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Challenging the Boundaries of Reasoning: An Olympiad-Level Math Benchmark for Large Language Models

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Table of Contents

2503.21380
Haoxiang Sun et el.
🤗 2025-03-28

↗ arXiv ↗ Hugging Face

TL;DR
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Existing benchmarks for evaluating LLMs’ mathematical reasoning are becoming saturated. The rapid development of LLMs necessitates more challenging and rigorous evaluations to identify their limitations and guide future improvements. Current benchmarks struggle to differentiate capabilities and multilingual assessment is also lacking.

This paper introduces OlymMATH, a new Olympiad-level math benchmark to rigorously test LLMs’ reasoning. OlymMATH has 200 problems with varying difficulty, manually verified and available in both English and Chinese. Experiments show that models like DeepSeek-R1 struggle with OlymMATH, highlighting its difficulty. The benchmark facilitates bilingual assessment and the analysis helps understand limitations in solving complex math problems.

Key Takeaways
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Why does it matter?
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This paper introduces a new Olympiad-level math benchmark for LLMs, addressing limitations in existing benchmarks. It offers rigorous evaluations, bilingual assessment, and insights into models’ reasoning abilities, opening new research avenues to improve LLMs in mathematical problem-solving.


Visual Insights
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🔼 This figure compares the performance of several large language models (LLMs) on various Olympiad-level mathematics benchmarks, including the OlymMATH benchmark (English version) introduced in this paper. The x-axis shows the different models tested, and the y-axis displays their accuracy (or percentile) on each benchmark. The bars are grouped by benchmark: OlymMATH-EN-EASY, OlymMATH-EN-HARD (the new benchmark proposed in the paper), and other existing Olympiad-level benchmarks like AIME 2024, AIME 2025, and HMMT 202502. The figure highlights that the OlymMATH-HARD subset is significantly more challenging than existing benchmarks, with even the state-of-the-art models showing considerably lower accuracy on this subset.

read the captionFigure 1: Performance comparisons of mainstream reasoning models between our OlymMATH (English version) and other Olympiad-level mathematical benchmarks. OlymMATH-HARD emerges as the most challenging, with significantly higher difficulty than existing evaluation benchmarks.
Problem:Given that two vertices of an equilateral triangle are on the parabolay2=4xsuperscript𝑦24𝑥y^{2}=4xitalic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 4 italic_x, and the third vertex is on the directrix of the parabola, and the distance from the center of the triangle to the directrix equals1919\frac{1}{9}divide start_ARG 1 end_ARG start_ARG 9 end_ARGof the perimeter. Find the area of the triangle.Subject:Geometry

🔼 This table compares various mathematical reasoning benchmarks used to evaluate large language models (LLMs). It lists the name of each benchmark, the number of problems it contains, the number of mathematical fields covered, the languages supported (English, Chinese, or both), the evaluation method used (rule-based or LLM-based), and the difficulty level of the problems (grade school, competition, university, or Olympiad). This allows for a comparison of benchmark scale, scope, and difficulty.

read the captionTable 1: Comparison of different math evaluation benchmarks. EN and ZH denote English and Chinese, respectively. “Univ.” denotes university-level problems.

In-depth insights
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OlymMATH Intro
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The introduction of OlymMATH addresses a critical gap in evaluating LLMs’ mathematical reasoning. Existing benchmarks are becoming saturated, failing to challenge advanced models effectively. OlymMATH, an Olympiad-level benchmark, aims to rigorously test complex reasoning. Its key features include meticulously curated problems, manual verification, parallel English and Chinese versions for comprehensive bilingual assessment, and two distinct difficulty tiers: AIME-level (easy) and significantly more challenging (hard). The problems span four core mathematical fields, each with verifiable numerical solutions for objective evaluation. Initial results show state-of-the-art models struggle with OlymMATH, particularly the hard subset, indicating its higher difficulty and potential to push the boundaries of mathematical reasoning evaluation.

LLM Reasoning Gap
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LLM reasoning gaps signify the disparities between expected and actual performance in complex tasks. These gaps often manifest in areas requiring multi-step inference, nuanced understanding of context, and the ability to handle ambiguity. Current LLMs, while excelling in pattern recognition, often struggle with causal reasoning, analogical thinking, and counterfactual simulations, exhibiting limitations in systematic generalization and transfer learning. Addressing these gaps necessitates innovations in model architectures, training methodologies, and evaluation benchmarks to better assess and enhance true reasoning capabilities. The focus should be on developing models that can robustly handle novel situations and exhibit more human-like reasoning proficiency.

Bilingual Analysis
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A bilingual analysis of mathematical reasoning models offers a potent lens for evaluating their capabilities. By assessing performance across both English and Chinese, we can discern the influence of language on problem-solving efficacy. Differences in linguistic structure, cultural context embedded in problems, and the nuances of mathematical terminology across languages can all contribute to variations in a model’s accuracy and efficiency. Furthermore, a bilingual approach helps mitigate biases inherent in datasets predominantly available in one language, fostering more robust and equitable evaluations. Observing performance gaps highlights the need for enhanced multilingual training, ensuring that models generalize effectively across diverse linguistic landscapes and achieve true cross-cultural reasoning proficiency.

Dataset Details
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In analyzing the dataset details, several key aspects emerge. First, the paper emphasizes the high-quality and original nature of the Olympiad-level problems, meticulously curated from printed resources to minimize data contamination, with problems manually sourced from specialized magazines, textbooks, and official competition materials. The dataset spans four primary mathematical fields—algebra, geometry, number theory, and combinatorics—to ensure comprehensive coverage of reasoning skills. Problems are categorized into two difficulty levels: easy and hard, with the former challenging standard prompting approaches and the latter testing advanced reasoning capabilities, such as slow-thinking modes. The dataset adheres to a format aligning with the MATH dataset, ensuring seamless integration with existing benchmarks and evaluation pipelines. It restricts answers to real numbers and intervals and also deliberately excludes other answer formats such as set operations, variables, complex numbers, and free-form text for easy evaluation. Furthermore, it specifically curates problems that yield challenging numerical answers and provides a summary of all potential outcomes, such as sums, sums of squares, and more.

New benchmark
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When introducing a new benchmark, it’s crucial to establish its uniqueness and value proposition compared to existing ones. This involves highlighting specific gaps it addresses, like handling complex reasoning or multilingual capabilities. A new benchmark’s design should minimize biases and ensure reliability through rigorous problem curation and validation. Additionally, its difficulty level must be calibrated to challenge current models without being insurmountable. The benchmark’s format and evaluation metrics should be clearly defined and easily reproducible, facilitating broader adoption and comparison. Also, the diversity of problem types is key to ensure a variety of scenarios. Finally, releasing a new benchmark should be accompanied by an analysis of its performance against state-of-the-art models, demonstrating its effectiveness in differentiating model capabilities and guiding future research.

More visual insights
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More on tables
Original problem:If the distances from the eight vertices of a cube to a certain plane are00,1111,2222,3333,4444,5555,6666,7777respectively, what is the possible edge length of this cube?After transformation:If the distances from the eight vertices of a cube to a certain plane are00,1111,2222,3333,4444,5555,6666,7777respectively, consider all possible edge lengths of this cube. Assuming the possible edge lengths form a setS𝑆Sitalic_S, find the sum of squares of all elements inS𝑆Sitalic_S.

🔼 This table presents a geometry problem from the OlymMATH dataset. The problem is described entirely in text, illustrating the dataset’s focus on textual problem representation, even for subjects like geometry that traditionally rely on diagrams. This exemplifies the OlymMATH design principle of ensuring that all problems are amenable to evaluation via language models, facilitating a fair and consistent assessment of LLMs’ reasoning capabilities.

read the captionFigure 3: A geometry problem described precisely in text from OlymMATH.
Problem (EN):Let $a_1, a_2, a_3, a_4, a_5 \in [0, 1]$, find the maximum value of $\prod_{1 \le i < j \le 5} |a_i - a_j|$.Answer (EN):\frac{3\sqrt{21}}{38416}Subject (EN):Algebra
Problem (ZH):{CJK}UTF8gbsn 设 $a_1,a_2,a_3,a_4,a_5 \in [0, 1]$, 求 $\prod_{1\le i<j \le 5} |a_i-a_j|$ 的最大值.Answer (ZH):\frac{3\sqrt{21}}{38416}Subject (ZH):{CJK}UTF8gbsn 代数

🔼 This table shows the distribution of the 200 problems in the OlymMATH benchmark across four mathematical categories (Algebra, Geometry, Number Theory, and Combinatorics) and two difficulty levels (Easy and Hard). It provides the counts of problems for each category and difficulty level combination, and the total number of problems in each category.

read the captionTable 2: The distribution of contest problems by category.
ModelsPass@1Cons@10
Alg.Geo.Num.Com.Avg.Alg.Geo.Num.Com.Avg.
The Hard Subset
Qwen2.5-32B-R1D17.218.817.60.813.636.020.024.00.020.0
QwQ-32B25.618.424.85.218.536.012.024.04.019.0
DeepSeek-R132.828.819.24.021.248.036.024.04.028.0
o3-mini (high)24.032.744.720.030.3/////
The Easy Subset
Qwen2.5-32B-R1D73.669.188.548.366.792.097.092.358.684.0
QwQ-32B84.879.783.867.978.196.0100.092.379.392.0
DeepSeek-R188.484.890.862.480.0100.097.092.375.991.0
o3-mini (high)90.089.4100.085.189.7/////

🔼 This table details the acceptable and unacceptable answer formats for the OlymMATH benchmark. Acceptable formats are limited to single real numbers, intervals (ranges), and sets of numbers. Unacceptable formats include set operations (union, intersection, etc.), variables (symbols representing unknown values), complex numbers, and any text-based answers. This restriction ensures consistency and ease of automated evaluation.

read the captionTable 3: The included and excluded formats of the final answer.
ModelsPass@1Cons@10
Alg.Geo.Num.Com.Avg.Alg.Geo.Num.Com.Avg.
The Hard Subset
Qwen2.5-32B-R1D5.66.88.01.25.44.08.012.00.06.0
QwQ-32B19.612.415.21.212.116.012.024.00.013.0
DeepSeek-R121.624.416.81.616.120.036.020.00.019.0
o3-mini (high)22.026.048.712.027.2/////
The Easy Subset
Qwen2.5-32B-R1D45.240.354.623.438.556.057.661.531.050.0
QwQ-32B76.873.387.748.668.992.097.092.362.185.0
DeepSeek-R179.274.893.150.771.392.090.9100.065.585.0
o3-mini (high)88.092.496.279.988.2/////

🔼 This figure displays a sample problem from the OlymMATH-HARD subset of the benchmark. It’s designed to evaluate whether large language models (LLMs) can correctly identify and provide all possible solutions to a complex mathematical problem, rather than just finding a single valid answer. The problem’s complexity and the requirement to find all possible answers make this a rigorous test of the model’s true reasoning capabilities.

read the captionFigure 4: An example from OlymMATH-HARD designed to assess whether the model can identify all possible answers.
Benchmarkso3-mini (high)DeepSeek-R1QwQ-32BQwen2.5-32B-R1D
MATH-500/97.3/94.3
AIME 202487.379.879.572.6
AIME 202586.570.066.059.0
HMMT 20250267.541.747.533.3
OlymMATH-EN-HARD30.321.218.513.6
OlymMATH-EN-EASY89.780.078.166.7
OlymMATH-ZH-HARD27.216.112.15.4
OlymMATH-ZH-EASY88.271.368.938.5

🔼 This table shows an example of the structured format used to present problems and answers in the OlymMATH dataset. It demonstrates the parallel presentation of problems and solutions in both English and Chinese, illustrating the bilingual nature of the dataset. The table includes problem statements, answers, and subject categories, highlighting the consistency in data structure across languages.

read the captionFigure 5: Example of the format used in OlymMATH.
DatasetAvg. Output LengthAvg. Reasoning Length
CorrectIncorrectTotalCorrectIncorrectTotal
AIME 2024159117071617199585083626820
AIME 2025154716251575233505300633894
OlymMATH-EN-HARD160816101610390244385042827
OlymMATH-EN-EASY158815941589291324600832507

🔼 Table 4 presents the performance of various large language models (LLMs) on the English version of the OlymMATH benchmark. The benchmark contains problems from four mathematical areas: Algebra, Geometry, Number Theory, and Combinatorics. The table shows the Pass@1 and Cons@10 scores for each model and category. Pass@1 indicates the accuracy of the top-ranked answer provided by the model, while Cons@10 represents the accuracy of the majority consensus answer among the top 10 model responses. Due to limitations with the 03-mini model’s API, only three responses per question were sampled, preventing a Cons@10 score from being reported for that model.

read the captionTable 4: Model performance on the English version of OlymMATH. The abbreviations “Alg.”, “Geo.”, “Num.”, and “Com.” represent the four problem categories: Algebra, Geometry, Number Theory, and Combinatorics, respectively. Due to o3-mini’s API instability, we sampled only three responses per question and cannot report its Cons@10 performance.
Problem:In the triangleABC𝐴𝐵𝐶\triangle ABC△ italic_A italic_B italic_C, the lengths of the three sides area,b,c𝑎𝑏𝑐a,b,citalic_a , italic_b , italic_c, satisfying the condition2a2+b2+c2=42superscript𝑎2superscript𝑏2superscript𝑐242a^{2}+b^{2}+c^{2}=42 italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_b start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_c start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 4. Find the maximum value of the area ofABC𝐴𝐵𝐶\triangle ABC△ italic_A italic_B italic_C.Answer:5555\frac{\sqrt{5}}{5}divide start_ARG square-root start_ARG 5 end_ARG end_ARG start_ARG 5 end_ARG. (Whena=25𝑎25a=\frac{2}{\sqrt{5}}italic_a = divide start_ARG 2 end_ARG start_ARG square-root start_ARG 5 end_ARG end_ARG,b=c=65𝑏𝑐65b=c=\sqrt{\frac{6}{5}}italic_b = italic_c = square-root start_ARG divide start_ARG 6 end_ARG start_ARG 5 end_ARG end_ARG)
Model Output:We wish to maximize the area ofABC𝐴𝐵𝐶\triangle ABC△ italic_A italic_B italic_C, subject to2a2+b2+c2=4.2superscript𝑎2superscript𝑏2superscript𝑐242a^{2}+b^{2}+c^{2}=4.2 italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_b start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_c start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 4 .Because the condition is symmetric inb𝑏bitalic_bandc𝑐citalic_c(but not ina𝑎aitalic_a), it is reasonable to look for an optimum among triangles havingb=c𝑏𝑐b=citalic_b = italic_c. (Later one can check that assumingb=c𝑏𝑐b=citalic_b = italic_cindeed gives the maximum.)The problem involves maximizing the area of an isosceles triangle with sidesa,b,b𝑎𝑏𝑏a,b,bitalic_a , italic_b , italic_bunder the constraint2a2+b2+c2=42superscript𝑎2superscript𝑏2superscript𝑐242a^{2}+b^{2}+c^{2}=42 italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_b start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_c start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 4. By assumingb=c𝑏𝑐b=citalic_b = italic_c, we derive a formula for the area in terms ofb𝑏bitalic_band find the maximum area by expressing it as a function ofu=b2𝑢superscript𝑏2u=b^{2}italic_u = italic_b start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT. We then determine the critical points of this quadratic function, identify the maximum area, and finally recover the side lengths to verify that they satisfy the original constraints. The maximum area achieved is5555\frac{\sqrt{5}}{5}divide start_ARG square-root start_ARG 5 end_ARG end_ARG start_ARG 5 end_ARG.

🔼 Table 5 presents the performance of several large language models (LLMs) on the Chinese version of the OlymMATH benchmark. The benchmark consists of 200 Olympiad-level math problems categorized into four areas: Algebra, Geometry, Number Theory, and Combinatorics. The table shows the models’ Pass@1 (accuracy of the top prediction) and Cons@10 (accuracy of the consensus answer across 10 predictions) scores for each category and overall. Due to limitations in the o3-mini model’s API, only three responses were sampled per question for this model, and its Cons@10 performance is therefore not reported.

read the captionTable 5: Model performance on the Chinese version of OlymMATH. The abbreviations “Alg.”, “Geo.”, “Num.”, and “Com.” represent the four problem categories: Algebra, Geometry, Number Theory, and Combinatorics, respectively. Due to o3-mini’s API instability, we sampled only three responses per question and cannot report its Cons@10 performance.
Problem:Let the sequence of rationalsx1,x2,subscript𝑥1subscript𝑥2x_{1},x_{2},\dotsitalic_x start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_x start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , …be defined such thatx1=2511subscript𝑥12511x_{1}=\frac{25}{11}italic_x start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT = divide start_ARG 25 end_ARG start_ARG 11 end_ARGandxk+1=13(xk+1xk1).subscript𝑥𝑘113subscript𝑥𝑘1subscript𝑥𝑘1x_{k+1}=\frac{1}{3}\left(x_{k}+\frac{1}{x_{k}}-1\right).italic_x start_POSTSUBSCRIPT italic_k + 1 end_POSTSUBSCRIPT = divide start_ARG 1 end_ARG start_ARG 3 end_ARG ( italic_x start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT + divide start_ARG 1 end_ARG start_ARG italic_x start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT end_ARG - 1 ) .x2025subscript𝑥2025x_{2025}italic_x start_POSTSUBSCRIPT 2025 end_POSTSUBSCRIPTcan be expressed asmn𝑚𝑛\frac{m}{n}divide start_ARG italic_m end_ARG start_ARG italic_n end_ARGfor relatively prime positive integersm𝑚mitalic_mandn𝑛nitalic_n. Find the remainder whenm+n𝑚𝑛m+nitalic_m + italic_nis divided by1000100010001000.Answer:248248248248.
Model Output:We will show that ifx1=2511subscript𝑥12511x_{1}=\frac{25}{11}italic_x start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT = divide start_ARG 25 end_ARG start_ARG 11 end_ARGandxk+1=13(xk+1xk1),subscript𝑥𝑘113subscript𝑥𝑘1subscript𝑥𝑘1x_{k+1}=\frac{1}{3}\Bigl{(}x_{k}+\frac{1}{x_{k}}-1\Bigr{)},italic_x start_POSTSUBSCRIPT italic_k + 1 end_POSTSUBSCRIPT = divide start_ARG 1 end_ARG start_ARG 3 end_ARG ( italic_x start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT + divide start_ARG 1 end_ARG start_ARG italic_x start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT end_ARG - 1 ) ,then after “many” iterations the answer tox2025=mnsubscript𝑥2025𝑚𝑛x_{2025}=\frac{m}{n}italic_x start_POSTSUBSCRIPT 2025 end_POSTSUBSCRIPT = divide start_ARG italic_m end_ARG start_ARG italic_n end_ARG(in lowest terms) satisfiesm+n248(mod1000).𝑚𝑛annotated248pmod1000m+n\equiv 248\pmod{1000}.italic_m + italic_n ≡ 248 start_MODIFIER ( roman_mod start_ARG 1000 end_ARG ) end_MODIFIER .In what follows we briefly describe one way (via a “miraculous” change-of–variables) to “solve” the recurrence; the details need not be reproduced fully here.A good hint is that the recurrencexk+1=13(xk+1xk1)subscript𝑥𝑘113subscript𝑥𝑘1subscript𝑥𝑘1x_{k+1}=\frac{1}{3}\Bigl{(}x_{k}+\frac{1}{x_{k}}-1\Bigr{)}italic_x start_POSTSUBSCRIPT italic_k + 1 end_POSTSUBSCRIPT = divide start_ARG 1 end_ARG start_ARG 3 end_ARG ( italic_x start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT + divide start_ARG 1 end_ARG start_ARG italic_x start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT end_ARG - 1 )has the curious symmetryT(x)=13(x+1x1)=T(1x),𝑇𝑥13𝑥1𝑥1𝑇1𝑥T(x)=\frac{1}{3}\Bigl{(}x+\frac{1}{x}-1\Bigr{)}=T\Bigl{(}\frac{1}{x}\Bigr{)},italic_T ( italic_x ) = divide start_ARG 1 end_ARG start_ARG 3 end_ARG ( italic_x + divide start_ARG 1 end_ARG start_ARG italic_x end_ARG - 1 ) = italic_T ( divide start_ARG 1 end_ARG start_ARG italic_x end_ARG ) ,so the expressionx+1x𝑥1𝑥x+\frac{1}{x}italic_x + divide start_ARG 1 end_ARG start_ARG italic_x end_ARG(which is “symmetric” inx𝑥xitalic_xand its reciprocal) naturally appears. In fact, one may rewrite the recurrence in the form3xk+1+1=xk+1xk.3subscript𝑥𝑘11subscript𝑥𝑘1subscript𝑥𝑘3x_{k+1}+1=x_{k}+\frac{1}{x_{k}}.3 italic_x start_POSTSUBSCRIPT italic_k + 1 end_POSTSUBSCRIPT + 1 = italic_x start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT + divide start_ARG 1 end_ARG start_ARG italic_x start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT end_ARG .A useful idea is then to “complete the square” by writing(xk1xk)2=xk+1xk2superscriptsubscript𝑥𝑘1subscript𝑥𝑘2subscript𝑥𝑘1subscript𝑥𝑘2\left(\sqrt{x_{k}}-\frac{1}{\sqrt{x_{k}}}\right)^{2}=x_{k}+\frac{1}{x_{k}}-2( square-root start_ARG italic_x start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT end_ARG - divide start_ARG 1 end_ARG start_ARG square-root start_ARG italic_x start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT end_ARG end_ARG ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_x start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT + divide start_ARG 1 end_ARG start_ARG italic_x start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT end_ARG - 2, so that3xk+1+1=(xk1xk)2+2.3subscript𝑥𝑘11superscriptsubscript𝑥𝑘1subscript𝑥𝑘223x_{k+1}+1=\left(\sqrt{x_{k}}-\frac{1}{\sqrt{x_{k}}}\right)^{2}+2.3 italic_x start_POSTSUBSCRIPT italic_k + 1 end_POSTSUBSCRIPT + 1 = ( square-root start_ARG italic_x start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT end_ARG - divide start_ARG 1 end_ARG start_ARG square-root start_ARG italic_x start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT end_ARG end_ARG ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + 2 .Defininguk=xk1xksubscript𝑢𝑘subscript𝑥𝑘1subscript𝑥𝑘u_{k}=\sqrt{x_{k}}-\frac{1}{\sqrt{x_{k}}}italic_u start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT = square-root start_ARG italic_x start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT end_ARG - divide start_ARG 1 end_ARG start_ARG square-root start_ARG italic_x start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT end_ARG end_ARG, one obtains the “linear-looking” relationuk2=3xk+11.superscriptsubscript𝑢𝑘23subscript𝑥𝑘11u_{k}^{2}=3x_{k+1}-1.italic_u start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 3 italic_x start_POSTSUBSCRIPT italic_k + 1 end_POSTSUBSCRIPT - 1 .With some additional work it turns out that if we writexk=mknksubscript𝑥𝑘subscript𝑚𝑘subscript𝑛𝑘x_{k}=\frac{m_{k}}{n_{k}}italic_x start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT = divide start_ARG italic_m start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT end_ARG start_ARG italic_n start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT end_ARG(in lowest terms), then one may prove by induction thatmk+nk=3612(2k11).subscript𝑚𝑘subscript𝑛𝑘36superscript12superscript2𝑘11m_{k}+n_{k}=36\cdot 12^{(2^{\,k-1}-1)}.italic_m start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT + italic_n start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT = 36 ⋅ 12 start_POSTSUPERSCRIPT ( 2 start_POSTSUPERSCRIPT italic_k - 1 end_POSTSUPERSCRIPT - 1 ) end_POSTSUPERSCRIPT .(In our case one checks thatm1+n1=25+11=36subscript𝑚1subscript𝑛1251136m_{1}+n_{1}=25+11=36italic_m start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT + italic_n start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT = 25 + 11 = 36,m2+n2=432subscript𝑚2subscript𝑛2432m_{2}+n_{2}=432italic_m start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT + italic_n start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT = 432,m3+n3=62208,subscript𝑚3subscript𝑛362208m_{3}+n_{3}=62208,italic_m start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT + italic_n start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT = 62208 ,and indeed36,432,62208364326220836,432,6220836 , 432 , 62208equal36120, 36121, 3612336superscript12036superscript12136superscript12336\cdot 12^{0},\,36\cdot 12^{1},\,36\cdot 12^{3}36 ⋅ 12 start_POSTSUPERSCRIPT 0 end_POSTSUPERSCRIPT , 36 ⋅ 12 start_POSTSUPERSCRIPT 1 end_POSTSUPERSCRIPT , 36 ⋅ 12 start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPTrespectively.)Thus, fork=2025𝑘2025k=2025italic_k = 2025we havem+n=3612(220241).𝑚𝑛36superscript12superscript220241m+n=36\cdot 12^{\left(2^{2024}-1\right)}.italic_m + italic_n = 36 ⋅ 12 start_POSTSUPERSCRIPT ( 2 start_POSTSUPERSCRIPT 2024 end_POSTSUPERSCRIPT - 1 ) end_POSTSUPERSCRIPT .Even though this number is astronomically large, we are only asked to computem+n(mod1000).annotated𝑚𝑛pmod1000m+n\pmod{1000}.italic_m + italic_n start_MODIFIER ( roman_mod start_ARG 1000 end_ARG ) end_MODIFIER .Using modular reductions, we computem+n𝑚𝑛m+nitalic_m + italic_nmodulo 1000. First, modulo 8, we find3612(220241)0(mod8)36superscript12superscript220241annotated0pmod836\cdot 12^{(2^{2024}-1)}\equiv 0\pmod{8}36 ⋅ 12 start_POSTSUPERSCRIPT ( 2 start_POSTSUPERSCRIPT 2024 end_POSTSUPERSCRIPT - 1 ) end_POSTSUPERSCRIPT ≡ 0 start_MODIFIER ( roman_mod start_ARG 8 end_ARG ) end_MODIFIER. Modulo 125, Euler’s theorem helps reduce12Nsuperscript12𝑁12^{N}12 start_POSTSUPERSCRIPT italic_N end_POSTSUPERSCRIPTto1215118(mod125)superscript1215annotated118pmod12512^{15}\equiv 118\pmod{125}12 start_POSTSUPERSCRIPT 15 end_POSTSUPERSCRIPT ≡ 118 start_MODIFIER ( roman_mod start_ARG 125 end_ARG ) end_MODIFIER, giving3612N123(mod125)36superscript12𝑁annotated123pmod12536\cdot 12^{N}\equiv 123\pmod{125}36 ⋅ 12 start_POSTSUPERSCRIPT italic_N end_POSTSUPERSCRIPT ≡ 123 start_MODIFIER ( roman_mod start_ARG 125 end_ARG ) end_MODIFIER. Using the Chinese Remainder Theorem, solving125k+1230(mod8)125𝑘123annotated0pmod8125k+123\equiv 0\pmod{8}125 italic_k + 123 ≡ 0 start_MODIFIER ( roman_mod start_ARG 8 end_ARG ) end_MODIFIERgivesk1(mod8)𝑘annotated1pmod8k\equiv 1\pmod{8}italic_k ≡ 1 start_MODIFIER ( roman_mod start_ARG 8 end_ARG ) end_MODIFIER, leading to the final answer:248248\boxed{248}248.

🔼 This table compares the performance of four large language models (LLMs) across five different mathematical reasoning benchmarks: MATH-500, AIME 2024, AIME 2025, HMMT 202502, and OlymMATH (both easy and hard versions in English and Chinese). The Pass@1 metric indicates the percentage of problems correctly solved by the model in a single attempt. Results marked with an asterisk (*) were taken from the original publications or the MathArena website, not directly from the authors’ experiments. The table highlights the relative difficulty of each benchmark and how the models’ performance varies across them.

read the captionTable 6: Model performance on different mathematical benchmarks (Pass@1). The results with (*) denote that they are copied from their reports or the third-party website MathArena.
Source:2016 IMO Shortlist A7Problem:Find all functionsf::𝑓f:\mathbb{R}\rightarrow\mathbb{R}italic_f : blackboard_R → blackboard_Rsuch thatf(0)0𝑓00f(0)\neq 0italic_f ( 0 ) ≠ 0and for allx,y𝑥𝑦x,y\in\mathbb{R}italic_x , italic_y ∈ blackboard_R,f(x+y)2=2f(x)f(y)+max{f(x2+y2),f(x2)+f(y2)}.𝑓superscript𝑥𝑦22𝑓𝑥𝑓𝑦𝑓superscript𝑥2superscript𝑦2𝑓superscript𝑥2𝑓superscript𝑦2f(x+y)^{2}=2f(x)f(y)+\max\left\{f(x^{2}+y^{2}),f(x^{2})+f(y^{2})\right\}.italic_f ( italic_x + italic_y ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 2 italic_f ( italic_x ) italic_f ( italic_y ) + roman_max { italic_f ( italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) , italic_f ( italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) + italic_f ( italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) } .Answer:f(x)=1 and f(x)=x1.𝑓𝑥1 and 𝑓𝑥𝑥1f(x)=-1\text{ and }f(x)=x-1.italic_f ( italic_x ) = - 1 and italic_f ( italic_x ) = italic_x - 1 .Solution:(Provided in Omni-MATH) To find all functionsf::𝑓f:\mathbb{R}\rightarrow\mathbb{R}italic_f : blackboard_R → blackboard_Rsatisfying the given functional equation, we start with the condition:f(0)0,𝑓00f(0)\neq 0,italic_f ( 0 ) ≠ 0 ,and the functional equation for allx,y𝑥𝑦x,y\in\mathbb{R}italic_x , italic_y ∈ blackboard_R:f(x+y)2=2f(x)f(y)+max{f(x2+y2),f(x2)+f(y2)}.𝑓superscript𝑥𝑦22𝑓𝑥𝑓𝑦𝑓superscript𝑥2superscript𝑦2𝑓superscript𝑥2𝑓superscript𝑦2f(x+y)^{2}=2f(x)f(y)+\max\left\{f(x^{2}+y^{2}),f(x^{2})+f(y^{2})\right\}.italic_f ( italic_x + italic_y ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 2 italic_f ( italic_x ) italic_f ( italic_y ) + roman_max { italic_f ( italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) , italic_f ( italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) + italic_f ( italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) } .Step 1: Test Simple FunctionsLet’s start by testing the functions given in the reference answer:f(x)=1𝑓𝑥1f(x)=-1italic_f ( italic_x ) = - 1andf(x)=x1𝑓𝑥𝑥1f(x)=x-1italic_f ( italic_x ) = italic_x - 1.Case 1:f(x)=1𝑓𝑥1f(x)=-1italic_f ( italic_x ) = - 1. Substitutef(x)=1𝑓𝑥1f(x)=-1italic_f ( italic_x ) = - 1into the equation:\cdotsCase 2:f(x)=x1𝑓𝑥𝑥1f(x)=x-1italic_f ( italic_x ) = italic_x - 1. Substitutingf(x)=x1𝑓𝑥𝑥1f(x)=x-1italic_f ( italic_x ) = italic_x - 1:\cdotsconfirmingf(x)=x1𝑓𝑥𝑥1f(x)=x-1italic_f ( italic_x ) = italic_x - 1is another valid solution.ConclusionAfter verifying the two functions, we conclude that the functions satisfying the original functional equation are:f(x)=1andf(x)=x1.𝑓𝑥1and𝑓𝑥𝑥1f(x)=-1\quad\text{and}\quad f(x)=x-1.italic_f ( italic_x ) = - 1 and italic_f ( italic_x ) = italic_x - 1 .Thus, the solutions are:f(x)=1 and f(x)=x1.𝑓𝑥1 and 𝑓𝑥𝑥1\boxed{f(x)=-1\text{ and }f(x)=x-1}.start_ARG italic_f ( italic_x ) = - 1 and italic_f ( italic_x ) = italic_x - 1 end_ARG .
Model Output:We will show that the only functionsf::𝑓f:\mathbb{R}\to\mathbb{R}italic_f : blackboard_R → blackboard_Rwithf(0)0𝑓00f(0)\neq 0italic_f ( 0 ) ≠ 0, satisfyingf(x+y)2=2f(x)f(y)+max{f(x2+y2),f(x2)+f(y2)}for all x,y,formulae-sequence𝑓superscript𝑥𝑦22𝑓𝑥𝑓𝑦𝑓superscript𝑥2superscript𝑦2𝑓superscript𝑥2𝑓superscript𝑦2for all 𝑥𝑦f(x+y)^{2}=2f(x)f(y)+\max\{\,f(x^{2}+y^{2}),\,f(x^{2})+f(y^{2})\,\}\quad\text{% for all }x,y\in\mathbb{R},italic_f ( italic_x + italic_y ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 2 italic_f ( italic_x ) italic_f ( italic_y ) + roman_max { italic_f ( italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) , italic_f ( italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) + italic_f ( italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) } for all italic_x , italic_y ∈ blackboard_R ,aref(x)=1for all xorf(x)=x1.\boxed{f(x)=-1\quad\text{for all }x\quad\text{or}\quad f(x)=x-1.}italic_f ( italic_x ) = - 1 for all italic_x or italic_f ( italic_x ) = italic_x - 1 .Below is an outline of one way to arrive at this result.Step 1. Plug iny=0𝑦0y=0italic_y = 0Lety=0𝑦0y=0italic_y = 0in the given equation:f(x)2=2f(x)f(0)+max{f(x2),f(x2)+f(0)}.𝑓superscript𝑥22𝑓𝑥𝑓0𝑓superscript𝑥2𝑓superscript𝑥2𝑓0f(x)^{2}=2f(x)f(0)+\max\{f(x^{2}),\,f(x^{2})+f(0)\}.italic_f ( italic_x ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 2 italic_f ( italic_x ) italic_f ( 0 ) + roman_max { italic_f ( italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) , italic_f ( italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) + italic_f ( 0 ) } .Writec=f(0)0𝑐𝑓00c=f(0)\neq 0italic_c = italic_f ( 0 ) ≠ 0. Notice that the two arguments inside the maximum differ byf(0)𝑓0f(0)italic_f ( 0 ); therefore, the larger one depends on the sign ofc𝑐citalic_c:- **Ifc>0𝑐0c>0italic_c > 0:**  Thenf(x2)+c>f(x2)𝑓superscript𝑥2𝑐𝑓superscript𝑥2f(x^{2})+c>f(x^{2})italic_f ( italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) + italic_c > italic_f ( italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT )so the equation becomesf(x)2=2cf(x)+f(x2)+c.𝑓superscript𝑥22𝑐𝑓𝑥𝑓superscript𝑥2𝑐f(x)^{2}=2c\,f(x)+f(x^{2})+c.italic_f ( italic_x ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 2 italic_c italic_f ( italic_x ) + italic_f ( italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) + italic_c .- **Ifc<0𝑐0c<0italic_c < 0:**  Thenf(x2)+c<f(x2)𝑓superscript𝑥2𝑐𝑓superscript𝑥2f(x^{2})+c<f(x^{2})italic_f ( italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) + italic_c < italic_f ( italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT )so we havef(x)2=2cf(x)+f(x2).𝑓superscript𝑥22𝑐𝑓𝑥𝑓superscript𝑥2f(x)^{2}=2c\,f(x)+f(x^{2}).italic_f ( italic_x ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 2 italic_c italic_f ( italic_x ) + italic_f ( italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) .Thus, in each case one obtains a relationship betweenf(x)𝑓𝑥f(x)italic_f ( italic_x )andf(x2)𝑓superscript𝑥2f(x^{2})italic_f ( italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ).Step 2. Check constant solutions\cdotsThus, the constant solution isf(x)=1for all x𝑓𝑥1for all 𝑥f(x)=-1\quad\text{for all }xitalic_f ( italic_x ) = - 1 for all italic_x.Step 3. Look for non-constant solutionsA natural next guess is to try a linear function.\cdotsA quick check verifies thatf(x)=x1𝑓𝑥𝑥1f(x)=x-1italic_f ( italic_x ) = italic_x - 1satisfies the original equation.\cdotswhich equals(x+y1)2superscript𝑥𝑦12(x+y-1)^{2}( italic_x + italic_y - 1 ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT, matching the left-hand side.ConclusionThe only functionsf::𝑓f:\mathbb{R}\to\mathbb{R}italic_f : blackboard_R → blackboard_Rwithf(0)0𝑓00f(0)\neq 0italic_f ( 0 ) ≠ 0that satisfy\cdotsfor all real numbersx𝑥xitalic_xandy𝑦yitalic_yare:f(x)=1for all xorf(x)=x1.\boxed{f(x)=-1\quad\text{for all }x\quad\text{or}\quad f(x)=x-1.}italic_f ( italic_x ) = - 1 for all italic_x or italic_f ( italic_x ) = italic_x - 1 .These are the complete solutions.

🔼 This table presents a quantitative analysis of the DeepSeek-R1 model’s response length across four different mathematical reasoning benchmarks: AIME 2024, AIME 2025, OlymMATH-EN-HARD, and OlymMATH-EN-EASY. For each benchmark, it shows the average length of the model’s final answer (Output Length) and the average length of its reasoning process (Reasoning Length). These lengths are further broken down into the average lengths for correct and incorrect answers, providing insight into the relationship between response length and solution accuracy. The data reveals how the model’s output and thought processes vary in length based on the complexity and nature of the problems within each benchmark.

read the captionTable 7: The average length of the responses generated from DeepSeek-R1. “Output Length” refers to the length of the final solution provided by the model, while “Reasoning Length” indicates the length of model’s thought process.
Problem:Leta,b,c𝑎𝑏𝑐a,b,c\in\mathbb{R}italic_a , italic_b , italic_c ∈ blackboard_R,a3b+b3c+c3a=3superscript𝑎3𝑏superscript𝑏3𝑐superscript𝑐3𝑎3a^{3}b+b^{3}c+c^{3}a=3italic_a start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT italic_b + italic_b start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT italic_c + italic_c start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT italic_a = 3, find the minimum value of the expressionf(a,b,c)=(a4)4+1000a2b2𝑓𝑎𝑏𝑐superscriptsuperscript𝑎441000superscript𝑎2superscript𝑏2f(a,b,c)=(\sum a^{4})^{4}+1000\sum a^{2}b^{2}italic_f ( italic_a , italic_b , italic_c ) = ( ∑ italic_a start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT ) start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT + 1000 ∑ italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_b start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT.Reference Answer:2625262526252625. (Equality condition:a=±47sin24π7,b=±47sin22π7,c=a=\pm\frac{4}{\sqrt{7}}\sin^{2}\frac{4\pi}{7},b=\pm\frac{4}{\sqrt{7}}\sin^{2}% \frac{2\pi}{7},c=italic_a = ± divide start_ARG 4 end_ARG start_ARG square-root start_ARG 7 end_ARG end_ARG roman_sin start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT divide start_ARG 4 italic_π end_ARG start_ARG 7 end_ARG , italic_b = ± divide start_ARG 4 end_ARG start_ARG square-root start_ARG 7 end_ARG end_ARG roman_sin start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT divide start_ARG 2 italic_π end_ARG start_ARG 7 end_ARG , italic_c =±47sin2π7plus-or-minus47superscript2𝜋7\pm\frac{4}{\sqrt{7}}\sin^{2}\frac{\pi}{7}± divide start_ARG 4 end_ARG start_ARG square-root start_ARG 7 end_ARG end_ARG roman_sin start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT divide start_ARG italic_π end_ARG start_ARG 7 end_ARGand its cyclic permutations)
Model Output:We wish to minimizef(a,b,c)=(a4+b4+c4)4+1000(a2b2+b2c2+c2a2)𝑓𝑎𝑏𝑐superscriptsuperscript𝑎4superscript𝑏4superscript𝑐441000superscript𝑎2superscript𝑏2superscript𝑏2superscript𝑐2superscript𝑐2superscript𝑎2f(a,b,c)=\left(a^{4}+b^{4}+c^{4}\right)^{4}+1000\left(a^{2}b^{2}+b^{2}c^{2}+c^% {2}a^{2}\right)italic_f ( italic_a , italic_b , italic_c ) = ( italic_a start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT + italic_b start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT + italic_c start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT ) start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT + 1000 ( italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_b start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_b start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_c start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_c start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT )subject to the (cyclic) constrainta3b+b3c+c3a=3,superscript𝑎3𝑏superscript𝑏3𝑐superscript𝑐3𝑎3a^{3}b+b^{3}c+c^{3}a=3,italic_a start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT italic_b + italic_b start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT italic_c + italic_c start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT italic_a = 3 ,witha,b,c𝑎𝑏𝑐a,b,c\in\mathbb{R}italic_a , italic_b , italic_c ∈ blackboard_R.A natural idea is to look for a "symmetric-like" solution. (Even though the constraint is cyclic rather than fully symmetric, the very "even" form of the objectivef𝑓fitalic_f— all terms are even — suggests that the optimum might occur when the "building blocks" have equal absolute value.) One obvious candidate is to trya=b=c=t.𝑎𝑏𝑐𝑡a=b=c=t.italic_a = italic_b = italic_c = italic_t .Then the constraint becomesa3b+b3c+c3a=3t4=3t4=1.superscript𝑎3𝑏superscript𝑏3𝑐superscript𝑐3𝑎3superscript𝑡43superscript𝑡41a^{3}b+b^{3}c+c^{3}a=3t^{4}=3\implies t^{4}=1.italic_a start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT italic_b + italic_b start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT italic_c + italic_c start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT italic_a = 3 italic_t start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT = 3 ⟹ italic_t start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT = 1 .Thust=1𝑡1t=1italic_t = 1ort=1𝑡1t=-1italic_t = - 1. (In\mathbb{R}blackboard_Rthe equationt4=1superscript𝑡41t^{4}=1italic_t start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT = 1has the two solutionst=1𝑡1t=1italic_t = 1andt=1𝑡1t=-1italic_t = - 1.)Now compute the value off𝑓fitalic_fata=b=c=t𝑎𝑏𝑐𝑡a=b=c=titalic_a = italic_b = italic_c = italic_t. Notice thata4+b4+c4=t4+t4+t4=3,superscript𝑎4superscript𝑏4superscript𝑐4superscript𝑡4superscript𝑡4superscript𝑡43a^{4}+b^{4}+c^{4}=t^{4}+t^{4}+t^{4}=3,italic_a start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT + italic_b start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT + italic_c start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT = italic_t start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT + italic_t start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT + italic_t start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT = 3 ,a2b2+b2c2+c2a2=t4+t4+t4=3superscript𝑎2superscript𝑏2superscript𝑏2superscript𝑐2superscript𝑐2superscript𝑎2superscript𝑡4superscript𝑡4superscript𝑡43a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}=t^{4}+t^{4}+t^{4}=3italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_b start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_b start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_c start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_c start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_t start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT + italic_t start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT + italic_t start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT = 3(sincet4=1superscript𝑡41t^{4}=1italic_t start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT = 1).Thenf=(3)4+10003=81+3000=3081.𝑓superscript34100038130003081f=(3)^{4}+1000\cdot 3=81+3000=3081.italic_f = ( 3 ) start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT + 1000 ⋅ 3 = 81 + 3000 = 3081 .Before concluding that3081308130813081is the minimum, one may ask: Could there be a choice ofa,b,c𝑎𝑏𝑐a,b,citalic_a , italic_b , italic_csatisfyinga3b+b3c+c3a=3superscript𝑎3𝑏superscript𝑏3𝑐superscript𝑐3𝑎3a^{3}b+b^{3}c+c^{3}a=3italic_a start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT italic_b + italic_b start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT italic_c + italic_c start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT italic_a = 3for which(a4+b4+c4)<3or(a2b2+b2c2+c2a2)<3?superscript𝑎4superscript𝑏4superscript𝑐43orsuperscript𝑎2superscript𝑏2superscript𝑏2superscript𝑐2superscript𝑐2superscript𝑎23?\left(a^{4}+b^{4}+c^{4}\right)<3\quad\text{or}\quad\left(a^{2}b^{2}+b^{2}c^{2}% +c^{2}a^{2}\right)<3?( italic_a start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT + italic_b start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT + italic_c start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT ) < 3 or ( italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_b start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_b start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_c start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_c start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) < 3 ?Sincea4,b4,c4superscript𝑎4superscript𝑏4superscript𝑐4a^{4},b^{4},c^{4}italic_a start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT , italic_b start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT , italic_c start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPTare nonnegative and the constraint (a homogeneous form of degree4444) "forces" the variables to be large enough in a certain sense, one may show by a careful (often inequality-based) argument that ifa3b+b3c+c3a=3superscript𝑎3𝑏superscript𝑏3𝑐superscript𝑐3𝑎3a^{3}b+b^{3}c+c^{3}a=3italic_a start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT italic_b + italic_b start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT italic_c + italic_c start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT italic_a = 3, then one must havea4+b4+c43anda2b2+b2c2+c2a23.superscript𝑎4superscript𝑏4superscript𝑐43andsuperscript𝑎2superscript𝑏2superscript𝑏2superscript𝑐2superscript𝑐2superscript𝑎23a^{4}+b^{4}+c^{4}\geq 3\quad\text{and}\quad a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}% \geq 3.italic_a start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT + italic_b start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT + italic_c start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT ≥ 3 and italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_b start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_b start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_c start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_c start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ≥ 3 .In fact, equality in these "loss-inequalities" forces|a|=|b|=|c|=1𝑎𝑏𝑐1|a|=|b|=|c|=1| italic_a | = | italic_b | = | italic_c | = 1, and it turns out (by checking the signs in the constraint) that the only possibility for equality isa=b=c=1𝑎𝑏𝑐1a=b=c=1italic_a = italic_b = italic_c = 1ora=b=c=1𝑎𝑏𝑐1a=b=c=-1italic_a = italic_b = italic_c = - 1.Also, one might try "extremal" cases such as letting one variable vanish (say,c=0𝑐0c=0italic_c = 0). In that case, the constraint reduces toa3b=3superscript𝑎3𝑏3a^{3}b=3italic_a start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT italic_b = 3so thata𝑎aitalic_aandb𝑏bitalic_bcannot be arbitrarily small (or close in absolute value to1111), and a short calculation shows that the value off𝑓fitalic_fthen exceeds3081308130813081.Thus, the candidatea=b=c=1𝑎𝑏𝑐1a=b=c=1italic_a = italic_b = italic_c = 1(ora=b=c=1𝑎𝑏𝑐1a=b=c=-1italic_a = italic_b = italic_c = - 1) gives the minimal value. In conclusion, the minimum value off𝑓fitalic_fis3081,30813081,3081 ,and it is attained exactly whena,b,c𝑎𝑏𝑐a,b,citalic_a , italic_b , italic_care all equal with|a|=1𝑎1|a|=1| italic_a | = 1and the signs arranged so thata3b=1superscript𝑎3𝑏1a^{3}b=1italic_a start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT italic_b = 1(that is,a=b=c=1𝑎𝑏𝑐1a=b=c=1italic_a = italic_b = italic_c = 1ora=b=c=1𝑎𝑏𝑐1a=b=c=-1italic_a = italic_b = italic_c = - 1).

🔼 This figure displays an example problem encountered during the OlymMATH dataset creation. The problem involves finding the maximum area of a triangle given a constraint on its side lengths. The model, 03-mini (high), arrived at the correct answer; however, its solution process lacked mathematical rigor. Due to space limitations, only a concise summary of the model’s solution steps is presented, highlighting the model’s non-rigorous approach.

read the captionFigure 7: An example during our data collection. o3-mini (high) found the correct answer without rigorously solving the problem. Due to length constraints, we provide only a summary of its original solution steps.

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