2025B AMC 10 真题（P版）

1 答案 B

每杯需要 \(20\) 克，因此 \(\frac{350}{20} = 17.5\)。半杯不能算作一杯，所以向下取整，最多能冲泡 \(17\) 大杯，正确答案是 B。

English

Each mug needs \(20\) grams, so we divide: \(\frac{350}{20} = 17.5.\) A half mug isn’t a mug, so round down. That leaves \(17.\) Thus, B is the correct answer.

2 答案 D

\(n^2\) 的个位数字只取决于 \(n\) 的个位数字，因此序列每 \(10\) 项循环：\(1, 4, 9, 6, 5, 6, 9, 4, 1, 0\)。一个循环的和为 \(45\)。又 \(2025 = 202 \cdot 10 + 5\)，所以共有 \(202\) 个完整循环，再加上前五项，总和为 \(202 \cdot 45\) \(+ (1 + 4 + 9 + 6 + 5)\) \(= 9090\) \(+ 25 = 9115\)。因此正确答案是 D。

English

The ones digit of \(n^2\) depends only on the ones digit of \(n,\) so the list repeats every \(10\) terms: \(1, 4, 9, 6, 5, 6, 9, 4, 1, 0.\) One block sums to \(45.\) Now \(2025 = 202 \cdot 10 + 5,\) so we get \(202\) full blocks plus the first five terms: \(202 \cdot 45\) \(+ (1 + 4 + 9 + 6 + 5)\) \(= 9090\) \(+ 25 = 9115.\) Therefore, the answer is D.

3 答案 D

设 \(S_n\) 为第 \(n\) 行的和。第 \(n-1\) 行的每个数都会贡献到下一行的两个位置，而固定的边界数 \(10\) 和 \(1\) 正好补上边缘缺少的项。因此 \(S_n = 2 S_{n-1}\)（当 \(n \ge 3\)）。由 \(S_2 = 11\)，得到 \(S_n = 11 \cdot 2^{n-2}\)，所以 \(S_{11} = 11 \cdot 2^9 = 5632\)。其各位数字之和为 \(5 + 6 + 3 + 2 = 16\)。因此正确答案是 D。

English

Let \(S_n\) be the sum of row \(n.\) Each entry in row \(n-1\) feeds the two entries just below it, and the fixed border numbers \(10\) and \(1\) exactly make up for the terms lost at the edges. So \(S_n = 2 S_{n-1}\) for \(n \ge 3.\) With \(S_2 = 11,\) this gives \(S_n = 11 \cdot 2^{n-2},\) so \(S_{11} = 11 \cdot 2^9 = 5632.\) Its digits sum to \(5 + 6 + 3 + 2 = 16.\) Thus, D is the correct answer.

4 答案 A

按位值，七进制的 \(\underline{a}\,\underline{b}\) 是 \(7a + b\)，九进制的 \(\underline{b}\,\underline{a}\) 是 \(9b + a\)。令两者相等：\(7a + b = 9b + a\)，所以 \(6a = 8b\)，即 \(3a = 4b\)。因此 \(a = 4t\)，且 \(b = 3t\)。因为它们是两位数，\(t \ne 0\)；又因为 \(a\) 是七进制数字，所以 \(t \le 1\)。因此 \(a = 4\)、\(b = 3\)，所以 \(a + b = 7\)。所以答案是 A。

English

By place value, \(\underline{a}\,\underline{b}\) in base seven is \(7a + b,\) and \(\underline{b}\,\underline{a}\) in base nine is \(9b + a.\) Set them equal: \(7a + b = 9b + a,\) so \(6a = 8b,\) that is \(3a = 4b.\) Thus \(a = 4t\) and \(b = 3t.\) Because these are two-digit numerals, \(t \ne 0;\) because \(a\) is a base-seven digit, \(t \le 1.\) Hence \(a = 4\) and \(b = 3,\) so \(a + b = 7.\) Therefore, the answer is A.

5 答案 C

因为 \(O\) 是外心，\(OA = OB = OC\)。圆周角 \(\angle B = 130^\circ\) 对应弧 \(AC\)，且由于 \(B\) 是钝角，对应的小圆心角为 \(\angle AOC = 360^\circ\) \(- 2 \cdot 130^\circ = 100^\circ\)。三角形 \(OAC\) 是等腰三角形，所以 \(\angle CAO = \tfrac{180^\circ - 100^\circ}{2} = 40^\circ\)。边长 \(AB\) 和 \(AC\) 实际上不需要用到。因此正确答案是 C。

English

Since \(O\) is the circumcenter, \(OA = OB = OC.\) The inscribed angle \(\angle B = 130^\circ\) subtends arc \(AC,\) and because \(B\) is obtuse, the central angle is \(\angle AOC = 360^\circ\) \(- 2 \cdot 130^\circ = 100^\circ.\) Triangle \(OAC\) is isosceles, so \(\angle CAO = \tfrac{180^\circ - 100^\circ}{2} = 40^\circ.\) (The lengths \(AB\) and \(AC\) never enter.) Thus, C is the correct answer.

6 答案 C

下方区域的面积为 \(\int_0^2\left(\tfrac{x}{3} + 1\right)dx = \tfrac{2}{3} + 2 = \tfrac{8}{3}\)。\(0 \le x \le a\) 的部分是面积为 \(a + \tfrac{a^2}{6}\) 的梯形。它应等于总面积的一半，即 \(\tfrac{4}{3}\)，所以 \(a^2 + 6a - 8 = 0\)，从而 \(a = -3 + \sqrt{17}\)。因此 \(s = 17\)，\(t = 3\)，\(s + t = 20\)。因此正确答案是 C。

English

The lower region has area \(\int_0^2\left(\tfrac{x}{3} + 1\right)dx = \tfrac{2}{3} + 2 = \tfrac{8}{3}.\) The slice with \(0 \le x \le a\) is a trapezoid of area \(a + \tfrac{a^2}{6}.\) We want that to be half the total, namely \(\tfrac{4}{3},\) so \(a^2 + 6a - 8 = 0\) and \(a = -3 + \sqrt{17}.\) Then \(s = 17,\) \(t = 3,\) and \(s + t = 20.\) Therefore, the answer is C.

7 答案 C

把栅栏放在 \(x\)-轴上，上锁的门在原点，Frances 在 \((0, -15)\)。则盒子在 \((x, 0)\)，东门在 \((x + 9, 0)\)，西门在 \((-8, 0)\)。东侧路线为 \(\sqrt{(x + 9)^2 + 15^2} + 9\)；西侧路线为 \(\sqrt{8^2 + 15^2}\) \(+ (x + 8) = 17 + x + 8\)。令二者相等，得 \(\sqrt{(x + 9)^2 + 225} = x + 16\)。平方并化简：\(18x + 306 = 32x + 256\)，所以 \(x = \tfrac{50}{14} = \tfrac{25}{7} = 3\tfrac{4}{7}\)。因此正确答案是 C。

English

Put the fence on the \(x\)-axis, the locked gate at the origin, and Frances at \((0, -15).\) Then the box is at \((x, 0),\) the east gate at \((x + 9, 0),\) and the west gate at \((-8, 0).\) The east route is \(\sqrt{(x + 9)^2 + 15^2} + 9;\) the west route is \(\sqrt{8^2 + 15^2}\) \(+ (x + 8) = 17 + x + 8.\) Set them equal: \(\sqrt{(x + 9)^2 + 225} = x + 16.\) Square and simplify to get \(18x + 306 = 32x + 256,\) so \(x = \tfrac{50}{14} = \tfrac{25}{7} = 3\tfrac{4}{7}.\) Thus, C is the correct answer.

8 答案 C

以美分为单位，总价是 \(10000A + 1000B + 100B\) \(+ 10B + A\) \(= 10001A + 1110B\)。它要平均分给 \(36\) 件运动衫，所以能被 \(36\) 整除。因为 \(10001 \equiv 29\) 且 \(1110 \equiv 30 \pmod{36}\)，所以需要 \(29A + 30B \equiv 0\)，也就是 \(7A + 6B \equiv 0 \pmod{36}\)。在 \(A \ne 0\) 的数字解中，唯一可行的是 \(A = 6, B = 5\)，因为 \(7 \cdot 6 + 6 \cdot 5 = 72\)。这对应 \(\$655.56\)，也就是每件 \(\$18.21\)，所以 \(A + B = 11\)。因此正确答案是 C。

English

In cents the total is \(10000A + 1000B + 100B\) \(+ 10B + A\) \(= 10001A + 1110B.\) Split evenly among \(36\) shirts, so it’s divisible by \(36.\) Now \(10001 \equiv 29\) and \(1110 \equiv 30 \pmod{36},\) so we need \(29A + 30B \equiv 0,\) which reduces to \(7A + 6B \equiv 0 \pmod{36}.\) The only digit solution with \(A \ne 0\) is \(A = 6, B = 5,\) since \(7 \cdot 6 + 6 \cdot 5 = 72.\) That’s \(\$655.56,\) or \(\$18.21\) a shirt, so \(A + B = 11.\) Therefore, the answer is C.

9 答案 C

令 \(p = -x + y + z\)，\(q = x - y + z\)，\(r = x + y - z\)。后三个不等式给出 \(p, q, r \le 2\)，第一个不等式给出 \(x + y + z \ge 2\)，并且 \(p + q + r = x + y + z\)。又因为 \(x = \tfrac{q + r}{2}\) 等，所以 \(p, q, r\) 必须同奇偶。现在数每项 \(\le 2\)、同奇偶、总和在 \([2, 6]\) 中的三元组。偶数情形有 \((2,2,2)\)、\((2,2,0)\) 的排列、\((2,0,0)\) 的排列，以及 \((2,2,-2)\) 的排列，共 \(10\) 个。奇数情形只有 \((1,1,1)\)。总共 \(11\) 个，并且每个都对应唯一的 \((x, y, z)\)。因此正确答案是 C。

English

Let \(p = -x + y + z,\) \(q = x - y + z,\) \(r = x + y - z.\) The last three inequalities say \(p, q, r \le 2,\) the first says \(x + y + z \ge 2,\) and \(p + q + r = x + y + z.\) Since \(x = \tfrac{q + r}{2}\) and so on, \(p, q, r\) must all share the same parity. Now count triples with each part \(\le 2,\) equal parity, and sum in \([2, 6].\) The even ones are \((2,2,2),\) the permutations of \((2,2,0),\) of \((2,0,0),\) and of \((2,2,-2),\) giving \(10.\) The only odd one is \((1,1,1).\) That’s \(11\) in all, and each yields a unique \((x, y, z).\) Thus, C is the correct answer.

10 答案 A

分解因式：\(f(n) = (n + 1)(n - 2)(n - 4)\)，\(g(n) = (n + 1)(n - 3)(n - 4)\)。避开 \(n \in \{-1, 3, 4\}\) 这些使 \(g\) 为零或出现 \(\tfrac{0}{0}\) 的值后，可约去公因式：\(\dfrac{f(n)}{g(n)} = \dfrac{n - 2}{n - 3} = 1 + \dfrac{1}{n - 3}\)。这只有在 \(n - 3 = \pm 1\) 时才是整数，所以 \(n = 2\) 或 \(n = 4\)。但 \(n = 4\) 使 \(g\) 为零，因此只有 \(n = 2\)，和为 \(2\)。因此正确答案是 A。

English

Factor both cubics: \(f(n) = (n + 1)(n - 2)(n - 4)\) and \(g(n) = (n + 1)(n - 3)(n - 4).\) Away from \(n \in \{-1, 3, 4\},\) where \(g\) vanishes or the ratio is \(\tfrac{0}{0},\) the common factors cancel and \(\dfrac{f(n)}{g(n)} = \dfrac{n - 2}{n - 3} = 1 + \dfrac{1}{n - 3}.\) That’s an integer only when \(n - 3 = \pm 1,\) so \(n = 2\) or \(n = 4.\) But \(n = 4\) kills \(g,\) so only \(n = 2\) survives, and the sum is \(2.\) Therefore, the answer is A.

11 答案 B

每天的分配都是把 \(6\) 名学生分给 \(6\) 名导师的一个排列。比较两天时，保持同一导师的学生数就是 \(\tau = \pi_{\text{Tue}}^{-1}\pi_{\text{Mon}}\) 的不动点个数，而这个置换是 \(6\) 个元素上的均匀随机排列。要恰有 \(2\) 个不动点，先选出这 \(2\) 个，有 \(\binom{6}{2}\) 种；其余 \(4\) 个要错排，有 \(D_4 = 9\) 种。因此概率为 \(\dfrac{\binom{6}{2} D_4}{6!} = \dfrac{15 \cdot 9}{720} = \dfrac{3}{16}\)。所以正确答案是 B。

English

Each day’s assignment is a permutation of the \(6\) students among the \(6\) tutors. Comparing the two days, the number who keep the same tutor is the number of fixed points of \(\tau = \pi_{\text{Tue}}^{-1}\pi_{\text{Mon}},\) itself a uniformly random permutation of \(6\) elements. We want exactly \(2\) fixed points, so choose those \(2\) in \(\binom{6}{2}\) ways and derange the other \(4,\) where \(D_4 = 9.\) The probability is \(\dfrac{\binom{6}{2} D_4}{6!} = \dfrac{15 \cdot 9}{720} = \dfrac{3}{16}.\) Thus, B is the correct answer.

12 答案 C

设三角形边长为 \(s\)。三个半径为 \(r\) 的圆盘给出 \(s = 2r(1 + \sqrt3)\)，所以 \(T = \dfrac{3\pi r^2}{(\frac{\sqrt3}{4})s^2}\) \(= \dfrac{(2\sqrt3 - 3)\pi}{2} \approx 0.73\)。对于边长为 \(a\) 的菱形，两个圆盘位于长对角线上，满足 \(a\sqrt3 = 6r\)，所以 \(r = \tfrac{a}{2\sqrt3}\)，且 \(R = \dfrac{2\pi r^2}{(\frac{a^2\sqrt3}{2})} = \dfrac{\pi\sqrt3}{9} \approx 0.60\)。对于边长为 \(a\) 的正六边形，六个圆盘中每个都与一条边在中点处相切，同样有 \(r = \tfrac{a}{2\sqrt3}\)，且 \(H = \dfrac{6\pi r^2}{(\frac{3\sqrt3}{2})a^2}\) \(= \dfrac{\pi\sqrt3}{9} \approx 0.60\)。因此 \(H = R \lt T\)，正确答案是 C。

English

Take the triangle with side \(s.\) Three disks of radius \(r\) give \(s = 2r(1 + \sqrt3),\) so \(T = \dfrac{3\pi r^2}{(\frac{\sqrt3}{4})s^2}\) \(= \dfrac{(2\sqrt3 - 3)\pi}{2} \approx 0.73.\) For the rhombus with side \(a,\) the two disks sit on the long diagonal \(a\sqrt3 = 6r,\) so \(r = \tfrac{a}{2\sqrt3}\) and \(R = \dfrac{2\pi r^2}{(\frac{a^2\sqrt3}{2})} = \dfrac{\pi\sqrt3}{9} \approx 0.60.\) For the hexagon with side \(a,\) each of the six disks touches a side at its midpoint, again giving \(r = \tfrac{a}{2\sqrt3}\) and \(H = \dfrac{6\pi r^2}{(\frac{3\sqrt3}{2})a^2}\) \(= \dfrac{\pi\sqrt3}{9} \approx 0.60.\) So \(H = R \lt T.\) Therefore, the answer is C.

13 答案 A

令直角顶点 \(C = (0,0)\)，短边 \(CB = 1\)，其中 \(B = (1, 0)\)，长边 \(CA = \sqrt3\)，其中 \(A = (0, \sqrt3)\)。从 \(C\) 到斜边 \(AB\) 的高的垂足为 \(H = \left(\tfrac34, \tfrac{\sqrt3}{4}\right)\)，且这条高在直线 \(x = \sqrt3\,y\) 上。从 \(A\) 到 \(CB\) 中点 \(\left(\tfrac12, 0\right)\) 的中线与这条高相交于 \(\left(\tfrac37, \tfrac{\sqrt3}{7}\right)\)。这把 \(CH\)（长度 \(\tfrac{\sqrt3}{2}\)）分成 \(\tfrac{4\sqrt3}{14}\) 和 \(\tfrac{3\sqrt3}{14}\)，所以 \(x = \tfrac{3\sqrt3}{14}\)，于是 \(\dfrac{x}{x + y} = \dfrac{\frac{3\sqrt3}{14}}{\frac{\sqrt3}{2}} = \dfrac{3}{7}\)。因此正确答案是 A。

English

Place the right angle at \(C = (0,0),\) the short leg \(CB = 1\) with \(B = (1, 0),\) and the long leg \(CA = \sqrt3\) with \(A = (0, \sqrt3).\) The altitude from \(C\) to hypotenuse \(AB\) has foot \(H = \left(\tfrac34, \tfrac{\sqrt3}{4}\right)\) and runs along \(x = \sqrt3\,y.\) The median from \(A\) to the midpoint \(\left(\tfrac12, 0\right)\) of \(CB\) meets that altitude at \(\left(\tfrac37, \tfrac{\sqrt3}{7}\right).\) This cuts \(CH\) (length \(\tfrac{\sqrt3}{2}\)) into \(\tfrac{4\sqrt3}{14}\) and \(\tfrac{3\sqrt3}{14},\) so \(x = \tfrac{3\sqrt3}{14}\) and \(\dfrac{x}{x + y} = \dfrac{\frac{3\sqrt3}{14}}{\frac{\sqrt3}{2}} = \dfrac{3}{7}.\) Thus, A is the correct answer.

14 答案 C

三名队长正好是最高的三人，当且仅当这三人进入三个不同的组；这样每人都是自己组中最高的。把这三人依次放入 \(9\) 个位置（每组 \(3\) 个）中。第二高的人避开第一高所在组的概率是 \(\tfrac{6}{8}\)，第三高的人避开前两人所在组的概率是 \(\tfrac{3}{7}\)。所以所求概率为 \(\tfrac{6}{8} \cdot \tfrac{3}{7} = \tfrac{9}{28}\)。因此正确答案是 C。

English

The captains are the three tallest exactly when those three land in three different groups, since each is then the tallest of its own group. Drop them into the \(9\) slots one at a time (\(3\) per group). The second tallest misses the first’s group with probability \(\tfrac{6}{8},\) and the third misses both with probability \(\tfrac{3}{7}.\) So the probability is \(\tfrac{6}{8} \cdot \tfrac{3}{7} = \tfrac{9}{28}.\) Therefore, the answer is C.

15 答案 D

分解 \(k^3 + 6k^2 + 8k = k(k + 2)(k + 4)\)，并作部分分式分解：\(\dfrac{1}{k(k + 2)(k + 4)} = \dfrac{\frac{1}{8}}{k}\) \(- \dfrac{\frac{1}{4}}{k + 2}\) \(+ \dfrac{\frac{1}{8}}{k + 4}\)。对所有 \(k\) 求和时，\(\tfrac1n\) 在 \(n \ge 5\) 的系数都会相消，所以只剩前几项：\(\tfrac18\left(1 + \tfrac12 - \tfrac13 - \tfrac14\right)\) \(= \tfrac18 \cdot \tfrac{11}{12} = \tfrac{11}{96}\)。因此 \(a + b = 11 + 96 = 107\)，正确答案是 D。

English

Factor \(k^3 + 6k^2 + 8k = k(k + 2)(k + 4),\) then split into partial fractions: \(\dfrac{1}{k(k + 2)(k + 4)} = \dfrac{\frac{1}{8}}{k}\) \(- \dfrac{\frac{1}{4}}{k + 2}\) \(+ \dfrac{\frac{1}{8}}{k + 4}.\) Summing over all \(k,\) the coefficient of \(\tfrac1n\) cancels for \(n \ge 5,\) so only the first few terms survive: \(\tfrac18\left(1 + \tfrac12 - \tfrac13 - \tfrac14\right)\) \(= \tfrac18 \cdot \tfrac{11}{12} = \tfrac{11}{96}.\) So \(a + b = 11 + 96 = 107.\) Thus, D is the correct answer.

16 答案 D

六个大小不同的扇形构成一个由 \(6\) 个可区分位置组成的固定环，所以要计算 \(6\) 环的合法 \(3\) 色涂色，并要求每种颜色恰好使用两次。合法 \(3\) 色涂色共有 \((3-1)^6 + (3-1) = 66\) 种。其中有 \(6\) 种只使用两种颜色：缺少的颜色有 \(3\) 种选择，其余两种颜色有 \(2\) 种交替方式。另有 \(36\) 种的颜色计数为 \((3,2,1)\)：出现三次的颜色有 \(3\) 种选择，它所占的交替位置组有 \(2\) 种选择，出现两次的颜色有 \(2\) 种选择，再从其余 \(3\) 个位置中选择它的 \(2\) 个位置，有 \(3\) 种方法。剩下的 \(66 - 6 - 36 = 24\) 种恰好每种颜色使用两次。所以答案是 D。

English

The six unequal sectors form a fixed cycle of \(6\) distinguishable positions, so we want proper \(3\)-colorings of a \(6\)-cycle that use each color exactly twice. There are \((3-1)^6 + (3-1) = 66\) proper \(3\)-colorings altogether. Of these, \(6\) use only two colors: choose the missing color in \(3\) ways, then alternate the other two colors in \(2\) ways. Another \(36\) have color counts \((3,2,1)\): choose the color used three times in \(3\) ways, choose one of the \(2\) alternating sets of positions for it, choose the color used twice in \(2\) ways, and choose its \(2\) positions among the other \(3\) in \(3\) ways. The remaining \(66 - 6 - 36 = 24\) use each color exactly twice. Therefore, the answer is D.

17 答案 B

令 \(A_k\) 为前 \(k\) 项的平均数。则 \(A_3 = 2025\)，且 \(A_k = A_{k-1} - 1\)（当 \(k \ge 4\)），所以 \(A_k = 2028 - k\)。部分和为 \(S_k = k(2028 - k)\)，因此当 \(k \ge 4\) 时各项为 \(x_k = S_k - S_{k-1} = 2029 - 2k\)，也就是 \(x_4 = 2021, x_5 = 2019, \ldots\)。只要 \(2029 - 2k \gt 0\)，即 \(k \le 1014\)，这些项就保持为正，此时 \(x_{1014} = 1\)。这个上界可以达到：取 \((x_1,x_2,x_3) = (2030,2023,2022)\)，它是递减的，和为 \(6075\)，并且各项都大于 \(x_4 = 2021\)。所以最大可能值为 \(n = 1014\)，因此正确答案是 B。

English

Let \(A_k\) be the average of the first \(k\) terms. Then \(A_3 = 2025\) and \(A_k = A_{k-1} - 1\) for \(k \ge 4,\) so \(A_k = 2028 - k.\) The partial sum is \(S_k = k(2028 - k),\) and for \(k \ge 4\) the terms are \(x_k = S_k - S_{k-1} = 2029 - 2k,\) namely \(x_4 = 2021, x_5 = 2019, \ldots.\) These stay positive as long as \(2029 - 2k \gt 0,\) that is \(k \le 1014,\) with \(x_{1014} = 1.\) This bound is attainable: take \((x_1,x_2,x_3) = (2030,2023,2022),\) which is decreasing, sums to \(6075,\) and lies above \(x_4 = 2021.\) Thus the greatest possible value is \(n = 1014,\) so B is the correct answer.

18 答案 D

对每个 \(m\)，当 \(\lfloor\sqrt{n}\rfloor = m\) 时，对应的 \(2m + 1\) 个整数满足 \(m^2 \le n \le (m + 1)^2 - 1\)。因为 \(\sqrt{2025} = 45\)，所以 \(1 \le m \le 44\) 的项贡献 \(\sum_{m=1}^{44} m(2m + 1)\)，再加上 \(n = 2025\) 对应的 \(45\)。这个和为 \(\sum_{m=1}^{44}(2m^2 + m)\) \(= 2 \cdot \tfrac{44 \cdot 45 \cdot 89}{6}\) \(+ \tfrac{44 \cdot 45}{2} = 58740\) \(+ 990 = 59730\)，所以总和是 \(59775\)，个位数字为 \(5\)。因此正确答案是 D。

English

For each \(m,\) \(\lfloor\sqrt{n}\rfloor = m\) on the \(2m + 1\) integers \(m^2 \le n \le (m + 1)^2 - 1.\) Since \(\sqrt{2025} = 45,\) the terms with \(1 \le m \le 44\) contribute \(\sum_{m=1}^{44} m(2m + 1),\) and \(n = 2025\) tacks on \(45.\) That sum is \(\sum_{m=1}^{44}(2m^2 + m)\) \(= 2 \cdot \tfrac{44 \cdot 45 \cdot 89}{6}\) \(+ \tfrac{44 \cdot 45}{2} = 58740\) \(+ 990 = 59730,\) so the total is \(59775.\) Its ones digit is \(5.\) Therefore, the answer is D.

19 答案 D

这个容器是一个正方台：高度比例为 \(t\) 的水平截面边长为 \(1 + 2t\)。把侧面向上延长到顶点，则到边长为 \(w\) 的截面为止的体积按 \(w^3\) 缩放。因此整个容器有 \(3^3 - 1^3 = 26\) 份，到中线（边长 \(2\)）的部分有 \(2^3 - 1^3 = 7\) 份，剩余部分有 \(3^3 - 2^3 = 19\) 份。\(7\) 份用时 \(35\) 分钟，所以每份 \(5\) 分钟。剩下 \(19\) 份需要 \(95\) 分钟。因此正确答案是 D。

English

The container is a square frustum: a horizontal slice at height fraction \(t\) has side \(1 + 2t.\) Extend the sides up to their apex, and the volume out to where the side length is \(w\) scales as \(w^3.\) So the whole container is \(3^3 - 1^3 = 26\) parts, the piece up to the midline (side \(2\)) is \(2^3 - 1^3 = 7\) parts, and the rest is \(3^3 - 2^3 = 19\) parts. Those \(7\) parts take \(35\) minutes, so each part is \(5\) minutes. The remaining \(19\) parts take \(95\) minutes. Thus, D is the correct answer.

20 答案 A

设每个半圆半径为 \(\rho\)，圆心可取如 \((\rho, 0)\) 和 \((1, \rho)\)。相邻半圆相切，所以这两个圆心距离为 \(2\rho\)：\((1 - \rho)^2 + \rho^2 = 4\rho^2\)。这给出 \(2\rho^2 + 2\rho - 1 = 0\)，所以 \(\rho = \tfrac{\sqrt3 - 1}{2}\)。小圆半径为 \(t\)，圆心在 \(\left(\tfrac12, \tfrac12\right)\)。它与一个半圆相切时，到该半圆圆心的距离等于 \(\rho + t\)。这个距离是 \(\sqrt{2 - \sqrt3}\)，所以 \(t = \sqrt{2 - \sqrt3} - \rho\)。直径为 \(2t = \sqrt6 - \sqrt2\) \(- \sqrt3 + 1 = (\sqrt3 - 1)(\sqrt2 - 1)\)。因此 \(a + b + c + d = 3\) \(+ (-1) + 2 + (-1) = 3\)，正确答案是 A。

English

Let each semicircle have radius \(\rho,\) with centers like \((\rho, 0)\) and \((1, \rho).\) Adjacent semicircles are tangent, so these centers are \(2\rho\) apart: \((1 - \rho)^2 + \rho^2 = 4\rho^2.\) This gives \(2\rho^2 + 2\rho - 1 = 0,\) so \(\rho = \tfrac{\sqrt3 - 1}{2}.\) The small circle of radius \(t\) sits at \(\left(\tfrac12, \tfrac12\right),\) and it’s tangent to a semicircle when its distance to that center equals \(\rho + t.\) That distance is \(\sqrt{2 - \sqrt3},\) so \(t = \sqrt{2 - \sqrt3} - \rho,\) and the diameter is \(2t = \sqrt6 - \sqrt2\) \(- \sqrt3 + 1 = (\sqrt3 - 1)(\sqrt2 - 1).\) So \(a + b + c + d = 3\) \(+ (-1) + 2 + (-1) = 3.\) Therefore, the answer is A.

21 答案 C

先计算各位置可区分的方格涂色。固定中心方格为红色，并按循环顺序列出四个边中点的颜色。在旋转或反射意义下，边中点的唯一可能模式是 \(YRBR\) 和 \(YRBY\)。第一种模式有 \(4\) 种放置方式，以及 \(3\) 个可能的循环角落字符串 \(BRYB, BYRB\)，和 \(BYYB\)；第二种模式有 \(8\) 种放置方式，以及 \(2\) 个可能的角落字符串 \(BYBB\) 和 \(BYRB\)。所以红色中心时共有 \(4 \cdot 3 + 8 \cdot 2 = 28\) 种涂色。中心有 \(3\) 种颜色选择，所以带标号涂色共有 \(84\) 种。

现在应用伯恩赛德引理。恒等变换固定全部 \(84\) 种涂色。没有非恒等旋转能固定合法涂色。水平轴和竖直轴的两个反射各固定 \(6\) 种涂色，而两个对角线反射均不固定任何涂色。因此在旋转和反射意义下，涂色数为 \(\dfrac{84 + 6 + 6}{8} = 12\)。所以正确答案是 C。

English

First count colorings of a grid whose positions are distinguished. Fix the center square as red and list the four edge-middle colors cyclically. Up to a rotation or reflection, the only possible edge patterns are \(YRBR\) and \(YRBY.\) The first has \(4\) placements and \(3\) possible cyclic corner strings, \(BRYB, BYRB,\) and \(BYYB;\) the second has \(8\) placements and \(2\) possible corner strings, \(BYBB\) and \(BYRB.\) Thus there are \(4 \cdot 3 + 8 \cdot 2 = 28\) colorings with a red center. The center has \(3\) possible colors, so there are \(84\) labeled colorings.

Now apply Burnside’s lemma. The identity fixes all \(84\) colorings. No nonidentity rotation fixes a valid coloring. Each of the two reflections across a horizontal or vertical axis fixes \(6\) colorings, while each diagonal reflection fixes none. Therefore, the number of colorings up to rotations and reflections is \(\dfrac{84 + 6 + 6}{8} = 12.\) Thus, C is the correct answer.

22 答案 A

数字和 \(61 = 63 - 2\)，表示七个数字全为 \(9\) 后总亏缺为 \(2\)，这样的数有 \(\binom{2 + 6}{6} = 28\) 个。要能被 \(11\) 整除，需要 \(O - E \equiv 0 \pmod{11}\)，其中 \(O\) 是 \(4\) 个奇数位数字之和，\(E\) 是 \(3\) 个偶数位数字之和。设亏缺为 \(d_O + d_E = 2\)。则 \(O - E = 9 - d_O\) \(+ d_E = 11 - 2 d_O\)，只有当 \(d_O = 0\) 时才是 \(11\) 的倍数。因此全部亏缺 \(2\) 都在 \(3\) 个偶数位上，有 \(\binom{2 + 2}{2} = 6\) 种。概率为 \(\tfrac{6}{28} = \tfrac{3}{14}\)。因此正确答案是 A。

English

A digit sum of \(61 = 63 - 2\) means all seven digits are \(9\) except for a total deficit of \(2,\) which gives \(\binom{2 + 6}{6} = 28\) numbers. For divisibility by \(11\) we need \(O - E \equiv 0 \pmod{11},\) where \(O\) sums the \(4\) odd-position digits and \(E\) the \(3\) even ones. Write the deficits as \(d_O + d_E = 2.\) Then \(O - E = 9 - d_O\) \(+ d_E = 11 - 2 d_O,\) a multiple of \(11\) only when \(d_O = 0.\) So all of the deficit \(2\) falls on the \(3\) even positions, giving \(\binom{2 + 2}{2} = 6\) ways. The probability is \(\tfrac{6}{28} = \tfrac{3}{14}.\) Therefore, the answer is A.

23 答案 C

在第 \(i\) 行第 \(j\) 列，Horace 写的是 \(91(i - 1) + j\)，Vera 写的是 \(141(j - 1) + i\)。令二者相等并化简，得到 \(9i - 14j = -5\)，所以 \(i = \tfrac{14j - 5}{9}\)，恰好在 \(j \equiv 1 \pmod 9\) 时为整数。对于 \(j = 1, 10, 19, \ldots, 91\)，共有 \(11\) 个值，且对应的 \(i\) 为 \(1, 15, 29, \ldots, 141\)，都在范围内。因此有 \(11\) 个小正方形匹配，正确答案是 C。

English

At row \(i,\) column \(j,\) Horace writes \(91(i - 1) + j\) and Vera writes \(141(j - 1) + i.\) Set them equal and simplify to get \(9i - 14j = -5,\) so \(i = \tfrac{14j - 5}{9},\) an integer exactly when \(j \equiv 1 \pmod 9.\) For \(j = 1, 10, 19, \ldots, 91,\) that’s \(11\) values, and \(i\) runs \(1, 15, 29, \ldots, 141,\) all within range. So \(11\) squares match. Thus, C is the correct answer.

24 答案 E

令 \(p_n\) 为从位置 \(n\) 出发到达 \(4\) 的概率，且 \(p_4 = 1\)。规则给出 \(p_0 = \tfrac12 p_1\)、\(p_1 = \tfrac14 p_0 + \tfrac14 p_2\)、\(p_2 = \tfrac14 p_1 + \tfrac14 p_3\)、\(p_3 = \tfrac14 p_2 + \tfrac14\)。向上求解可得 \(p_1 = \tfrac27 p_2\) 和 \(p_2 = \tfrac{7}{26} p_3\)。继续代回得到 \(p_3 = \tfrac{26}{97}\)、\(p_2 = \tfrac{7}{97}\)、\(p_1 = \tfrac{2}{97}\)，最后 \(p_0 = \tfrac{1}{97}\)。因此正确答案是 E。

English

Let \(p_n\) be the probability of reaching \(4\) from position \(n,\) with \(p_4 = 1.\) The rules give \(p_0 = \tfrac12 p_1,\) \(p_1 = \tfrac14 p_0 + \tfrac14 p_2,\) \(p_2 = \tfrac14 p_1 + \tfrac14 p_3,\) and \(p_3 = \tfrac14 p_2 + \tfrac14.\) Work upward: \(p_1 = \tfrac27 p_2\) and \(p_2 = \tfrac{7}{26} p_3.\) These unwind to \(p_3 = \tfrac{26}{97},\) \(p_2 = \tfrac{7}{97},\) \(p_1 = \tfrac{2}{97},\) and finally \(p_0 = \tfrac{1}{97}.\) Therefore, the answer is E.

25 答案 B

令 \(A = (0,0)\)、\(B = (4,0)\)、\(C = (4,4)\)、\(D = (0,4)\)，则 \(P = \left(0, \tfrac85\right)\)、\(Q = \left(\tfrac{10}{3}, 4\right)\)。初始方向为 \(\left(\tfrac{10}{3}, \tfrac{12}{5}\right) \parallel (25, 18)\)。把台球路径展开成由反射正方形拼成的网格，并从 \(P\) 出发沿直线前进。在网格顶点 \((4a,4b)\) 处，由横坐标得 \(t = \tfrac{4a}{25}\)。于是纵坐标要求 \(10 + 18a = 25b\)，即 \(a \equiv 5 \pmod{25}\)。第一个正的可能值是 \(a = 5\)，由此得 \(b = 4\)，展开后的顶点为 \((20,16)\)。横向穿过 \(5\) 个小正方形（奇数），说明它落在 \(x = 4\) 这条边上；纵向穿过 \(4\) 个小正方形（偶数），说明它落在 \(y = 0\) 上。这个顶点就是 \((4,0) = B\)。因此正确答案是 B。

English

Place \(A = (0,0),\) \(B = (4,0),\) \(C = (4,4),\) \(D = (0,4),\) so \(P = \left(0, \tfrac85\right)\) and \(Q = \left(\tfrac{10}{3}, 4\right).\) The initial direction is \(\left(\tfrac{10}{3}, \tfrac{12}{5}\right) \parallel (25, 18).\) Unfold the billiard into a grid of reflected copies and follow the straight line from \(P.\) At a grid corner \((4a,4b),\) the horizontal coordinate gives \(t = \tfrac{4a}{25}.\) The vertical coordinate then requires \(10 + 18a = 25b,\) so \(a \equiv 5 \pmod{25}.\) The first positive possibility is \(a = 5,\) giving \(b = 4\) and the unfolded corner \((20,16).\) Crossing \(5\) cells across (odd) puts it on the side \(x = 4,\) and \(4\) cells up (even) puts it on \(y = 0.\) That’s vertex \((4,0) = B.\) Thus, B is the correct answer.