2025A AMC 12 真题（P版）

1 答案 E

设 \(t\) 为从 \(1{:}30\) 起经过的小时数。Andy 向北行驶了 \(8t\) 英里，而 Betsy 晚一小时出发，向东行驶了 \(12(t-1)\) 英里。

令两人的距离相等，得到 \[8t = 12(t-1)\text{，}\] 所以 \(4t = 12\)，\(t = 3\)。

从 \(1{:}30\) 算起，三小时后的时间是 \(4{:}30\)。

因此，正确答案是 E。

English

Let \(t\) be the number of hours since \(1{:}30.\) Andy has traveled \(8t\) miles north, and Betsy, who started an hour later, has traveled \(12(t-1)\) miles east.

Setting the distances equal, \[8t = 12(t-1),\] so \(4t = 12\) and \(t = 3.\)

Three hours after \(1{:}30\) is \(4{:}30.\)

Thus, the correct answer is E.

2 答案 B

第一盒中有 \(5\) 磅花生、\(2\) 磅腰果和 \(3\) 磅杏仁。加入 \(x\) 磅第二种混合物会增加 \(0.2x\) 磅花生和 \(0.4x\) 磅腰果。

新的花生比例是 \(40\%\)，所以 \[\frac{5+0.2x}{10+x}=0.4\text{。}\] 这给出 \(5+0.2x=4+0.4x\)，所以 \(x=5\)。

现在腰果总量为 \(2 + 0.4(5) = 4\) 磅。

因此，正确答案是 B。

English

The first box has \(5\) lb peanuts, \(2\) lb cashews, and \(3\) lb almonds. Adding \(x\) pounds of the second mix contributes \(0.2x\) lb peanuts and \(0.4x\) lb cashews.

The new peanut fraction is \(40\%,\) so \[\frac{5+0.2x}{10+x}=0.4.\] This gives \(5+0.2x=4+0.4x,\) so \(x=5.\)

The cashews now total \(2 + 0.4(5) = 4\) pounds.

Thus, the correct answer is B.

3 答案 A

设学生人数为 \(s\)，Ash 的年龄为 \(a\)。学生年龄总和为 \(12s\)，加入 Ash 后得到 \[ \begin{aligned} 12s + a &= 14(s+1) \\ &\implies a = 2s + 14 \end{aligned}\text{。}\]

老师有 \(15 - s\) 人，年龄总和为 \(55(15-s)\)，加入 Ash 后得到 \[ \begin{aligned} 55(15-s) + a &= 52(16-s) \\ &\implies a = 3s + 7 \end{aligned}\text{。}\]

令 \(2s+14 = 3s+7\)，得 \(s = 7\)，所以 \(a = 2(7)+14 = 28\)。

因此，正确答案是 A。

English

Let \(s\) be the number of students and \(a\) be Ash’s age. The students’ ages total \(12s,\) and adding Ash gives \[ \begin{aligned} 12s + a &= 14(s+1) \\ &\implies a = 2s + 14. \end{aligned} \]

There are \(15 - s\) teachers with ages totaling \(55(15-s),\) and adding Ash gives \[ \begin{aligned} 55(15-s) + a &= 52(16-s) \\ &\implies a = 3s + 7. \end{aligned} \]

Setting \(2s+14 = 3s+7\) gives \(s = 7,\) so \(a = 2(7)+14 = 28.\)

Thus, the correct answer is A.

4 答案 B

设 \(T\) 为真陈述的数量。这四个陈述分别断言 \(T \ge 1\)、\(T \ge 2\)、\(T \le 2\) 和 \(T \le 3\)。

检查 \(T = 3\)：条件 \(T\ge1\)、\(T\ge2\)、\(T\le3\) 成立，也就是第一、第二、第四个陈述为真；而 \(T\le2\) 不成立，也就是第三个陈述为假。恰好有 \(3\) 个陈述为真，与 \(T = 3\) 一致。

对 \(T=0,1,2,3,4\)，成立的条件个数依次为 \(2,3,4,3,2\)，所以没有其他值是自洽的。因此恰好有一个陈述是假的。

因此，正确答案是 B。

English

Let \(T\) be the number of true statements. The statements assert \(T \ge 1,\) \(T \ge 2,\) \(T \le 2,\) and \(T \le 3,\) respectively.

Testing \(T = 3\): the conditions \(T\ge1,\) \(T\ge2,\) \(T\le3\) hold (statements one, two, four) and \(T\le2\) fails (statement three). Exactly \(3\) statements are true, matching \(T = 3.\)

For \(T=0,1,2,3,4,\) the respective numbers of true conditions are \(2,3,4,3,2,\) so no other value is self-consistent. Therefore exactly one statement is false.

Thus, the correct answer is B.

5 答案 D

设外层正方形面积为 \(1\)。嵌套正方形的面积为 \(1, k^2, k^4, \ldots\)，因此第 \(n\) 个正方形与第 \((n+1)\) 个正方形之间的环带面积为 \(k^{2n}(1-k^2)\)。

阴影环带是交替的那些，即 \(n = 0, 2, 4, \ldots\)，总面积为 \[ \begin{aligned} \sum_{j=0}^{\infty} k^{4j}(1-k^2) &= \frac{1-k^2}{1-k^4} \\ &= \frac{1}{1+k^2} \end{aligned}\text{。}\]

令 \(\dfrac{1}{1+k^2} = \dfrac{16}{25}\)，得 \(1 + k^2 = \dfrac{25}{16}\)，所以 \(k^2 = \dfrac{9}{16}\)，\(k = \dfrac{3}{4}\)。

因此，正确答案是 D。

English

Let the outer square have area \(1.\) The nested squares have areas \(1, k^2, k^4, \ldots,\) so the ring between the \(n\)th and \((n+1)\)th squares has area \(k^{2n}(1-k^2).\)

The shaded rings are the alternate ones \(n = 0, 2, 4, \ldots,\) with total area \[ \begin{aligned} \sum_{j=0}^{\infty} k^{4j}(1-k^2) &= \frac{1-k^2}{1-k^4} \\ &= \frac{1}{1+k^2}. \end{aligned} \]

Setting \(\dfrac{1}{1+k^2} = \dfrac{16}{25}\) gives \(1 + k^2 = \dfrac{25}{16},\) so \(k^2 = \dfrac{9}{16}\) and \(k = \dfrac{3}{4}.\)

Thus, the correct answer is D.

6 答案 B

给学生选择 \(2\) 把椅子、给老师选择 \(2\) 把椅子，共有 \(\binom{6}{2}\binom{4}{2} = 15 \cdot 6 = 90\) 种等可能结果。

圆桌周围有 \(6\) 对相邻椅子。学生可以得到任意一对相邻椅子；在剩下的 \(4\) 把椅子中，老师正好有 \(3\) 对相邻椅子可选。因此有 \(6 \cdot 3 = 18\) 种有利结果。

所求概率为 \(\dfrac{18}{90} = \dfrac{1}{5}\)。

因此，正确答案是 B。

English

Choosing \(2\) chairs for the students and \(2\) for the teachers gives \(\binom{6}{2}\binom{4}{2} = 15 \cdot 6 = 90\) equally likely outcomes.

A round table has \(6\) adjacent pairs of chairs. Give the students any adjacent pair; among the remaining \(4\) chairs there are exactly \(3\) adjacent pairs for the teachers. That is \(6 \cdot 3 = 18\) favorable outcomes.

The probability is \(\dfrac{18}{90} = \dfrac{1}{5}.\)

Thus, the correct answer is B.

7 答案 C

取对数，\(\log v = \log k + a\log n + b\log m\)。

当 \(n = 5\) 时，式子变为 \(\log v = (\log k + a\log 5)\) \(+ b\log m\)，与 \(4 + 2\log m\) 对应，所以 \(b = 2\)，且 \(\log k + a\log 5 = 4\)。

当 \(m = 25\) 时，式子变为 \(\log v = (\log k + b\log 25)\) \(+ a\log n\)，与 \(4 + 4\log n\) 对应，所以 \(a = 4\)，且 \(\log k + 2\log 25 = 4\)。

因此 \(\log k = 4 - \log 625\) \(= \log\dfrac{10000}{625}\) \(= \log 16\)，所以 \(k = 16\)。于是 \(k + a + b = 16 + 4 + 2 = 22\)。

因此，正确答案是 C。

English

Taking logarithms, \(\log v = \log k + a\log n + b\log m.\)

With \(n = 5,\) this reads \(\log v = (\log k + a\log 5)\) \(+ b\log m,\) matching \(4 + 2\log m,\) so \(b = 2\) and \(\log k + a\log 5 = 4.\)

With \(m = 25,\) it reads \(\log v = (\log k + b\log 25)\) \(+ a\log n,\) matching \(4 + 4\log n,\) so \(a = 4\) and \(\log k + 2\log 25 = 4.\)

Then \(\log k = 4 - \log 625\) \(= \log\dfrac{10000}{625}\) \(= \log 16,\) so \(k = 16.\) Hence \(k + a + b = 16 + 4 + 2 = 22.\)

Thus, the correct answer is C.

8 答案 E

圆周角 \(\angle BEC = 30^\circ\) 对着弧 \(BC = 60^\circ\)，所以同样对着弧 \(BC\) 的 \(\angle BAC\) 等于 \(30^\circ\)。同理，\(\angle CAD = 30^\circ\)。

因此 \(AC\) 平分 \(\angle BAD = 60^\circ\)。在 \(\triangle ABD\) 中， \[ \begin{aligned} BD^2 &= 9^2 + 24^2 \\ &\quad {}- 2(9)(24)\cos 60^\circ \\ &= 657 - 216 = 441\text{，} \end{aligned} \] 所以 \(BD = 21\)。

因为 \(AF\)（沿着 \(AC\)）平分 \(\angle BAD\)， 由角平分线定理， \(\dfrac{BF}{FD} = \dfrac{AB}{AD} = \dfrac{9}{24} = \dfrac{3}{8}\)。因此 \(BF = \dfrac{3}{11}\cdot 21 = \dfrac{63}{11}\)。

因此，正确答案是 E。

English

The inscribed angle \(\angle BEC = 30^\circ\) subtends arc \(BC = 60^\circ,\) so \(\angle BAC,\) which also subtends arc \(BC,\) equals \(30^\circ.\) Likewise \(\angle CAD = 30^\circ.\)

Thus \(AC\) bisects \(\angle BAD = 60^\circ.\) In \(\triangle ABD,\) \[ \begin{aligned} BD^2 &= 9^2 + 24^2 \\ &\quad {}- 2(9)(24)\cos 60^\circ \\ &= 657 - 216 = 441, \end{aligned} \] so \(BD = 21.\)

Since \(AF\) (along \(AC\)) bisects \(\angle BAD,\) the Angle Bisector Theorem gives \(\dfrac{BF}{FD} = \dfrac{AB}{AD} = \dfrac{9}{24} = \dfrac{3}{8}.\) Hence \(BF = \dfrac{3}{11}\cdot 21 = \dfrac{63}{11}.\)

Thus, the correct answer is E.

9 答案 E

计算得 \(w^2 = (2+i)^2 = 3 + 4i\)，所以这些点是 \((2,1)\) 和 \((3,4)\)。

过它们的直线斜率为 \(\dfrac{4-1}{3-2} = 3\)，因而方程为 \(y = 3x - 5\)。令 \(y = 0\) 得到 \(x = \dfrac{5}{3}\)。

所以 \(r = \dfrac{5}{3}\)。

因此，正确答案是 E。

English

Compute \(w^2 = (2+i)^2 = 3 + 4i,\) so the points are \((2,1)\) and \((3,4).\)

The line through them has slope \(\dfrac{4-1}{3-2} = 3,\) giving \(y = 3x - 5.\) Setting \(y = 0\) yields \(x = \dfrac{5}{3}.\)

So \(r = \dfrac{5}{3}.\)

Thus, the correct answer is E.

10 答案 B

设 \(R_1 = OA = OD\)，\(R_2 = OB = OC\)，并设 \(\alpha = \angle AOD = \angle BOC\)（两条射线重合）。劣弧 \(BC\) 的长度为 \(R_2\alpha = 2\pi\)，而优弧 \(AD\) 是其反向的大弧，所以 \(R_1(2\pi - \alpha) = 2\pi\)。

每条线段 \(AB = CD = R_2 - R_1 = 2\pi\)。

由前两个方程，\(R_2 = \dfrac{2\pi}{\alpha}\)，\(R_1 = \dfrac{2\pi}{2\pi - \alpha}\)。代入 \(R_2 - R_1 = 2\pi\) 并两边除以 \(2\pi\)，得到 \[\frac{1}{\alpha} - \frac{1}{2\pi - \alpha} = 1\text{，}\] 它可化简为 \(\alpha^2 - 2(1+\pi)\alpha + 2\pi = 0\)。

较大的根大于 \(2\pi\)，不可能是这段劣弧的圆心角，所以 \(\alpha = (1+\pi) - \sqrt{1+\pi^2}\)。于是 \[ \begin{aligned} R_1 &= \frac{2\pi}{2\pi - \alpha} \\ &= \frac{2\pi}{\pi - 1 + \sqrt{1+\pi^2}} \\ &= 1 - \pi + \sqrt{1+\pi^2} \end{aligned}\text{，}\] 其中最后一步作了有理化（分母乘以 \(\sqrt{1+\pi^2} - (\pi-1)\) 等于 \(2\pi\)）。

因此，正确答案是 B。

English

Let \(R_1 = OA = OD\) and \(R_2 = OB = OC,\) and let \(\alpha = \angle AOD = \angle BOC\) (the rays coincide). The minor arc \(BC\) has length \(R_2\alpha = 2\pi,\) and the major arc \(AD\) is the reflex arc, so \(R_1(2\pi - \alpha) = 2\pi.\)

Each segment \(AB = CD = R_2 - R_1 = 2\pi.\)

From the first two equations, \(R_2 = \dfrac{2\pi}{\alpha}\) and \(R_1 = \dfrac{2\pi}{2\pi - \alpha}.\) Substituting into \(R_2 - R_1 = 2\pi\) and dividing by \(2\pi\) gives \[\frac{1}{\alpha} - \frac{1}{2\pi - \alpha} = 1,\] which simplifies to \(\alpha^2 - 2(1+\pi)\alpha + 2\pi = 0.\)

The larger root is greater than \(2\pi,\) so it cannot be an angle of this minor arc. Thus \(\alpha = (1+\pi) - \sqrt{1+\pi^2}.\) Then \[ \begin{aligned} R_1 &= \frac{2\pi}{2\pi - \alpha} \\ &= \frac{2\pi}{\pi - 1 + \sqrt{1+\pi^2}} \\ &= 1 - \pi + \sqrt{1+\pi^2}, \end{aligned} \] after rationalizing (the denominator times \(\sqrt{1+\pi^2} - (\pi-1)\) equals \(2\pi\)).

Thus, the correct answer is B.

11 答案 A

由于 \(B\) 和 \(C\) 都在直线 \(y = 27\) 上，边 \(BC\) 是水平的，所以从 \(A\) 作的高是竖直直线 \(x = 2\)。

边 \(AC\) 的斜率为 \(\dfrac{27 - 31}{18 - 2} = -\dfrac{1}{4}\)，所以从 \(B\) 作的高斜率为 \(4\)：\[y - 27 = 4(x - 8)\text{。}\]

当 \(x = 2\) 时，\(y = 27 + 4(2 - 8) = 3\)。垂心为 \((2, 3)\)，坐标和为 \(5\)。

因此，正确答案是 A。

English

Since \(B\) and \(C\) both have \(y = 27,\) side \(BC\) is horizontal and the altitude from \(A\) is the vertical line \(x = 2.\)

Side \(AC\) has slope \(\dfrac{27 - 31}{18 - 2} = -\dfrac{1}{4},\) so the altitude from \(B\) has slope \(4\): \[y - 27 = 4(x - 8).\]

At \(x = 2,\) \(y = 27 + 4(2 - 8) = 3.\) The orthocenter is \((2, 3),\) with coordinate sum \(5.\)

Thus, the correct answer is A.

12 答案 B

每个因式 \(kx^2 - 4x - 3\) 的判别式为 \(16 + 12k \gt 0\)，所以它有两个实根；总共有 \(4050\) 个根。

对 \(kx^2 - 4x - 3\) 的根，倒数之和为 \(\dfrac{\text{和}}{\text{积}} = \dfrac{\frac{4}{k}}{-\frac{3}{k}} = -\dfrac{4}{3}\)，与 \(k\) 无关。

对全部 \(2025\) 个因式求和，\(\displaystyle\sum \frac{1}{r} = 2025\left(-\frac{4}{3}\right) = -2700\)。调和平均数为 \[\frac{4050}{-2700} = -\frac{3}{2}\text{。}\]

因此，正确答案是 B。

English

Each factor \(kx^2 - 4x - 3\) has discriminant \(16 + 12k \gt 0,\) so it has two real roots; there are \(4050\) roots in all.

For the roots of \(kx^2 - 4x - 3,\) the sum of reciprocals is \(\dfrac{\text{sum}}{\text{product}} = \dfrac{\frac{4}{k}}{-\frac{3}{k}} = -\dfrac{4}{3},\) independent of \(k.\)

Summing over all \(2025\) factors, \(\displaystyle\sum \frac{1}{r} = 2025\left(-\frac{4}{3}\right) = -2700.\) The harmonic mean is \[\frac{4050}{-2700} = -\frac{3}{2}.\]

Thus, the correct answer is B.

13 答案 D

要避免五个连续整数，只需去掉两个元素（例如 \(5\) 和 \(10\)），而去掉一个元素不可能同时击中两个不相交的区块 \(\{1,\ldots,5\}\) 和 \(\{9,\ldots,13\}\)。因此 \(N = 11\)。

从 \(13\) 个元素中选 \(11\) 个，等价于去掉 \(2\) 个，共有 \(\binom{13}{2} = 78\) 种。所选集合不含五个连续整数，当且仅当两个被去掉的元素合起来与每个窗口 \(\{t, t+1, t+2, t+3, t+4\}\) 相交，其中 \(t = 1, \ldots, 9\)。

这迫使一个被去掉的元素在 \(\{1,\ldots,5\}\) 中，另一个在 \(\{9,\ldots,13\}\) 中，并且两者相距不超过 \(5\)。有效的去法为 \(\{4,9\}\)、\(\{5,9\}\) 和 \(\{5,10\}\)，共 \(3\) 种。

概率为 \(\dfrac{3}{78} = \dfrac{1}{26}\)。

因此，正确答案是 D。

English

To avoid five consecutive integers, it suffices to remove two elements (for example \(5\) and \(10\)), and no single removal can hit both disjoint blocks \(\{1,\ldots,5\}\) and \(\{9,\ldots,13\}.\) Thus \(N = 11.\)

Choosing \(11\) of \(13\) elements is the same as removing \(2,\) which can be done in \(\binom{13}{2} = 78\) ways. The chosen set avoids five consecutive integers exactly when the two removed elements together intersect every window \(\{t, t+1, t+2, t+3, t+4\}\) for \(t = 1, \ldots, 9.\)

This forces one removed element in \(\{1,\ldots,5\},\) the other in \(\{9,\ldots,13\},\) and the two within \(5\) of each other. The valid removals are \(\{4,9\},\) \(\{5,9\},\) and \(\{5,10\},\) giving \(3\) of them.

The probability is \(\dfrac{3}{78} = \dfrac{1}{26}.\)

Thus, the correct answer is D.

14 答案 D

离心率相同，则 \(b = a\sqrt{1 - e^2}\)，所以面积 \(\pi a b \propto a^2\)。面积之比为 \(2025\)，因此 \(\dfrac{a_1}{a_2} = \sqrt{2025} = 45\)，其中 \(a_1, a_2\) 是两个半长轴。

两个椭圆共用焦点 \(G\)。在大椭圆中，\(G\) 是右焦点，所以右顶点位于 \(G\) 右侧 \(a_1 - c_1\) 处。在小椭圆中，\(G\) 是左焦点，所以右顶点位于 \(G\) 右侧 \(a_2 + c_2\) 处。内切使这两个点重合：\[a_1 - c_1 = a_2 + c_2\text{。}\]

利用 \(c = ea\)，得 \(a_1(1 - e) = a_2(1 + e)\)，所以 \(45(1 - e) = 1 + e\)，从而 \(46e = 44\)，\(e = \dfrac{22}{23}\)。

因此，正确答案是 D。

English

With the same eccentricity, \(b = a\sqrt{1 - e^2},\) so the area \(\pi a b \propto a^2.\) The area ratio \(2025\) gives \(\dfrac{a_1}{a_2} = \sqrt{2025} = 45,\) where \(a_1, a_2\) are the semi-major axes.

Both ellipses share focus \(G.\) On the large ellipse \(G\) is the right focus, so its right vertex lies \(a_1 - c_1\) to the right of \(G.\) On the small ellipse \(G\) is the left focus, so its right vertex lies \(a_2 + c_2\) to the right of \(G.\) Internal tangency makes these coincide: \[a_1 - c_1 = a_2 + c_2.\]

Using \(c = ea,\) \(a_1(1 - e) = a_2(1 + e),\) so \(45(1 - e) = 1 + e,\) giving \(46e = 44\) and \(e = \dfrac{22}{23}.\)

Thus, the correct answer is D.

15 答案 C

集合 \(\{11, 12, \ldots, 20\}\) 有 \(10\) 个元素，并且是无和集，因为其中任意两个元素之和至少为 \(22 \gt 20\)。

为了证明上界，设 \(a_1 \lt a_2 \lt \cdots \lt a_k\) 是一个无和子集。对每个 \(i \lt k\)，差 \(a_k - a_i\) 不能在 \(S\) 中，因为若在，则 \((a_k - a_i) + a_i = a_k \in S\)，违反无和性。

这 \(k - 1\) 个差互不相同，属于 \(\{1, \ldots, 19\}\)，并且与 \(S\) 的 \(k\) 个元素不相交。所以 \(k + (k - 1) \le 20\)，得到 \(k \le 10\)。

因此，正确答案是 C。

English

The set \(\{11, 12, \ldots, 20\}\) has \(10\) elements and is sum-free, since any two elements sum to at least \(22 \gt 20.\)

For the upper bound, let \(a_1 \lt a_2 \lt \cdots \lt a_k\) be a sum-free subset. Each difference \(a_k - a_i\) for \(i \lt k\) cannot lie in \(S,\) because \((a_k - a_i) + a_i = a_k \in S\) would violate sum-freeness.

These \(k - 1\) differences are distinct, lie in \(\{1, \ldots, 19\},\) and are disjoint from the \(k\) elements of \(S.\) So \(k + (k - 1) \le 20,\) giving \(k \le 10.\)

Thus, the correct answer is C.

16 答案 D

由余弦定理，\[ \begin{aligned} \cos B &= \frac{80^2 + 45^2 - 75^2}{2 \cdot 80 \cdot 45} \\ &= \frac{2800}{7200} = \frac{7}{18} \end{aligned}\text{。}\]

到 \(AB\) 的高是从 \(C\) 作出的，其垂足在边 \(AB\) 上，距 \(B\) 的距离为 \(BC\cos B = 45 \cdot \dfrac{7}{18} = 17.5\)。

沿着从 \(B\) 出发的角平分线，平行于 \(AB\) 的分量是 \(BP\cos\dfrac{B}{2}\)，它必须到达高的垂足：\(BP\cos\dfrac{B}{2} = 17.5\)。

因为 \(\cos\dfrac{B}{2} = \sqrt{\dfrac{1 + \frac{7}{18}}{2}}\) \(= \sqrt{\dfrac{25}{36}}\) \(= \dfrac{5}{6}\)，所以 \(BP = \dfrac{17.5}{\frac{5}{6}} = 21\)。

因此，正确答案是 D。

English

By the Law of Cosines, \[ \begin{aligned} \cos B &= \frac{80^2 + 45^2 - 75^2}{2 \cdot 80 \cdot 45} \\ &= \frac{2800}{7200} = \frac{7}{18}. \end{aligned} \]

The altitude to \(AB\) is drawn from \(C,\) and its foot is at distance \(BC\cos B = 45 \cdot \dfrac{7}{18} = 17.5\) from \(B\) along \(AB.\)

Along the bisector from \(B,\) the component parallel to \(AB\) is \(BP\cos\dfrac{B}{2},\) which must reach the altitude’s foot: \(BP\cos\dfrac{B}{2} = 17.5.\)

Since \(\cos\dfrac{B}{2} = \sqrt{\dfrac{1 + \frac{7}{18}}{2}}\) \(= \sqrt{\dfrac{25}{36}}\) \(= \dfrac{5}{6},\) we get \(BP = \dfrac{17.5}{\frac{5}{6}} = 21.\)

Thus, the correct answer is D.

17 答案 A

根的和为 \(-6i\)，所以重心为 \(-2i\)。代入 \(z = u - 2i\)，得 \[ \begin{gathered} (u - i)(u)(u + i) + 10 \\ = u(u^2 + 1) + 10 \\ = u^3 + u + 10 \end{gathered}\text{。}\]

因为 \(u = -2\) 是一个根，\(u^3 + u + 10\) \(= (u + 2)(u^2 - 2u + 5)\)，所以根为 \(u = -2\) 和 \(u = 1 \pm 2i\)。

这些点是 \((-2, 0)\)、\((1, 2)\)、\((1, -2)\)。\((1, 2)\) 与 \((1, -2)\) 之间的底边长为 \(4\)，到 \((-2, 0)\) 的水平距离为 \(3\)，所以面积为 \(\dfrac{1}{2}(4)(3) = 6\)。平移不改变面积。

因此，正确答案是 A。

English

The sum of the roots is \(-6i,\) so the centroid is \(-2i.\) Substituting \(z = u - 2i,\) \[ \begin{gathered} (u - i)(u)(u + i) + 10 \\ = u(u^2 + 1) + 10 \\ = u^3 + u + 10. \end{gathered} \]

Since \(u = -2\) is a root, \(u^3 + u + 10\) \(= (u + 2)(u^2 - 2u + 5),\) giving roots \(u = -2\) and \(u = 1 \pm 2i.\)

These are the points \((-2, 0),\) \((1, 2),\) \((1, -2).\) The base between \((1, 2)\) and \((1, -2)\) has length \(4,\) at horizontal distance \(3\) from \((-2, 0),\) so the area is \(\dfrac{1}{2}(4)(3) = 6.\) Translation does not change the area.

Thus, the correct answer is A.

18 答案 C

如果某个变量为 \(0\)，例如 \(z = 0\)，那么 \(zx = 0 \gt y\) 不可能成立。因此 \(x, y, z \in \{1, \ldots, 8\}\) 是互不相同的正整数。

条件是对称的。对互异的数值 \(a \lt b \lt c\) 有 \(ac \gt b\) 和 \(bc \gt a\) 自动成立，所以唯一真正的限制是 \(ab \gt c\)。只要它成立，所有 \(6\) 种排列都可行。

对每个可能的最小值 \(a\)，满足 \(ab\gt c\) 的数对 \(a\lt b\lt c\le8\) 的个数为 \[ \begin{array}{c|rrrrrr} a&1&2&3&4&5&6\\ \hline \text{个数}&0&11&10&6&3&1 \end{array}\text{。}\] 对固定的 \(a,b\)，这个个数来自在 \(b+1\) 与 \(\min(8,ab-1)\) 之间选取 \(c\)。因此共有 \(31\) 个无序三元组，而且每个三元组的全部 \(6\) 种排序都可行。答案为 \(6\cdot31=186\)。

因此，正确答案是 C。

English

If any variable is \(0,\) say \(z = 0,\) then \(zx = 0 \gt y\) is impossible. So \(x, y, z \in \{1, \ldots, 8\}\) are distinct positive integers.

The conditions are symmetric. For distinct values \(a \lt b \lt c,\) we have \(ac \gt b\) and \(bc \gt a\) automatically, so the only real constraint is \(ab \gt c.\) When it holds, all \(6\) orderings work.

For each possible smallest value \(a,\) the numbers of pairs \(a\lt b\lt c\le8\) satisfying \(ab\gt c\) are \[ \begin{array}{c|rrrrrr} a&1&2&3&4&5&6\\ \hline \text{count}&0&11&10&6&3&1. \end{array} \] For fixed \(a,b,\) this count comes from choosing \(c\) between \(b+1\) and \(\min(8,ab-1).\) Thus there are \(31\) unordered triples, and all \(6\) orderings of each work. The answer is \(6\cdot31=186.\)

Thus, the correct answer is C.

19 答案 E

由韦达定理，\(a + b + c = 0\)， \(ab + bc + ca = k\)， 且 \(abc = -1\)。

将原和分组为 \[ \begin{aligned} &a^2b^2(a + b) + b^2c^2(b + c) \\ &\quad {}+ c^2a^2(c + a)\text{。} \end{aligned} \] 因为 \(a + b + c = 0\)，所以 \(a + b = -c\)、\(b + c = -a\)、\(c + a = -b\)。

因此原和等于 \[ \begin{gathered} -a^2b^2 c - ab^2c^2 - a^2bc^2 \\ = -abc(ab + bc + ca) \\ = -(-1)(k) = k\text{。} \end{gathered} \]

因此，正确答案是 E。

English

By Vieta’s formulas, \(a + b + c = 0,\) \(ab + bc + ca = k,\) and \(abc = -1.\)

Group the sum as \[ \begin{aligned} &a^2b^2(a + b) + b^2c^2(b + c) \\ &\quad {}+ c^2a^2(c + a). \end{aligned} \] Since \(a + b + c = 0,\) we have \(a + b = -c,\) \(b + c = -a,\) \(c + a = -b.\)

So the sum equals \[ \begin{gathered} -a^2b^2 c - ab^2c^2 - a^2bc^2 \\ = -abc(ab + bc + ca) \\ = -(-1)(k) = k. \end{gathered} \]

Thus, the correct answer is E.

20 答案 C

顶部是一条长为 \(7\) 的棱，居中位于底面上方某个高度 \(h\)。它的两个端点分别在 \(13 \times 8\) 底面的 \((3, 4)\) 和 \((10, 4)\) 上方。从端点到一个底面顶点的斜边长为 \(\sqrt{3^2 + 4^2 + h^2} = 13\)，所以 \(h = 12\)。

在高度 \(z\) 处，水平截面是一个矩形，尺寸为 \(\left(13 - \dfrac{z}{2}\right)\) 乘 \(\left(8 - \dfrac{2z}{3}\right)\)。当 \(z = 0\) 时面积为 \(104\)；当 \(z = 6\) 时面积为 \(10 \cdot 4 = 40\)；当 \(z = 12\) 时顶棱面积为 \(0\)。

由棱台公式，\[ \begin{aligned} V &= \frac{12}{6}\left(104 + 4 \cdot 40 + 0\right) \\ &= 2(264) = 528 \end{aligned}\text{。}\]

因此，正确答案是 C。

English

The top is a ridge of length \(7,\) centered above the base at some height \(h.\) Its endpoints sit above \((3, 4)\) and \((10, 4)\) of the \(13 \times 8\) base. A slant edge to a base corner has length \(\sqrt{3^2 + 4^2 + h^2} = 13,\) so \(h = 12.\)

At height \(z,\) the horizontal cross-section is a rectangle measuring \(\left(13 - \dfrac{z}{2}\right)\) by \(\left(8 - \dfrac{2z}{3}\right).\) At \(z = 0\) its area is \(104\); at \(z = 6\) it is \(10 \cdot 4 = 40\); at \(z = 12\) the ridge has area \(0.\)

By the prismatoid formula, \[ \begin{aligned} V &= \frac{12}{6}\left(104 + 4 \cdot 40 + 0\right) \\ &= 2(264) = 528. \end{aligned} \]

Thus, the correct answer is C.

21 答案 A

如果 \(k=0\) 或 \(m=0\)，原比值就是 \(2^a\)，不可能等于 \(964\)。所以 \(k,m\gt0\)。

对等比级数求和，分子为 \(4^a\dfrac{4^{k(m+1)} - 1}{4^k - 1}\)，分母为 \(2^a\dfrac{2^{k(m+1)} - 1}{2^k - 1}\)。利用 \(4^N - 1 = (2^N - 1)(2^N + 1)\)，比值化简为 \[2^a \cdot \frac{2^{k(m+1)} + 1}{2^k + 1} = 964\text{。}\]

左边的分数是奇数除以奇数，所以它的 \(2\) 的指数为 \(0\)。因为 \(964=2^2\cdot241\)，必须有 \(a=2\)，并且 \[ 2^{k(m+1)}+1=241(2^k+1)\text{。}\]将此式模 \(2^k\) 化简，得到 \(240\) 是 \(2^k\) 的倍数，所以 \(k\le4\)。当 \(k=1,2,3\) 时，所需的 \(2\) 的幂将分别为 \(722,1204,2168\)，而它们都不是 \(2\) 的幂。然而当 \(k=4\) 时，\(241(2^4+1)=4097=2^{12}+1\)，所以 \(k(m+1)=12\) 且 \(m=2\)。这也证明了解的唯一性。

于是 \(a + k + m = 2 + 4 + 2 = 8\)。

因此，正确答案是 A。

English

If \(k=0\) or \(m=0,\) the original ratio is just \(2^a,\) which cannot equal \(964.\) Hence \(k,m\gt0.\)

Summing the geometric series, the numerator is \(4^a\dfrac{4^{k(m+1)} - 1}{4^k - 1}\) and the denominator is \(2^a\dfrac{2^{k(m+1)} - 1}{2^k - 1}.\) Using \(4^N - 1 = (2^N - 1)(2^N + 1),\) the ratio simplifies to \[2^a \cdot \frac{2^{k(m+1)} + 1}{2^k + 1} = 964.\]

The fraction on the left is odd over odd, so its power of \(2\) is \(0.\) Since \(964=2^2\cdot241,\) we must have \(a=2,\) and \[ 2^{k(m+1)}+1=241(2^k+1). \] Reducing this equation modulo \(2^k\) shows that \(240\) is divisible by \(2^k,\) so \(k\le4.\) For \(k=1,2,3,\) the required powers of \(2\) would be \(722,1204,2168,\) none of which is a power of \(2.\) For \(k=4,\) however, \(241(2^4+1)=4097=2^{12}+1,\) so \(k(m+1)=12\) and \(m=2.\) This also proves uniqueness.

Then \(a + k + m = 2 + 4 + 2 = 8.\)

Thus, the correct answer is A.

22 答案 E

将数按大小排列为 \(x_1 \gt x_2 \gt x_3\)；顺序统计量在这个区域上的联合密度为 \(6\)。所求事件为 \(x_1 \gt 2x_2\)。

对 \(x_3\) 从 \(0\) 到 \(x_2\) 积分，会贡献因子 \(x_2\)。于是 \[ \begin{aligned} P &= 6\int_0^{\frac{1}{2}} x_2\int_{2x_2}^{1} dx_1\, dx_2 \\ &= 6\int_0^{\frac{1}{2}} x_2(1 - 2x_2)\, dx_2 \end{aligned}\text{。}\]

这等于 \(6\left(\dfrac{1}{8} - \dfrac{1}{12}\right) = 6 \cdot \dfrac{1}{24} = \dfrac{1}{4}\)。

因此，正确答案是 E。

English

Order the values as \(x_1 \gt x_2 \gt x_3\); the joint density of the order statistics is \(6\) on this region. The event is \(x_1 \gt 2x_2.\)

Integrating \(x_3\) from \(0\) to \(x_2\) contributes a factor of \(x_2.\) Then \[ \begin{aligned} P &= 6\int_0^{\frac{1}{2}} x_2\int_{2x_2}^{1} dx_1\, dx_2 \\ &= 6\int_0^{\frac{1}{2}} x_2(1 - 2x_2)\, dx_2. \end{aligned} \]

This equals \(6\left(\dfrac{1}{8} - \dfrac{1}{12}\right) = 6 \cdot \dfrac{1}{24} = \dfrac{1}{4}.\)

Thus, the correct answer is E.

23 答案 C

数字互不相同，并且来自 \(\{1, \ldots, 9\}\)，条件“没有数字同时与两个更大的数字相邻”等价于没有内部数字小于它的两个相邻数字。

对固定的 \(n\) 个数字，从大到小插入来构造排列；每个新的较小数字必须放在当前排列的两个端点之一，因此有 \(2^{n-1}\) 个有效排列。

对所有非空数字子集求和，\[ \begin{gathered} \sum_{n=1}^{9}\binom{9}{n}2^{n-1} \\ = \frac{1}{2}\sum_{n=1}^{9}\binom{9}{n}2^{n} \\ = \frac{3^9 - 1}{2} = \frac{19682}{2} = 9841 \end{gathered}\text{。}\]

因此，正确答案是 C。

English

The digits are distinct and drawn from \(\{1, \ldots, 9\},\) and “no digit adjacent to two greater digits” means no interior digit is smaller than both neighbors.

For a fixed set of \(n\) digits, build the arrangement by inserting digits from largest to smallest; each new (smaller) digit must go to one of the two ends, giving \(2^{n-1}\) valid arrangements.

Summing over all nonempty digit subsets, \[ \begin{gathered} \sum_{n=1}^{9}\binom{9}{n}2^{n-1} \\ = \frac{1}{2}\sum_{n=1}^{9}\binom{9}{n}2^{n} \\ = \frac{3^9 - 1}{2} = \frac{19682}{2} = 9841. \end{gathered} \]

Thus, the correct answer is C.

24 答案 C

\(12\) 个外圆的圆心位于半径为 \(r + 1\) 的圆上，形成一个正 \(12\) 边形。相邻圆心距离为 \(2\)（两个圆半径都为 \(1\)），它们之间的圆心角为 \(30^\circ\)。

因此 \(2(r + 1)\sin 15^\circ = 2\)，所以 \(r + 1 = \dfrac{1}{\sin 15^\circ}\)。由于 \(\sin 15^\circ = \dfrac{\sqrt{6} - \sqrt{2}}{4}\)，\[r + 1 = \frac{4}{\sqrt{6} - \sqrt{2}} = \sqrt{6} + \sqrt{2}\text{。}\]

于是 \(r = \sqrt{6} + \sqrt{2} - 1\)，所以 \(a + b + c = 6 + 2 - 1 = 7\)。

因此，正确答案是 C。

English

The centers of the \(12\) outer circles lie on a circle of radius \(r + 1,\) forming a regular \(12\)-gon. Adjacent centers are \(2\) apart (both circles have radius \(1\)), and the central angle between them is \(30^\circ.\)

Thus \(2(r + 1)\sin 15^\circ = 2,\) so \(r + 1 = \dfrac{1}{\sin 15^\circ}.\) Since \(\sin 15^\circ = \dfrac{\sqrt{6} - \sqrt{2}}{4},\) \[r + 1 = \frac{4}{\sqrt{6} - \sqrt{2}} = \sqrt{6} + \sqrt{2}.\]

Then \(r = \sqrt{6} + \sqrt{2} - 1,\) so \(a + b + c = 6 + 2 - 1 = 7.\)

Thus, the correct answer is C.

25 答案 E

本题已作废：按题面作答，正确答案不在选项中。下面给出计数。

对于 \(\{f \le 0\} = [a, b] \cup (c, d)\)，闭区间端点 \(a, b\) 必须是 \(P\) 的零点，且在这些点 \(Q\ne0\)，而 \(c,d\) 必须是极点。因此符号的变化必须与下面这个函数一致： \[ g(x)=\frac{(x-a)(x-b)}{(x-c)(x-d)}\text{。}\] \(P\) 中未使用的第三个因子必须与 \(Q\) 中未使用的第三个因子相同；否则会多出零点、极点或符号变化。

如果这个公共因子是 \(a,b,c,d\) 之外的第五个值，它不能在任一区间内部形成空点。在五个数从小到大的排列中，它只能位于 \(a\) 之前、\(b\) 与 \(c\) 之间，或 \(d\) 之后，因此给出 \(3\) 个函数。

另一种可能是公共因子等于 \(c\) 或 \(d\)。对 \(a,b,c,d\) 的每一种选择（共 \(\binom54=5\) 种），这两个多项式对化简为同一个公式 \(g\)，且定义域相同（都排除 \(c,d\)），所以只定义一个函数。这再给出 \(5\) 个函数，按题面字面理解总数为 \(3+5=8\)。

如果在后五种情形中把两个多项式对分别计数，就会得到 \(3+2\cdot5=13\)，即临时答案 (E)，但这没有回答题目所问的函数个数。因此印刷的选项都不正确。

English

This problem was voided: as written, its answer is not among the choices. Here is the count.

For \(\{f \le 0\} = [a, b] \cup (c, d),\) the endpoints \(a, b\) of the closed interval must be zeros of \(P\) at which \(Q\ne0,\) while \(c,d\) must be poles. The required sign pattern is therefore that of \[ g(x)=\frac{(x-a)(x-b)}{(x-c)(x-d)}. \] The unused third factor of \(P\) must match the unused third factor of \(Q;\) otherwise there would be an extra zero, pole, or sign change.

If the common factor is the fifth value not among \(a,b,c,d,\) it cannot make a hole inside either interval. In the ordered list of five values it may therefore occur before \(a,\) between \(b\) and \(c,\) or after \(d,\) giving \(3\) functions.

Alternatively, the common factor can equal \(c\) or \(d.\) For each of the \(\binom54=5\) choices of \(a,b,c,d,\) these two polynomial pairs simplify to the same formula \(g\) and have the same domain (both omit \(c,d\)), so they define only one function. This gives \(5\) more functions, for a literal total of \(3+5=8.\)

Counting the two polynomial pairs separately in each of the last five cases gives \(3+2\cdot5=13,\) the provisional answer (E), but that does not answer the stated question about functions. Thus none of the printed choices is correct.