2024B AMC 12 真题（P版）

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.

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.

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.

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.

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.

\(N\) 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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

\(3\)-\(4\)-\(5\) 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.

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.

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.

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.

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.