2024B AMC 12 真题（P版）

1 答案 B

目标人物左边有 \(1013 - 1 = 1012\) 人，右边有 \(1010 - 1 = 1009\) 人。包括此人自己，总人数为 \(1012 + 1009 + 1 = 2022\)。

所以正确答案是 B。

English

The target person has \(1013 - 1 = 1012\) people to the left and \(1010 - 1 = 1009\) people to the right. Including the person themself, the line has \(1012 + 1009 + 1 = 2022\) people.

Thus, the correct answer is B.

2 答案 B

注意 \(6! = 720 = 8 \cdot 9 \cdot 10\)。因此 \[7! \cdot 6! = 7! \cdot (8 \cdot 9 \cdot 10) = 10!\text{。}\] 所以 \(10! - 7! \cdot 6! = 10! - 10! = 0\)。

所以正确答案是 B。

English

Note that \(6! = 720 = 8 \cdot 9 \cdot 10.\) Therefore \[7! \cdot 6! = 7! \cdot (8 \cdot 9 \cdot 10) = 10!.\] So \(10! - 7! \cdot 6! = 10! - 10! = 0.\)

Thus, the correct answer is B.

3 答案 E

不等式 \(|2x| \le 7\pi\) 等价于 \(|x| \le \dfrac{7\pi}{2} \approx 10.99\)。满足条件的整数从 \(-10\) 到 \(10\)，共有 \(10 + 10 + 1 = 21\) 个。

所以正确答案是 E。

English

The inequality \(|2x| \le 7\pi\) is equivalent to \(|x| \le \dfrac{7\pi}{2} \approx 10.99.\) The integers satisfying this run from \(-10\) to \(10,\) which is \(10 + 10 + 1 = 21\) values.

Thus, the correct answer is E.

4 答案 D

第 \(k\) 步放入 \(k\) 个球，所以第 \(k\) 步后共放入 \(\dfrac{k(k+1)}{2}\) 个球。又 \(\dfrac{63 \cdot 64}{2} = 2016\) 而 \(\dfrac{64 \cdot 65}{2} = 2080\)，所以球 \(2024\) 在第 \(64\) 步。

步骤按 \(A, B, C, D, E\) 循环，所以第 \(k\) 步使用位置 \((k-1) \bmod 5\) 的箱。这里 \((64 - 1) \bmod 5 = 63 \bmod 5 = 3\)，即第四个箱 \(D\)。

所以正确答案是 D。

English

Step \(k\) deposits \(k\) balls, so after step \(k\) a total of \(\dfrac{k(k+1)}{2}\) balls have been placed. Since \(\dfrac{63 \cdot 64}{2} = 2016\) and \(\dfrac{64 \cdot 65}{2} = 2080,\) ball \(2024\) falls in step \(64.\)

The steps cycle through the bins \(A, B, C, D, E,\) so step \(k\) uses position \((k-1) \bmod 5.\) Here \((64 - 1) \bmod 5 = 63 \bmod 5 = 3,\) which is the fourth bin, \(D.\)

Thus, the correct answer is D.

5 答案 B

原表达式是前 \(50\) 个正奇数之和，等于 \(50^2 = 2500\)。把值为 \(v\) 的项从 \(+\) 改为 \(-\)，总和会减少 \(2v\)，所以要使结果为负，翻号项的和必须大于 \(\dfrac{2500}{2} = 1250\)。

为使项数尽量少，应从 \(99, 97, 95, \ldots\) 开始翻号。最大的 \(k\) 个奇数之和为 \(k(100 - k)\)。当 \(k = 14\) 时，\(14 \cdot 86 = 1204 \le 1250\)；当 \(k = 15\) 时，\(15 \cdot 85 = 1275 \gt 1250\)。所以 \(15\) 个足够，而 \(14\) 个不够。

所以正确答案是 B。

English

The original expression sums the first \(50\) odd numbers, giving \(50^2 = 2500.\) Changing a term of value \(v\) from \(+\) to \(-\) decreases the total by \(2v,\) so to make the result negative the flipped terms must total more than \(\dfrac{2500}{2} = 1250.\)

To use as few terms as possible, flip the largest odd numbers \(99, 97, 95, \ldots\) The largest \(k\) of them sum to \(k(100 - k).\) With \(k = 14\) this is \(14 \cdot 86 = 1204 \le 1250,\) but with \(k = 15\) it is \(15 \cdot 85 = 1275 \gt 1250.\) So \(15\) sign changes suffice and \(14\) do not.

Thus, the correct answer is B.

6 答案 B

\(N\) 的 \(5\) 进制位数是 \(\lfloor \log_5 N \rfloor + 1\)。对于 \(N = 5 \cdot 10^{13}\)，有 \[\log_{10} N = 13 + \log_{10} 5 = 13.7\text{。}\] 换底可得 \(\log_5 N\) \(= \dfrac{13.7}{\log_{10} 5}\) \(= \dfrac{13.7}{0.7}\) \(= 19.57\ldots\)。

因此位数为 \(\lfloor 19.57 \rfloor + 1 = 19 + 1 = 20\)。

所以正确答案是 B。

English

The number of digits of \(N\) in base \(5\) is \(\lfloor \log_5 N \rfloor + 1.\) With \(N = 5 \cdot 10^{13},\) \[\log_{10} N = 13 + \log_{10} 5 = 13.7.\] Converting bases, \(\log_5 N\) \(= \dfrac{13.7}{\log_{10} 5}\) \(= \dfrac{13.7}{0.7}\) \(= 19.57\ldots\)

Thus the number of digits is \(\lfloor 19.57 \rfloor + 1 = 19 + 1 = 20.\)

Thus, the correct answer is B.

7 答案 C

设 \(X = (0,0)\)、\(W = (0,4)\)、\(Y = (8,0)\)、\(Z = (8,4)\)，其中 \(M = (m, 0)\) 在 \(\overline{XY}\) 上，\(A = (8, a)\) 在 \(\overline{YZ}\) 上。

因为 \(\angle WMA = 90^\circ\)，所以 \(\overrightarrow{MW} \cdot \overrightarrow{MA}\) \(= (-m)(8-m)\) \(+ 4a = 0\)，即 \(4a = m(8-m)\)。面积为 \([\triangle WXM] = \tfrac12 \cdot 4 \cdot m = 2m\) 而 \([\triangle WAZ]\) \(= \tfrac12 \cdot 8 \cdot (4 - a)\) \(= 4(4 - a)\)。面积相等给出 \(m = 8 - 2a\)。

把 \(a = \tfrac{8-m}{2}\) 代入 \(4a = m(8-m)\)，得 \(2(8-m) = m(8-m)\)，所以 \(m = 2\)，\(a = 3\)。另一个代数根 \(m=8\) 会导致 \(M=A=Y\)，此时 \(\angle WMA\) 没有定义，所以不成立。于是取 \(W = (0,4)\)、\(M = (2,0)\)、\(A = (8,3)\)，得 \[ \begin{aligned} [\triangle WMA] &= \tfrac12\,\bigl|2(3 - 4) + 8(4 - 0)\bigr| \\ &= \tfrac12 (30) = 15 \end{aligned}\text{。}\]

所以正确答案是 C。

English

Set \(X = (0,0),\) \(W = (0,4),\) \(Y = (8,0),\) \(Z = (8,4),\) with \(M = (m, 0)\) on \(\overline{XY}\) and \(A = (8, a)\) on \(\overline{YZ}.\)

Since \(\angle WMA = 90^\circ,\) \(\overrightarrow{MW} \cdot \overrightarrow{MA}\) \(= (-m)(8-m)\) \(+ 4a = 0,\) so \(4a = m(8-m).\) The areas give \([\triangle WXM] = \tfrac12 \cdot 4 \cdot m = 2m\) and \([\triangle WAZ]\) \(= \tfrac12 \cdot 8 \cdot (4 - a)\) \(= 4(4 - a).\) Setting these equal yields \(m = 8 - 2a.\)

Substituting \(a = \tfrac{8-m}{2}\) into \(4a = m(8-m)\) gives \(2(8-m) = m(8-m),\) so \(m = 2\) and \(a = 3.\) The other algebraic root, \(m=8,\) gives \(M=A=Y\) and no defined angle \(\angle WMA,\) so it is invalid. Then with \(W = (0,4),\) \(M = (2,0),\) \(A = (8,3),\) \[ \begin{aligned} [\triangle WMA] &= \tfrac12\,\bigl|2(3 - 4) + 8(4 - 0)\bigr| \\ &= \tfrac12 (30) = 15. \end{aligned} \]

Thus, the correct answer is C.

8 答案 C

分子分母同除以 \(\log_2 x \cdot \log_3 x\)，左边变为 \[ \begin{gathered} \frac{1}{\dfrac{1}{\log_2 x} + \dfrac{1}{\log_3 x}} \\ = \frac{1}{\log_x 2 + \log_x 3} \\ = \frac{1}{\log_x 6} \end{gathered}\text{。}\] 因此 \(\dfrac{1}{\log_x 6} = 2\)，即 \(\log_x 6 = \dfrac12\)，也就是 \(x^{\frac{1}{2}} = 6\)。

因此 \(x = 36\)。

所以正确答案是 C。

English

Dividing top and bottom by \(\log_2 x \cdot \log_3 x,\) the left side becomes \[ \begin{gathered} \frac{1}{\dfrac{1}{\log_2 x} + \dfrac{1}{\log_3 x}} \\ = \frac{1}{\log_x 2 + \log_x 3} \\ = \frac{1}{\log_x 6}. \end{gathered} \] So \(\dfrac{1}{\log_x 6} = 2,\) meaning \(\log_x 6 = \dfrac12,\) i.e. \(x^{\frac{1}{2}} = 6.\)

Therefore \(x = 36.\)

Thus, the correct answer is C.

9 答案 B

飞镖靶 \(|x| + |y| \le 8\) 是一个对角线长为 \(16\) 的正方形，面积为 \(\tfrac12 \cdot 16 \cdot 16 = 128\)。目标条件 \((x^2 + y^2 - 25)^2 \le 49\) 等价于 \(-7 \le x^2 + y^2 - 25 \le 7\)，即 \(18 \le x^2 + y^2 \le 32\)，这是一个面积为 \(\pi(32 - 18) = 14\pi\) 的圆环。

原点到正方形一条边（如 \(x + y = 8\)）的距离为 \(\dfrac{8}{\sqrt2} = \sqrt{32}\)，正好等于环形区域外半径，所以整个环形区域都在 \(B\) 内。概率为 \(\dfrac{14\pi}{128} = \dfrac{7}{64}\pi\)，因此 \(m + n = 7 + 64 = 71\)。

所以正确答案是 B。

English

The dartboard \(|x| + |y| \le 8\) is a square with diagonals \(16,\) so its area is \(\tfrac12 \cdot 16 \cdot 16 = 128.\) The target condition \((x^2 + y^2 - 25)^2 \le 49\) means \(-7 \le x^2 + y^2 - 25 \le 7,\) i.e. \(18 \le x^2 + y^2 \le 32,\) an annulus of area \(\pi(32 - 18) = 14\pi.\)

The distance from the origin to a side of the square (for instance \(x + y = 8\)) is \(\dfrac{8}{\sqrt2} = \sqrt{32},\) exactly the annulus’s outer radius. So the annulus is tangent to the square and lies entirely within \(B.\) The probability is \(\dfrac{14\pi}{128} = \dfrac{7}{64}\pi,\) giving \(m + n = 7 + 64 = 71.\)

Thus, the correct answer is B.

10 答案 C

六个固定数之和为 \(24.8\)，取值范围为 \([1,7]\)。整个列表的最小值与最大值有三种可能的安排。

若最小值与最大值为 \(0\) 和 \(7\)，则 \(x=0\) 且 \(z\le7\)。此时平均数为整数只能取 \(3\) 或 \(4\)，分别要求 \(y+z=2.2\) 或 \(11.2\)。前者使中位数变成 \(2.2\)。后者若要中位数为整数，必须 \(y=5\)，从而 \(z=6.2\)。这给出 \((0,5,6.2)\)。

若最小值与最大值为 \(1\) 和 \(8\)，则 \(z=8\)，而平均数为整数要求 \(x+y=3.2\) 或 \(12.2\)。前者使中位数变成 \(3.2\)；后者只有在 \(x=6,\ y=6.2\) 时中位数才是整数。这给出 \((6,6.2,8)\)。

最后，若两个极值都由新数给出，写成 \(x=t,\ z=t+7\)，其中 \(0\lt t\lt1\)。平均数必须为 \(4\)，所以 \(y=4.2-2t\)。中位数是 \(1,2.2,3.2,5.2,6.2,7,y\) 中的第四个数；只有当 \(y=4\) 时它才是整数，此时 \(t=0.1\)。于是第三个三元组是 \((0.1,4,7.1)\)，总共恰有 \(3\) 个。

所以正确答案是 C。

English

The six fixed numbers sum to \(24.8\) and span \([1,7].\) There are three possible arrangements of the overall extremes.

If the extremes are \(0\) and \(7,\) then \(x=0\) and \(z\le7.\) The only possible integer means are \(3\) and \(4,\) requiring \(y+z=2.2\) or \(11.2.\) The first makes the median \(2.2.\) In the second, an integer median forces \(y=5,\) hence \(z=6.2.\) This gives \((0,5,6.2).\)

If the extremes are \(1\) and \(8,\) then \(z=8\) and the integer mean forces \(x+y=3.2\) or \(12.2.\) The first makes the median \(3.2;\) the second has an integer median only for \(x=6,\ y=6.2.\) This gives \((6,6.2,8).\)

Finally, if both extremes are new, write \(x=t,\ z=t+7\) with \(0\lt t\lt1.\) The mean must be \(4,\) so \(y=4.2-2t.\) The median is the fourth number among \(1,2.2,3.2,5.2,6.2,7,y;\) it is an integer only when \(y=4,\) giving \(t=0.1.\) Thus the third triple is \((0.1,4,7.1),\) and there are exactly \(3\) in all.

Thus, the correct answer is C.

11 答案 E

由 \(\sin^2\theta = \dfrac{1 - \cos 2\theta}{2}\)，可得 \[ \begin{aligned} \sum_{n=1}^{90} \sin^2(n^\circ) &= \frac{90}{2} \\ &\quad {}- \frac12 \sum_{n=1}^{90}\cos(2n^\circ)\text{。} \end{aligned} \] 在余弦和中，\(n\) 与 \(90 - n\) 的项满足 \(\cos(2n^\circ) + \cos(180^\circ - 2n^\circ) = 0\)，且 \(\cos 90^\circ = 0\)，最后只剩 \(\cos 180^\circ = -1\)。

所以总和为 \(45 - \tfrac12(-1) = 45.5\)，平均数为 \(\dfrac{45.5}{90} = \dfrac{91}{180}\)。

所以正确答案是 E。

English

Using \(\sin^2\theta = \dfrac{1 - \cos 2\theta}{2},\) \[ \begin{aligned} \sum_{n=1}^{90} \sin^2(n^\circ) &= \frac{90}{2} \\ &\quad {}- \frac12 \sum_{n=1}^{90}\cos(2n^\circ). \end{aligned} \] In the cosine sum, the terms for \(n\) and \(90 - n\) satisfy \(\cos(2n^\circ) + \cos(180^\circ - 2n^\circ) = 0,\) and \(\cos 90^\circ = 0,\) so everything cancels except \(\cos 180^\circ = -1.\)

Hence the sum is \(45 - \tfrac12(-1) = 45.5,\) and the mean is \(\dfrac{45.5}{90} = \dfrac{91}{180}.\)

Thus, the correct answer is E.

12 答案 D

对顶点 \(0, z, z^2, z^3\)，鞋带公式给出面积 \[ \begin{aligned} &\tfrac12\,\bigl|\operatorname{Im}(\bar z z^2 + \overline{z^2}\,z^3)\bigr| \\ &= \tfrac12\,\bigl|\operatorname{Im}\bigl((|z|^2 + |z|^4)z\bigr)\bigr| \\ &= \tfrac12(|z|^2 + |z|^4)\operatorname{Im}(z) \end{aligned}\text{。}\]

由 \(|z| = 2\)，面积为 \(\tfrac12(4 + 16)\operatorname{Im}(z) = 10\operatorname{Im}(z)\)。令 \(10\operatorname{Im}(z) = 15\)，得 \(\operatorname{Im}(z) = \dfrac32\)。又 \(\operatorname{Re}(z) = \sqrt{4 - \tfrac94} = \tfrac{\sqrt7}{2} \gt 1\)，满足条件。

所以正确答案是 D。

English

For vertices \(0, z, z^2, z^3\) the shoelace formula gives area \[ \begin{aligned} &\tfrac12\,\bigl|\operatorname{Im}(\bar z z^2 + \overline{z^2}\,z^3)\bigr| \\ &= \tfrac12\,\bigl|\operatorname{Im}\bigl((|z|^2 + |z|^4)z\bigr)\bigr| \\ &= \tfrac12(|z|^2 + |z|^4)\operatorname{Im}(z). \end{aligned} \]

With \(|z| = 2,\) this is \(\tfrac12(4 + 16)\operatorname{Im}(z) = 10\operatorname{Im}(z).\) Setting \(10\operatorname{Im}(z) = 15\) gives \(\operatorname{Im}(z) = \dfrac32.\) (Then \(\operatorname{Re}(z) = \sqrt{4 - \tfrac94} = \tfrac{\sqrt7}{2} \gt 1,\) as required.)

Thus, the correct answer is D.

13 答案 C

两式相加，得 \[ \begin{aligned} h + k &= 2x^2 + 2y^2 - 16x - 4y \\ &= 2(x - 4)^2 \\ &\quad {}+ 2(y - 1)^2 - 34\text{。} \end{aligned} \] 两个平方项都非负，所以最小值在 \(x = 4\)、\(y = 1\) 时取得，为 \(h + k = -34\)。

所以正确答案是 C。

English

Adding the equations, \[ \begin{aligned} h + k &= 2x^2 + 2y^2 - 16x - 4y \\ &= 2(x - 4)^2 \\ &\quad {}+ 2(y - 1)^2 - 34. \end{aligned} \] Both squared terms are nonnegative, so the minimum occurs at \(x = 4,\) \(y = 1,\) giving \(h + k = -34.\)

Thus, the correct answer is C.

14 答案 B

若 \(n\) 与 \(5\) 互质，则由 \(\varphi(125) = 100\) 和 Euler 定理可得 \(n^{100} \equiv 1 \pmod{125}\)。若 \(n\) 是 \(5\) 的倍数，则 \(n^{100}\) 被 \(5^{100}\) 整除，因而也被 \(125\) 整除，余数为 \(0\)。

因此只可能出现余数 \(0\) 和 \(1\)，共 \(2\) 种。

所以正确答案是 B。

English

If \(n\) is coprime to \(5,\) then since \(\varphi(125) = 100,\) Euler’s theorem gives \(n^{100} \equiv 1 \pmod{125}.\) If \(n\) is a multiple of \(5,\) then \(n^{100}\) is divisible by \(5^{100},\) hence by \(125,\) leaving remainder \(0.\)

So the only possible remainders are \(0\) and \(1,\) which is \(2\) distinct values.

Thus, the correct answer is B.

15 答案 B

顶点为 \(A = (0, 1)\)、\(B = (\log_2 3, 2)\)、\(C = (\log_2 7, 3)\)。由鞋带公式，\[ \begin{gathered} [\triangle ABC]\\ {}=\tfrac12\,\bigl|2\log_2 3-\log_2 7\bigr| \end{gathered}\text{。}\]

该式等于 \(\tfrac12\log_2 \dfrac{9}{7} = \log_2 \sqrt{\tfrac{9}{7}} = \log_2 \dfrac{3}{\sqrt7}\)。

所以正确答案是 B。

English

The vertices are \(A = (0, 1),\) \(B = (\log_2 3, 2),\) \(C = (\log_2 7, 3).\) By the shoelace formula, \[ \begin{gathered} [\triangle ABC]\\ {}=\tfrac12\,\bigl|2\log_2 3-\log_2 7\bigr|. \end{gathered} \]

This equals \(\tfrac12\log_2 \dfrac{9}{7} = \log_2 \sqrt{\tfrac{9}{7}} = \log_2 \dfrac{3}{\sqrt7}.\)

Thus, the correct answer is B.

16 答案 A

将 \(16\) 人分成 \(4\) 个不可区分的 \(4\) 人组有 \(\dfrac{16!}{(4!)^4\, 4!}\) 种。每个委员会选主席和秘书有 \(4 \cdot 3 = 12\) 种选择，贡献因子 \(12^4\)。所以总数为 \(\dfrac{16!}{(4!)^4\,4!}\cdot 12^4\)。

只计数因子 \(3\)：\(16!\) 贡献 \(\lfloor \frac{16}{3}\rfloor + \lfloor \frac{16}{9}\rfloor = 6\) 个；分母 \((4!)^4\,4!\) 贡献 \(4 + 1 = 5\) 个；而 \(12^4 = (2^2\cdot 3)^4\) 贡献 \(4\) 个。因此 \(r = 6 - 5 + 4 = 5\)。

所以正确答案是 A。

English

The number of ways to split \(16\) people into \(4\) indistinguishable groups of \(4\) is \(\dfrac{16!}{(4!)^4\, 4!}.\) Each committee then chooses a chairperson and a secretary in \(4 \cdot 3 = 12\) ways, contributing \(12^4.\) So the total is \(\dfrac{16!}{(4!)^4\,4!}\cdot 12^4.\)

Counting factors of \(3:\) \(16!\) contributes \(\lfloor \frac{16}{3}\rfloor + \lfloor \frac{16}{9}\rfloor = 6.\) The denominator \((4!)^4\,4!\) contributes \(4 + 1 = 5.\) And \(12^4 = (2^2\cdot 3)^4\) contributes \(4.\) Thus \(r = 6 - 5 + 4 = 5.\)

Thus, the correct answer is A.

17 答案 C

集合中有 \(21\) 个整数，所以有 \(21 \cdot 20 = 420\) 个有序选择 \((a, b)\)。若多项式有不同整数根 \(p, q, r\)，则 \(pqr = -6\)，\(a = -(p + q + r)\)，\(b = pq + qr + rp\)。

乘积为 \(-6\) 的不同整数根三元组为 \(\{1, 2, -3\}\)，\(\{1, -2, 3\}\)，\(\{-1, 2, 3\}\)，\(\{-1, -2, -3\}\)，以及 \(\{1, -1, 6\}\)。它们给出 \((a, b) = (0, -7)\)，\((-2, -5)\)，\((-4, 1)\)，\((6, 11)\)，和 \((-6, -1)\)。第四组有 \(b = 11 \gt 10\)，无效；其余四组有效且互不相同。

所求概率为 \(\dfrac{4}{420} = \dfrac{1}{105}\)。

所以正确答案是 C。

English

The set has \(21\) integers, so there are \(21 \cdot 20 = 420\) ordered choices of \((a, b).\) If the polynomial has distinct integer roots \(p, q, r,\) then \(pqr = -6,\) \(a = -(p + q + r),\) and \(b = pq + qr + rp.\)

The triples of distinct integers with product \(-6\) are \(\{1, 2, -3\},\) \(\{1, -2, 3\},\) \(\{-1, 2, 3\},\) \(\{-1, -2, -3\},\) and \(\{1, -1, 6\}.\) These give \((a, b) = (0, -7),\) \((-2, -5),\) \((-4, 1),\) \((6, 11),\) and \((-6, -1).\) The fourth has \(b = 11 \gt 10,\) so it is invalid; the other four are valid and distinct.

The probability is \(\dfrac{4}{420} = \dfrac{1}{105}.\)

Thus, the correct answer is C.

18 答案 B

因为 \(F_{2k} = F_k L_k\)，其中 \(L_k\) 为第 \(k\) 个 Lucas 数，所以 \(\dfrac{F_{2k}}{F_k} = L_k\)。所求和为 \[ \begin{gathered} L_1 + L_2 + \cdots + L_{10} \\ = 1 + 3 + 4 + 7 + 11 + 18 \\ {}+ 29 + 47 + 76 + 123 \\ = 319\text{。} \end{gathered} \] 等价地，\(L_1 + \cdots + L_{10}\) \(= L_{12} - 3\) \(= 322 - 3 = 319\)。

所以正确答案是 B。

English

Since \(F_{2k} = F_k L_k\) where \(L_k\) is the \(k\)th Lucas number, each term \(\dfrac{F_{2k}}{F_k} = L_k.\) The sum is \[ \begin{gathered} L_1 + L_2 + \cdots + L_{10} \\ = 1 + 3 + 4 + 7 + 11 + 18 \\ {}+ 29 + 47 + 76 + 123 \\ = 319. \end{gathered} \] (Equivalently, \(L_1 + \cdots + L_{10}\) \(= L_{12} - 3\) \(= 322 - 3 = 319.\))

Thus, the correct answer is B.

19 答案 B

六个顶点都在外接圆上，半径 \(R = \dfrac{14}{\sqrt3}\)，所以 \(R^2 = \dfrac{196}{3}\)。圆心角在 \(\theta\) 与 \(120^\circ - \theta\) 之间交替。六边形面积为 \[ \begin{aligned} &\tfrac12 R^2\bigl(3\sin\theta + 3\sin(120^\circ - \theta)\bigr) \\ &= 98\bigl(\sin\theta + \sin(120^\circ - \theta)\bigr) \end{aligned}\text{。}\]

由和差化积，\(\sin\theta + \sin(120^\circ - \theta)\) \(= 2\sin 60^\circ\cos(\theta - 60^\circ)\) \(= \sqrt3\cos(60^\circ - \theta)\)。令面积为 \(91\sqrt3\)，得 \(\sqrt3\cos(60^\circ - \theta)\) \(= \dfrac{91\sqrt3}{98} = \dfrac{13\sqrt3}{14}\)，所以 \(\cos(60^\circ - \theta) = \dfrac{13}{14}\)，且 \(\sin(60^\circ - \theta) = \dfrac{3\sqrt3}{14}\)。

于是 \(\tan(60^\circ - \theta) = \dfrac{3\sqrt3}{13}\)，并且 \[ \begin{aligned} &\tan\theta = \tan\bigl(60^\circ - (60^\circ - \theta)\bigr) \\ &= \frac{\sqrt3 - \frac{3\sqrt3}{13}}{1 + \sqrt3\cdot\frac{3\sqrt3}{13}} \\ &= \frac{\frac{10\sqrt3}{13}}{\frac{22}{13}} \\ &= \frac{5\sqrt3}{11} \end{aligned}\text{。}\]

所以正确答案是 B。

English

The six vertices lie on the circumcircle of radius \(R = \dfrac{14}{\sqrt3},\) so \(R^2 = \dfrac{196}{3}.\) Going around, the central angles alternate between \(\theta\) (three times) and \(120^\circ - \theta\) (three times). The cyclic-hexagon area is \[ \begin{aligned} &\tfrac12 R^2\bigl(3\sin\theta + 3\sin(120^\circ - \theta)\bigr) \\ &= 98\bigl(\sin\theta + \sin(120^\circ - \theta)\bigr). \end{aligned} \]

By sum-to-product, \(\sin\theta + \sin(120^\circ - \theta)\) \(= 2\sin 60^\circ\cos(\theta - 60^\circ)\) \(= \sqrt3\cos(60^\circ - \theta).\) Setting the area to \(91\sqrt3\) gives \(\sqrt3\cos(60^\circ - \theta)\) \(= \dfrac{91\sqrt3}{98} = \dfrac{13\sqrt3}{14},\) so \(\cos(60^\circ - \theta) = \dfrac{13}{14}\) and \(\sin(60^\circ - \theta) = \dfrac{3\sqrt3}{14}.\)

Then \(\tan(60^\circ - \theta) = \dfrac{3\sqrt3}{13},\) and \[ \begin{aligned} &\tan\theta = \tan\bigl(60^\circ - (60^\circ - \theta)\bigr) \\ &= \frac{\sqrt3 - \frac{3\sqrt3}{13}}{1 + \sqrt3\cdot\frac{3\sqrt3}{13}} \\ &= \frac{\frac{10\sqrt3}{13}}{\frac{22}{13}} \\ &= \frac{5\sqrt3}{11}. \end{aligned} \]

Thus, the correct answer is B.

20 答案 C

设 \(a = BC\)。中线长度满足 \(x^2\) \(= \dfrac{2\cdot 1600 + 2\cdot 1764 - a^2}{4}\) \(= \dfrac{6728 - a^2}{4}\)。三角形不等式要求 \(2 \lt a \lt 82\)，即 \(4 \lt a^2 \lt 6724\)，从而 \(1 \lt x \lt 41\)。所以 \((p, q) = (1, 41)\)。

在 \(AB = 40\)、\(AC = 42\) 固定时，面积为 \(\tfrac12\cdot 40\cdot 42\sin A\)，当 \(\angle A = 90^\circ\) 时最大，得 \(r = 840\)。此时 \(a^2 = 40^2 + 42^2 = 3364\)，\(x^2 = \dfrac{6728 - 3364}{4} = 841\)，所以 \(s = 29\)。

因此 \(p + q + r + s\) \(= 1 + 41 + 840 + 29\) \(= 911\)。

所以正确答案是 C。

English

Let \(a = BC.\) The median length gives \(x^2\) \(= \dfrac{2\cdot 1600 + 2\cdot 1764 - a^2}{4}\) \(= \dfrac{6728 - a^2}{4}.\) The triangle inequality requires \(2 \lt a \lt 82,\) i.e. \(4 \lt a^2 \lt 6724,\) which translates to \(1 \lt x \lt 41.\) So \((p, q) = (1, 41).\)

With \(AB = 40\) and \(AC = 42\) fixed, the area \(\tfrac12\cdot 40\cdot 42\sin A\) is largest when \(\angle A = 90^\circ,\) giving \(r = 840.\) Then \(a^2 = 40^2 + 42^2 = 3364,\) so \(x^2 = \dfrac{6728 - 3364}{4} = 841,\) i.e. \(s = 29.\)

Thus \(p + q + r + s\) \(= 1 + 41 + 840 + 29\) \(= 911.\)

Thus, the correct answer is C.

21 答案 C

\(3\)-\(4\)-\(5\) 与 \(5\)-\(12\)-\(13\) 三角形的最小角 \(\alpha, \beta\) 满足 \(\tan\alpha = \tfrac34\)，\(\tan\beta = \tfrac{5}{12}\)。\[ \begin{aligned} &\tan(\alpha + \beta) = \frac{\tfrac34 + \tfrac{5}{12}}{1 - \tfrac34\cdot\tfrac{5}{12}} \\ &= \frac{\tfrac{14}{12}}{\tfrac{33}{48}} \\ &= \frac{56}{33} \end{aligned}\text{。}\]

第三个最小角 \(\gamma\) 满足 \(\gamma = 90^\circ - (\alpha + \beta)\)，所以 \(\tan\gamma = \dfrac{33}{56}\)。直角边为 \(33\)、\(56\) 的直角三角形斜边为 \(\sqrt{33^2 + 56^2} = \sqrt{4225} = 65\)，构成本原三元组，周长为 \(33 + 56 + 65 = 154\)。

所以正确答案是 C。

English

The smallest angles \(\alpha, \beta\) of the \(3\)-\(4\)-\(5\) and \(5\)-\(12\)-\(13\) triangles have \(\tan\alpha = \tfrac34\) and \(\tan\beta = \tfrac{5}{12}.\) By the tangent addition formula, \[ \begin{aligned} &\tan(\alpha + \beta) = \frac{\tfrac34 + \tfrac{5}{12}}{1 - \tfrac34\cdot\tfrac{5}{12}} \\ &= \frac{\tfrac{14}{12}}{\tfrac{33}{48}} \\ &= \frac{56}{33}. \end{aligned} \]

The third smallest angle \(\gamma\) satisfies \(\gamma = 90^\circ - (\alpha + \beta),\) so \(\tan\gamma = \dfrac{33}{56}.\) The right triangle with legs \(33\) and \(56\) has hypotenuse \(\sqrt{33^2 + 56^2} = \sqrt{4225} = 65,\) a primitive triple. Its perimeter is \(33 + 56 + 65 = 154.\)

Thus, the correct answer is C.

22 答案 C

当 \(\angle B = 2\angle A\) 时，边长满足 \(b^2 = a(a + c)\)，其中 \(a = BC\)、\(b = CA\)、\(c = AB\)。所以 \(c = \dfrac{b^2 - a^2}{a}\) 必须是正整数，且三边必须构成非退化三角形。

又因为 \(\angle B=2\angle A\gt\angle A\)，所以 \(b\gt a\)。周长为 \(a+b+c=b+\dfrac{b^2}{a}\gt2b\)，因此周长小于 \(15\) 就要求 \(b\le6\)。对 \(b=2,3,4,5,6\)，\(b^2\) 的满足 \(a\lt b\) 的因数给出下列可能的数对 \[ \begin{gathered} (b,a)=(2,1),(3,1),\\ (4,1),(4,2),(5,1),\\ (6,1),(6,2),(6,3),(6,4) \end{gathered}\text{。}\] 逐一代入可知，除了 \((a,b)=(4,6)\) 给出 \(c=5\) 以外，其余都得到退化或不合法的三角形。因此 \((4,5,6)\) 是第一个合法的三角形，其周长为 \(15\)。

所以正确答案是 C。

English

When \(\angle B = 2\angle A,\) the side lengths satisfy \(b^2 = a(a + c),\) where \(a = BC,\) \(b = CA,\) \(c = AB.\) So \(c = \dfrac{b^2 - a^2}{a}\) must be a positive integer, and the sides must form a valid triangle.

Also \(b\gt a,\) because \(\angle B=2\angle A\gt\angle A.\) The perimeter is \(a+b+c=b+\dfrac{b^2}{a}\gt2b.\) Therefore a perimeter below \(15\) would require \(b\le6.\) For \(b=2,3,4,5,6,\) the divisors \(a\lt b\) of \(b^2\) give the possible pairs \[ \begin{gathered} (b,a)=(2,1),(3,1),\\ (4,1),(4,2),(5,1),\\ (6,1),(6,2),(6,3),(6,4). \end{gathered} \] Substitution gives a degenerate or invalid triangle in every case except \((a,b)=(4,6),\) which gives \(c=5.\) Thus \((4,5,6)\) is the first valid triangle, and its perimeter is \(15.\)

Thus, the correct answer is C.

23 答案 B

令 \(R\) 为正八边形外接圆半径，\(L\) 为每条侧棱长，则 \(L^2 = h^2 + R^2\)。由于 \(\angle AVD = 90^\circ\)，有 \(AD^2 = 2L^2\)。

顶点 \(A\) 与 \(D\) 相隔三步，圆心角为 \(135^\circ\)，所以 \(AD^2\) \(= 2R^2(1 - \cos 135^\circ)\) \(= R^2(2 + \sqrt2)\)。结合上式得 \(R^2(2 + \sqrt2) = 2(h^2 + R^2)\)，即 \(2h^2 = R^2\sqrt2\)。

边长为 \(1\) 的正八边形满足 \(R^2 = \dfrac{1}{2\sin^2(22.5^\circ)} = \dfrac{2 + \sqrt2}{2}\)。因此 \(h^2\) \(= \dfrac{R^2\sqrt2}{2}\) \(= \dfrac{(2 + \sqrt2)\sqrt2}{4}\) \(= \dfrac{2\sqrt2 + 2}{4}\) \(= \dfrac{1 + \sqrt2}{2}\)。

所以正确答案是 B。

English

Let \(R\) be the circumradius of the octagon and \(L\) the length of each lateral edge, so \(L^2 = h^2 + R^2.\) Since \(\angle AVD = 90^\circ,\) \(AD^2 = 2L^2.\)

Vertices \(A\) and \(D\) are three steps apart, a central angle of \(135^\circ,\) so \(AD^2\) \(= 2R^2(1 - \cos 135^\circ)\) \(= R^2(2 + \sqrt2).\) Setting \(R^2(2 + \sqrt2) = 2(h^2 + R^2)\) gives \(2h^2 = R^2\sqrt2.\)

For a regular octagon of side \(1,\) \(R^2 = \dfrac{1}{2\sin^2(22.5^\circ)} = \dfrac{2 + \sqrt2}{2}.\) Therefore \(h^2\) \(= \dfrac{R^2\sqrt2}{2}\) \(= \dfrac{(2 + \sqrt2)\sqrt2}{4}\) \(= \dfrac{2\sqrt2 + 2}{4}\) \(= \dfrac{1 + \sqrt2}{2}.\)

Thus, the correct answer is B.

24 答案 B

将每条边写成 \(\dfrac{2[\triangle]}{h}\)，则半周长为 \([\triangle]\bigl(\tfrac1a + \tfrac1b + \tfrac1c\bigr)\)。由 \(r = \dfrac{[\triangle]}{s}\)，可得 \(\dfrac1r = \dfrac1a + \dfrac1b + \dfrac1c\)。我们需要该和等于正整数 \(r\) 的倒数 \(\dfrac1r\)。边长与 \(\tfrac1a, \tfrac1b, \tfrac1c\) 成比例，所以非退化条件要求 \(\tfrac1a \lt \tfrac1b + \tfrac1c\)。

因为 \(a\le b\le c\le9\)，倒数之和至少为 \(\frac{3}{c}\ge\frac{1}{3}\)，所以整数 \(r\) 只能是 \(1,2,3\) 之一。又 \(\frac{1}{a}\lt\frac{1}{r}\le\frac{3}{a}\)，所以 \(r\lt a\le3r\)。对这少数几组 \(r,a\)，把 \(b=a,a+1,\ldots,9\) 代入 \[ c=\frac{abr}{ab-ar-br}\text{。}\] 只保留满足 \(b\le c\le9\) 的整数 \(c\)，就得到完整的列表 \[ \begin{array}{c|l} r& (a,b,c)\\ \hline 1&(2,3,6),(2,4,4),(3,3,3)\\ 2&(4,8,8),(6,6,6)\\ 3&(9,9,9) \end{array}\text{。}\] 其中三元组 \((2,3,6),(2,4,4)\) 和 \((4,8,8)\) 满足 \(\tfrac1a=\tfrac1b+\tfrac1c\)，因而给出退化三角形。剩下的三元组是 \((3,3,3),(6,6,6)\) 和 \((9,9,9)\)，所以答案是 \(3\)。

所以正确答案是 B。

English

Writing each side as \(\dfrac{2[\triangle]}{h},\) the semiperimeter is \([\triangle]\bigl(\tfrac1a + \tfrac1b + \tfrac1c\bigr),\) so the inradius \(r = \dfrac{[\triangle]}{s}\) satisfies \(\dfrac1r = \dfrac1a + \dfrac1b + \dfrac1c.\) We need this to be \(\dfrac1r\) for a positive integer \(r,\) with the sides (proportional to \(\tfrac1a, \tfrac1b, \tfrac1c\)) forming a non-degenerate triangle, requiring \(\tfrac1a \lt \tfrac1b + \tfrac1c.\)

Because \(a\le b\le c\le9,\) the reciprocal sum is at least \(\frac{3}{c}\ge\frac{1}{3},\) so the integer \(r\) is one of \(1,2,3.\) Also \(\frac{1}{a}\lt\frac{1}{r}\le\frac{3}{a},\) so \(r\lt a\le3r.\) For each of these few values of \(r,a,\) substitute \(b=a,a+1,\ldots,9\) into \[ c=\frac{abr}{ab-ar-br}. \] Keeping only integral \(c\) with \(b\le c\le9\) gives the complete list \[ \begin{array}{c|l} r& (a,b,c)\\ \hline 1&(2,3,6),(2,4,4),(3,3,3)\\ 2&(4,8,8),(6,6,6)\\ 3&(9,9,9). \end{array} \] The triples \((2,3,6),(2,4,4),\) and \((4,8,8)\) have \(\tfrac1a=\tfrac1b+\tfrac1c\) and therefore give degenerate triangles. The remaining triples are \((3,3,3),(6,6,6),\) and \((9,9,9),\) so the answer is \(3.\)

Thus, the correct answer is B.

25 答案 A

每个球独立地属于四种等可能类型之一：红色条纹、红色圆点、蓝色条纹、蓝色圆点。设 \(6\) 个球中有 \(k\) 个红色条纹球，红球总数为 \(R\)，条纹球总数为 \(S\)。对 Frida 的均匀随机选择，\(P(\text{红}) = \tfrac{R}{6}\)，\(P(\text{条纹}) = \tfrac{S}{6}\)，且 \(P(\text{红且条纹}) = \tfrac{k}{6}\)。独立性意味着 \(\dfrac{k}{6} = \dfrac{R}{6}\cdot\dfrac{S}{6}\)，即 \(6k = RS\)。

先计数满足 \(R\in\{0,6\}\) 或 \(S\in\{0,6\}\) 的分配。这四个条件各给另一种属性留下 \(2^6=64\) 种选择，而它们两两交集中的四个分配被重复计算。因此并集贡献 \(4\cdot64-4=252\)。

当 \(1\le R,S\le5\) 时，\(RS\) 是 \(6\) 的倍数这一条件只留下 \((R,S)=(2,3),(3,2)\) 或 \((3,4),(4,3)\)。每种情形都有 \(k=\frac{RS}{6}\)，四种类型的数量恰是 \((1,1,2,2)\) 的一个排列。因此每个数对贡献 \(\dfrac{6!}{1!1!2!2!}=180\) 种分配。有利分配数为 \(252+4\cdot180=972\)。在 \(4^6=4096\) 个等可能分配中，概率为 \(\dfrac{972}{4096}=\dfrac{243}{1024}\)，所以 \(m=243\)。

因此，正确答案是 A。

English

Each ball is independently one of four equally likely types: red-striped, red-dotted, blue-striped, blue-dotted. Suppose among the \(6\) balls there are \(k\) red-striped, with \(R\) red and \(S\) striped in total. For Frida’s uniform pick, \(P(\text{red}) = \tfrac{R}{6},\) \(P(\text{striped}) = \tfrac{S}{6},\) and \(P(\text{red and striped}) = \tfrac{k}{6}.\) Independence means \(\dfrac{k}{6} = \dfrac{R}{6}\cdot\dfrac{S}{6},\) i.e. \(6k = RS.\)

First count assignments with \(R\in\{0,6\}\) or \(S\in\{0,6\}.\) Each of these four conditions leaves \(2^6=64\) choices for the other attribute, and the four assignments at their pairwise intersections have been counted twice. Their union therefore contributes \(4\cdot64-4=252.\)

For \(1\le R,S\le5,\) the condition that \(RS\) be divisible by \(6\) leaves only \((R,S)=(2,3),(3,2)\) or \((3,4),(4,3).\) In each case \(k=\frac{RS}{6},\) and the four type counts are a permutation of \((1,1,2,2).\) Thus each pair contributes \(\dfrac{6!}{1!1!2!2!}=180\) assignments. The favorable count is therefore \(252+4\cdot180=972.\) Out of \(4^6=4096\) equally likely assignments, the probability is \(\dfrac{972}{4096}=\dfrac{243}{1024},\) so \(m=243.\)

Thus, the correct answer is A.