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

Yafu Li, Runzhe Zhan, Haoran Zhang, Shunkai Zhang, Yizhuo Li, Zhilin Wang, Jiacheng Chen, Futing Wang, Xuyang Hu, Yuchen Fan, Bangjie Xu, Yucheng Su, Xinmiao Han, Chenxi Li, Haodi Lei, Yufeng Zhao, Zejin Lin, Qianjia Cheng, Tong Zhu, Xiaoye Qu, Ganqu Cui, Peng Ye, Yun Luo, Zhouchen Lin, Yu Qiao, Bowen Zhou, Ning Ding, Yu Cheng

arXiv:2605.13301v1cs.AIcs.CL

TL;DR

The paper asks whether a compact reasoning backbone can reach olympiad-level mathematical and scientific performance through a unified recipe. It combines proof-oriented SFT, staged RL, and test-time scaling, producing SU-01, which achieves gold-medal-level olympiad performance and transfers to broader scientific reasoning.

  • Problem

    It is unclear whether a compact reasoning backbone can reach olympiad-level performance through one domain-unified pipeline across mathematical and scientific problems.

    Source: §1 · p. 2

  • Method

    The recipe combines reverse-perplexity curriculum SFT, staged RL with verifiable and proof-level rewards, and test-time scaling for self-verification and refinement.

    Source: §1 · p. 2, §8 · p. 13

  • Results

    SU-01 reaches gold-medal-level performance on mathematical and physical olympiads, sustains reasoning trajectories beyond 100K tokens, and shows scientific-reasoning transfer beyond olympiad benchmarks.

    Source: §1 · pp. 2–3, §8 · p. 13

  • Takeaways & Limitations

    A compact 30B-A3B backbone can be driven toward expert-level proof reasoning while retaining meaningful scientific transfer.

    Source: §1 · p. 3

    SU-01 still benefits from additional search and self-correction on brittle proof attempts, with harder problems remaining unsolved in a single pass.

    Source: USAMO 2026 · p. 10

Abstract

from arXiv · show

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 a 30B-A3B backbone with SFT on around 340K sub-8K-token trajectories followed by 200 RL steps. The resulting model, SU-01, supports stable reasoning on difficult problems with trajectories exceeding 100K tokens, while achieving gold-medal-level performance on mathematical and physical olympiad competitions, including IMO 2025/USAMO 2026 and IPhO 2024/2025. It also demonstrates strong generalization of scientific reasoning to domains beyond mathematics and physics.

1 Introduction

The paper presents a compact, unified pipeline that converts a broadly capable 30B-A3B reasoning backbone into an olympiad-level solver through rigorous SFT, staged RL, and test-time scaling. SU-01 achieves strong proof, competition, and scientific-reasoning results while sustaining verification-rich trajectories beyond 100K tokens.

Supervised fine-tuning

  • 338K sub-8K-token trajectories and reverse-perplexity ordering instill proof search, self-checking, and repair while helping preserve and recover the post-trained model’s capabilities.The SFT mixture draws from mathematical, scientific, coding, and instruction-following sources, with each pass beginning from trajectories most mismatched to the initial policy.Source: §1 · p. 2

Reinforcement learning

  • Refined RL shifts optimization from answer correctness to proof quality using generative proof rewards, critique-and-repair training, and experience replay for rare successful trajectories.Coarse RL first uses reliable binary rewards from verifiable prompts, and test-time scaling repeatedly verifies and repairs candidate proofs.Source: §1 · p. 2

Benchmark results

  • 57.6% on IMO-ProofBench with direct generation and 70.2% with test-time scaling substantially outperform similar-size models, while SU-01 nearly matches Qwen3.6-35B-A3B on answer-verifiable benchmarks.SU-01 also achieves the best similar-size overall score on FrontierScience-Research, indicating transfer toward research-style scientific reasoning.Source: §1 · p. 2

Competition results

  • 35 points on both IMO 2025 and USAMO 2026 with test-time scaling meet the IMO gold line and exceed the USAMO gold line by 10 points.On USAMO 2026, this matches the highest reported human total among 340 competitors; direct SU-01 also exceeds IPhO 2024/2025 gold lines and clears IMO/USAMO bronze lines.Source: §1 · pp. 2–3

Training and inference pipeline

  • The unified recipe reshapes a 30B-A3B backbone with rigorous long-form SFT, scales behaviors through coarse and refined RL, and applies test-time verification and refinement.The stages align behavior shaping, reward design, experience replay, and self-verification across mathematical and scientific problems.Source: §1 · p. 2, Figure 2 · p. 3

Emergent reasoning behavior

  • SU-01 sustains reasoning trajectories beyond 100K tokens, conditions on drafts and error analyses, and repeatedly verifies and repairs candidate proofs while retaining scientific transfer.These results support specializing a broadly capable post-trained model toward expert-level proof reasoning rather than building a narrow olympiad solver.Source: §1 · pp. 2–3

2 Instilling Rigorous Reasoning via SFT

SFT specializes a post-trained reasoning backbone toward explicit, rigorous proof-search while preserving broad scientific competence. The data mixture combines diverse reasoning sources with self-verification and self-refinement, and reverse-perplexity ordering stabilizes long-CoT adaptation.

2 Instilling Rigorous Reasoning via SFT

  • SFT reshapes P1-30B-A3B’s reasoning toward explicit, disciplined proof-search while preserving its existing mathematical, physical, and broader scientific capabilities.Starting from a post-trained model lets SFT modify reasoning patterns rather than rebuild instruction-following and problem-solving abilities, providing a stronger policy for later RL.Source: §2 · p. 3

2.1 SFT Data Curation

  • The curated mixture spans mathematical, scientific, instruction-following, and coding prompts, with contamination and noisy generations filtered and trajectories capped at 8,192 tokens.The resulting filtered SFT set contains 338K trajectories; Figure 3 groups its data into direct-generation and self-improvement sources.Source: §2.1 · pp. 3–4, Figure 3 · p. 4
  • Mathematical examples include verification and refinement traces, teaching the model to check proof validity and repair flaws in its arguments.These self-improvement examples complement direct solution trajectories in the final filtered mixture.Source: §2.1 · p. 4

2.2 Reverse-Perplexity Curriculum for SFT

  • For 8,192-token-capped trajectories, four epochs usually recover most capability after the initial behavioral shift, while validation truncation below 5% indicates sufficient adaptation.Insufficient adaptation produces shallow, repetitive, or endless reasoning and can substantially degrade performance; recovery also depends on the data mixture and learning rate.Source: §2.2 · p. 4
  • Reverse-perplexity curriculum orders examples from high to low perplexity, exposing the model first to teacher trajectories most mismatched with its current policy.The ordering is intended to promote behavioral adaptation to unfamiliar proof-search patterns before consolidating more familiar examples.Source: §2.2 · p. 4

3 Boosting Reasoning Capability with RL

The RL stage uses a two-level pipeline: coarse RL strengthens answer-seeking behavior with verifiable rewards, while refined RL optimizes rigorous, complete proofs with process-level feedback. Refined RL further uses self-refinement and targeted replay to turn failures into repair tasks and preserve rare successful proofs.

3 Boosting Reasoning Capability with RL

  • RL first applies coarse optimization for answer-seeking, then refined optimization for complete, auditable proof construction.Coarse RL uses reliable outcome rewards to improve search, coverage, and direct solving, whereas refined RL uses finer feedback to encourage rigor and self-refinement.Source: §3 · p. 5

3.1 RL Data Curation

  • The RL pool contains 8,967 verifiable prompts and 16,287 non-verifiable prompts after deduplication, decontamination, rejection sampling, and noise filtering.The verifiable set supports reliably checked answers or structured outputs, while the non-verifiable set contains proof-oriented or open-ended prompts requiring softer judgment.Source: §3.1 · p. 5

3.2 Coarse RL

  • Coarse RL trains on verifiable prompts with RLVR and GSPO, assigning binary correctness rewards at the complete-response level.GSPO aligns reward assignment and policy clipping with outcome-reward training, while layered verification prioritizes canonical matching and Math-Verify before generative verification by gptoss-120b (OpenAI, 2025).Source: §3.2 · pp. 5–6

3.3 Refined RL

  • Refined RL evaluates full solutions with a binary proof-quality reward from DeepSeekMath-V2, targeting validity, rigor, and completeness rather than final-answer matching alone.Anti-hack preprocessing replaces malformed, delimiter-corrupted, or severely repetitive generations with a safe fallback before reward-model evaluation.Source: §3.3 · p. 6
  • Self-refinement converts groups with average proof reward below τref = 0.5 into critique-and-repair prompts mixed into later batches at ηref = 0.2.The repair prompts include the original problem and failed solution, and ask the policy to locate gaps, fix errors, supply justifications, and produce a complete proof.Source: §3.3 · p. 6
  • Targeted replay preserves rare valid proofs when 0 < n+(q) < 2, selects the lowest-entropy stored trajectory, replays it at ρ = 0.25, and retires queries once n+(q) ≥4.Because refined-RL success is judged on the full proof, rare successful trajectories are more likely to encode reusable proof patterns than answer-only lucky hits; replay is mixed with fresh training rather than run separately.Source: §3.3 · pp. 6–7

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

Test-time scaling addresses olympiad proofs’ need for substantial search, verification, and revision beyond single-pass or best-of-many generation. SU-01 implements this through repeated solve–refine–verify–verdict cycles, sustaining coherent inference trajectories beyond 100K tokens.

Motivation

  • Structured verification-and-refinement outperforms single-pass and best-of-many inference for rigorous IMO-style proofs, which can fail despite correct conclusions when they contain hidden gaps or fallacies.The motivation is that complete proofs require checking logical validity, not merely selecting a correct final answer.Source: §4 · p. 7
  • Repeated solve–verify–refine stages allocate additional computation to exploration, lemma checking, counterexample search, and exposition repair while keeping each step focused and auditable.After training, SU-01 sustains coherent reasoning trajectories longer than 100K tokens during inference.Source: §4 · p. 7

Test-time scaling procedure

  • SU-01 generates a rigorous draft, repairs weak steps, produces a structured bug report, and uses a verdict to accept, reject, or trigger further refinement until verification passes or the budget is exhausted.Multiple independent runs may execute in parallel or serial, and candidates are selected only after repeated verification establishes proof stability.Source: §4 · p. 7

5 Experimental Results

SU-01 achieves gold-medal-level performance on official mathematics and physics olympiad problems while remaining competitive on answer-verifiable, proof-oriented, and broader scientific reasoning benchmarks. Test-time scaling substantially improves proof and competition performance, while revealing failures on especially delicate structural arguments.

5.1 Benchmarks

  • The evaluation covers answer-verifiable tasks, proof-oriented and non-verifiable scientific problems, and official IMO, USAMO, and IPhO competitions.Answer-verifiable benchmarks use final-answer checks or high-confidence automatic verifiers, while proof-oriented tasks assess rigorous reasoning and scientific problem solving beyond answer matching.Source: §5 · p. 7, §5.1 · pp. 7–8

5.2 Verifiable Problems

  • 77.3% average on five answer-verifiable benchmarks nearly matches Qwen3.6-35B-A3B at 77.4%, with SU-01 leading similar-size results on AMO-Bench and AIME 2025/2026.SU-01 scores 59.8% on AMO-Bench, 94.6%/93.3% on AIME 2025/2026, and transfers to Chemistry at 69.4% and Biology at 25.0% despite math-and-physics-only RL signals.Source: §5.2 · p. 8, Table 1 · p. 8

5.3 Non-verifiable Problems

  • 70.2% on IMO-ProofBench with test-time scaling, up from 57.6% in direct generation, includes 91.0% on basic and 49.5% on advanced problems.These results indicate that self-verification and refinement matter when correctness depends on complete proofs rather than final-answer matching.Source: §5.3 · p. 8, Table 3 · p. 9

5.4 Olympiad Competition Problems

  • 23.5 and 20.3 points without test-time scaling exceed the IPhO 2024/2025 gold lines, while scaling raises them to 25.3 and 21.7.SU-01 also reaches 11.7% on FrontierScience-Research, the best similar-size overall score, despite RL using only mathematics and physics signals.Source: §5.4 · p. 9, Table 2 · p. 9, §5.3 · p. 9
  • 35 points on both IMO 2025 and USAMO 2026 let SU-01 meet the IMO gold line and exceed the USAMO gold line by 10 points with test-time scaling.It solves five of six USAMO problems for full credit and five IMO problems for full credit, while remaining unresolved on IMO P6 and USAMO P2.Source: USAMO 2026 · p. 10, Table 4

6 Analysis and Discussion

The analysis shows that staged training progressively converts answer-seeking ability into rigorous proof performance, while test-time scaling further strengthens proof verification and repair. It also finds that reverse-perplexity ordering improves capability retention and stability, supporting strong reasoning with a compact post-training setup.

6.1 Progressive Rigorous Reasoning

  • 91.0 on ProofBench-Basic and 49.5 on ProofBench-Advanced after test-time scaling, with the largest trained-model gain occurring on Advanced proofs.Refined RL improves ProofBench-Advanced from 25.2 to 38.1 while leaving AnswerBench at 77.5 and ProofBench-Basic at 77.1; it uses proof-level rewards, self-refinement, and experience replay.Source: Figure 4 · p. 10, §6.1 · p. 11

6.2 Characterizing Inference Scaling

  • Long refinement traces show the model can sustain reasoning beyond 100K tokens while localizing flaws, preserving useful arguments, and synthesizing corrected proofs.Refinement conditions on a candidate solution and structured verifier feedback or a bug report, rather than generating only from the problem statement.Source: §6.2 · p. 11

6.3 Reverse-Perplexity Ordering

  • 55.8 on AnswerBench and 40.0 on AMO-Bench with descending-PPL ordering, versus 39.5 and 31.0 for random ordering, while truncation falls to 0.3% and 0.0%.Low-PPL-first is weakest, reaching 24.3 on AnswerBench and 15.0 on AMO-Bench; descending-PPL ordering preserves capability while reshaping reasoning behavior.Source: Figure 6 · p. 11, §6.3 · pp. 11–12

6.4 Cost Analysis

  • SU-01 trains a 30B-A3B backbone with 338K sub-8K-token SFT trajectories and 25K RL prompts over 200 steps.Compared with DeepSeek-V3.2 and Nemotron-Cascade 2, the setup is presented as compact and unified while preserving scientific transfer.Source: §6.4 · p. 12

7 Related Work

Prior work has advanced reasoning through post-training, symbolic and natural-language olympiad systems, verification and refinement, and test-time search. This work proposes a simple, unified post-training recipe that gives a 30B-A3B model rigorous proof behavior and olympiad-level performance.

Post-training for Large Reasoning Models

  • Post-training methods improve reasoning through self-improvement, instruction tuning, preference optimization, reinforcement learning, curricula, replay, entropy control, sequence-level optimization, and tool-oriented training.Representative systems include DeepSeekMath, Qwen2.5-Math, DeepSeek-R1, Kimi k1.5, GSPO, MiniMax-M2.5, Kimi-K2.5, and GLM-5.Source: §7 · p. 12

Toward Olympiad-Level Reasoning

  • Olympiad reasoning requires complete, rigorous, and gap-resistant solutions, motivating symbolic deduction, formal search, natural-language reasoning, test-time search, and verification-and-refinement pipelines.AlphaGeometry, AlphaProof/AlphaGeometry 2, Gemini Deep Think, model-agnostic verification and refinement, DeepSeekMath-V2, and Nemotron-Cascade 2 represent these directions.Source: §7 · p. 12
  • The proposed simple, unified recipe enables a 30B-A3B model to develop rigorous proof behavior and achieve olympiad-level performance through self-verification, refinement, and test-time scaling.The contribution combines post-training with these mechanisms rather than relying on a single-pass answer.Source: §7 · p. 13

8 Conclusion · Appendix

SU-01 presents a unified recipe that strengthens a compact post-trained reasoning model through behavior shaping, scalable reward feedback, proof-level specialization, and inference-time repair. It achieves gold-medal-level olympiad performance, supports trajectories beyond 100K tokens, and transfers to broader scientific domains.

8 Conclusion

  • SU-01 combines reverse-perplexity curriculum SFT, outcome-verified coarse RL, proof-level RL, self-refinement, experience replay, and test-time scaling.These stages target behavior shaping, scalable reward feedback, proof-level specialization, and inference-time repair.Source: §8 · p. 13
  • SU-01 reaches gold-medal-level performance on mathematical and physical olympiad competitions.Source: §8 · p. 13
  • SU-01 sustains reasoning trajectories beyond 100K tokens during inference.Source: §8 · p. 13
  • SU-01 transfers scientific reasoning to domains beyond its main mathematics and physics training signals.Source: §8 · p. 13
  • The recipe turns a broadly capable 30B-A3B backbone into a stronger mathematical and scientific reasoner.Source: §8 · p. 13
  • The results support viewing compact reasoning models as specializable generalists that can retain meaningful scientific transfer.Source: §8 · p. 13

A Implementation and Evaluation Details · B Problem-Solving Prompt

The appendix specifies the implementation and evaluation settings needed to interpret and reproduce SU-01’s training, inference, serving, reward, and grading procedures. It also defines the fixed prompt used across RL training and inference, requiring complete, rigorous, LaTeX-formatted solutions.

A Implementation and Evaluation Details

  • The appendix consolidates settings needed to reproduce the reported training, inference, reward-model serving, and evaluation procedures.It focuses on settings affecting objectives, data semantics, inference behavior, reward definitions, and evaluation criteria.Source: A Implementation and Evaluation Details · p. 17
  • The implementation summary covers effective modeling, optimization, decoding, serving, and grading settings extracted from launch scripts and evaluation protocols.Source: A Implementation and Evaluation Details · p. 17
  • Infrastructure-only commands are omitted when they do not affect training objectives, data semantics, inference behavior, reward definitions, or evaluation criteria.Source: A Implementation and Evaluation Details · p. 17

B Problem-Solving Prompt

  • Unless otherwise specified, SU-01 uses one fixed problem-solving prompt throughout all RL training and inference stages.Source: B Problem-Solving Prompt · p. 17
  • The prompt instructs the model to solve olympiad problems with complete reasoning and proof, using LaTeX for variables and formulas.Source: Problem-Solving Prompt · p. 17
  • For value-finding tasks, the prompt requires final answers in \boxed{}, while proof tasks require a clear and rigorous argument.Source: Problem-Solving Prompt · p. 17

C SFT Training Details

The SFT stage uses reverse-perplexity-ordered trajectories to provide controlled behavioral adaptation while preserving the backbone’s existing mathematical and scientific competence. Training preserves curriculum order and applies a specified four-epoch optimization configuration on eight GPUs.

C SFT Training Details

  • SFT is initialized from P1-30B and trains on eight GPUs with the reverse-perplexity curriculum, preserving its order by disabling rollout shuffling.The curriculum is implemented with slime.Source: C SFT Training Details · p. 18
  • The objective is controlled behavioral adaptation: exposing the backbone to stable rigorous long-form trajectories while preserving its existing mathematical and scientific competence.SFT is not optimized directly for benchmark performance.Source: C SFT Training Details · p. 18
  • Training runs for four epochs with batch size 128, Adam, learning rate 1 × 10−5, cosine decay to 1 × 10−6, 0.1 warmup fraction, 0.1 weight decay, and β1 = 0.9, β2 = 0.95.Source: C SFT Training Details · p. 18

D RL Training Details · E Inference and Reward-Model Serving Details · F Compared Models

The paper specifies a two-stage, 200-step GSPO RL implementation with replay and self-refinement, followed by iterative verification-and-refinement inference under detailed serving constraints. It also defines the model families and sourcing used for benchmark comparisons.

D RL Training Details

  • 200 RL steps use GSPO on 64 GPUs, progressing from 96 coarse-RL steps to 104 refined-RL steps over balanced verifiable, proof-reward, self-refinement, and replayed data.Rollouts use 128 prompts, 8 samples per prompt, 160K-token responses, four policy updates, dynamic sampling, partial rollout, and replay filtering.Source: D RL Training Details · p. 18
  • Self-refinement samples 20% of queries, targeting rollout groups with average reward below 0.5, while replay admits queries below 25% correctness and retires them at 50%.The implementation uses one replayed trajectory per query, hard-positive admission when only one rollout succeeds, entropy-based selection, and Adam with learning rate 1 × 10−6.Source: D RL Training Details · p. 18

E Inference and Reward-Model Serving Details

  • Inference follows an iterative solve–verify–refine loop in which structured verifier critiques guide proof revision until acceptance or budget exhaustion.SGLang is the default serving backend, while Nemotron-Cascade-2 and DeepSeek-V3.2-Speciale use vLLM; the setup follows Huang & Yang (2025).Source: E Inference and Reward-Model Serving… · p. 18
  • Generation defaults to 160,000 tokens at temperature 1.0 and top-p 0.95, with benchmark-specific limits of 131K or 256K tokens and API caps of 128,000 or 65,535 tokens.Nemotron uses top-p 1.0; GPT-5.5 API and Gemini 3.1-Pro impose the 128,000- and 65,535-token limits, respectively.Source: E Inference and Reward-Model Serving… · pp. 18–19
  • TTS permits up to 10 independent runs per problem, with 30 exploration rounds per run and stopping after five consecutive verification passes or ten consecutive failures.These stopping rules align test-time proof repair with the self-refinement workflow used during refined RL.Source: E Inference and Reward-Model Serving… · p. 19
  • Proof-reward evaluation deploys the DeepSeekMath-V2 server on 32 GPUs with 32K context and uses three-step speculative decoding with one EAGLE draft candidate and four draft tokens per step.The configuration is intended to improve verifier efficiency while preserving evaluation correctness.Source: E Inference and Reward-Model Serving… · p. 19

F Compared Models

  • The comparison set combines larger models DeepSeek-V3.2, GPT-5.5, and Gemini 3.1 Pro Thinking with similar-size models including GLM-4.7-Flash, Nemotron-Cascade-2-30B-A3B, Qwen3.6-35B-A3B, Gemma-4-31B, and SU-01.The same active families are reused in the non-verifiable benchmark table, with missing entries marked by –.Source: F Compared Models · p. 19
  • Comparison results are sourced from public technical reports, official model cards, and official benchmark reports, with model variants and provenance specified when reports are unavailable.GPT-5.5-High is treated as GPT-5.5’s high-reasoning variant, while Gemini 3.1 Pro Thinking and DeepSeek V3.2 retain their stated source families.Source: F Compared Models · p. 19

G Evaluation Details … Solution

This section details layered evaluation procedures and presents model solutions for geometric and functional-equation problems. It develops analytic and geometric constructions, proves a tangency result, and classifies admissible function-value triples into two families.

G Evaluation Details

  • For answer-verifiable tasks, grading uses rule-based checks first and gpt-oss-120b generative verification when unresolved, with a dedicated FrontierScience-O prompt.Rule-based checks include canonicalized answer matching and symbolic or expression-level verification with Math-Verify16.Source: G Evaluation Details · p. 19
  • For proof-oriented tasks, official benchmark protocols are followed, including four-level {0, 1, 6, 7} scoring for ProofBench and expert-conservative scoring for IMO and USAMO.ProofBench uses Gemini-2.5-Pro, FrontierScience-R uses GPT-5-high, and IMO/USAMO final scores are the worst of three independent gold-medal experts’ scores.Source: G Evaluation Details · p. 19

1. Constructions for k = 0, 1, 3

  • Explicit constructions attain k = 0 using horizontal lines, k = 1 using n −1 horizontal lines plus one sunny line, and k = 3 by reducing the remaining six points to a translated P3 configuration.For k = 3, three sunny lines cover the translated P3 points and combine with n −3 horizontal lines.Source: §1 · pp. 20–21

2. Upper bound on points per sunny line

  • A sunny line contains at most floor((n + 1)/2) points of Pn, using rational-slope parametrization and bounds on x, y, and x + y.Irrational-slope or integer-point-free lines contain at most one integer point; for rational slopes, the sunny constraints force max{q, |p|, |p + q|} ≥2.Source: §2 · p. 21

3. No covering by all sunny lines for n ≥4

  • For n ≥4, no set of n distinct sunny lines covers Pn: even n fails by incidence counting, while odd n would force every line into one of only three extremal sunny lines.The odd-n equality case implies pairwise disjoint lines, but n ≥5 lines cannot all be among the three classified extremal lines.Source: §3 · pp. 21–23

4. Induction on n

  • Strong induction shows every covering contains at least one non-sunny boundary line y = 1, x = 1, or x + y = n + 1, whose removal reduces the problem to Pn−1 without changing k.If none is present, all lines would be sunny, contradicting the no-all-sunny-lines lemma; removing any present boundary line preserves the sunny-line count.Source: §4 · p. 23

5. Conclusion

  • The IMO 2025 Problem 1 solution proves that for every n ≥3, the only possible numbers of sunny lines in an n-line covering are 0, 1, and 3, and all three values are attainable.The model-solutions section presents solutions used for human-expert scoring, with a USAMO 2026 Problem 2 excerpt shown because the full response is substantially longer.Source: H Model Solutions for IMO 2025 and… · p. 20, §5 · p. 24

Solution

  • The proof uses analytic geometry: by similarity invariance, place M and N on the x-axis, scale MN = 1, and let intersecting circles have radii 0 < r < R satisfying R − r < 1 < R + r.With M = (0, 0), N = (1, 0), and P = (xP, yP), the altitude from P is x = xP, while the altitude from M is perpendicular to PN.Source: Solution · p. 24, §1 · pp. 24–25, §8 · p. 27

9. Circumcenter O′ of △BEF

  • The circumcenter O′ of △BEF is determined by enforcing equal distances from B, E, and F, using the perpendicular bisector of EF and vector equations.Since EF is collinear with v⃗, the perpendicular-bisector condition is v⃗ · (O′ − M_EF) = 0; equating distances to B and E supplies the second equation.Source: §9 · pp. 27–28

10. Computation of |⃗v|2

  • The computation simplifies the direction-vector norm to |v⃗|2 = SRr/(R + r − 1), using p2 − q2 = 4Rr and p2 − 1 = (R + r − 1)S.This identity is substituted into the later distance calculations.Source: §10 · pp. 28–29

11. Simplify y from (4)

  • Substituting the vector-norm identity into the circumcenter equations simplifies the y-coordinate, while the remaining equation yields x = 1/2 because R ≠ r.The x-coordinate follows from 2(R − r)x + (R + r) = 2R.Source: §11 · p. 29, §12 · p. 29

13. Distance from O′ to the line through H parallel to⃗v

  • The distance from O′ to the line through H parallel to v⃗ is computed via the planar cross product, with the numerator reduced using p2 − q2 = 4Rr.The resulting squared distance is 4h2(R + r − 1)/(SRr), equivalently (R − r)2/[SRr(R + r − 1)].Source: §13 · pp. 29–30, §14 · p. 31

14. Squared distance from O′ to ℓ

  • The circumradius Rc of △BEF is computed from |O′ − B| and simplified to the same expression as the distance from O′ to ℓ.The comparison establishes equality between the center-to-line distance and the circumcircle radius.Source: §15 · p. 31, §16 · p. 31

16. Conclusion

  • Because the distance from O′ to ℓ equals the radius of its circumcircle, ℓ is tangent to the circumcircle of △BEF.Here ℓ is the line through H parallel to AP.Source: §16 · p. 31

Solution

  • For bonza functions, preliminary divisibility lemmas show f(1) = 1 and f(a) | a^a, so every prime divisor of f(a) divides a.These facts restrict f(3) to {1, 3, 9, 27} and f(5) to powers 5^δ with 0 ≤ δ ≤ 5.Source: Solution · p. 32
  • The only admissible triples (f(2), f(3), f(5)) are (1, 1, 1), (2, 1, 1), (4, 1, 1), and (2, 3, 5), forming Families A and B.Family A consists of the first three triples with f(3) = 1; Family B is the triple (2, 3, 5).Source: Solution · pp. 33–34

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

For Family A with f(3)=1, the argument forces f(q)=1 for every odd prime q and shows that every f(n) is a power of 2. Computing v2(3^n−1) then gives explicit parity-dependent bounds on these powers.

Prime values

  • Every odd prime q satisfies f(q)=1 when f(3)=1.From (q,q), write f(q)=q^k; applying (q,3) and Fermat’s little theorem rules out k≥1, so k=0.Source: §3 · p. 34

Power-of-two structure

  • For every positive integer n, f(n) is a power of 2.If an odd prime divided f(n), Lemma 2 would imply it divides n; applying (n,r) then yields the contradiction r | 3^n−1 and r | 3^n−1.Source: §3 · p. 34
  • For all n, f(n) divides 3^n−1, so the exponent of 2 in f(n) is bounded by v2(3^n−1).Writing f(n)=2^e(n) and applying the defining relation to (n,3) gives 2^e(n) | 3^n−1.Source: §3 · p. 34

2-adic valuation

  • The valuation is v2(3^n−1)=1 for odd n and v2(3^n−1)=v2(n)+2 for even n.The odd case follows from factoring 3^n−1 into 2 times an odd sum; the even case is proved by induction on v2(n).Source: §3 · pp. 34–35

4. Analysis of Family B (f(2) = 2, f(3) = 3, f(5) = 5) … Alice prevents Bazza from winning

This section establishes the behavior of Family B, including the identity characterization and sharpness of the universal bound, then develops fixed-point and orbit lemmas and applies them to characterize admissible starting values and analyze winning, preventing, and drawing strategies in the inekoalaty game.

4. Analysis of Family B (f(2) = 2, f(3) = 3, f(5) = 5)

  • The induction establishes f(p)=p by excluding k=0, k=2, and every k≥3 when f(p)=pk.The cases k=0 and k=2 contradict Bertrand’s postulate, while k≥3 yields incompatible p-adic valuations.Source: §4 · pp. 35, 37
  • Family B satisfies f(n)=n for every positive integer n, and therefore f(n)≤4n.Strong induction proves f(p)=p for every prime p; divisibility by every prime then forces n−f(n)=0.Source: §4 · p. 37

5. Upper bound summary

  • Together, Families A and B imply f(n)≤4n for every bonza function and every positive integer n.The supplied upper-bound summary states the Family A bound and combines it with the exact Family B identity.Source: §5 · p. 37

Verification

  • The function h is bonza and attains h(n)=4n for every n=2^t with t≥2, including h(4)=16.The verification checks odd, exceptional, and even cases using divisibility and 2-adic valuations.Source: Verification · p. 39
  • Because h(4)=16>c·4 for every c<4, no universal constant smaller than 4 can replace the upper bound.Thus 4 is both attainable and minimal.Source: Verification · p. 39

7. Conclusion

  • The conclusion is that 4 is the smallest real constant satisfying f(n)≤cn for every bonza function f and every n∈N.The result follows from the upper bound and the function h attaining equality.Source: §7 · p. 39

Solution

  • For IMO 2025 Problem 4, the solution defines f(N) as the sum of the three largest proper divisors and characterizes fixed points through 1/a+1/b+1/c=1.The unique divisor triple is (2,3,6), yielding the fixed-point condition d2=2, d3=3, and d4=6.Source: §7 · p. 39, Solution · p. 39, §2 · p. 40

4. Lemma on odd numbers

  • The orbit analysis shows that odd terms strictly decrease, while an even term not divisible by 12 can lie on an infinite orbit only if it is a fixed point.The odd-number lemma rules out infinite odd tails, and the even-number case analysis forces membership in the fixed-point set F.Source: §4 · p. 40, §5 · pp. 40–42

7. Sufficiency - every such number works

  • For sufficiency, induction reduces a1 = 12^tK to 13K ∈ F, preserving infinitude of the sequence at every step.The base case is a fixed point, while the inductive step uses 13K ∈ F.Source: §7 · p. 42

8. Final characterization

  • The possible starting values are exactly a1 = 12^t · K with t ≥ 0, 2 ∥ K, 3 | K, and 5 ∤ K.Necessity and sufficiency together completely solve the sequence problem.Source: §8 · pp. 42–43

The decisive move

  • Alice can force a finite win by playing zeros on earlier odd turns and then choosing a decisive value that makes Bazza’s next move impossible.The construction proves the resulting sum of squares exceeds Bazza’s bound before an even turn.Source: Two elementary lemmas · p. 44, The decisive move · p. 45

Alice loses

  • Bazza can force a win with the maximal even-turn strategy, eventually making Alice’s sum constraint fail even if she chooses zero.His strategy maintains the required inductive bounds until Alice has no legal odd-turn move.Source: The decisive move · pp. 45–46, Alice loses · p. 46
  • At the boundary case, Bazza’s maximal strategy always leaves Alice a legal move, so Alice cannot force a win.Bazza may reply with zero or any move preserving the square-sum constraint.Source: Alice loses · p. 47

Alice prevents Bazza from winning

  • Conversely, Alice can prevent Bazza from winning by playing zero on every odd turn, yielding a draw that can continue indefinitely.The accumulated sum remains within Alice’s constraint, so she never loses and Bazza cannot force a win.Source: Alice prevents Bazza from winning · p. 47

Conclusion … Solution

The paper classifies the game outcome for every positive λ and solves the IMO 2025 tiling problem exactly, reducing it to a diagonal configuration and proving an optimum of 2n − 2 tiles.

Conclusion

  • For every positive λ, Alice wins exactly when λ > √2, Bazza wins exactly when λ < √2, and λ = √2 produces no winner.Source: Conclusion · pp. 47–48

Solution

  • For n = 2025, the uncovered squares form a permutation matrix, while the remaining n^2 − n squares must be covered by non-overlapping axis-aligned rectangles.Source: Solution · p. 48

1. Reduction to the diagonal configuration

  • Relabeling columns transforms any uncovered permutation matrix into the main diagonal without changing tile count or validity, so the diagonal case suffices.Source: §1 · p. 48

2. A construction using 2n −2 tiles

  • A valid diagonal tiling uses 2n − 2 rectangles: n − 1 vertical tiles cover the lower triangle and n − 1 horizontal tiles cover the upper triangle.Source: §2 · p. 48

3. Lower bound: at least 2n −2 tiles

  • Any tile leaving the diagonal uncovered lies wholly in either the lower or upper triangle, preventing tiles from covering both regions.Source: §3 · pp. 48–49
  • The boundary cells immediately below and above the diagonal require at least n − 1 tiles each, establishing a lower bound of 2n − 2.Source: §3 · p. 49

4. Minimum number of tiles

  • The IMO 2025 Problem 6 solution therefore determines the minimum number of tiles as exactly 2n − 2.Substituting n = 2025 gives the contest instance’s exact optimum, as stated after matching the construction and lower bound.Source: §4 · p. 49

H.2 USAMO 2026

  • USAMO 2026 Problem 1 asks which real x maximizes a floor-function expression for integer n > 1 and what maximal value it attains.Source: §H.2 · p. 50

Step 2. Value on the interval · Solution

The section establishes a maximum value of H_n−1, attained exactly when {x} ∈ [1−1/n, 1), and presents an inductive strategy for Annie’s game. The strategy targets a final value F ≤ 4·2^k and handles the difficult sum-increase case through a stronger auxiliary lemma.

Step 2. Value on the interval

  • On [1−1/n, 1), the expression is constant and equals H_n−1; the global upper bound follows by decomposing θ through N = ⌊nθ⌋ and a remainder parameter.The decomposition reduces the problem to proving A(N) ≤ H_n−1 for every N ∈ {0, …, n−1}.Source: Step 2. Value on the interval · pp. 50–51
  • Writing d = gcd(N,n) and applying the rearrangement inequality to the residue classes yields A(N) ≤ H_n−1, with equality requiring d = 1.The proof uses the permutation of residues modulo n_1 and decreasing weights w_j.Source: Step 2. Value on the interval · pp. 51–53
  • Equality in the global bound forces N = n−1, and this condition is equivalent to the fractional-part interval [1−1/n, 1).For N = n−1, the remainder conditions also give equality in the preceding bound.Source: Step 2. Value on the interval · pp. 53–54
  • The expression’s maximum is H_n−1, attained exactly when the fractional part satisfies {x} ∈ [1−1/n, 1).Here H_n = 1 + 1/2 + ··· + 1/n.Source: Step 2. Value on the interval · p. 54

Solution

  • Annie can always win by strong induction: for a total sum S with 2^k ≤ S < 2^(k+1), some sequence ends with F ≤ 4·2^k, implying F < 4S.If S is a power of two, the proof claims the sharper bound F ≤ 2S.Source: Solution · p. 54

Case 1: There exists an adjacent equal pair

  • When adjacent equal powers exist, replacing them by their double preserves S and k, so the induction hypothesis directly proves the target bound.The replacement uses 2a, the smallest allowed power of two greater than the equal entries.Source: Case 1: There exists an adjacent equal… · p. 54
  • With no adjacent equal pair, merging a carefully chosen unequal pair either preserves k or invokes a stronger lemma controlling each remaining block by 2^(k+1).The lemma applies when a block sum lies between 2^L and 2^(L+1) and is at most 3·2^(L−1)−1; then subsequent merges remain bounded by 4·2^k.Source: Case 1: There exists an adjacent equal… · pp. 54–55

Proof of the Lemma … 3. Case I: N is prime

This section develops the geometry and number-theoretic arguments underlying the result, including a counterexample to the stated lemma, a complex-coordinate analysis of the relevant centers and radii, and an infinite-descent classification in the prime case.

Proof of the Lemma

  • The stated lemma is false: for L = 3, the list [2, 8, 1] has allowed sum 11 but every strategy ends at 32, exceeding 2L+1 = 16.The attempted induction fails in the three-element case containing 2L; merging either adjacent pair produces [16, 1] or [2, 16], whose final merge is at least 32.Source: Proof of the Lemma · p. 56

1. Complex numbers on the unit circle

  • The geometry solution places the circumcircle as the unit circle, represents A, B, C by unit complex numbers, and uses the line equation z + uv z̄ = u + v.This establishes the complex-coordinate framework for subsequent intersections and identities.Source: §1 · p. 57

2. Equations of the relevant lines

  • The constructed lines satisfy ℓC: z + c^2z̄ = −c and N^2 = bc, while intersecting these lines with AC and AB yields algebraic coordinates for Y and Z.The derivation introduces S = a + b + c and D1 = ac − b^2 for the coordinate formulas.Source: §2 · pp. 57–58

4. A convenient rotation

  • After rotating so N = 1, the coordinates become b = e^iα, c = e^−iα, with k = cos α and α = ∠A, enabling explicit expressions for S, T, Q, and Δ.The assumptions imply α lies between 0° and 90° and α ≠ 60°.Source: §4 · p. 58

5. A key algebraic identity

  • The key identity (a + 2k)^2 = QT + Δ supports identifying the circumcenter O of △AYZ, and equality of the relevant distances confirms O is the circumcenter.The circumradius is then obtained from the common distance to A, Y, and Z.Source: §5 · pp. 58–59, Equality of the distances · p. 60

7. Cartesian description of triangle R

  • In Cartesian coordinates, ℓB and ℓC are symmetric lines and the tangent at N = 1 is x = 1, so these three lines bound triangle R.The construction uses the rotated coordinates b = e^iα and c = e^−iα.Source: §7 · p. 60

8. Vertices and interior of R

  • The vertices and interior of R are obtained by pairwise line intersections, with interior inequalities 2, x cos α − y sin α > −1/2, and x < 1.The triangle is symmetric about the x-axis, placing its incenter at I = (p, 0).Source: §8 · pp. 60–61, §9 · p. 61

9. Incenter I and inradius r of R

  • Equal distances from I to the three sides give p = 1/2(1 + cos α) and inradius r = (1 + 2k)/(2(1 + k)); the subsequent OI calculation sets U = aQT + pΔ and OI = |U|/|Δ|.The algebra expands the coefficients of U in the rotated complex coordinates.Source: §9 · p. 61, §10 · p. 62

10. Distance OI

  • The algebra gives L = r(2 cos 3α − t), and 2 cos 3α − t = ε|∆|, where ε = ±1 depending on the sign of t − 2 cos 3α.Source: §10 · p. 64

Final step

  • The distance between O and I equals either R + r or |R − r|, so the circumcircle and incircle are tangent.This follows by substituting the preceding identity into the distance formula.Source: Final step · p. 64

12. Remarks on non-vanishing denominators

  • The proof verifies ∆ ≠ 0 and a + 2k ≠ 0, while cos α > 0, preventing degenerate cases and justifying the algebra.The excluded equal-side and α = 60° cases establish the required nonvanishing conditions.Source: §12 · p. 64

3. Case I: N is prime

  • When N is prime, φ(N)=ab, so the condition becomes a^2+b^2+1=kab; descent forces k=3 and hence a^2−3ab+b^2=−1.The descent reduces solutions to a=1, where the discriminant equation forces k=3.Source: §3 · p. 74, §3.1 · p. 75

3.1. The value of k

  • These generated pairs equal (F_{2n−1},F_{2n+1}), so all prime-case solutions consist of consecutive odd-index Fibonacci numbers.The identification follows by induction from A_0=A_1=1, A_{n+2}=3A_{n+1}−A_n and the Fibonacci recurrence.Source: §3.2 · p. 75
  • Every positive solution of a^2−3ab+b^2=−1 is generated from (1,1) and (1,2) by (x,y)↦(y,3y−x), yielding (1,1),(1,2),(2,5),(5,13),….The quadratic-root descent reduces any solution to a minimal one, while the inverse transformation generates all solutions.Source: §3.2 · p. 75

4. Case II: N is composite and odd (N > 2) · 5. Conclusion

The composite odd case is reduced to N = 9, yielding only (a,b) = (1,8) and (8,1); combined with the even and prime cases, this proves that a and b are Fibonacci numbers.

4. Case II: N is composite and odd (N > 2)

  • The composite odd case satisfies a^2 + b^2 + 1 ≡ 2 (mod 4), which constrains the possible factorization of N.Source: §4 · p. 76

4.1. N cannot have two distinct odd prime factors

  • N cannot have two distinct odd prime divisors, because they would make 4 divide φ(N), contradicting φ(N) | a^2 + b^2 + 1 ≡ 2 (mod 4).Therefore, N must be a prime power.Source: §4.1 · p. 76

4.2. The prime p cannot be ≡1 (mod 4)

  • The prime base p of the prime power cannot satisfy p ≡ 1 (mod 4), since then 4 | φ(N) would again contradict the modulo-4 condition.Source: §4.2 · p. 76

4.3. Analysis modulo 3

  • Modulo 3, p ≡ 1 is impossible because 3 | φ(N) would force 3 | S, while p ≡ 2 leads to a nonexistent solution of y^2 + y + 1 ≡ 0 (mod p).Thus the only remaining prime base is p = 3.Source: §4.3 · p. 76
  • For p = 3, exponents e ≥ 3 imply 9 | φ(N) | S, but S ≡ 3 (mod 9), so e must equal 2 and N = 9.The contradiction uses ab ≡ −1 (mod 9), a + b ≡ 0 (mod 3), and S = (a + b)^2 − 2ab + 1.Source: §4.3 · pp. 76–77
  • For N = 9, the divisibility condition holds only for (a,b) = (1,8) and (8,1), and both entries are Fibonacci numbers.The pair (1,8) gives S = 66 with 66/6 = 11, whereas (2,4) gives 21/6 = 3.5 and fails.Source: §4.3 · p. 77

5. Conclusion

  • Under φ(ab + 1) | a^2 + b^2 + 1, the complete classification forces a and b to be Fibonacci numbers, with the composite case contributing (1,8) and (8,1).The even case gives N = 2 and (1,1), while the prime case yields Fibonacci pairs through a^2 − 3ab + b^2 = −1.Source: §5 · p. 77
  • The conclusion identifies the prime-case solutions as (F_{2n−1}, F_{2n+1}) and symmetric pairs, while (1,8) also fits the Fibonacci pattern.Fibonacci numbers are defined by F1 = F2 = 1 and F_{n+2} = F_{n+1} + F_n.Source: §5 · p. 77

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