2025A AMC 10 真题（P版）

1 答案 E

设 \(t\) 为从 \(1{:}30\) 起经过的小时数。Andy 向北走了 \(8t\) 英里。Betsy 晚一小时出发，所以她向东走了 \(12(t-1)\) 英里。要求两人的距离相等，即 \(8t = 12(t-1)\)。因此 \(4t = 12\)，\(t = 3\)。\(1{:}30\) 之后三小时是 \(4{:}30\)。因此正确答案是 E。

English

Let \(t\) be the hours since \(1{:}30.\) Andy has gone \(8t\) miles north. Betsy starts an hour later, so she’s gone \(12(t-1)\) miles east. We want these equal: \(8t = 12(t-1).\) That gives \(4t = 12,\) so \(t = 3.\) Three hours past \(1{:}30\) is \(4{:}30.\) Thus, E is the correct answer.

2 答案 B

原来的 \(10\) 磅混合物中有 \(5\) 磅花生和 \(2\) 磅腰果。设加入 \(x\) 磅第二种混合物，其中 \(20\%\) 是花生。新的花生比例要为 \(40\%\)，所以 \(\frac{5 + 0.2x}{10 + x} = 0.4\)。这给出 \(5 + 0.2x = 4 + 0.4x\)，\(x = 5\)。这 \(5\) 磅新混合物带来 \(0.4 \cdot 5 = 2\) 磅腰果，所以盒子里现在有 \(2 + 2 = 4\) 磅腰果。因此正确答案是 B。

English

The starting \(10\)-pound mix holds \(5\) pounds of peanuts and \(2\) pounds of cashews. Add \(x\) pounds of the second mix, which is \(20\%\) peanuts. We want the new peanut fraction to be \(40\%,\) so \(\frac{5 + 0.2x}{10 + x} = 0.4.\) This means \(5 + 0.2x = 4 + 0.4x,\) giving \(x = 5.\) Those \(5\) pounds bring \(0.4 \cdot 5 = 2\) more pounds of cashews, so the box now has \(2 + 2 = 4.\) Therefore, the answer is B.

3 答案 D

分成两种情况。若有两条边都等于 \(2025\)，第三边可以是从 \(1\) 到 \(2025\) 的任意整数，共有 \(2025\) 个三角形。现在假设 \(2025\) 是唯一的最长边。两条相等的腰长 \(s\) 必须满足三角不等式 \(2s \gt 2025\)，并且 \(s \le 2024\)。因此 \(s\) 从 \(1013\) 到 \(2024\)，共有 \(1012\) 个三角形。总数为 \(2025 + 1012 = 3037\)。因此正确答案是 D。

English

Split into two cases. Say two sides both equal \(2025.\) Then the third side can be any integer from \(1\) to \(2025,\) which is \(2025\) triangles. Now suppose \(2025\) is the unique longest side. The two equal legs \(s\) must satisfy \(2s \gt 2025\) by the triangle inequality, and \(s \le 2024.\) So \(s\) runs from \(1013\) to \(2024,\) giving \(1012\) triangles. Adding up, \(2025 + 1012 = 3037.\) Thus, D is the correct answer.

4 答案 A

设学生人数为 \(s\)。如果 Ash 加入学生队，他的年龄为 \(14(s+1) - 12s = 2s + 14\)。如果他加入老师队，老师有 \(15 - s\) 人，他的年龄为 \(52(16 - s)\) \(- 55(15 - s) = 3s + 7\)。两式描述的是同一个 Ash，所以 \(2s + 14 = 3s + 7\)。解得 \(s = 7\)，Ash 的年龄为 \(2 \cdot 7 + 14 = 28\)。因此正确答案是 A。

English

Let \(s\) be the number of students. If Ash joins them, his age is \(14(s+1) - 12s = 2s + 14.\) If he joins the teachers instead (there are \(15 - s\) of them), his age is \(52(16 - s)\) \(- 55(15 - s) = 3s + 7.\) Both describe the same Ash, so \(2s + 14 = 3s + 7.\) That gives \(s = 7,\) and Ash is \(2 \cdot 7 + 14 = 28.\) Therefore, the answer is A.

5 答案 E

按块分组。第 \(k\) 块为 \(k, k-1, \ldots, 2, 1, 2, \ldots, k-1, k\)，共有 \(2k - 1\) 项，并以 \(k\) 结束。累计到第 \(k\) 块末尾共用了 \(1 + 3 + \cdots + (2k-1) = k^2\) 项。注意 \(2025 = 45^2\)。这正好是第 \(45\) 块的末尾，而该块的最后一项为 \(45\)。因此正确答案是 E。

English

Group the sequence into blocks. Block \(k\) reads \(k, k-1, \ldots, 2, 1, 2, \ldots, k-1, k,\) which is \(2k - 1\) terms and ends on \(k.\) So after block \(k\) we’ve used \(1 + 3 + \cdots + (2k-1) = k^2\) terms. Notice \(2025 = 45^2.\) That’s exactly the end of block \(45,\) whose last term is \(45.\) Thus, E is the correct answer.

6 答案 C

设等边三角形为 \(ABC\)。每个 \(60^\circ\) 角被分成三个 \(20^\circ\) 部分。取从 \(A\) 和 \(B\) 出发的外侧三等分线：它们形成的三角形底角都是 \(\tfrac23 \cdot 60^\circ = 40^\circ\)，所以对应的六边形顶角为 \(180^\circ - 2\cdot 40^\circ = 100^\circ\)。从 \(A\) 和 \(B\) 出发的内侧三等分线形成底角为 \(20^\circ\) 的三角形，顶角为 \(180^\circ - 2\cdot 20^\circ = 140^\circ\)，由对顶角可得到相对的六边形角。因此六个角交替为 \(100^\circ\) 和 \(140^\circ\)。最小角为 \(100^\circ\)。因此正确答案是 C。

English

Label the equilateral triangle \(ABC.\) Each \(60^\circ\) angle splits into three \(20^\circ\) pieces. Take the outermost trisectors from \(A\) and \(B\): they meet at base angles \(\tfrac23 \cdot 60^\circ = 40^\circ,\) so the hexagon vertex there has angle \(180^\circ - 2\cdot 40^\circ = 100^\circ.\) The innermost trisectors from \(A\) and \(B\) meet at base angles \(20^\circ,\) giving apex \(180^\circ - 2\cdot 20^\circ = 140^\circ,\) and by vertical angles that’s the opposite hexagon angle. So the six angles alternate \(100^\circ\) and \(140^\circ.\) The smallest is \(100^\circ.\) Therefore, the answer is C.

7 答案 E

由余数定理，直接代入即可。我们有 \(p(1) = 1 + 1 + a + b = 4\)，所以 \(a + b = 2\)。又有 \(p(2) = 8 + 4 + 2a + b = 6\)，所以 \(2a + b = -6\)。第二式减去第一式，得 \(a = -8\)，从而 \(b = 10\)。因此 \(b - a = 10 - (-8) = 18\)，所以正确答案是 E。

English

By the Remainder Theorem, just plug in. We get \(p(1) = 1 + 1 + a + b = 4,\) so \(a + b = 2.\) And \(p(2) = 8 + 4 + 2a + b = 6,\) so \(2a + b = -6.\) Subtract the first from the second: \(a = -8,\) hence \(b = 10.\) Then \(b - a = 10 - (-8) = 18.\) Thus, E is the correct answer.

8 答案 B

给四个陈述编号：\((1)\) 至少一个真，\((2)\) 至少两个真，\((3)\) 至少两个假，\((4)\) 至少一个假。假设 \((3)\) 为真，则至少有两个陈述为假。但这时 \((1)\)、\((2)\) 和 \((4)\) 都会是真的，因此最多只有一个假陈述，矛盾。所以 \((3)\) 必须为假。现在 \((1)\)、\((2)\) 和 \((4)\) 都为真，而且这与只有一个假陈述完全一致。所以恰好有 \(1\) 个陈述是假的。因此正确答案是 B。

English

Number them: \((1)\) at least one true, \((2)\) at least two true, \((3)\) at least two false, \((4)\) at least one false. Suppose \((3)\) is true. Then at least two statements are false. But then \((1),\) \((2),\) and \((4)\) all read as true, which leaves at most one false statement. That’s a contradiction, so \((3)\) must be false. Now \((1),\) \((2),\) and \((4)\) are all true, and each matches reality with just one false statement. So exactly \(1\) statement is false. Therefore, the answer is B.

9 答案 C

图像经过 \((1,25)\)，当且仅当 \(f(1 - a) = 25\)。令 \(t = 1 - a\)，于是只需要数 \(f(t) = 25\) 的解的个数。分解得 \(f(x) = 100x(x-1)(x-2)\)，根为 \(0, 1, 2\)。函数在 \((0,1)\) 上为正，且 \(f(0.5) = 37.5 \gt 25\)，所以由连续性可知 \(0.5\) 两侧各有一个解。函数在 \((1,2)\) 上为负，而当 \(x \gt 2\) 时从 \(0\) 单调增加到无穷大，因此还有一个解。三次方程最多有 \(3\) 个实根，所以这些就是全部的解。每个解给出唯一的 \(a\)，因此共有 \(3\) 个值。正确答案是 C。

English

The graph passes through \((1,25)\) exactly when \(f(1 - a) = 25.\) Let \(t = 1 - a,\) so we count solutions of \(f(t) = 25.\) Factor \(f(x) = 100x(x-1)(x-2),\) with roots \(0, 1, 2.\) On \((0,1),\) the function is positive and \(f(0.5) = 37.5 \gt 25,\) so continuity gives one root on each side of \(0.5.\) On \((1,2)\) the function is negative, while for \(x \gt 2\) it increases from \(0\) to infinity, giving one more root. A cubic equation has at most \(3\) real roots, so these are all the solutions. Each gives one \(a,\) so there are \(3\) values. Thus, C is the correct answer.

10 答案 C

设 \(O\) 是 \(AB\) 上的圆心，\(P\) 是弦 \(CD\) 的中点。令 \(r = OP\) 为小半圆半径，\(R = OD\) 为大半圆半径。因为 \(PD = 8\)，在直角三角形 \(OPD\) 中由勾股定理得 \(R^2 - r^2 = 64\)。阴影面积等于大半圆面积减去小半圆面积：\(\tfrac12\pi R^2 - \tfrac12\pi r^2\) \(= \tfrac12\pi(R^2 - r^2)\) \(= 32\pi\)。因此正确答案是 C。

English

Let \(O\) be the center on \(AB\) and \(P\) the midpoint of chord \(CD.\) Set \(r = OP\) for the small radius and \(R = OD\) for the large one. Since \(PD = 8,\) the Pythagorean theorem in triangle \(OPD\) gives \(R^2 - r^2 = 64.\) The shaded area is the big semicircle minus the small one: \(\tfrac12\pi R^2 - \tfrac12\pi r^2\) \(= \tfrac12\pi(R^2 - r^2)\) \(= 32\pi.\) Therefore, the answer is C.

11 答案 E

从等差数列可知 \(z = 1 + 3d\)，所以 \(z \equiv 1 \pmod 3\)。从等比数列可知 \(z = p^3\)，其中整数公比 \(p \ge 2\)。为了让 \(z\) 最小，测试 \(p = 2, 3, 4\)。只有 \(p = 4\) 可行，因为 \(p^3 = 64 \equiv 1 \pmod 3\)。于是 \(d = 21\)，两个序列为 \(1, 22, 43, 64\) 和 \(1, 4, 16, 64\)。因此 \(x + y + z + p + q\) \(= 22 + 43 + 64 + 4 + 16\) \(= 149\)。所以正确答案是 E。

English

From the arithmetic sequence, \(z = 1 + 3d,\) so \(z \equiv 1 \pmod 3.\) From the geometric one, \(z = p^3\) for some integer ratio \(p \ge 2.\) We want the smallest such \(z,\) so test \(p = 2, 3, 4.\) Only \(p = 4\) works, since \(p^3 = 64 \equiv 1 \pmod 3.\) That forces \(d = 21,\) and the sequences are \(1, 22, 43, 64\) and \(1, 4, 16, 64.\) So \(x + y + z + p + q\) \(= 22 + 43 + 64 + 4 + 16\) \(= 149.\) Thus, E is the correct answer.

12 答案 D

没有数字是 \(0\)，所以可用数字为 \(1\) 到 \(9\)。先固定唯一的偶数数字在第一个位置，最后再乘以 \(4\) 来选择它的位置。按这个偶数数字分类。若它是质数 \(2\)，则三个奇数数字都必须不是质数，所以每个只能是 \(1\) 或 \(9\)，共有 \(2^3 = 8\) 种。否则偶数数字是 \(4, 6\) 或 \(8\)，有 \(3\) 种；三个奇数位置中恰好一个是质数，可选 \(3, 5\) 或 \(7\)，有 \(3\) 种，并可放在 \(3\) 个奇数位置之一；另外两个奇数来自 \(\{1, 9\}\)，有 \(2^2\) 种。这部分共有 \(3 \cdot 3 \cdot 3 \cdot 4 = 108\) 种。总数为 \(4(8 + 108) = 464\)。因此正确答案是 D。

English

No digit is \(0,\) so digits run from \(1\) to \(9.\) Put the single even digit in the first slot for now and multiply by \(4\) at the end to place it. Split on that even digit. If it’s the prime \(2,\) then the three odd digits all have to be non-prime, so each is \(1\) or \(9,\) giving \(2^3 = 8\) ways. Otherwise the even digit is \(4, 6,\) or \(8\) (\(3\) choices), and exactly one of the odd digits is prime, worth \(3, 5,\) or \(7\) (\(3\) choices) in one of the \(3\) odd positions, while the other two odds come from \(\{1, 9\}\) (\(2^2\) ways). That’s \(3 \cdot 3 \cdot 3 \cdot 4 = 108.\) Altogether, \(4(8 + 108) = 464.\) Therefore, the answer is D.

13 答案 D

设外层正方形边长为 \(1\)。各正方形边长依次为 \(1, k, k^2, \ldots\)，而阴影环带交替出现，所以阴影面积为 \(1 - k^2 + k^4 - k^6 + \cdots = \frac{1}{1 + k^2}\)。题目给出这等于 \(64\% = \frac{16}{25}\)，所以 \(1 + k^2 = \frac{25}{16}\)。因此 \(k^2 = \frac{9}{16}\)，得到 \(k = \frac{3}{4}\)。所以正确答案是 D。

English

Let the outer side be \(1.\) The squares have sides \(1, k, k^2, \ldots,\) and the shaded rings alternate, so the shaded area is \(1 - k^2 + k^4 - k^6 + \cdots = \frac{1}{1 + k^2}.\) We’re told this equals \(64\% = \frac{16}{25},\) so \(1 + k^2 = \frac{25}{16}.\) Then \(k^2 = \frac{9}{16},\) giving \(k = \frac{3}{4}.\) Thus, D is the correct answer.

14 答案 B

先让第一名学生坐任意位置。第二名学生坐到相邻椅子的概率为 \(\tfrac{2}{5}\)，因为剩下五把椅子中有两把与第一名学生相邻。接着老师占据剩下四把椅子中的两把。在 \(\binom{4}{2} = 6\) 种选法中，恰好有 \(3\) 对是相邻椅子。因此所求概率为 \(\tfrac{2}{5} \cdot \tfrac{3}{6} = \tfrac{1}{5}\)。所以正确答案是 B。

English

Seat the first student anywhere. The second student lands next to them with probability \(\tfrac{2}{5},\) since two of the remaining five chairs are adjacent. Now the teachers fill two of the four leftover chairs. Of the \(\binom{4}{2} = 6\) ways to do that, exactly \(3\) are adjacent pairs. So the probability is \(\tfrac{2}{5} \cdot \tfrac{3}{6} = \tfrac{1}{5}.\) Therefore, the answer is B.

15 答案 A

设 \(x = BC\)。因为 \(ABEF\) 是矩形，\(AB = 1\)、\(AF = 7\)，且 \(AD = 5\)，所以 \(AC = \sqrt{1 + x^2}\)，\(CE = 7 - x\)，\(CD = 5 - \sqrt{1 + x^2}\)。三角形 \(\triangle ABC\) 和 \(\triangle EDC\) 相似，因此 \(\frac{7 - x}{\sqrt{1 + x^2}} = \frac{5 - \sqrt{1 + x^2}}{x}\)。清除分母并平方，可得 \(24x^2 + 14x - 24 = 0\)，它分解为 \((4x - 3)(3x + 4) = 0\)。正根为 \(x = \tfrac{3}{4}\)。面积为 \(\tfrac12 \cdot \tfrac34 \cdot 1 = \tfrac{3}{8}\)。因此正确答案是 A。

English

Let \(x = BC.\) Since \(ABEF\) is a rectangle with \(AB = 1\) and \(AF = 7,\) and \(AD = 5,\) we get \(AC = \sqrt{1 + x^2},\) \(CE = 7 - x,\) and \(CD = 5 - \sqrt{1 + x^2}.\) The triangles \(\triangle ABC\) and \(\triangle EDC\) are similar, so \(\frac{7 - x}{\sqrt{1 + x^2}} = \frac{5 - \sqrt{1 + x^2}}{x}.\) Clear denominators and square to get \(24x^2 + 14x - 24 = 0,\) which factors as \((4x - 3)(3x + 4) = 0.\) The positive root is \(x = \tfrac{3}{4}.\) So the area is \(\tfrac12 \cdot \tfrac34 \cdot 1 = \tfrac{3}{8}.\) Thus, A is the correct answer.

16 答案 D

共有 \(3^3 = 27\) 种等可能的放法。其中 \(3\) 种把所有硬币放在同一个罐子中，此时最大值为 \(3\)；\(6\) 种把三个罐子各放一枚硬币，此时最大值为 \(1\)。其余 \(18\) 种为 \(2\)-\(1\) 分布，此时最大值为 \(2\)。所以最大硬币数的期望为 \(\frac{3 \cdot 3 + 2 \cdot 18 + 1 \cdot 6}{27} = \frac{51}{27} = \frac{17}{9}\)。因此正确答案是 D。

English

There are \(3^3 = 27\) equally likely placements. Of these, \(3\) pile all coins into one jar (max \(3\)), and \(6\) put one coin in each jar (max \(1\)). The other \(18\) split \(2\)-\(1\) (max \(2\)). So the expected maximum is \(\frac{3 \cdot 3 + 2 \cdot 18 + 1 \cdot 6}{27} = \frac{51}{27} = \frac{17}{9}.\) Therefore, the answer is D.

17 答案 E

减去余数。\(273420 = 273436 - 16\) 和 \(272745 = 272760 - 15\) 都是 \(N\) 的倍数，所以它们的差 \(675\) 也是倍数。又有 \(272745 = 404 \cdot 675 + 45\)，因此 \(45\) 也是 \(N\) 的倍数。余数 \(16\) 意味着 \(N \gt 16\)，而 \(45\) 的大于 \(16\) 的因数只有 \(45\) 本身。所以 \(N = 45\)，十位数字是 \(4\)。因此正确答案是 E。

English

Subtract off the remainders. Both \(273420 = 273436 - 16\) and \(272745 = 272760 - 15\) are multiples of \(N,\) so their difference \(675\) is too. Now \(272745 = 404 \cdot 675 + 45,\) which makes \(45\) a multiple of \(N\) as well. The remainder \(16\) means \(N \gt 16,\) and the only divisor of \(45\) bigger than \(16\) is \(45\) itself. So \(N = 45,\) and its tens digit is \(4.\) Thus, E is the correct answer.

18 答案 B

看其中一个因式 \(kx^2 - 4x - 3\)。它的判别式 \(16 + 12k\) 为正，所以它有两个不同的实根。来自不同因式的根也互不相同：若 \(x\) 是下标为 \(k_1\) 和 \(k_2\) 的两个因式的公共根，就会有 \((k_1-k_2)x^2=0\)，但 \(x\ne0\)。由韦达定理，同一个因式的两个根的倒数之和为 \(\frac{x_1 + x_2}{x_1 x_2} = \frac{\frac{4}{k}}{-\frac{3}{k}} = -\tfrac{4}{3}\)。对全部 \(2025\) 个因式求和，这些倒数的总和为 \(2025 \cdot \left(-\tfrac{4}{3}\right) = -2700\)。实根共有 \(4050\) 个，所以调和平均数为 \(\frac{4050}{-2700} = -\tfrac{3}{2}\)。因此正确答案是 B。

English

Look at one factor \(kx^2 - 4x - 3.\) Its discriminant \(16 + 12k\) is positive, so it has two distinct real roots. Roots from different factors are also distinct: a common root \(x\) for indices \(k_1\) and \(k_2\) would satisfy \((k_1-k_2)x^2=0,\) but \(x\ne0.\) By Vieta, the two reciprocals from one factor sum to \(\frac{x_1 + x_2}{x_1 x_2} = \frac{\frac{4}{k}}{-\frac{3}{k}} = -\tfrac{4}{3}.\) Summing over all \(2025\) factors, the reciprocals total \(2025 \cdot \left(-\tfrac{4}{3}\right) = -2700.\) There are \(4050\) real roots in all, so the harmonic mean is \(\frac{4050}{-2700} = -\tfrac{3}{2}.\) Therefore, the answer is B.

19 答案 A

每个内部项会贡献给下一行的两个项，而端点 \(-1\) 和 \(1\) 在总和中相互抵消。因此每一行的总和都是上一行的两倍。顶行总和为 \(3\)，且 \(12{,}288 = 3 \cdot 2^{12}\)，所以这是第 \(12\) 行（顶行算第 \(0\) 行）。从左侧追踪对角线。第二条对角线为 \(3, 2, 1, 0, -1, \ldots\)，每行减少 \(1\)。第三条对角线是这些数的累加：当 \(n \ge 4\) 时，它等于 \(7 - (0 + 1 + \cdots + (n-4))\) \(= 7 - \frac{(n-4)(n-3)}{2}\)。代入 \(n = 12\)，得到 \(7 - \frac{8 \cdot 9}{2} = -29\)。因此正确答案是 A。

English

Each interior entry feeds two entries below, and the end values \(-1\) and \(1\) cancel in the sum. So every row’s total doubles the one above. The top row sums to \(3,\) and \(12{,}288 = 3 \cdot 2^{12},\) so this is the \(12\)th row (counting the top as row \(0\)). Track the diagonals from the left. The second diagonal is \(3, 2, 1, 0, -1, \ldots,\) dropping by \(1\) each row. The third diagonal adds these up: for \(n \ge 4\) it equals \(7 - (0 + 1 + \cdots + (n-4))\) \(= 7 - \frac{(n-4)(n-3)}{2}.\) Plug in \(n = 12\): \(7 - \frac{8 \cdot 9}{2} = -29.\) Thus, A is the correct answer.

20 答案 A

将筒仓中心放在原点，半径为 \(10\)。则 MacDonald 在 \(D = (-20, -15)\)，McGregor 在 \(G = (20, -g)\)。从 \(D\) 到切点的切线长为 \(DT = \sqrt{DS^2 - 10^2}\) \(= \sqrt{25^2 - 100}\) \(= \sqrt{525}\)，从 \(G\) 到切点的切线长为 \(TG = \sqrt{g^2 + 20^2 - 10^2}\)。切点 \(T\) 在两人之间，所以 \(DG = DT + TG\)。另一方面 \(DG = \sqrt{40^2 + (15 - g)^2}\)。两次平方并化简，得到 \(3g^2 + 150g - 925 = 0\)。它的正根满足原来的切线长方程，即 \(g = \frac{20\sqrt{21} - 75}{3}\)。所以 \(a + b + c + d\) \(= 20 + 21 + 75 + 3\) \(= 119\)。因此正确答案是 A。

English

Put the silo’s center at the origin with radius \(10.\) Then MacDonald is at \(D = (-20, -15)\) and McGregor at \(G = (20, -g).\) The tangent length from \(D\) is \(DT = \sqrt{DS^2 - 10^2}\) \(= \sqrt{25^2 - 100}\) \(= \sqrt{525},\) and from \(G\) it is \(TG = \sqrt{g^2 + 20^2 - 10^2}.\) The tangent point \(T\) sits between the two men, so \(DG = DT + TG.\) But also \(DG = \sqrt{40^2 + (15 - g)^2}.\) Squaring twice and simplifying gives \(3g^2 + 150g - 925 = 0.\) Its positive root, which satisfies the original tangent-length equation, is \(g = \frac{20\sqrt{21} - 75}{3}.\) Therefore \(a + b + c + d\) \(= 20 + 21 + 75 + 3\) \(= 119.\) Thus, A is the correct answer.

21 答案 C

我们可以达到 \(10\)。所有奇数组成的集合 \(\{1, 3, 5, \ldots, 19\}\) 是无和集，因为两个奇数之和为偶数。\(\{11, 12, \ldots, 20\}\) 也是无和集，因为其中任意两个数之和都超过 \(20\)。它们各有 \(10\) 个元素。现在设 \(m\) 是任意一个无和子集的最大元素。对于 \(1\le i<\frac{m}{2}\)，\(\{i,m-i\}\) 中最多只能选一个，因为这两个数之和为 \(m\)。若 \(m\) 是偶数，则 \(\frac{m}{2}\) 也不能选，因为它可以被用两次，而 \(\frac{m}{2}+\frac{m}{2}=m\)。于是除 \(m\) 之外最多还能选 \(\lfloor\frac{m-1}{2}\rfloor\) 个元素，总数最多为 \(\lfloor\frac{m-1}{2}\rfloor+1\le10\)。因此最大可能的元素个数是 \(10\)。所以正确答案是 C。

English

We can reach \(10.\) The odds \(\{1, 3, 5, \ldots, 19\}\) are sum-free, since two odds sum to an even. So is \(\{11, 12, \ldots, 20\},\) since any two of those sum past \(20.\) Each has \(10\) elements. Now let \(m\) be the largest element of any sum-free subset. For \(1\le i<\frac{m}{2},\) at most one member of \(\{i,m-i\}\) can be chosen, because the two sum to \(m.\) If \(m\) is even, \(\frac{m}{2}\) cannot be chosen either, since it can be used twice and \(\frac{m}{2}+\frac{m}{2}=m.\) Thus besides \(m\) there are at most \(\lfloor\frac{m-1}{2}\rfloor\) chosen elements, for a total of at most \(\lfloor\frac{m-1}{2}\rfloor+1\le10.\) Therefore the greatest possible size is \(10.\) Thus, C is the correct answer.

22 答案 B

三个外圆圆心 \(A, B, C\) 的两两距离分别为 \(AB = 1 + 2 = 3\)，\(AC = 1 + 3 = 4\)，\(BC = 2 + 3 = 5\)，构成 \(3\)-\(4\)-\(5\) 直角三角形。现在对四个相互相切的圆应用笛卡尔圆定理，曲率分别为 \(1, \tfrac12, \tfrac13\)，以及 \(\tfrac1r\)：\(\frac1r = 1 + \tfrac12 + \tfrac13\) \(+ 2\sqrt{\tfrac12 + \tfrac16 + \tfrac13}\) \(= \tfrac{11}{6} + 2\sqrt{1}\) \(= \tfrac{23}{6}\)。取倒数得 \(r = \tfrac{6}{23}\)。因此正确答案是 B。

English

The three outer centers \(A, B, C\) are pairwise \(AB = 1 + 2 = 3,\) \(AC = 1 + 3 = 4,\) and \(BC = 2 + 3 = 5\) apart, a \(3\)-\(4\)-\(5\) right triangle. Now apply Descartes’ Circle Theorem with curvatures \(1, \tfrac12, \tfrac13,\) and \(\tfrac1r,\) all mutually tangent: \(\frac1r = 1 + \tfrac12 + \tfrac13\) \(+ 2\sqrt{\tfrac12 + \tfrac16 + \tfrac13}\) \(= \tfrac{11}{6} + 2\sqrt{1}\) \(= \tfrac{23}{6}.\) Inverting, \(r = \tfrac{6}{23}.\) Therefore, the answer is B.

23 答案 D

设 \(\angle B\) 的角平分线交 \(AC\) 于 \(D\)。由角平分线定理，\(\frac{AD}{DC} = \frac{AB}{BC} = \frac{80}{45}\)，又因为 \(AC = 75\)，得到 \(AD = 48\)、\(CD = 27\)。三角形 \(BCD\) 与 \(ACB\) 共有 \(\angle C\)，夹角两边的比例都为 \(\tfrac{45}{75}=\tfrac{27}{45}=\tfrac35\)，所以由边角边相似。于是 \(BD=\tfrac35\cdot80=48=AD\)，故 \(\triangle ADB\) 是等腰三角形。设 \(\angle DAB=\angle DBA=\theta\)。因为过 \(C\) 的高垂直于 \(AB\)，所以 \(\angle DPC\) 和 \(\angle DCP\) 都等于 \(90^\circ-\theta\)。因此 \(\triangle CDP\) 是等腰三角形，故 \(PD=CD=27\)。由于 \(P\) 位于 \(BD\) 上，所以 \(BP=BD-PD\) \(=48-27=21\)。所以正确答案是 D。

English

Let the bisector of \(\angle B\) hit \(AC\) at \(D.\) By the Angle Bisector Theorem, \(\frac{AD}{DC} = \frac{AB}{BC} = \frac{80}{45},\) and since \(AC = 75,\) we get \(AD = 48\) and \(CD = 27.\) Triangles \(BCD\) and \(ACB\) share \(\angle C,\) with adjacent sides in the common ratio \(\tfrac{45}{75}=\tfrac{27}{45}=\tfrac35,\) so they are similar by SAS. Hence \(BD=\tfrac35\cdot80=48=AD,\) making \(\triangle ADB\) isosceles. Put \(\angle DAB=\angle DBA=\theta.\) Because the altitude through \(C\) is perpendicular to \(AB,\) both \(\angle DPC\) and \(\angle DCP\) equal \(90^\circ-\theta.\) Thus \(\triangle CDP\) is isosceles, so \(PD=CD=27.\) Since \(P\) lies on \(BD,\) we get \(BP=BD-PD\) \(=48-27=21.\) Thus, D is the correct answer.

24 答案 C

在一个公平数中，数字必须先上升到最大数字 \(m\)，再下降。事实上，任何一次下降之后的第一次上升，都会起始于一个比左右两个邻居都小的数字。对于 \(k\) 位数，从 \(1\) 到 \(9\) 中选择数字集合，有 \(\binom{9}{k}\) 种。比 \(m\) 小的那 \(k-1\) 个数字各自独立地放在递增的左侧或递减的右侧，之后它们的位置就被唯一确定。因此总数为 \(\sum_{k=1}^{9}\binom{9}{k}2^{k-1}\) \(= \tfrac12\big((1+2)^9-1\big)\) \(= \tfrac{3^9-1}{2}\) \(=9841\)。因此正确答案是 C。

English

A fair number’s digits must increase up to its largest digit \(m\) and then decrease. Indeed, the first ascent after any descent would begin at a digit smaller than both of its neighbors. For \(k\) digits, choose the digit set from \(1\) to \(9\) in \(\binom{9}{k}\) ways. Each of the \(k-1\) digits below \(m\) independently goes on the increasing left side or the decreasing right side, after which its position is forced. Hence the total is \(\sum_{k=1}^{9}\binom{9}{k}2^{k-1}\) \(= \tfrac12\big((1+2)^9-1\big)\) \(= \tfrac{3^9-1}{2}\) \(=9841.\) Therefore, the answer is C.

25 答案 A

将 \(A = (0,0)\)、\(B = (1,0)\) 放在单位正方形中。\(AP\) 是中间长度，当且仅当 \(BP \lt AP \lt AB\) 或 \(AB \lt AP \lt BP\)。这两个条件给出的区域由以 \(A\) 为圆心、半径为 \(1\) 的圆（其中 \(AP = AB\)）和垂直平分线 \(x = \tfrac12\)（其中 \(AP = BP\)）围出。设 \(S\) 是该圆与 \(x = \tfrac12\) 的交点，则 \(\triangle ABS\) 为等边三角形，所以相关扇形角为 \(\angle BAS = 60^\circ\)。计算这些区域面积，较大区域为 \(\frac{4\pi - 3\sqrt3}{24}\)，较小区域为 \(\frac{12 - 2\pi - 3\sqrt3}{24}\)。相加得概率 \(\frac{6 + \pi - 3\sqrt3}{12}\)。因此 \(a + b + c + d + e\) \(= 6 + 1 + 3 + 3 + 12\) \(= 25\)，所以正确答案是 A。

English

Place \(A = (0,0)\) and \(B = (1,0)\) on the unit square. \(AP\) is the middle length when \(BP \lt AP \lt AB\) or \(AB \lt AP \lt BP.\) These regions are bounded by the circle centered at \(A\) with radius \(1\) (where \(AP = AB\)) and the line \(x = \tfrac12\) (where \(AP = BP\)). Let \(S\) be where the circle meets \(x = \tfrac12.\) Then \(\triangle ABS\) is equilateral, so \(\angle BAS = 60^\circ.\) The larger region has area \(\frac{4\pi - 3\sqrt3}{24}\) and the smaller has area \(\frac{12 - 2\pi - 3\sqrt3}{24}.\) They add to \(\frac{6 + \pi - 3\sqrt3}{12}.\) Thus \(a + b + c + d + e\) \(= 6 + 1 + 3 + 3 + 12\) \(= 25.\) Thus, A is the correct answer.