\neq 0, soΔ≠0\\Delta\\neq 0\.

⊳\\trianglerighta\+2​k=0a\+2k=0would givea=−2​ka=\-2kreal; with\|a\|=1\|a\|=1we get2​k=12k=1i\.e\.cos⁡α=1/2\\cos\\alpha=1/2,α=60∘\\alpha=60^\{\\circ\}, which is excluded\. Thusa\+2​k≠0a\+2k\\neq 0, ensuringR\>0R\>0\.

⊳\\trianglerightcos⁡α\>0\\cos\\alpha\>0becauseα\\alphais acute\.

All algebraic manipulations are therefore justified, and the proof is complete\.

#### USAMO 2026 Problem 4\.

A positive integernnis called*solitary*if, for any nonnegative integersaaandbbsuch thata\+b=na\+b=n, eitheraaorbbcontains the digit “1”\. Determine, with proof, the number of solitary integers less than10202610^\{2026\}\.

Solution

1\. Equivalent formulation

Let

S=\{x∈ℕ0∣the decimal representation of​x​contains no digit​1\}\.S=\\\{x\\in\\mathbb\{N\}\_\{0\}\\mid\\text\{the decimal representation of \}x\\text\{ contains no digit \}1\\\}\.\(Note that0∈S0\\in Sbecause its representation ”0” has no digit11\.\)

Ifn=a\+bn=a\+bwitha,b∈Sa,b\\in S, then the pair\(a,b\)\(a,b\)shows thatnnisnotsolitary \(bothaaandbblack a digit11\)\. Conversely, if for every representationn=a\+bn=a\+bat least one ofa,ba,bcontains a digit11, then certainly no representation with both inSSexists\. Hence

n​is solitary⟺n∉S\+S,where​S\+S=\{x\+y∣x,y∈S\}\.n\\text\{ is solitary \}\\Longleftrightarrow\\;n\\notin S\+S,\\qquad\\text\{where \}S\+S=\\\{x\+y\\mid x,y\\in S\\\}\.We need to count thepositiveintegersn<102026n<10^\{2026\}that are not inS\+SS\+S\.

Set

We will first count how many numbers0≤n<10N0\\leq n<10^\{N\}belong toS\+SS\+S; then subtract to get the number of positive solitary integers\.

2\. Digit\-wise analysis and carries

Every integermmwith0≤m<10N0\\leq m<10^\{N\}can be written uniquely as

m=∑i=0N−1di​10i,m=\\sum\_\{i=0\}^\{N\-1\}d\_\{i\}10^\{i\},where each digitdid\_\{i\}is in\{0,1,…,9\}\\\{0,1,\\dots,9\\\}\. To have a uniform treatment, we pad the representation with leading zeros so that every suchmmuses exactlyNNdigits\.

Ifa,b∈Sa,b\\in S, then each of their digits belongs to

D=\{0,2,3,4,5,6,7,8,9\}\(all digits except​1\)\.D=\\\{0,2,3,4,5,6,7,8,9\\\}\\qquad\(\\text\{all digits except \}1\)\.Writea=∑ai​10ia=\\sum a\_\{i\}10^\{i\},b=∑bi​10ib=\\sum b\_\{i\}10^\{i\}withai,bi∈Da\_\{i\},b\_\{i\}\\in D\. The additiona\+b=na\+b=nproceeds digit by digit with carriesc0,c1,…,cNc\_\{0\},c\_\{1\},\\dots,c\_\{N\}:

ai\+bi\+ci=ni\+10​ci\+1,c0=0,a\_\{i\}\+b\_\{i\}\+c\_\{i\}=n\_\{i\}\+10\\,c\_\{i\+1\},\\qquad c\_\{0\}=0,where eachcic\_\{i\}is either0or11\(since the maximum sum is9\+9\+1=199\+9\+1=19\)\. Becausen<10Nn<10^\{N\}, the final carry must becN=0c\_\{N\}=0\.

3\. Possible sums of two digits fromDD

Define

Σ=\{ai\+bi∣ai,bi∈D\}\.\\Sigma=\\\{a\_\{i\}\+b\_\{i\}\\mid a\_\{i\},b\_\{i\}\\in D\\\}\.Since the only digit missing fromDDis11, the sums that can be obtained are

Σ=\{0\}∪\{2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\}\.\\Sigma=\\\{0\\\}\\cup\\\{2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\\\}\.\(Indeed,11cannot be expressed as a sum of two digits fromDD; all other integers from0to1818can\.\)

4\. Transition setsT​\(s,d\)T\(s,d\)

For a fixed carryins∈\{0,1\}s\\in\\\{0,1\\\}and a target digitd∈\{0,…,9\}d\\in\\\{0,\\dots,9\\\}, let

T​\(s,d\)=\{t∈\{0,1\}\|∃ai,bi∈D​with​ai\+bi\+s=d\+10​t\}\.T\(s,d\)=\\bigl\\\{t\\in\\\{0,1\\\}\\;\\big\|\\;\\exists a\_\{i\},b\_\{i\}\\in D\\text\{ with \}a\_\{i\}\+b\_\{i\}\+s=d\+10t\\bigr\\\}\.UsingΣ\\Sigma, we can computeT​\(s,d\)T\(s,d\)\.

Cases=0s=0

Here the total sum before splitting is simplys0s\_\{0\}withs0∈Σs\_\{0\}\\in\\Sigma\.

⊳\\trianglerightIfs0≤9s\_\{0\}\\leq 9, we may taket=0t=0and the digit iss0s\_\{0\}\.

⊳\\trianglerightIfs0≥10s\_\{0\}\\geq 10, we may taket=1t=1and the digit iss0−10s\_\{0\}\-10\.

Thus:

dT​\(0,d\)0\{0,1\}\(totals​0​and​10\)1\{1\}\(total​11\)2\{0,1\}\(totals​2​and​12\)3\{0,1\}\(totals​3​and​13\)4\{0,1\}\(totals​4​and​14\)5\{0,1\}\(totals​5​and​15\)6\{0,1\}\(totals​6​and​16\)7\{0,1\}\(totals​7​and​17\)8\{0,1\}\(totals​8​and​18\)9\{0\}\(total​9\)\\begin\{array\}\[\]\{c\|l\}d&T\(0,d\)\\\\ \\hline\\cr 0&\\\{0,1\\\}\\quad\(\\text\{totals \}0\\text\{ and \}10\)\\\\ 1&\\\{1\\\}\\quad\(\\text\{total \}11\)\\\\ 2&\\\{0,1\\\}\\quad\(\\text\{totals \}2\\text\{ and \}12\)\\\\ 3&\\\{0,1\\\}\\quad\(\\text\{totals \}3\\text\{ and \}13\)\\\\ 4&\\\{0,1\\\}\\quad\(\\text\{totals \}4\\text\{ and \}14\)\\\\ 5&\\\{0,1\\\}\\quad\(\\text\{totals \}5\\text\{ and \}15\)\\\\ 6&\\\{0,1\\\}\\quad\(\\text\{totals \}6\\text\{ and \}16\)\\\\ 7&\\\{0,1\\\}\\quad\(\\text\{totals \}7\\text\{ and \}17\)\\\\ 8&\\\{0,1\\\}\\quad\(\\text\{totals \}8\\text\{ and \}18\)\\\\ 9&\\\{0\\\}\\quad\(\\text\{total \}9\)\\\\ \\end\{array\}In words:

⊳\\trianglerightd=9d=9gives onlyt=0t=0\.

⊳\\trianglerightd=1d=1gives onlyt=1t=1\.

⊳\\trianglerightThe other eight digits \(0,2,3,4,5,6,7,80,2,3,4,5,6,7,8\) allow botht=0t=0andt=1t=1\.

Cases=1s=1

Now the total sum iss0\+1s\_\{0\}\+1withs0∈Σs\_\{0\}\\in\\Sigma\. Hence the attainable totals are

Σ\+1=\{1\}∪\{3,4,5,…,19\},\\Sigma\+1=\\\{1\\\}\\cup\\\{3,4,5,\\dots,19\\\},i\.e\., all integers from11to1919except22\.

Proceeding analogously:

dT​\(1,d\)0\{1\}\(total​19\)1\{0,1\}\(totals​1​and​11\)2\{1\}\(total​12\)3\{0,1\}\(totals​3​and​13\)4\{0,1\}\(totals​4​and​14\)5\{0,1\}\(totals​5​and​15\)6\{0,1\}\(totals​6​and​16\)7\{0,1\}\(totals​7​and​17\)8\{0,1\}\(totals​8​and​18\)9\{0,1\}\(totals​9​and​19\)\\begin\{array\}\[\]\{c\|l\}d&T\(1,d\)\\\\ \\hline\\cr 0&\\\{1\\\}\\quad\(\\text\{total \}19\)\\\\ 1&\\\{0,1\\\}\\quad\(\\text\{totals \}1\\text\{ and \}11\)\\\\ 2&\\\{1\\\}\\quad\(\\text\{total \}12\)\\\\ 3&\\\{0,1\\\}\\quad\(\\text\{totals \}3\\text\{ and \}13\)\\\\ 4&\\\{0,1\\\}\\quad\(\\text\{totals \}4\\text\{ and \}14\)\\\\ 5&\\\{0,1\\\}\\quad\(\\text\{totals \}5\\text\{ and \}15\)\\\\ 6&\\\{0,1\\\}\\quad\(\\text\{totals \}6\\text\{ and \}16\)\\\\ 7&\\\{0,1\\\}\\quad\(\\text\{totals \}7\\text\{ and \}17\)\\\\ 8&\\\{0,1\\\}\\quad\(\\text\{totals \}8\\text\{ and \}18\)\\\\ 9&\\\{0,1\\\}\\quad\(\\text\{totals \}9\\text\{ and \}19\)\\\\ \\end\{array\}Thus:

⊳\\trianglerightd=0d=0andd=2d=2give onlyt=1t=1\.

⊳\\trianglerightThe other eight digits \(1,3,4,5,6,7,8,91,3,4,5,6,7,8,9\) allow botht=0t=0andt=1t=1\.

5\. State of the possible carries

For a givennnwith digitsn0,n1,…,nN−1n\_\{0\},n\_\{1\},\\dots,n\_\{N\-1\}\(least significant first\), consider all sequences of carriesc0,c1,…,cNc\_\{0\},c\_\{1\},\\dots,c\_\{N\}that can arise from some choices ofai,bi∈Da\_\{i\},b\_\{i\}\\in D\. Define

Ri=\{ci∣there exist​aj,bj∈D​\(j<i\)​such that the carry after processing digits​0,…,i−1​is​ci\}\.R\_\{i\}=\\\{\\,c\_\{i\}\\mid\\text\{there exist \}a\_\{j\},b\_\{j\}\\in D\\ \(j<i\)\\ \\text\{such that the carry after processing digits \}0,\\dots,i\-1\\text\{ is \}c\_\{i\}\\,\\\}\.By definition,R0=\{0\}R\_\{0\}=\\\{0\\\}\. Fori≥0i\\geq 0,

Ri\+1=⋃s∈RiT\(s,ni\)\.\(1\)R\_\{i\+1\}=\\bigcup\_\{s\\in R\_\{i\}\}T\(s,n\_\{i\}\)\.\\quad\\text\{\(1\)\}EachT​\(s,d\)T\(s,d\)is a subset of\{0,1\}\\\{0,1\\\}, and a simple check shows it is never empty\. Hence everyRiR\_\{i\}is a non\-empty subset of\{0,1\}\\\{0,1\\\}\. Consequently,RiR\_\{i\}can only be one of three types:

⊳\\trianglerightAiA\_\{i\}:Ri=\{0\}R\_\{i\}=\\\{0\\\}\(only zero carry possible\),

⊳\\trianglerightBiB\_\{i\}:Ri=\{1\}R\_\{i\}=\\\{1\\\}\(only one carry possible\),

⊳\\trianglerightCiC\_\{i\}:Ri=\{0,1\}R\_\{i\}=\\\{0,1\\\}\(both carries possible\)\.

Let

Ai\\displaystyle A\_\{i\}=\#​\{n∈\[0,10i\)∣Ri=\{0\}\},\\displaystyle=\\\#\\\{\\,n\\in\[0,0^\{i\}\)\\mid R\_\{i\}=\\\{0\\\}\\,\\\},Bi\\displaystyle B\_\{i\}=\#​\{n∈\[0,10i\)∣Ri=\{1\}\},\\displaystyle=\\\#\\\{\\,n\\in\[0,0^\{i\}\)\\mid R\_\{i\}=\\\{1\\\}\\,\\\},Ci\\displaystyle C\_\{i\}=\#​\{n∈\[0,10i\)∣Ri=\{0,1\}\}\.\\displaystyle=\\\#\\\{\\,n\\in\[0,0^\{i\}\)\\mid R\_\{i\}=\\\{0,1\\\}\\,\\\}\.
6\. Transition counts

We now determine, given the current stateRiR\_\{i\}, for which next digitsd=nid=n\_\{i\}we obtain each possible next stateRi\+1R\_\{i\+1\}\.

⊳\\trianglerightIfRi=\{0\}R\_\{i\}=\\\{0\\\}: thenRi\+1=T​\(0,d\)R\_\{i\+1\}=T\(0,d\)\. From the table fors=0s=0:

⊳\\trianglerightd=9d=9gives\{0\}\\\{0\\\};

⊳\\trianglerightd=1d=1gives\{1\}\\\{1\\\};

⊳\\trianglerightthe other eight digits \(0,2,3,4,5,6,7,80,2,3,4,5,6,7,8\) give\{0,1\}\\\{0,1\\\}\.

Hence:

⊳\\triangleright1 digit yields state\{0\}\\\{0\\\},

⊳\\triangleright1 digit yields state\{1\}\\\{1\\\},

⊳\\triangleright8 digits yield state\{0,1\}\\\{0,1\\\}\.

⊳\\trianglerightIfRi=\{1\}R\_\{i\}=\\\{1\\\}:Ri\+1=T​\(1,d\)R\_\{i\+1\}=T\(1,d\)\. From the table fors=1s=1:

⊳\\trianglerightd=0d=0andd=2d=2give\{1\}\\\{1\\\};

⊳\\trianglerightthe other eight digits \(1,3,4,5,6,7,8,91,3,4,5,6,7,8,9\) give\{0,1\}\\\{0,1\\\}\.

Hence:

⊳\\triangleright2 digits yield state\{1\}\\\{1\\\},

⊳\\triangleright8 digits yield state\{0,1\}\\\{0,1\\\}\.

⊳\\trianglerightIfRi=\{0,1\}R\_\{i\}=\\\{0,1\\\}:Ri\+1=T​\(0,d\)∪T​\(1,d\)R\_\{i\+1\}=T\(0,d\)\\cup T\(1,d\)\. Checking each digit individually shows that this union is always\{0,1\}\\\{0,1\\\}\. Indeed:

⊳\\trianglerightd=0d=0:T​\(0,0\)=\{0,1\}T\(0,0\)=\\\{0,1\\\},T​\(1,0\)=\{1\}T\(1,0\)=\\\{1\\\}→\\tounion =\{0,1\}\\\{0,1\\\};

⊳\\trianglerightd=1d=1:T​\(0,1\)=\{1\}T\(0,1\)=\\\{1\\\},T​\(1,1\)=\{0,1\}T\(1,1\)=\\\{0,1\\\}→\\tounion =\{0,1\}\\\{0,1\\\};

⊳\\trianglerightd=2d=2:T​\(0,2\)=\{0,1\}T\(0,2\)=\\\{0,1\\\},T​\(1,2\)=\{1\}T\(1,2\)=\\\{1\\\}→\\tounion =\{0,1\}\\\{0,1\\\};

⊳\\trianglerightd=3,…,8d=3,\\dots,8: both sets contain0and11;

⊳\\trianglerightd=9d=9:T​\(0,9\)=\{0\}T\(0,9\)=\\\{0\\\},T​\(1,9\)=\{0,1\}T\(1,9\)=\\\{0,1\\\}→\\tounion =\{0,1\}\\\{0,1\\\}\.

Therefore,all 10 digitskeep the state as\{0,1\}\\\{0,1\\\}\.

Summarising the transitions in a matrix:

from\\to\{0\}\{1\}\{0,1\}\{0\}118\{1\}028\{0,1\}0010\\begin\{array\}\[\]\{c\|ccc\}\\text\{from\}\\backslash\\text\{to\}&\\\{0\\\}&\\\{1\\\}&\\\{0,1\\\}\\\\ \\hline\\cr\\\{0\\\}&1&1&8\\\\ \\\{1\\\}&0&2&8\\\\ \\\{0,1\\\}&0&0&10\\end\{array\}
7\. Recurrence relations

A number withi\+1i\+1digits \(i\.e\., an integer in\[0,10i\+1\)\[0,10^\{i\+1\}\)\) is obtained by taking a numbermmin\[0,10i\)\[0,10^\{i\}\)\(which yields a certain state\) and appending a new most significant digitdd\(which becomesnin\_\{i\}\)\. The number of ways to reach each state for lengthi\+1i\+1is therefore:

Ai\+1=1⋅Ai\+0⋅Bi\+0⋅Ci=Ai,Bi\+1=1⋅Ai\+2⋅Bi\+0⋅Ci=Ai\+2​Bi,Ci\+1=8⋅Ai\+8⋅Bi\+10⋅Ci\.\(2\)\\begin\{aligned\} A\_\{i\+1\}&=1\\cdot A\_\{i\}\+0\\cdot B\_\{i\}\+0\\cdot C\_\{i\}=A\_\{i\},\\\\\[4\.0pt\] B\_\{i\+1\}&=1\\cdot A\_\{i\}\+2\\cdot B\_\{i\}\+0\\cdot C\_\{i\}=A\_\{i\}\+2B\_\{i\},\\\\\[4\.0pt\] C\_\{i\+1\}&=8\\cdot A\_\{i\}\+8\\cdot B\_\{i\}\+10\\cdot C\_\{i\}\.\\end\{aligned\}\\quad\\text\{\(2\)\}Initial conditions: Fori=0i=0we have processed no digits\. The only possible ”number” is0, and the only possible carry isc0=0c\_\{0\}=0\. Hence

A0=1,B0=0,C0=0\.A\_\{0\}=1,\\qquad B\_\{0\}=0,\\qquad C\_\{0\}=0\.
8\. Solving the recurrences

8\.1\.AiA\_\{i\}

FromAi\+1=AiA\_\{i\+1\}=A\_\{i\}andA0=1A\_\{0\}=1, we immediately obtain

Ai=1for all​i≥0\.\\boxed\{A\_\{i\}=1\\quad\\text\{for all \}i\\geq 0\}\.
8\.2\.BiB\_\{i\}

SubstitutingAi=1A\_\{i\}=1into the recurrence forBiB\_\{i\}:

Bi\+1=1\+2​Bi,B0=0\.B\_\{i\+1\}=1\+2B\_\{i\},\\qquad B\_\{0\}=0\.We claim

Bi=2i−1for all​i≥0\.\\boxed\{B\_\{i\}=2^\{i\}\-1\\quad\\text\{for all \}i\\geq 0\}\.*Proof by induction\.*

⊳\\trianglerighti=0i=0:20−1=02^\{0\}\-1=0, true\.

⊳\\trianglerightAssumeBi=2i−1B\_\{i\}=2^\{i\}\-1\. Then

Bi\+1=1\+2​\(2i−1\)=1\+2i\+1−2=2i\+1−1,B\_\{i\+1\}=1\+2\(2^\{i\}\-1\)=1\+2^\{i\+1\}\-2=2^\{i\+1\}\-1,which completes the induction\.

8\.3\.CiC\_\{i\}

Now substituteAi=1A\_\{i\}=1andBi=2i−1B\_\{i\}=2^\{i\}\-1into the recurrence forCi\+1C\_\{i\+1\}:

Ci\+1\\displaystyle C\_\{i\+1\}=8⋅1\+8⋅\(2i−1\)\+10​Ci\\displaystyle=8\\cdot 1\+8\\cdot\(2^\{i\}\-1\)\+0\\,C\_\{i\}=8\+8⋅2i−8\+10​Ci\\displaystyle=8\+8\\cdot 2^\{i\}\-8\+0\\,C\_\{i\}=8⋅2i\+10​Ci\.\\displaystyle=8\\cdot 2^\{i\}\+0\\,C\_\{i\}\.WithC0=0C\_\{0\}=0\. We claim

Ci=10i−2ifor all​i≥0\.\\boxed\{C\_\{i\}=10^\{i\}\-2^\{i\}\\quad\\text\{for all \}i\\geq 0\}\.*Proof by induction\.*

⊳\\trianglerighti=0i=0:100−20=1−1=010^\{0\}\-2^\{0\}=1\-1=0, true\.

⊳\\trianglerightAssumeCi=10i−2iC\_\{i\}=10^\{i\}\-2^\{i\}\. Then

Ci\+1\\displaystyle C\_\{i\+1\}=8⋅2i\+10​\(10i−2i\)\\displaystyle=8\\cdot 2^\{i\}\+0\(0^\{i\}\-2^\{i\}\)=8⋅2i\+10i\+1−10⋅2i\\displaystyle=8\\cdot 2^\{i\}\+0^\{i\+1\}\-0\\cdot 2^\{i\}=10i\+1−2⋅2i\\displaystyle=0^\{i\+1\}\-2\\cdot 2^\{i\}=10i\+1−2i\+1,\\displaystyle=0^\{i\+1\}\-2^\{i\+1\},which establishes the claim\.

9\. Representable numbers

After processing allNNdigits, a numbern∈\[0,10N\)n\\in\[0,10^\{N\}\)can be written asa\+ba\+bwitha,b∈Sa,b\\in Siffthere exists a carry sequence withcN=0c\_\{N\}=0\. In terms of the sets, this is equivalent to0∈RN0\\in R\_\{N\}\.

The possible values ofRNR\_\{N\}are\{0\}\\\{0\\\},\{1\}\\\{1\\\}, or\{0,1\}\\\{0,1\\\}\. Among these,0∈RN0\\in R\_\{N\}exactly whenRN=\{0\}R\_\{N\}=\\\{0\\\}orRN=\{0,1\}R\_\{N\}=\\\{0,1\\\}\. Therefore

\|S\+S∩\[0,10N\)\|=AN\+CN\.\|S\+S\\cap\[0,10^\{N\}\)\|=A\_\{N\}\+C\_\{N\}\.Plugging in the formulas:

\|S\+S∩\[0,10N\)\|=1\+\(10N−2N\)=10N−2N\+1\.\|S\+S\\cap\[0,10^\{N\}\)\|=1\+\(10^\{N\}\-2^\{N\}\)=10^\{N\}\-2^\{N\}\+1\.This count includesn=0n=0\(since0=0\+00=0\+0and0∈S0\\in S\)\.

10\. Positive solitary numbers

We are interested inpositiveintegersn<10Nn<10^\{N\}that are solitary, i\.e\.,n\>0n\>0andn∉S\+Sn\\notin S\+S\.

Total positive integers less than10N10^\{N\}are

The number of positive integers that belong toS\+SS\+S\(hence arenotsolitary\) is

\|\(S\+S\)∖\{0\}\|=\(10N−2N\+1\)−1=10N−2N\.\|\(S\+S\)\\setminus\\\{0\\\}\|=\(10^\{N\}\-2^\{N\}\+1\)\-1=10^\{N\}\-2^\{N\}\.Thus the number of positive solitary integers is

\(10N−1\)−\(10N−2N\)=2N−1\.\(10^\{N\}\-1\)\-\(10^\{N\}\-2^\{N\}\)=2^\{N\}\-1\.
11\. Final answer

ForN=2026N=2026, we obtain

22026−1\.\\boxed\{2^\{2026\}\-1\}\.
12\. Small\-case verification \(optional\)

⊳\\trianglerightN=1N=1: numbers1,…,91,\\dots,9\. The formula gives21−1=12^\{1\}\-1=1solitary number\. Indeed,n=1n=1is solitary because every pair\(a,b\)\(a,b\)witha\+b=1a\+b=1has eithera=0,b=1a=0,b=1ora=1,b=0a=1,b=0; in both cases the number11contains a digit11\. All othernn\(e\.g\.,22\) have a representation with both numbers lacking digit11\(e\.g\.,2=2\+02=2\+0\), so they are not solitary\.

⊳\\trianglerightN=2N=2: numbers1,…,991,\\dots,99\. The formula gives22−1=32^\{2\}\-1=3solitary numbers\. One can check that they are1,19,211,19,21\. This matches the known pattern\.

The solution is complete\.

#### USAMO 2026 Problem 5\.

LetA​B​CABCbe a triangle\. PointsDD,EE, andFFlie on sidesB​CBC,C​ACA, andA​BAB, respectively, such that∠​A​F​E=∠​B​D​F=∠​C​E​D\.\\angle AFE=\\angle BDF=\\angle CED\.LetOAO\_\{A\},OBO\_\{B\}, andOCO\_\{C\}be the circumcenters of trianglesA​F​EAFE,B​D​FBDF, andC​E​DCED, respectively\. LetMM,NN, andOObe the circumcenters of trianglesA​B​CABC,D​E​FDEF, andOA​OB​OCO\_\{A\}O\_\{B\}O\_\{C\}, respectively\. Prove thatO​M=O​NOM=ON\.

Solution

We are given triangleA​B​CABCwith pointsD∈B​CD\\in BC,E∈C​AE\\in CA,F∈A​BF\\in ABsuch that

∠AFE=∠BDF=∠CED=:θ\.\\angle AFE=\\angle BDF=\\angle CED=:\\theta\.We shall use complex numbers\. Without loss of generality we may place the triangle in the complex plane and denote bya,b,ca,b,cthe complex numbers corresponding toA,B,CA,B,C\.

1\. Angle Chase

Denote the angles of△​A​B​C\\triangle ABCbyα=∠​B​A​C\\alpha=\\angle BAC,β=∠​A​B​C\\beta=\\angle ABC,γ=∠​A​C​B\\gamma=\\angle ACB\.

At vertexFF

PointsA,F,BA,F,Bare collinear, so∠​A​F​B=180∘\\angle AFB=180^\{\\circ\}\. The three angles aroundFFon the side ofA​BABthat contains the triangle are∠​A​F​E\\angle AFE,∠​E​F​D\\angle EFD,∠​D​F​B\\angle DFBand they sum to180∘180^\{\\circ\}:

∠AFE\+∠EFD\+∠DFB=180∘\.\(1\)\\angle AFE\+\\angle EFD\+\\angle DFB=180^\{\\circ\}\.\\quad\\text\{\(1\)\}In△​B​D​F\\triangle BDFwe have∠​B​D​F=θ\\angle BDF=\\thetaand∠​D​B​F=β\\angle DBF=\\beta\(sinceDDlies onB​CBCandFFonA​BAB\)\. Hence

∠​B​F​D=180∘−θ−β,\\angle BFD=180^\{\\circ\}\-\\theta\-\\beta,so∠​D​F​B=180∘−θ−β\\angle DFB=180^\{\\circ\}\-\\theta\-\\beta\.

Inserting∠​A​F​E=θ\\angle AFE=\\thetainto \(1\) gives

θ\+∠​E​F​D\+\(180∘−θ−β\)=180∘⟹∠​E​F​D=β\.\\theta\+\\angle EFD\+\(180^\{\\circ\}\-\\theta\-\\beta\)=180^\{\\circ\}\\quad\\Longrightarrow\\quad\\angle EFD=\\beta\.Thus

∠DFE=β\.\(2\)\\angle DFE=\\beta\.\\quad\\text\{\(2\)\}
At vertexDD

PointsB,D,CB,D,Care collinear, so∠​B​D​C=180∘\\angle BDC=180^\{\\circ\}\. The angles atDDare∠​B​D​F\\angle BDF,∠​F​D​E\\angle FDE,∠​E​D​C\\angle EDCand they sum to180∘180^\{\\circ\}:

∠BDF\+∠FDE\+∠EDC=180∘\.\(3\)\\angle BDF\+\\angle FDE\+\\angle EDC=180^\{\\circ\}\.\\quad\\text\{\(3\)\}Given∠​B​D​F=θ\\angle BDF=\\theta\.

In△​C​E​D\\triangle CED,∠​C​E​D=θ\\angle CED=\\thetaand∠​E​C​D=γ\\angle ECD=\\gamma, so

∠​C​D​E=180∘−θ−γ,\\angle CDE=180^\{\\circ\}\-\\theta\-\\gamma,i\.e\.,∠​E​D​C=180∘−θ−γ\\angle EDC=180^\{\\circ\}\-\\theta\-\\gamma\.

Substituting into \(3\):

θ\+∠​F​D​E\+\(180∘−θ−γ\)=180∘⟹∠​F​D​E=γ\.\\theta\+\\angle FDE\+\(180^\{\\circ\}\-\\theta\-\\gamma\)=180^\{\\circ\}\\quad\\Longrightarrow\\quad\\angle FDE=\\gamma\.Thus

∠EDF=γ\.\(4\)\\angle EDF=\\gamma\.\\quad\\text\{\(4\)\}
At vertexEE

PointsC,E,AC,E,Aare collinear, so∠​C​E​A=180∘\\angle CEA=180^\{\\circ\}\. The angles atEEare∠​C​E​D\\angle CED,∠​D​E​F\\angle DEF,∠​F​E​A\\angle FEAand they sum to180∘180^\{\\circ\}:

∠CED\+∠DEF\+∠FEA=180∘\.\(5\)\\angle CED\+\\angle DEF\+\\angle FEA=180^\{\\circ\}\.\\quad\\text\{\(5\)\}Given∠​C​E​D=θ\\angle CED=\\theta\.

In△​A​F​E\\triangle AFE,∠​A​F​E=θ\\angle AFE=\\thetaand∠​F​A​E=α\\angle FAE=\\alpha, so

∠​A​E​F=180∘−θ−α,\\angle AEF=180^\{\\circ\}\-\\theta\-\\alpha,i\.e\.,∠​F​E​A=180∘−θ−α\\angle FEA=180^\{\\circ\}\-\\theta\-\\alpha\.

Substituting into \(5\):

θ\+∠​D​E​F\+\(180∘−θ−α\)=180∘⟹∠​D​E​F=α\.\\theta\+\\angle DEF\+\(180^\{\\circ\}\-\\theta\-\\alpha\)=180^\{\\circ\}\\quad\\Longrightarrow\\quad\\angle DEF=\\alpha\.Thus

∠DEF=α\.\(6\)\\angle DEF=\\alpha\.\\quad\\text\{\(6\)\}From \(2\), \(4\), \(6\) we obtain

∠​D​F​E=β,∠​E​D​F=γ,∠​D​E​F=α\.\\angle DFE=\\beta,\\quad\\angle EDF=\\gamma,\\quad\\angle DEF=\\alpha\.Therefore△​D​E​F\\triangle DEFhas anglesα,β,γ\\alpha,\\beta,\\gamma; it issimilarto△​A​B​C\\triangle ABC\. The vertex correspondence is

E⟷A,F⟷B,D⟷C\.\(7\)E\\longleftrightarrow A,\\qquad F\\longleftrightarrow B,\\qquad D\\longleftrightarrow C\.\\quad\\text\{\(7\)\}
2\. Complex Representation of the Similarity

The similarity \(7\) may be either orientation\-preserving \(direct\) or orientation\-reversing \(opposite\)\. The statementO​M=O​NOM=ONinvolves only distances, which are invariant under reflection\. Hence we may reflect the whole configuration if necessary and assume that the similarity isdirect\. Consequently there exist a non\-zero complex numberkkand a complex numberttsuch that

e=ka\+t,f=kb\+t,d=kc\+t\.\(8\)e=ka\+t,\\qquad f=kb\+t,\\qquad d=kc\+t\.\\quad\\text\{\(8\)\}\(Ifk=0k=0, thene=f=te=f=t, making△​A​F​E\\triangle AFEdegenerate, contrary to the existence of its circumcenterOAO\_\{A\}\.\)

3\. Eliminating the Translation

Assume for contradiction thatk=1k=1\. Then \(8\) becomes

e=a\+t,f=b\+t,d=c\+t\.e=a\+t,\\quad f=b\+t,\\quad d=c\+t\.BecauseFFlies onA​BAB, the pointsAA,FF,BBare collinear; hence

f−ab−a=b\+t−ab−a=1\+tb−a∈ℝ\.\\frac\{f\-a\}\{b\-a\}=\\frac\{b\+t\-a\}\{b\-a\}=1\+\\frac\{t\}\{b\-a\}\\in\\mathbb\{R\}\.Similarly,DDlies onB​CBC, so

d−bc−b=c\+t−bc−b=1\+tc−b∈ℝ,\\frac\{d\-b\}\{c\-b\}=\\frac\{c\+t\-b\}\{c\-b\}=1\+\\frac\{t\}\{c\-b\}\\in\\mathbb\{R\},thustb−a∈ℝ\\frac\{t\}\{b\-a\}\\in\\mathbb\{R\}andtc−b∈ℝ\\frac\{t\}\{c\-b\}\\in\\mathbb\{R\}\.

The vectorsb−ab\-aandc−bc\-bare not parallel \(they are sides of a non\-degenerate triangle\), so the only complex number that is a real multiple of both is0\. Hencet=0t=0, which givese=ae=a,f=bf=b,d=cd=c\. Then△​A​F​E\\triangle AFEdegenerates to the segmentA​FAF, contradicting the existence ofOAO\_\{A\}\. Thereforek≠1k\\neq 1\.

Consider the fixed point of the spiral similarityz↦k​z\+tz\\mapsto kz\+t\(whenk≠1k\\neq 1\):

Translate the plane so thatXXbecomes the origin\. \(Translation is an isometry, so all distances and circumcenters are preserved up to the same translation; we keep the same letters for the new coordinates\.\) After this translation we have

e=ka,f=kb,d=kc\.\(9\)e=ka,\\qquad f=kb,\\qquad d=kc\.\\quad\\text\{\(9\)\}\(We also note thata,b,c≠0a,b,c\\neq 0; otherwise, e\.g\.,a=0a=0would implyA=XA=X, and thene=k⋅0=0e=k\\cdot 0=0, soAAandEEcoincide, making△​A​F​E\\triangle AFEdegenerate \- impossible\. Hencea,b,ca,b,care non\-zero\.\)

4\. The Origin Lies on the Three Circles

We now show that the origin0belongs to the circumcircles of△​A​F​E\\triangle AFE,△​B​D​F\\triangle BDF, and△​C​E​D\\triangle CED\.

Lemma \(Cross Ratio and Concyclicity\)\.Four distinct pointsz1,z2,z3,z4z\_\{1\},z\_\{2\},z\_\{3\},z\_\{4\}in the complex plane lie on a common circle or line if and only if their cross ratio

\(z1,z2;z3,z4\)=\(z1−z3\)​\(z2−z4\)\(z1−z4\)​\(z2−z3\)\(z\_\{1\},z\_\{2\};z\_\{3\},z\_\{4\}\)=\\frac\{\(z\_\{1\}\-z\_\{3\}\)\(z\_\{2\}\-z\_\{4\}\)\}\{\(z\_\{1\}\-z\_\{4\}\)\(z\_\{2\}\-z\_\{3\}\)\}is a real number\.

*Proof\.*The Möbius transformation

T​\(z\)=\(z−z1\)​\(z2−z3\)\(z−z3\)​\(z2−z1\)T\(z\)=\\frac\{\(z\-z\_\{1\}\)\(z\_\{2\}\-z\_\{3\}\)\}\{\(z\-z\_\{3\}\)\(z\_\{2\}\-z\_\{1\}\)\}mapsz1,z2,z3z\_\{1\},z\_\{2\},z\_\{3\}to0,1,∞0,1,\\inftyrespectively\. UnderTT, the circle/line throughz1,z2,z3z\_\{1\},z\_\{2\},z\_\{3\}is mapped to the real line\. Hencez4z\_\{4\}lies on that circle/line exactly whenT​\(z4\)T\(z\_\{4\}\)is real, andT​\(z4\)T\(z\_\{4\}\)equals the cross ratio\(z1,z2;z3,z4\)\(z\_\{1\},z\_\{2\};z\_\{3\},z\_\{4\}\)\.

Circle\(A​F​E\)\(AFE\)

Takez1=az\_\{1\}=a,z2=fz\_\{2\}=f,z3=ez\_\{3\}=e,z4=0z\_\{4\}=0\. Using \(9\):

a−e=a−k​a=a​\(1−k\),f−0=k​b,a−0=a,f−e=k​b−k​a=k​\(b−a\)\.a\-e=a\-ka=a\(1\-k\),\\quad f\-0=kb,\\quad a\-0=a,\\quad f\-e=kb\-ka=k\(b\-a\)\.Thus

\(a,f;e,0\)=\(a−e\)​\(f−0\)\(a−0\)​\(f−e\)=a​\(1−k\)⋅k​ba⋅k​\(b−a\)=\(1−k\)​bb−a\.\(10\)\(a,f;e,0\)=\\frac\{\(a\-e\)\(f\-0\)\}\{\(a\-0\)\(f\-e\)\}=\\frac\{a\(1\-k\)\\cdot kb\}\{a\\cdot k\(b\-a\)\}=\\frac\{\(1\-k\)b\}\{b\-a\}\.\\quad\\text\{\(10\)\}BecauseA,F,BA,F,Bare collinear, the ratiof−ab−a\\dfrac\{f\-a\}\{b\-a\}is real\. Compute

f−ab−a=k​b−ab−a\.\\frac\{f\-a\}\{b\-a\}=\\frac\{kb\-a\}\{b\-a\}\.Now

\(1−k\)​bb−a=1−k​b−ab−a,\\frac\{\(1\-k\)b\}\{b\-a\}=1\-\\frac\{kb\-a\}\{b\-a\},which is therefore real\. The pointsa,f,ea,f,eare not collinear \(otherwise△​A​F​E\\triangle AFEwould be degenerate\)\. By the lemma,0lies on the circle througha,f,ea,f,e, i\.e\.,

0∈\(AFE\)\.\(11\)0\\in\(AFE\)\.\\quad\\text\{\(11\)\}
Circle\(B​D​F\)\(BDF\)

Takez1=bz\_\{1\}=b,z2=dz\_\{2\}=d,z3=fz\_\{3\}=f,z4=0z\_\{4\}=0\. Using \(9\):

b−d=b−k​c,f−0=k​b,b−0=b,f−d=k​b−k​c=k​\(b−c\)\.b\-d=b\-kc,\\quad f\-0=kb,\\quad b\-0=b,\\quad f\-d=kb\-kc=k\(b\-c\)\.Hence

\(b,d;f,0\)=\(b−d\)​\(f−0\)\(b−0\)​\(f−d\)=\(b−k​c\)⋅k​bb⋅k​\(b−c\)=b−k​cb−c\.\(12\)\(b,d;f,0\)=\\frac\{\(b\-d\)\(f\-0\)\}\{\(b\-0\)\(f\-d\)\}=\\frac\{\(b\-kc\)\\cdot kb\}\{b\\cdot k\(b\-c\)\}=\\frac\{b\-kc\}\{b\-c\}\.\\quad\\text\{\(12\)\}Collinearity ofB,D,CB,D,Cgivesd−bc−b∈ℝ\\dfrac\{d\-b\}\{c\-b\}\\in\\mathbb\{R\}:

k​c−bc−b∈ℝ\.\\frac\{kc\-b\}\{c\-b\}\\in\\mathbb\{R\}\.Observe that

b−k​cb−c=−\(k​c−b\)−\(c−b\)=k​c−bc−b,\\frac\{b\-kc\}\{b\-c\}=\\frac\{\-\(kc\-b\)\}\{\-\(c\-b\)\}=\\frac\{kc\-b\}\{c\-b\},so the cross ratio is real\. Sinceb,d,fb,d,fare non\-collinear \(triangleB​D​FBDFis non\-degenerate\), the lemma yields

0∈\(BDF\)\.\(13\)0\\in\(BDF\)\.\\quad\\text\{\(13\)\}
Circle\(C​E​D\)\(CED\)

Takez1=cz\_\{1\}=c,z2=ez\_\{2\}=e,z3=dz\_\{3\}=d,z4=0z\_\{4\}=0\. Using \(9\):

c−e=c−k​a,d−0=k​c,c−0=c,d−e=k​c−k​a=k​\(c−a\)\.c\-e=c\-ka,\\quad d\-0=kc,\\quad c\-0=c,\\quad d\-e=kc\-ka=k\(c\-a\)\.Thus

\(c,e;d,0\)=\(c−e\)​\(d−0\)\(c−0\)​\(d−e\)=\(c−k​a\)⋅k​cc⋅k​\(c−a\)=c−k​ac−a\.\(14\)\(c,e;d,0\)=\\frac\{\(c\-e\)\(d\-0\)\}\{\(c\-0\)\(d\-e\)\}=\\frac\{\(c\-ka\)\\cdot kc\}\{c\\cdot k\(c\-a\)\}=\\frac\{c\-ka\}\{c\-a\}\.\\quad\\text\{\(14\)\}Collinearity ofC,E,AC,E,Agivese−ca−c∈ℝ\\dfrac\{e\-c\}\{a\-c\}\\in\\mathbb\{R\}:

k​a−ca−c∈ℝ\.\\frac\{ka\-c\}\{a\-c\}\\in\\mathbb\{R\}\.Since

c−k​ac−a=−\(k​a−c\)−\(a−c\)=k​a−ca−c,\\frac\{c\-ka\}\{c\-a\}=\\frac\{\-\(ka\-c\)\}\{\-\(a\-c\)\}=\\frac\{ka\-c\}\{a\-c\},the cross ratio is real\. Non\-collinearity ofc,e,dc,e,dgives

0∈\(CED\)\.\(15\)0\\in\(CED\)\.\\quad\\text\{\(15\)\}
5\.kkis Not Real

Suppose, for contradiction, thatk∈ℝk\\in\\mathbb\{R\}\. Then from \(9\) we havee=k​ae=ka, so points0,aa,eeare collinear \(they lie on the line through the origin andaa\)\. However, by \(11\) all three points0,a,e0,a,elie on the circle\(A​F​E\)\(AFE\)\. A non\-degenerate circle cannot contain three distinct collinear points\. This contradiction forces

k∉ℝ\.\(16\)k\\notin\\mathbb\{R\}\.\\quad\\text\{\(16\)\}\(Recallk≠0,1k\\neq 0,1; therefore0,1,k0,1,kare distinct and not collinear\.\)

6\. CircumcentersOA,OB,OCO\_\{A\},O\_\{B\},O\_\{C\}

Since0lies on\(A​F​E\)\(AFE\), the circle througha,f,ea,f,ecoincides with the circle through0,a,e0,a,e\(because three non\-collinear points determine a circle, and0,a,e0,a,eare distinct and, becausek∉ℝk\\notin\\mathbb\{R\}, they are not collinear\)\. HenceOAO\_\{A\}is the circumcenter of△​\(0,a,e\)\\triangle\(0,a,e\)\.

Now consider the triangle with vertices0,1,k0,1,k\. Ask∉ℝk\\notin\\mathbb\{R\}andk≠0,1k\\neq 0,1, these points are distinct and non\-collinear\. Letμ\\mube its circumcenter; thus

\|μ−0\|=\|μ−1\|=\|μ−k\|\.\(17\)\|\\mu\-0\|=\|\\mu\-1\|=\|\\mu\-k\|\.\\quad\\text\{\(17\)\}We claim thatOA=a​μO\_\{A\}=a\\mu\. Indeed, the mapz↦a​zz\\mapsto az\(a spiral similarity about0\) sends

0↦0,1↦a,k↦k​a=e\.0\\mapsto 0,\\quad 1\\mapsto a,\\quad k\\mapsto ka=e\.Under a similarity, the image of the circumcenter of a triangle is the circumcenter of the image triangle\. Thereforea​μa\\muis the circumcenter of△​\(0,a,e\)\\triangle\(0,a,e\), i\.e\.,OA=a​μO\_\{A\}=a\\mu\.

One may also verify directly: The distances froma​μa\\muto0,a,e0,a,eare

\|a​μ−0\|=\|a\|​\|μ\|,\|a​μ−a\|=\|a\|​\|μ−1\|,\|a​μ−e\|=\|a\|​\|μ−k\|,\|a\\mu\-0\|=\|a\|\\,\|\\mu\|,\\quad\|a\\mu\-a\|=\|a\|\\,\|\\mu\-1\|,\\quad\|a\\mu\-e\|=\|a\|\\,\|\\mu\-k\|,and by \(17\) these three numbers are equal, soa​μa\\muis equidistant from0,a,e0,a,e\. Because0,a,e0,a,eare not collinear,a​μa\\muis the unique point with that property, hence it is the circumcenter\.

Analogously, applying the same reasoning to the other two circles, we obtain

OB=bμ,OC=cμ\.\(18\)O\_\{B\}=b\\mu,\\qquad O\_\{C\}=c\\mu\.\\quad\\text\{\(18\)\}
7\. CircumcentersMM,NN,OO

LetMMbe the circumcenter of△​A​B​C\\triangle ABC, so

\|M−a\|=\|M−b\|=\|M−c\|=:R\.\(19\)\|M\-a\|=\|M\-b\|=\|M\-c\|=:R\.\\quad\\text\{\(19\)\}
CircumcenterOOof△​OA​OB​OC\\triangle O\_\{A\}O\_\{B\}O\_\{C\}

The vertices areOA=a​μO\_\{A\}=a\\mu,OB=b​μO\_\{B\}=b\\mu,OC=c​μO\_\{C\}=c\\mu\. Consider the pointμ​M\\mu M:

\|μ​M−OA\|=\|μ​M−a​μ\|=\|μ\|​\|M−a\|=\|μ\|​R,\|\\mu M\-O\_\{A\}\|=\|\\mu M\-a\\mu\|=\|\\mu\|\\,\|M\-a\|=\|\\mu\|R,\|μ​M−OB\|=\|μ​M−b​μ\|=\|μ\|​\|M−b\|=\|μ\|​R,\|\\mu M\-O\_\{B\}\|=\|\\mu M\-b\\mu\|=\|\\mu\|\\,\|M\-b\|=\|\\mu\|R,\|μ​M−OC\|=\|μ​M−c​μ\|=\|μ\|​\|M−c\|=\|μ\|​R\.\|\\mu M\-O\_\{C\}\|=\|\\mu M\-c\\mu\|=\|\\mu\|\\,\|M\-c\|=\|\\mu\|R\.Thusμ​M\\mu Mis equidistant fromOA,OB,OCO\_\{A\},O\_\{B\},O\_\{C\}\. Since△​OA​OB​OC\\triangle O\_\{A\}O\_\{B\}O\_\{C\}is non\-degenerate \(it is the image of△​A​B​C\\triangle ABCunder the non\-zero similarityz↦μ​zz\\mapsto\\mu z\),μ​M\\mu Mis the circumcenter:

O=μM\.\(20\)O=\\mu M\.\\quad\\text\{\(20\)\}
CircumcenterNNof△​D​E​F\\triangle DEF

The vertices areD=k​cD=kc,E=k​aE=ka,F=k​bF=kb\. Consider the pointk​MkM:

\|k​M−D\|=\|k​M−k​c\|=\|k\|​\|M−c\|=\|k\|​R,\|kM\-D\|=\|kM\-kc\|=\|k\|\\,\|M\-c\|=\|k\|R,\|k​M−E\|=\|k​M−k​a\|=\|k\|​\|M−a\|=\|k\|​R,\|kM\-E\|=\|kM\-ka\|=\|k\|\\,\|M\-a\|=\|k\|R,\|k​M−F\|=\|k​M−k​b\|=\|k\|​\|M−b\|=\|k\|​R\.\|kM\-F\|=\|kM\-kb\|=\|k\|\\,\|M\-b\|=\|k\|R\.Hencek​MkMis equidistant fromD,E,FD,E,F, and because△​D​E​F\\triangle DEFis non\-degenerate \(it is similar to△​A​B​C\\triangle ABCand, under the given configuration, non\-degenerate\), we have

N=kM\.\(21\)N=kM\.\\quad\\text\{\(21\)\}\(IfM=0M=0, thenO=0O=0,N=0N=0, and the equalities hold trivially\.\)

8\. Proof ofO​M=O​NOM=ON

Using \(20\) and \(21\),

O​M=\|O−M\|=\|μ​M−M\|=\|M\|​\|μ−1\|,OM=\|O\-M\|=\|\\mu M\-M\|=\|M\|\\,\|\\mu\-1\|,O​N=\|O−N\|=\|μ​M−k​M\|=\|M\|​\|μ−k\|\.ON=\|O\-N\|=\|\\mu M\-kM\|=\|M\|\\,\|\\mu\-k\|\.Nowμ\\muis the circumcenter of△​\(0,1,k\)\\triangle\(0,1,k\), so by \(17\)

\|μ−1\|=\|μ−k\|\.\|\\mu\-1\|=\|\\mu\-k\|\.Consequently,

O​M=\|M\|​\|μ−1\|=\|M\|​\|μ−k\|=O​N\.OM=\|M\|\\,\|\\mu\-1\|=\|M\|\\,\|\\mu\-k\|=ON\.
9\. Conclusion

We have shown that under the given conditions the distancesO​MOMandO​NONare equal\.

#### USAMO 2026 Problem 6\.

Letaaandbbbe positive integers such thatφ​\(a​b\+1\)\\varphi\(ab\+1\)dividesa2\+b2\+1a^\{2\}\+b^\{2\}\+1\. Prove thataaandbbare Fibonacci numbers\.

Solution

Letaaandbbbe positive integers such that

φ​\(a​b\+1\)∣a2\+b2\+1\.\\varphi\(ab\+1\)\\mid a^\{2\}\+b^\{2\}\+1\.SetN=a​b\+1N=ab\+1and define the positive integer

k=a2\+b2\+1φ​\(N\)\.k=\\frac\{a^\{2\}\+b^\{2\}\+1\}\{\\varphi\(N\)\}\.We will prove that bothaaandbbmust be Fibonacci numbers\.

1\. Preliminary bounds

SinceN\>1N\>1, we haveφ​\(N\)≤N−1=a​b\\varphi\(N\)\\leq N\-1=ab\(with equality exactly whenNNis prime\)\. Hence

a2\+b2\+1=kφ\(N\)≤kab\.\(1\)a^\{2\}\+b^\{2\}\+1=k\\,\\varphi\(N\)\\leq k\\,ab\.\\quad\\text\{\(1\)\}By the quadratic\-mean\-arithmetic\-mean inequality,a2\+b2≥2​a​ba^\{2\}\+b^\{2\}\\geq 2ab, so

a2\+b2\+1≥2ab\+1\.\(2\)a^\{2\}\+b^\{2\}\+1\\geq 2ab\+1\.\\quad\\text\{\(2\)\}Combining \(1\) and \(2\) yields

2​a​b\+1≤k​a​b⟹k≥2\+1a​b\.2ab\+1\\leq k\\,ab\\quad\\Longrightarrow\\quad k\\geq 2\+\\frac\{1\}\{ab\}\.Becausekkis an integer, ifa​b\>1ab\>1we obtaink≥3k\\geq 3\. Ifa​b=1ab=1then necessarilya=b=1a=b=1,N=2N=2,φ​\(2\)=1\\varphi\(2\)=1, and thenk=3k=3\. Thus in every case

k≥3\.\(3\)\\boxed\{k\\geq 3\}\.\\quad\\text\{\(3\)\}
2\. Parity and the even case

Lemma 2\.1\.IfNNis even andN\>2N\>2, then the divisibility condition cannot hold\.

*Proof\.*ForN\>2N\>2,φ​\(N\)\\varphi\(N\)is even \(a standard fact:φ​\(n\)\\varphi\(n\)is even for everyn\>2n\>2\)\. BecauseNNeven impliesa​b=N−1ab=N\-1odd, bothaaandbbare odd\. Thena2\+b2\+1a^\{2\}\+b^\{2\}\+1is odd \(odd\+odd\+1\)\. An even number cannot divide an odd number\. The only even possibility isN=2N=2, which givesa=b=1a=b=1and indeedφ​\(2\)=1\\varphi\(2\)=1divides33\.

Consequently, the only evenNNthat can occur isN=2N=2, producing\(a,b\)=\(1,1\)\(a,b\)=\(1,1\)\- both Fibonacci numbers\. From now on we assume

N​is odd and​N\>2\.N\\ \\text\{is odd and\}\\ N\>2\.Thena​b=N−1ab=N\-1is even, so at least one ofa,ba,bis even\. If both were even,a2\+b2\+1a^\{2\}\+b^\{2\}\+1would be odd, while for oddN\>2N\>2we still haveφ​\(N\)\\varphi\(N\)even\. Hence exactly one ofa,ba,bis even and the other is odd\. In particular,

a2\+b2\+1≡0\+1\+1≡2\(mod4\)\.\(4\)a^\{2\}\+b^\{2\}\+1\\equiv 0\+1\+1\\equiv 2\\pmod\{4\}\.\\quad\\text\{\(4\)\}
3\. Case I:NNis prime

AssumeN=a​b\+1N=ab\+1is prime\. Thenφ​\(N\)=N−1=a​b\\varphi\(N\)=N\-1=ab, and \(1\) becomes

a2\+b2\+1=kab\.\(5\)a^\{2\}\+b^\{2\}\+1=k\\,ab\.\\quad\\text\{\(5\)\}We first determinekk\.

3\.1\. The value ofkk

Without loss of generality, order the variables so thata≤ba\\leq b\. Write \(5\) as a quadratic inbb:

b2−\(k​a\)​b\+\(a2\+1\)=0\.b^\{2\}\-\(ka\)\\,b\+\(a^\{2\}\+1\)=0\.Ifbbis a root, the other root is

b′=a2\+1b=k​a−b,b^\{\\prime\}=\\frac\{a^\{2\}\+1\}\{b\}=ka\-b,which is a positive integer\.

Claim\.Ifa≥2a\\geq 2, thenb′<bb^\{\\prime\}<b\.

*Proof\.*Supposeb′≥bb^\{\\prime\}\\geq b\. Thena2\+1b≥b\\frac\{a^\{2\}\+1\}\{b\}\\geq b, i\.e\.a2\+1≥b2a^\{2\}\+1\\geq b^\{2\}\. Becauseb≥ab\\geq a, we havea2≤b2≤a2\+1a^\{2\}\\leq b^\{2\}\\leq a^\{2\}\+1\. Hence eitherb2=a2b^\{2\}=a^\{2\}orb2=a2\+1b^\{2\}=a^\{2\}\+1\.

⊳\\trianglerightIfb2=a2b^\{2\}=a^\{2\}, thenb=ab=a\. Substitutingb=ab=ainto \(5\) gives2​a2\+1=k​a22a^\{2\}\+1=ka^\{2\}⇒\\Rightarrowk=2\+1a2k=2\+\\frac\{1\}\{a^\{2\}\}, which is not an integer fora≥2a\\geq 2\. Contradiction\.

⊳\\trianglerightIfb2=a2\+1b^\{2\}=a^\{2\}\+1, thena2\+1a^\{2\}\+1is a perfect square\. Fora≥2a\\geq 2, this is impossible because betweena2a^\{2\}and\(a\+1\)2=a2\+2​a\+1\(a\+1\)^\{2\}=a^\{2\}\+2a\+1there is no other square\.

Thus whenevera≥2a\\geq 2, we can replace\(a,b\)\(a,b\)by\(a,b′\)\(a,b^\{\\prime\}\)\(which is also a solution of \(5\)\) witha\+b′<a\+ba\+b^\{\\prime\}<a\+b\. Repeating this descent \(the suma\+ba\+bstrictly decreases each step\) we eventually reach a solution witha=1a=1\(the minimal sum cannot havea≥2a\\geq 2\)\.

Now considera=1a=1\. Equation \(5\) becomes

1\+b2\+1=k⋅1⋅b⟹b2−kb\+2=0\.\(6\)1\+b^\{2\}\+1=k\\cdot 1\\cdot b\\quad\\Longrightarrow\\quad b^\{2\}\-kb\+2=0\.\\quad\\text\{\(6\)\}For integerbb, the discriminantΔ=k2−8\\Delta=k^\{2\}\-8must be a perfect square, sayd2d^\{2\}\. Then

k2−d2=8⟹\(k−d\)​\(k\+d\)=8\.k^\{2\}\-d^\{2\}=8\\quad\\Longrightarrow\\quad\(k\-d\)\(k\+d\)=8\.Both factors are positive integers of the same parity\. The factor pairs of88are\(1,8\)\(1,8\)and\(2,4\)\(2,4\)\.

⊳\\triangleright\(1,8\)\(1,8\)givesk−d=1k\-d=1,k\+d=8k\+d=8⇒\\Rightarrow2​k=92k=9⇒\\Rightarrowk=9/2k=9/2, not integer\.

⊳\\triangleright\(2,4\)\(2,4\)givesk−d=2k\-d=2,k\+d=4k\+d=4⇒\\Rightarrow2​k=62k=6⇒\\Rightarrowk=3k=3,d=1d=1\.

Hencek=3k=3is forced\. Withk=3k=3, equation \(6\) isb2−3​b\+2=0b^\{2\}\-3b\+2=0, whose roots areb=1b=1andb=2b=2\. Thus the only solutions witha=1a=1are\(1,1\)\(1,1\)and\(1,2\)\(1,2\)\(and by symmetry,\(2,1\)\(2,1\)if we had not assumeda≤ba\\leq b\)\. Since the reduction process preserveskk, every solution of \(5\) must havek=3k=3\.

Therefore, whenNNis prime,

a2\+b2\+1=3ab⟺a2−3ab\+b2=−1\.\(7\)a^\{2\}\+b^\{2\}\+1=3ab\\quad\\Longleftrightarrow\\quad a^\{2\}\-3ab\+b^\{2\}=\-1\.\\quad\\text\{\(7\)\}
3\.2\. Solvinga2−3​a​b\+b2=−1a^\{2\}\-3ab\+b^\{2\}=\-1

Equation \(7\) is symmetric; we may assumea≤ba\\leq b\.

Define a sequence\(An\)n≥0\(A\_\{n\}\)\_\{n\\geq 0\}by

A0=1,A1=1,An\+2=3​An\+1−An\(n≥0\)\.A\_\{0\}=1,\\qquad A\_\{1\}=1,\\qquad A\_\{n\+2\}=3A\_\{n\+1\}\-A\_\{n\}\\ \\ \(n\\geq 0\)\.One verifies by induction that\(An,An\+1\)\(A\_\{n\},A\_\{n\+1\}\)satisfies \(7\): Forn=0n=0,12−3⋅1⋅1\+12=−11^\{2\}\-3\\cdot 1\\cdot 1\+1^\{2\}=\-1\. Assuming it holds fornn, using the recurrence one checks that it also holds forn\+1n\+1\. \(The computation is straightforward and can be filled in similarly to the proof in Lemma 4\.1 of the original draft\.\)

Now let\(a,b\)\(a,b\)be any positive integer solution of \(7\) witha≤ba\\leq b\. View \(7\) as a quadratic inbb:

b2−3​a​b\+\(a2\+1\)=0\.b^\{2\}\-3a\\,b\+\(a^\{2\}\+1\)=0\.Its two roots arebband

b′=a2\+1b=3​a−b,b^\{\\prime\}=\\frac\{a^\{2\}\+1\}\{b\}=3a\-b,which is an integer\. A similar argument as in 3\.1 shows that ifa≥2a\\geq 2, thenb′<bb^\{\\prime\}<b\. Consequently, by infinite descent ona\+ba\+b, we can reduce any solution to one witha=1a=1\. The only solutions witha=1a=1are\(1,1\)\(1,1\)and\(1,2\)\(1,2\)\(and their symmetric versions\)\. Therefore every solution can be obtained from these minimal ones by the inverse transformation: if\(x,y\)\(x,y\)is a solution withx≤yx\\leq y, then\(y,3​y−x\)\(y,\\,3y\-x\)is also a solution\. Starting from\(1,1\)\(1,1\)and repeatedly applying this transformation yields the increasing sequence

\(1,1\),\(1,2\),\(2,5\),\(5,13\),\(13,34\),…\(1,1\),\\ \(1,2\),\\ \(2,5\),\\ \(5,13\),\\ \(13,34\),\\ \\dotswhich exactly corresponds to the pairs\(An,An\+1\)\(A\_\{n\},A\_\{n\+1\}\)forn≥0n\\geq 0\.

It remains to identify theseAnA\_\{n\}with Fibonacci numbers\. Define the Fibonacci numbers by

F1=1,F2=1,Fn\+2=Fn\+1\+Fn\.F\_\{1\}=1,\\quad F\_\{2\}=1,\\quad F\_\{n\+2\}=F\_\{n\+1\}\+F\_\{n\}\.We claim that forn≥1n\\geq 1,

An=F2​n−1\.A\_\{n\}=F\_\{2n\-1\}\.*Proof by induction\.*

⊳\\trianglerightBasen=1n=1:A1=1=F1A\_\{1\}=1=F\_\{1\}\.

⊳\\trianglerightn=2n=2:A2=3​A1−A0=3⋅1−1=2=F3A\_\{2\}=3A\_\{1\}\-A\_\{0\}=3\\cdot 1\-1=2=F\_\{3\}\.

⊳\\trianglerightInductive step: AssumeAn=F2​n−1A\_\{n\}=F\_\{2n\-1\}andAn\+1=F2​n\+1A\_\{n\+1\}=F\_\{2n\+1\}\. Then

An\+2=3​An\+1−An=3​F2​n\+1−F2​n−1\.A\_\{n\+2\}=3A\_\{n\+1\}\-A\_\{n\}=3F\_\{2n\+1\}\-F\_\{2n\-1\}\.Using the identities

F2​n\+1−F2​n−1=F2​n,andF2​n\+1\+F2​n=F2​n\+2,F\_\{2n\+1\}\-F\_\{2n\-1\}=F\_\{2n\},\\quad\\text\{and\}\\quad F\_\{2n\+1\}\+F\_\{2n\}=F\_\{2n\+2\},we compute

3​F2​n\+1−F2​n−1\\displaystyle 3F\_\{2n\+1\}\-F\_\{2n\-1\}=2​F2​n\+1\+\(F2​n\+1−F2​n−1\)\\displaystyle=2F\_\{2n\+1\}\+\(F\_\{2n\+1\}\-F\_\{2n\-1\}\)=2​F2​n\+1\+F2​n\\displaystyle=2F\_\{2n\+1\}\+F\_\{2n\}=F2​n\+1\+\(F2​n\+1\+F2​n\)\\displaystyle=F\_\{2n\+1\}\+\(F\_\{2n\+1\}\+F\_\{2n\}\)=F2​n\+1\+F2​n\+2\\displaystyle=F\_\{2n\+1\}\+F\_\{2n\+2\}=F2​n\+3\.\\displaystyle=F\_\{2n\+3\}\.ThusAn\+2=F2​n\+3=F2​\(n\+2\)−1A\_\{n\+2\}=F\_\{2n\+3\}=F\_\{2\(n\+2\)\-1\}, completing the induction\.

Hence all solutions of \(7\) are pairs of Fibonacci numbers \(specifically, consecutive odd\-index Fibonacci numbers\)\. This completesCase I\.

4\. Case II:NNis composite and odd \(N\>2N\>2\)

NowN=a​b\+1N=ab\+1is composite, odd, andN\>2N\>2\. From \(4\) we have

a2\+b2\+1≡2\(mod4\)\.a^\{2\}\+b^\{2\}\+1\\equiv 2\\pmod\{4\}\.
4\.1\.NNcannot have two distinct odd prime factors

SupposeNNhas at least two distinct odd prime divisorsppandqq\. Thenφ​\(N\)\\varphi\(N\)is divisible by\(p−1\)​\(q−1\)\(p\-1\)\(q\-1\)\. Sinceppandqqare odd, bothp−1p\-1andq−1q\-1are even, hence4​∣\(p−1\)​\(q−1\)∣​φ​\(N\)4\\mid\(p\-1\)\(q\-1\)\\mid\\varphi\(N\)\. Thus4∣φ​\(N\)4\\mid\\varphi\(N\)\. Sinceφ​\(N\)∣a2\+b2\+1\\varphi\(N\)\\mid a^\{2\}\+b^\{2\}\+1, we would have4∣a2\+b2\+14\\mid a^\{2\}\+b^\{2\}\+1, contradicting \(4\)\. ThereforeNNcannot possess two distinct odd prime factors\.

It follows thatNNmust be a prime power:

N=pe,e≥2​\(since​N​is composite\)\.N=p^\{e\},\\qquad e\\geq 2\\ \(\\text\{since \}N\\text\{ is composite\}\)\.
4\.2\. The primeppcannot be≡1\(mod4\)\\equiv 1\\pmod\{4\}

Ifp≡1\(mod4\)p\\equiv 1\\pmod\{4\}, thenp−1p\-1is divisible by44, so4∣φ​\(N\)4\\mid\\varphi\(N\)\. Again this forces4∣a2\+b2\+14\\mid a^\{2\}\+b^\{2\}\+1, which is impossible by \(4\)\. Hence

p≡3\(mod4\)\.\(8\)p\\equiv 3\\pmod\{4\}\.\\quad\\text\{\(8\)\}
4\.3\. Analysis modulo33

We now examine the residue ofppmodulo33\.

Subcasep≡1\(mod3\)p\\equiv 1\\pmod\{3\}\.Then3∣\(p−1\)3\\mid\(p\-1\), so3∣φ​\(N\)3\\mid\\varphi\(N\)\. Moreover,p≡1\(mod3\)p\\equiv 1\\pmod\{3\}impliespe≡1\(mod3\)p^\{e\}\\equiv 1\\pmod\{3\}, soa​b=pe−1≡0\(mod3\)ab=p^\{e\}\-1\\equiv 0\\pmod\{3\}\. Thus3∣a​b3\\mid ab\.

Now evaluateS=a2\+b2\+1S=a^\{2\}\+b^\{2\}\+1modulo33\. Since3∣a​b3\\mid ab, there are two possibilities:

⊳\\triangleright3∣a3\\mid aand3∣b3\\mid b: thena2≡0a^\{2\}\\equiv 0,b2≡0b^\{2\}\\equiv 0, soS≡1\(mod3\)S\\equiv 1\\pmod\{3\}\.

⊳\\trianglerightExactly one ofa,ba,bis divisible by33: thenS≡0\+1\+1=2\(mod3\)S\\equiv 0\+1\+1=2\\pmod\{3\}\.

In neither case isSSdivisible by33\. However,φ​\(N\)∣S\\varphi\(N\)\\mid Sand3∣φ​\(N\)3\\mid\\varphi\(N\)would force3∣S3\\mid S\. Contradiction\. Thereforep≢1\(mod3\)p\\not\\equiv 1\\pmod\{3\}\.

Subcasep≡2\(mod3\)p\\equiv 2\\pmod\{3\}\.Thenp−1≡1\(mod3\)p\-1\\equiv 1\\pmod\{3\}, so3∤\(p−1\)3\\nmid\(p\-1\)\. But becausee≥2e\\geq 2,φ​\(N\)=pe−1​\(p−1\)\\varphi\(N\)=p^\{e\-1\}\(p\-1\)contains the factorpe−1p^\{e\-1\}, so in particularp∣φ​\(N\)p\\mid\\varphi\(N\)\. Consequently,

p∣S\.\(9\)p\\mid S\.\\quad\\text\{\(9\)\}Also, froma​b=pe−1ab=p^\{e\}\-1we have

ab≡−1\(modp\)\.\(10\)ab\\equiv\-1\\pmod\{p\}\.\\quad\\text\{\(10\)\}In particular,ppdoes not divideaaorbb\(otherwise \(10\) would give0≡−1\(modp\)0\\equiv\-1\\pmod\{p\}, impossible\)\. Henceaaandbbare invertible modulopp\. Writeb≡−a−1\(modp\)b\\equiv\-a^\{\-1\}\\pmod\{p\}\. Substituting into \(9\):

a2\+\(−a−1\)2\+1≡0\(modp\),a^\{2\}\+\(\-a^\{\-1\}\)^\{2\}\+1\\equiv 0\\pmod\{p\},i\.e\.,

a2\+a−2\+1≡0\(modp\)\.a^\{2\}\+a^\{\-2\}\+1\\equiv 0\\pmod\{p\}\.Multiply bya2a^\{2\}\(invertible modulopp\):

a4\+a2\+1≡0\(modp\)\.\(11\)a^\{4\}\+a^\{2\}\+1\\equiv 0\\pmod\{p\}\.\\quad\\text\{\(11\)\}Sety=a2\(modp\)y=a^\{2\}\\pmod\{p\}; then \(11\) becomes

y2\+y\+1≡0\(modp\)\.\(12\)y^\{2\}\+y\+1\\equiv 0\\pmod\{p\}\.\\quad\\text\{\(12\)\}Equation \(12\) has a solution in𝔽p\\mathbb\{F\}\_\{p\}iff−3\-3is a quadratic residue modulopp\. Indeed, multiplying \(12\) by44gives\(2​y\+1\)2≡−3\(modp\)\(2y\+1\)^\{2\}\\equiv\-3\\pmod\{p\}\.

We need the following standard lemma:

Lemma\.For an odd primep≠3p\\neq 3,−3\-3is a quadratic residue moduloppif and only ifp≡1\(mod3\)p\\equiv 1\\pmod\{3\}\.

*Proof\.*The multiplicative group𝔽p×\\mathbb\{F\}\_\{p\}^\{\\times\}is cyclic of orderp−1p\-1\. The equationy2\+y\+1=0y^\{2\}\+y\+1=0is equivalent toy3=1y^\{3\}=1withy≠1y\\neq 1\(since\(y−1\)​\(y2\+y\+1\)=y3−1\(y\-1\)\(y^\{2\}\+y\+1\)=y^\{3\}\-1\. Hence a solution exists iff the group contains an element of order33, i\.e\., iff3∣p−13\\mid p\-1, which isp≡1\(mod3\)p\\equiv 1\\pmod\{3\}\. Conversely, ifp≡1\(mod3\)p\\equiv 1\\pmod\{3\}, then3∣p−13\\mid p\-1, so such an element exists\.

Sincep≡2\(mod3\)p\\equiv 2\\pmod\{3\}, we havep−1p\-1not divisible by33, so there is no element of order33in𝔽p×\\mathbb\{F\}\_\{p\}^\{\\times\}; therefore \(12\) has no solution modulopp\. This contradicts the existence ofaa\. Thusp≢2\(mod3\)p\\not\\equiv 2\\pmod\{3\}\.

The only remaining possibility for the odd primeppisp=3p=3\.

4\.4\. The exponentee

NowN=3eN=3^\{e\}withe≥2e\\geq 2\. Then

φ​\(N\)=3e−1⋅2\.\\varphi\(N\)=3^\{e\-1\}\\cdot 2\.Ife≥3e\\geq 3:Then3e−13^\{e\-1\}is at least32=93^\{2\}=9, so9∣φ​\(N\)9\\mid\\varphi\(N\)\. Sinceφ​\(N\)∣S\\varphi\(N\)\\mid S, we must have9∣S9\\mid S\.

But we will show thatS≡3\(mod9\)S\\equiv 3\\pmod\{9\}, which is not divisible by99, a contradiction\. Let us verify this congruence\.

Becausee≥3e\\geq 3,N=3eN=3^\{e\}is divisible by2727, in particularN≡0\(mod9\)N\\equiv 0\\pmod\{9\}\. Hence

ab=N−1≡−1\(mod9\)\.\(13\)ab=N\-1\\equiv\-1\\pmod\{9\}\.\\quad\\text\{\(13\)\}Moreover,N≡0\(mod3\)N\\equiv 0\\pmod\{3\}givesa​b≡−1≡2\(mod3\)ab\\equiv\-1\\equiv 2\\pmod\{3\}\. Hence3∤a3\\nmid aand3∤b3\\nmid b\(otherwisea​b≡0\(mod3\)ab\\equiv 0\\pmod\{3\}\)\. Thusaaandbbare each≡1\\equiv 1or2\(mod3\)2\\pmod\{3\}, and their product is2\(mod3\)2\\pmod\{3\}; the only possibilities are\{a,b\}≡\{1,2\}\(mod3\)\\\{a,b\\\}\\equiv\\\{1,2\\\}\\pmod\{3\}\. Consequently,

a\+b≡1\+2≡0\(mod3\)\.\(14\)a\+b\\equiv 1\+2\\equiv 0\\pmod\{3\}\.\\quad\\text\{\(14\)\}Now compute

S=a2\+b2\+1=\(a\+b\)2−2​a​b\+1\.S=a^\{2\}\+b^\{2\}\+1=\(a\+b\)^\{2\}\-2ab\+1\.From \(14\),a\+b=3​ta\+b=3tfor some integertt, so\(a\+b\)2=9​t2≡0\(mod9\)\(a\+b\)^\{2\}=9t^\{2\}\\equiv 0\\pmod\{9\}\. Using \(13\),a​b≡−1\(mod9\)ab\\equiv\-1\\pmod\{9\}, so−2​a​b≡−2​\(−1\)=2\(mod9\)\-2ab\\equiv\-2\(\-1\)=2\\pmod\{9\}\. Thus

S≡0\+2\+1=3\(mod9\)\.S\\equiv 0\+2\+1=3\\pmod\{9\}\.Thus9∤S9\\nmid S, contradicting9​∣φ​\(N\)∣​S9\\mid\\varphi\(N\)\\mid S\. Thereforeeecannot be≥3\\geq 3\.

The only remaining possibility ise=2e=2, i\.e\.,N=9N=9\. Thenφ​\(9\)=6\\varphi\(9\)=6anda​b=8ab=8\. The positive factor pairs of88are

\(1,8\),\(2,4\),\(4,2\),\(8,1\)\.\(1,8\),\\ \(2,4\),\\ \(4,2\),\\ \(8,1\)\.ComputeSSfor each:

⊳\\triangleright\(1,8\)\(1,8\):S=1\+64\+1=66S=1\+64\+1=66, and66/6=11∈ℤ66/6=11\\in\\mathbb\{Z\}\.

⊳\\triangleright\(2,4\)\(2,4\):S=4\+16\+1=21S=4\+16\+1=21,21/6=3\.5∉ℤ21/6=3\.5\\notin\\mathbb\{Z\}\.

Thus only\(1,8\)\(1,8\)and\(8,1\)\(8,1\)satisfy the divisibility condition\. Both11and88are Fibonacci numbers \(1=F11=F\_\{1\},8=F68=F\_\{6\}\)\.

5\. Conclusion

We have examined all possible values ofN=a​b\+1N=ab\+1:

⊳\\trianglerightEven case:N=2N=2gives\(a,b\)=\(1,1\)\(a,b\)=\(1,1\)\- Fibonacci numbers\.

⊳\\trianglerightPrime case:NNprime leads to the equationa2−3​a​b\+b2=−1a^\{2\}\-3ab\+b^\{2\}=\-1, whose complete set of positive integer solutions consists of pairs of Fibonacci numbers \(as shown in 3\.2\)\.

⊳\\trianglerightComposite case:NNcomposite forcesN=9N=9and yields\(a,b\)=\(1,8\)\(a,b\)=\(1,8\)or\(8,1\)\(8,1\), again Fibonacci numbers\.

Hence, under the hypothesisφ​\(a​b\+1\)∣a2\+b2\+1\\varphi\(ab\+1\)\\mid a^\{2\}\+b^\{2\}\+1, the integersaaandbbare necessarily Fibonacci numbers\.

*Remark\.*The Fibonacci numbers are defined byF1=F2=1F\_\{1\}=F\_\{2\}=1andFn\+2=Fn\+1\+FnF\_\{n\+2\}=F\_\{n\+1\}\+F\_\{n\}\. The solutions from the prime case are precisely\(F2​n−1,F2​n\+1\)\(F\_\{2n\-1\},F\_\{2n\+1\}\)\(and symmetric\) forn≥1n\\geq 1, together with the degenerate pair\(1,1\)\(1,1\)which can be seen as\(F1,F2\)\(F\_\{1\},F\_\{2\}\)or\(F2,F1\)\(F\_\{2\},F\_\{1\}\)\. The pair\(1,8\)\(1,8\)from the composite case also fits the Fibonacci sequence \(F1=1F\_\{1\}=1,F6=8F\_\{6\}=8\)\. This completes the proof\.