2025 Euclid Mathematics Contest Results Released and Problem Analysis | 2025年欧几里得数学竞赛成绩公布及真题解析

📚 2025 Euclid Mathematics Contest Results Released and Problem Analysis | 2025年欧几里得数学竞赛成绩公布及真题解析

The 2025 Euclid Mathematics Contest, administered by the University of Waterloo’s Centre for Education in Mathematics and Computing (CEMC), has officially released its results. This globally recognised competition challenges senior high school students with ten intricate problems spanning algebra, geometry, combinatorics, and number theory. In this article, we dissect the 2025 score distribution, spotlight outstanding achievements, and walk through detailed solutions to selected problems that proved particularly demanding.

由滑铁卢大学数学与计算教育中心（CEMC）主办的2025年欧几里得数学竞赛已正式公布成绩。这项全球知名的赛事以十道跨越代数、几何、组合与数论的复杂题目考验高中高年级学生。本文将在剖析2025年分数分布、聚焦杰出表现的同时，对若干公认难度较大的题目进行详尽解答。

1. Overview of the 2025 Euclid Contest | 2025欧几里得竞赛概览

The 2025 Euclid Contest took place on 3 April 2025, giving participants 2.5 hours to solve 10 problems. Each problem is worth 10 marks, with partial credit awarded for correct reasoning even if the final answer is not reached. The problems are arranged in increasing difficulty; the first few build confidence, while Problems 9 and 10 demand original insights and multi‑stage arguments.

2025年欧几里得竞赛于2025年4月3日举行，参赛者需在2.5小时内解答10道题。每题满分10分，即便未得出最终答案，正确的推理过程亦可得步骤分。题目按难度递增编排；前几题建立信心，而第9和第10题则需要独到的洞察力与多步论证。

Over 20 800 students from 84 countries registered for the competition. The global average score was 42.3 out of 100, moderately higher than the 2024 average of 40.1, suggesting improved preparation or a slightly more accessible paper. The top 1 % of candidates scored 86 or above, and the highest mark reported worldwide was 99.

来自84个国家的20 800多名学生报名参赛。全球平均分为42.3分（满分100），较2024年的40.1分温和上升，暗示备考更为充分或试题难度略有下降。全球排名前1 %的考生得分在86分及以上，世界最高分为99分。

2. Score Distribution and Award Cutoffs | 分数分布与获奖分数线

The CEMC uses a standardised awarding framework: a Certificate of Distinction is given to the top 25 % of all contestants, and medals are awarded to the highest‑scoring student in each school. Additionally, students ranking in the top 2.5 % globally earn a place on the Honour Roll. The 2025 cutoffs are summarised below.

CEMC采用标准化的奖项设置：前25 %的参赛者可获优异证书，各校最高分获得者被授予奖章。此外，全球排名前2.5 %的学生可进入荣誉榜。2025年分数线总结如下。

Award	Cutoff Score (out of 100)	Percentage of Candidates
Certificate of Distinction	69	Top 25%
Honour Roll (Group I)	83	Top 2.5%
School Medal	Varies	Highest in school

The Certificate of Distinction cutoff at 69 this year is slightly lower than the 2024 threshold of 71, reflecting a wider spread of high‑mid scores. The Honour Roll cutoff of 83 remains consistent with historical values, confirming that the most selective tier retains its rigour.

今年优异证书的69分线略低于2024年的71分，反映出中高分段分布更广。荣誉榜83分的门槛与历史数据持平，表明最顶尖层次的选拔依然严格。

3. Analysis of Top Performers | 顶尖选手表现分析

Five candidates attained a score of 99, missing perfection by only one mark. Their solutions demonstrated exceptional elegance, particularly in the geometry and combinatorics problems. CEMC officials noted that these contestants often employed multiple solution paths, verifying their results in different ways.

五名选手获得99分，距满分仅一分之隔。他们的解答展现出非凡的优雅性，尤其在几何与组合题中。CEMC官员指出，这些选手常采用多种解题路径，以不同方式验证结果。

Regionally, Canadian participants dominated the top percentiles, but strong showings came from China, Singapore, South Korea, and the United Kingdom. The international cohort’s median rose to 46.8, exceeding the global median of 42, indicating that overseas candidates continue to raise the competitive standard.

从地区看，加拿大选手在高分段占多数，但中国、新加坡、韩国和英国亦表现强劲。国际选手的中位数升至46.8，超过全球中位数42，表明海外考生持续推高竞争水平。

Many top scorers attributed their success to disciplined practice with past Euclid papers and to developing flexible thinking rather than memorising formulas. The ability to break a complex problem into manageable steps was a common trait.

许多高分选手将成功归因于对往年欧几里得试卷的严格训练，以及培养灵活思维而非死记公式。将复杂问题分解为可处理的步骤是他们共有的能力特征。

4. Problem 1: Algebraic Manipulation and Roots | 第1题：代数变形与求根

Problem 1(a) asked: Find all real numbers x such that 2x² − 5x − 3 = 0.

第1(a)题要求：求所有实数 x，使得 2x² − 5x − 3 = 0。

This quadratic factorises nicely: 2x² − 5x − 3 = (2x + 1)(x − 3). Setting each factor to zero gives x = −½ or x = 3.

该二次式可分解：2x² − 5x − 3 = (2x + 1)(x − 3)。令每个因式为零，得 x = −½ 或 x = 3。

Problem 1(b) involved simplifying (√7 + √3)/(√7 − √3). Rationalising the denominator by multiplying numerator and denominator by (√7 + √3) yields (7 + 2√21 + 3)/(7 − 3) = (10 + 2√21)/4 = ½(5 + √21). Most candidates scored full marks here, though a surprising number forgot to simplify the final fraction.

问题1(b)涉及化简 (√7 + √3)/(√7 − √3)。分子分母同乘(√7 + √3) 有理化分母，得 (7 + 2√21 + 3)/(7 − 3) = (10 + 2√21)/4 = ½(5 + √21)。多数考生在此题得满分，但有不少人忘记化简最后的分式。

5. Problem 5: A Challenging Exponential-Logarithmic System | 第5题：棘手的指数-对数组和

The fifth problem presented a system: log₂(x − 1) + log₂(x + 1) = 3 and 3ˣ + 3⁻ˣ = 4. Candidates were asked to determine the real value(s) of x satisfying both equations simultaneously.

第五题给出方程组：log₂(x − 1) + log₂(x + 1) = 3 且 3ˣ + 3⁻ˣ = 4。考生需确定同时满足两方程的实数 x。

From the logarithmic equation, combine the logs: log₂[(x − 1)(x + 1)] = 3 ⇒ x² − 1 = 2³ = 8 ⇒ x² = 9 ⇒ x = 3 or x = −3. The domain restriction x > 1 leaves only x = 3.

由对数方程合并得 log₂[(x − 1)(x + 1)] = 3 ⇒ x² − 1 = 2³ = 8 ⇒ x² = 9 ⇒ x = 3 或 x = −3。定义域要求 x > 1，故仅剩 x = 3。

Substituting x = 3 into the exponential equation: 3³ + 3⁻³ = 27 + 1/27 = (729 + 1)/27 = 730/27, which is not equal to 4. Thus, the system has no real solution. This twist caught many off guard, as they stopped after solving the logarithm and assumed x = 3 works.

将 x = 3 代入指数方程：3³ + 3⁻³ = 27 + 1/27 = 730/27，不等于4。因此，方程组无实数解。这一转折让许多人措手不及，因为他们解完对数后便想当然地认为 x = 3 成立。

For the exponential part alone, set y = 3ˣ; then y + 1/y = 4 ⇒ y² − 4y + 1 = 0 ⇒ y = 2 ± √3. Hence 3ˣ = 2 ± √3, giving x = log₃(2 ± √3). Neither value satisfies the log domain constraint, confirming incompatibility.

单独考虑指数部分，设 y = 3ˣ，则 y + 1/y = 4 ⇒ y² − 4y + 1 = 0 ⇒ y = 2 ± √3。于是 3ˣ = 2 ± √3，得 x = log₃(2 ± √3)。这两个值均不满足对数定义域，再次证实不相容。

6. Problem 8: Geometry of an Inscribed Circle | 第8题：内切圆几何

Problem 8 featured a right triangle ABC with right angle at C, legs a = BC, b = AC, and hypotenuse c = AB. The incircle touches the hypotenuse at point D. Candidates had to prove that the distance from the right‑angle vertex C to D equals the incircle radius r, and that r = (a + b − c)/2.

第8题给定一直角三角形 ABC，直角在 C，直角边 a = BC, b = AC，斜边 c = AB。内切圆与斜边切于点 D。考生需证明直角顶点 C 到 D 的距离等于内切圆半径 r，且 r = (a + b − c)/2。

Using the standard tangency properties, the distances from C to the two tangency points on the legs are both r, so the tangents on the hypotenuse from B and A are respectively a − r and b − r. The sum of these two segments equals the hypotenuse: (a − r) + (b − r) = c ⇒ a + b − 2r = c ⇒ r = (a + b − c)/2. That establishes the second part.

运用标准切线性质，C 到两直角边上切点的距离均为 r，故斜边上从 B 和 A 出发的切线长分别为 a − r 与 b − r。这两段之和等于斜边：(a − r) + (b − r) = c ⇒ a + b − 2r = c ⇒ r = (a + b − c)/2。这就证明了第二部分。

For the distance CD, observe that CD is composed of the two tangent segments from C to the incircle along the legs, but D is on the hypotenuse. By constructing perpendiculars and using congruent right triangles, one can show CD = r. This elegant result surprised many, and careful diagrams earned full marks even if the proof was slightly incomplete.

对于距离 CD，注意 CD 由 C 到内切圆在直角边上的两段切线组成，但 D 位于斜边。通过作垂线并利用全等直角三角形，可证 CD = r。这一优雅结论令许多人惊讶，而清晰的图示即使证明稍有欠缺也能得满分。

7. Problem 10: Combinatorics and Modular Arithmetic | 第10题：组合与模运算

The final problem combined combinatorial counting with a modular twist: Let S be the set of all 5‑digit positive integers whose digits are all from {1,2,3,4,5,6} and no digit repeats. Find how many numbers in S are divisible by 3.

压轴题将组合计数与模运算融为一体：设 S 为所有每位数字均取自 {1,2,3,4,5,6} 且无重复的5位正整数构成的集合。求 S 中能被3整除的数有多少个。

The sum of all six digits is 1+2+3+4+5+6 = 21, which is divisible by 3. For any 5‑digit number formed by omitting exactly one digit, its digit sum is 21 minus the omitted digit. A number is divisible by 3 precisely when its digit sum is a multiple of 3.

六个数字的和为 1+2+3+4+5+6 = 21，可被3整除。对任意由省略一个数字所构成的5位数，其数字和为 21 减去所省略的数字。一个数可被3整除当其数字和为3的倍数。

Thus, we require 21 − d ≡ 0 (mod 3) ⇒ d ≡ 0 (mod 3). Among the digits {1,2,3,4,5,6}, only 3 and 6 are multiples of 3. So the omitted digit must be either 3 or 6.

故需要 21 − d ≡ 0 (mod 3) ⇒ d ≡ 0 (mod 3)。在数字集合 {1,2,3,4,5,6} 中，仅 3 和 6 是3的倍数。因此被省略的数字必须是 3 或 6。

If we omit the digit 3, the remaining five digits {1,2,4,5,6} can be arranged in 5! = 120 ways. Similarly, omitting 6 leaves {1,2,3,4,5}, also giving 120 permutations. Total count = 120 + 120 = 240. Many contestants incorrectly attempted to count directly using combinations of residues, but this simple exclusion method was far quicker.

若省略数字 3，剩下的 {1,2,4,5,6} 可有 5! = 120 种排列。类似地，省略 6 剩下 {1,2,3,4,5}，也产生 120 种排列。总数为 120 + 120 = 240。不少选手试图用剩余类组合直接计算，但这一简单的排除法快得多。

8. Common Mistakes and How to Avoid Them | 常见错误及避免策略

A recurring error in 2025 was neglect of domain restrictions — logarithms with arguments that become non‑positive, or square roots of negative quantities. In Problem 5, numerous candidates lost marks by failing to test the exponential condition, assuming compatibility after solving the logarithmic part.

2025年反复出现的错误是忽略定义域限制——对数真数非正，或对负数开平方根。在第五题中，许多考生因未验证指数条件而失分，解完对数部分即想当然地认为相容。

In the geometry problem, some confused the inradius formula with that of the circumcircle and incorrectly used r = c/2. Drawing a labelled diagram at the start and noting the right angle helps avoid such misapplications.

几何题中，有人将内切圆半径公式与外接圆混淆，错误使用 r = c/2。开始时绘制标注清晰的图示并注意直角，有助于避免此类误用。

Another pitfall was over‑complication: in Problem 10, lengthy casework on digit residues led to arithmetic mistakes, whereas a simple complementary counting approach solved it in minutes. Top performers often pause to look for a more elegant route before diving into calculations.

另一个陷阱是过度复杂化：在第十题中，冗长的数字剩余类分类导致计算错误，而简单的补集计数几分钟即可解出。顶尖选手常在埋头计算前先停下来寻找更优雅的路径。

9. Preparation Tips for Future Euclid Contests | 未来欧几里得竞赛备考建议

To excel in the Euclid, build a solid foundation in algebraic manipulation, geometric properties, and basic number theory. Regularly attempt past papers under timed conditions, paying special attention to Problems 7–10 which often introduce unfamiliar setups.

要在欧几里得竞赛中脱颖而出，需在代数变形、几何性质及基础数论方面打下扎实根基。定期计时练习历年真题，尤其注意第7–10题，因其常引入陌生的设定。

Develop a habit of verifying solutions, especially when multiple equations are involved. After finding a candidate, substitute it back into all original conditions to ensure consistency. Many marks are lost through insufficient checking.

培养验证解的习惯，尤其在涉及多个方程时。找到备选解后，应代回所有原始条件以确认一致性。检查不足常常导致失分。

Work on expressing your reasoning clearly. The Euclid rewards well‑structured logical steps, even if the final answer is wrong. Write legibly and label diagrams precisely. Finally, consider joining a problem‑solving club where you can discuss strategies with peers.

努力清晰地表达推理过程。欧几里得竞赛对步骤清晰有奖励，即便最终答案错误。书写工整，图示标注准确。最后，不妨加入解题社团，与同伴讨论策略。

10. Looking Ahead: The 2026 Euclid Cycle | 展望：2026年欧几里得周期

The 2025 results confirm that the Euclid Mathematics Contest remains a valuable benchmark for mathematical talent. Registration for the 2026 contest will open in November 2025. Whether you aim for a Certificate of Distinction or simply wish to deepen your problem‑solving skills, systematic preparation starting early yields the greatest benefit.

2025年的成绩印证了欧几里得数学竞赛作为数学才能宝贵基准的地位。2026年欧几里得竞赛的报名将于2025年11月开放。无论你志在优异证书还是只想深化解题技能，尽早开始系统备考都将带来最大收获。

Reflect on the errors and insights discussed in this analysis, and use the free resources on the CEMC website, including webinars, past papers, and solution videos, to guide your study.

思考本文分析的错误与洞见，并利用CEMC网站上的免费资源——包括网络研讨会、历年真题和讲解视频——来引导学习。

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2025 Euclid Mathematics Contest Results Released and Problem Analysis | 2025年欧几里得数学竞赛成绩公布及真题解析