2020A AMC 12 真题（P版）

1 答案 C

Carlos 拿走 \(70\%\) 后，剩余部分为 \(30\%\)。

Maria 拿走其中三分之一，即 \(\tfrac13 \cdot 30\% = 10\%\)，所以还剩 \(30\% - 10\% = 20\%\)。

因此，正确答案是 C。

English

After Carlos takes \(70\%,\) the remaining portion is \(30\%.\)

Maria takes one third of this, namely \(\tfrac13 \cdot 30\% = 10\%,\) leaving \(30\% - 10\% = 20\%.\)

Thus, C is the correct answer.

2 答案 C

把每个字母拆成线段。字母 \(A\) 包含一条长为 \(2\sqrt{2}\) 的斜线、一条长为 \(2\) 的竖线，以及一条长为 \(1\) 的横线。

字母 \(M\) 有两条长为 \(2\) 的竖线和两条各长 \(\sqrt{2}\) 的斜线。字母 \(C\) 由三条长为 \(2\) 的边组成。

直线段的总长为 \(2 + 1 + 2 + 2 + 2 + 2 + 2 = 13\)，斜线段的总长为 \(2\sqrt2 + \sqrt2 + \sqrt2 = 4\sqrt2\)。

总和为 \(13 + 4\sqrt2\)。

因此，正确答案是 C。

English

Split each letter into its segments. The \(A\) is a diagonal of length \(2\sqrt{2},\) a vertical of length \(2,\) and a crossbar of length \(1.\)

The \(M\) has two verticals of length \(2\) and two diagonals of length \(\sqrt{2}\) each. The \(C\) is three sides of length \(2.\)

The straight pieces total \(2 + 1 + 2 + 2 + 2 + 2 + 2 = 13,\) and the diagonal pieces total \(2\sqrt2 + \sqrt2 + \sqrt2 = 4\sqrt2.\)

The sum is \(13 + 4\sqrt2.\)

Thus, C is the correct answer.

3 答案 E

\(2\) 小时内她行驶 \(120\) 英里，收入为 \(120 \cdot \$0.50 = \$60\)。

她用掉 \(120 \div 30 = 4\) 加仑汽油，花费 \(4 \cdot \$2.00 = \$8\)。

净收入为 \(\$60 - \$8 = \$52\)，因此每小时为 \(\$52 \div 2 = \$26\)。

因此，正确答案是 E。

English

In \(2\) hours she drives \(120\) miles, earning \(120 \cdot \$0.50 = \$60.\)

She uses \(120 \div 30 = 4\) gallons, costing \(4 \cdot \$2.00 = \$8.\)

Her net earnings are \(\$60 - \$8 = \$52,\) so her rate is \(\$52 \div 2 = \$26\) per hour.

Thus, E is the correct answer.

4 答案 B

要能被 \(5\) 整除，末位必须是 \(0\) 或 \(5\)，又因为所有数字都要是偶数，所以末位只能是 \(0\)。

首位是非零偶数：\(2, 4, 6, 8\)，共有 \(4\) 种选择。中间两位各可以是任意偶数 \(0, 2, 4, 6, 8\)，各有 \(5\) 种选择。

总数为 \(4 \cdot 5 \cdot 5 \cdot 1 = 100\)。

因此，正确答案是 B。

English

To be divisible by \(5\) the last digit is \(0\) or \(5,\) and to be even it must be \(0.\) So the units digit is fixed.

The leading digit is a nonzero even digit: \(2, 4, 6, 8\) give \(4\) choices. Each of the two middle digits is any even digit \(0, 2, 4, 6, 8,\) giving \(5\) choices each.

The total is \(4 \cdot 5 \cdot 5 \cdot 1 = 100.\)

Thus, B is the correct answer.

5 答案 C

这 \(25\) 个整数的和为 \(\dfrac{(-10 + 14) \cdot 25}{2} = 50\)。

五行的和都相同，并且合起来就是总和，所以每行的和为 \(50 \div 5 = 10\)。

因此，正确答案是 C。

English

The sum of the \(25\) integers is \(\dfrac{(-10 + 14) \cdot 25}{2} = 50.\)

The five rows each have the same sum and together account for the total, so each row sums to \(50 \div 5 = 10.\)

Thus, C is the correct answer.

6 答案 D

要同时具有这两种对称性，对称轴必须是 \(5\) 乘 \(4\) 网格的竖直和水平中心线。每个已涂色正方形都会迫使它关于每条轴反射得到的正方形也被涂色。

顶部的正方形不在中心线上，所以它的反射组共有 \(4\) 个正方形，需要再涂 \(3\) 个。中间的正方形在中央列上，所以它的反射组共有 \(2\) 个正方形，需要再涂 \(1\) 个。右下的正方形同样有一个大小为 \(4\) 的反射组，需要再涂 \(3\) 个。

至少还需涂色的正方形数为 \(3 + 1 + 3 = 7\)。

因此，正确答案是 D。

English

For both symmetries, the lines must be the vertical and horizontal center lines of the \(5\)-by-\(4\) grid. Every shaded square then forces the squares obtained by reflecting it across each line.

The top square lies off-center, so its reflection group has \(4\) squares, requiring \(3\) more. The middle square sits on the central column, so its group has \(2\) squares, requiring \(1\) more. The bottom-right square again has a group of \(4,\) requiring \(3\) more.

The least number of additional squares is \(3 + 1 + 3 = 7.\)

Thus, D is the correct answer.

7 答案 B

这些立方体的边长为 \(1, 2, \ldots, 7\)。边长为 \(k\) 的立方体的四个侧面贡献 \(4k^2\)，所以竖直侧面的总面积为 \(4(1^2 + 2^2 + \cdots + 7^2)\) \(= 4 \cdot 140 = 560\)。

从正上方看，每一块朝上的水平面都无重叠地投影到底部的 \(7 \times 7\) 正方形上，面积为 \(49\)。从下方看同理，再得到 \(49\)。

总表面积为 \(560 + 49 + 49 = 658\)。

因此，正确答案是 B。

English

The side lengths are \(1, 2, \ldots, 7.\) The four side faces of cube \(k\) contribute \(4k^2,\) so the vertical faces total \(4(1^2 + 2^2 + \cdots + 7^2)\) \(= 4 \cdot 140 = 560.\)

Viewed from directly above, every upward-facing horizontal patch projects onto the \(7 \times 7\) base without overlap, giving \(49.\) Viewed from below, the same is true, giving another \(49.\)

The total surface area is \(560 + 49 + 49 = 658.\)

Thus, B is the correct answer.

8 答案 C

中位数是第 \(2020\) 小和第 \(2021\) 小的值的平均。

不超过 \(2020\) 的完全平方数为 \(1^2, \ldots, 44^2\)（因为 \(44^2 = 1936\) 且 \(45^2 = 2025\)），所以共有 \(44\) 个。

在这个列表中，数值 \(\le 1976\) 的项包括 \(1976\) 个整数 \(1, \ldots, 1976\) 以及这 \(44\) 个平方数，总数为 \(1976 + 44 = 2020\)。

因此第 \(2020\) 个值是 \(1976\)，第 \(2021\) 个值是 \(1977\)，中位数为 \(\dfrac{1976 + 1977}{2} = 1976.5\)。

因此，正确答案是 C。

English

The median is the average of the \(2020\)th and \(2021\)st smallest values.

The perfect squares that are at most \(2020\) are \(1^2, \ldots, 44^2\) (since \(44^2 = 1936\) and \(45^2 = 2025\)), so there are \(44\) of them.

Among the list, the numbers \(\le 1976\) are the \(1976\) integers \(1, \ldots, 1976\) together with those \(44\) squares, totaling \(1976 + 44 = 2020.\)

Thus the \(2020\)th value is \(1976\) and the \(2021\)st value is \(1977,\) making the median \(\dfrac{1976 + 1977}{2} = 1976.5.\)

Thus, C is the correct answer.

9 答案 E

在 \([0, 2\pi]\) 上，\(\cos\left(\tfrac{x}{2}\right)\) 的图像是一条从 \(1\) 单调下降到 \(-1\) 的弧。

函数 \(\tan(2x)\) 的周期为 \(\tfrac{\pi}{2}\)，竖直渐近线位于 \(x = \tfrac{\pi}{4}, \tfrac{3\pi}{4}, \tfrac{5\pi}{4}, \tfrac{7\pi}{4}\)。它们将区间分成五个分支。在每个分支上，\(\tan(2x)\) 严格递增，而 \(\cos(\tfrac{x}{2})\) 递减，所以每个分支至多有一个交点。

中间三个分支都从 \(-\infty\) 变化到 \(+\infty\)，所以各有一个交点。在第一个分支上，\(\tan(0)=0\lt1=\cos(0)\)，而正切趋于 \(+\infty\)。在最后一个分支上，正切从 \(-\infty\) 开始，在终点为 \(0\gt-1=\cos\pi\)。因此两个外侧分支也各有一个交点，总计 \(5\) 个。

所以 E 是正确答案。

English

On \([0, 2\pi],\) the graph of \(\cos\left(\tfrac{x}{2}\right)\) is a single arc decreasing from \(1\) down to \(-1.\)

The function \(\tan(2x)\) has period \(\tfrac{\pi}{2}\) with vertical asymptotes at \(x = \tfrac{\pi}{4}, \tfrac{3\pi}{4}, \tfrac{5\pi}{4}, \tfrac{7\pi}{4}.\) These split the interval into five branches. On every branch \(\tan(2x)\) is strictly increasing, while \(\cos(\tfrac{x}{2})\) is decreasing, so there is at most one intersection per branch.

Each of the three interior branches runs from \(-\infty\) to \(+\infty,\) so each has one intersection. On the first branch, \(\tan(0)=0\lt1=\cos(0)\) and the tangent tends to \(+\infty.\) On the last, the tangent starts at \(-\infty\) and ends at \(0\gt-1=\cos\pi.\) Thus the two outer branches also have one intersection each, for \(5\) total.

Thus, E is the correct answer.

10 答案 E

因为 \(\log_{16} n = \tfrac12 \log_4 n\)，令 \(y = \log_4 n\)。方程变为 \(\log_2\left(\tfrac{y}{2}\right) = \log_4 y = \tfrac12 \log_2 y\)。

两边乘以 \(2\) 得到 \(\log_2\left(\tfrac{y}{2}\right)^2 = \log_2 y\)，所以 \(\left(\tfrac{y}{2}\right)^2 = y\)，从而 \(y = 4\)。

因此 \(\log_4 n = 4\)，所以 \(n = 4^4 = 256\)，各位数字之和为 \(2 + 5 + 6 = 13\)。

因此，正确答案是 E。

English

Since \(\log_{16} n = \tfrac12 \log_4 n,\) set \(y = \log_4 n.\) The equation becomes \(\log_2\left(\tfrac{y}{2}\right) = \log_4 y = \tfrac12 \log_2 y.\)

Multiplying by \(2\) gives \(\log_2\left(\tfrac{y}{2}\right)^2 = \log_2 y,\) so \(\left(\tfrac{y}{2}\right)^2 = y,\) which yields \(y = 4.\)

Then \(\log_4 n = 4,\) so \(n = 4^4 = 256,\) and the digit sum is \(2 + 5 + 6 = 13.\)

Thus, E is the correct answer.

11 答案 B

令 \(P(x, y)\) 为在竖直边上结束的概率。在竖直边上 \(P = 1\)，在水平边上 \(P = 0\)，在内部点处 \(P\) 是四个相邻点概率的平均。

由左右对称性，\(P(2, 2) = \tfrac12\)。令 \(a = P(1, 2)\)，\(b = P(1, 1) = P(1, 3)\)，且 \(c = P(2, 1) = P(2, 3)\)。则

\(a = \tfrac14\left(1 + \tfrac12 + 2b\right)\)，\(\;b = \tfrac14(1 + c + a)\)，且 \(c = \tfrac14\left(2b + \tfrac12\right)\)。

代入可得 \(b = \tfrac12\)，因此 \(a = \tfrac38 + \tfrac12 b = \tfrac58\)。

因此，正确答案是 B。

English

Let \(P(x, y)\) be the probability of ending on a vertical side. On a vertical side \(P = 1,\) on a horizontal side \(P = 0,\) and at an interior point \(P\) is the average of its four neighbors.

By left-right symmetry \(P(2, 2) = \tfrac12.\) Let \(a = P(1, 2),\) \(b = P(1, 1) = P(1, 3),\) and \(c = P(2, 1) = P(2, 3).\) Then

\(a = \tfrac14\left(1 + \tfrac12 + 2b\right),\) \(\;b = \tfrac14(1 + c + a),\) and \(c = \tfrac14\left(2b + \tfrac12\right).\)

Substituting gives \(b = \tfrac12,\) hence \(a = \tfrac38 + \tfrac12 b = \tfrac58.\)

Thus, B is the correct answer.

12 答案 B

注意 \((20, 20)\) 满足 \(3x - 5y + 40 = 0\)，所以它在 \(\ell\) 上，旋转后仍在 \(k\) 上。\(\ell\) 的斜率为 \(\tfrac{3}{5}\)。

旋转 \(45^\circ\) 后的斜率为 \(\dfrac{\tfrac35 + 1}{1 - \tfrac35} = \dfrac{\tfrac85}{\tfrac25} = 4\)。

直线 \(k\) 为 \(y - 20 = 4(x - 20)\)。令 \(y = 0\) 得到 \(-20 = 4(x - 20)\)，所以 \(x - 20 = -5\)，即 \(x = 15\)。

因此，正确答案是 B。

English

Note \((20, 20)\) satisfies \(3x - 5y + 40 = 0,\) so it is on \(\ell\) and remains on \(k.\) The slope of \(\ell\) is \(\tfrac{3}{5}.\)

Rotating by \(45^\circ\) gives slope \(\dfrac{\tfrac35 + 1}{1 - \tfrac35} = \dfrac{\tfrac85}{\tfrac25} = 4.\)

Line \(k\) is \(y - 20 = 4(x - 20).\) Setting \(y = 0\) gives \(-20 = 4(x - 20),\) so \(x - 20 = -5\) and \(x = 15.\)

Thus, B is the correct answer.

13 答案 B

左边等于 \(N\) 的 \(\dfrac{1}{a} + \dfrac{1}{ab} + \dfrac{1}{abc}\) 次幂，而这个指数必须等于 \(\dfrac{25}{36}\)。

因为 \(b,c\ge2\)，这个指数至多为 \(\dfrac{7}{4a}\)。如果 \(a\ge3\)，它至多为 \(\dfrac{7}{12}\lt\dfrac{25}{36}\)，所以 \(a=2\)。方程变为 \(\dfrac{1}{2b}+\dfrac{1}{2bc}=\dfrac{7}{36}\)。

清除分母得到 \(c(7b-18)=18\)。令 \(d=7b-18\)，则 \(d\) 是 \(18\) 的正因数。在 \(1,2,3,6,9,18\) 中，只有 \(d=3\) 能使 \(b=\dfrac{d+18}{7}\) 为整数。因此 \(b=3\) 且 \(c=6\)。

所以 B 是正确答案。

English

The left side equals \(N\) raised to the exponent \(\dfrac{1}{a} + \dfrac{1}{ab} + \dfrac{1}{abc},\) which must equal \(\dfrac{25}{36}.\)

Because \(b,c\ge2,\) this exponent is at most \(\dfrac{7}{4a}.\) If \(a\ge3,\) it is at most \(\dfrac{7}{12}\lt\dfrac{25}{36},\) so \(a=2.\) The equation becomes \(\dfrac{1}{2b}+\dfrac{1}{2bc}=\dfrac{7}{36}.\)

Clearing denominators gives \(c(7b-18)=18.\) Set \(d=7b-18,\) so \(d\) is a positive divisor of \(18.\) Among \(1,2,3,6,9,18,\) only \(d=3\) makes \(b=\dfrac{d+18}{7}\) an integer. Hence \(b=3\) and \(c=6.\)

Thus, B is the correct answer.

14 答案 B

四个顶点 \(A, C, E, G\) 构成一个正方形，因为它们是正八边形中每隔一个顶点取出的点。

取外接圆半径为一，八边形面积为 \(2\sqrt2\)，而正方形 \(ACEG\) 的对角线等于该圆的直径，因此面积为 \(2\)。

比值为 \(\dfrac{2}{2\sqrt2} = \dfrac{1}{\sqrt2} = \dfrac{\sqrt2}{2}\)。

因此，正确答案是 B。

English

The four vertices \(A, C, E, G\) form a square, since they are every other vertex of the regular octagon.

Taking a unit circumradius, the octagon’s area is \(2\sqrt2\) and the square \(ACEG\) has diagonal equal to the circle’s diameter, giving area \(2.\)

The ratio is \(\dfrac{2}{2\sqrt2} = \dfrac{1}{\sqrt2} = \dfrac{\sqrt2}{2}.\)

Thus, B is the correct answer.

15 答案 D

集合 \(A\) 由 \(8\) 的三个立方根组成：\(2\)、\(-1 + i\sqrt3\) 和 \(-1 - i\sqrt3\)。

分组因式分解，\(z^3 - 8z^2 - 8z + 64\) \(= z^2(z - 8) - 8(z - 8)\) \(= (z - 8)(z^2 - 8)\)，所以 \(B = \{8, 2\sqrt2, -2\sqrt2\}\)，全都是实数。

最大距离是从 \(-1 \pm i\sqrt3\) 到 \(8\) 的距离：\(\sqrt{(8 - (-1))^2 + (\sqrt3)^2}\) \(= \sqrt{81 + 3}\) \(= \sqrt{84} = 2\sqrt{21}\)。

因此，正确答案是 D。

English

The set \(A\) consists of the cube roots of \(8:\) \(2,\) \(-1 + i\sqrt3,\) and \(-1 - i\sqrt3.\)

Factoring by grouping, \(z^3 - 8z^2 - 8z + 64\) \(= z^2(z - 8) - 8(z - 8)\) \(= (z - 8)(z^2 - 8),\) so \(B = \{8, 2\sqrt2, -2\sqrt2\},\) all real.

The greatest distance is from \(-1 \pm i\sqrt3\) to \(8:\) \(\sqrt{(8 - (-1))^2 + (\sqrt3)^2}\) \(= \sqrt{81 + 3}\) \(= \sqrt{84} = 2\sqrt{21}.\)

Thus, D is the correct answer.

16 答案 B

由周期性，只需考虑一个四个角都是格点的单位格子。距离某个角不超过 \(d\) 的区域由四个半径为 \(d\) 的四分之一圆盘组成，合成一个面积为 \(\pi d^2\) 的完整圆盘。所得的值将小于 \(\tfrac12\)，所以这些四分之一圆盘互不重叠。

令 \(\pi d^2 = \tfrac12\) 得到 \(d = \sqrt{\dfrac{1}{2\pi}} \approx 0.399\)。

四舍五入到十分位，\(d = 0.4\)。

因此，正确答案是 B。

English

By periodicity it suffices to consider one unit cell with a lattice point at each corner. The region within \(d\) of a corner consists of four quarter-disks of radius \(d,\) forming one full disk of area \(\pi d^2.\) The resulting value will be less than \(\tfrac12,\) so these quarter-disks do not overlap.

Setting \(\pi d^2 = \tfrac12\) gives \(d = \sqrt{\dfrac{1}{2\pi}} \approx 0.399.\)

To the nearest tenth, \(d = 0.4.\)

Thus, B is the correct answer.

17 答案 D

设四个顶点的 \(x\) 坐标为 \(n, n+1, n+2, n+3\)，对应的 \(y\) 坐标为对这些数取 \(\ln\) 所得的值。用鞋带公式并化简，面积为 \(\ln\dfrac{(n+1)(n+2)}{n(n+3)}\)。

令 \(\dfrac{(n+1)(n+2)}{n(n+3)} = \dfrac{91}{90}\)，得到 \(1 + \dfrac{2}{n^2 + 3n} = \dfrac{91}{90}\)，所以 \(n^2 + 3n = 180\)。

因此 \((n - 12)(n + 15) = 0\)，所以 \(n = 12\)。

因此，正确答案是 D。

English

Let the vertices have \(x\)-coordinates \(n, n+1, n+2, n+3\) with \(y\)-coordinates \(\ln\) of those values. Applying the shoelace formula and simplifying, the area is \(\ln\dfrac{(n+1)(n+2)}{n(n+3)}.\)

Setting \(\dfrac{(n+1)(n+2)}{n(n+3)} = \dfrac{91}{90}\) gives \(1 + \dfrac{2}{n^2 + 3n} = \dfrac{91}{90},\) so \(n^2 + 3n = 180.\)

Then \((n - 12)(n + 15) = 0,\) so \(n = 12.\)

Thus, D is the correct answer.

18 答案 D

设 \(A = (0,0)\)，\(C = (20, 0)\)。因为 \(\angle ACD = 90^\circ\)，所以 \(D = (20, 30)\)；又因 \(AE = 5\)，所以 \(E = (5, 0)\)。

因为 \(\angle ABC = 90^\circ\)，点 \(B\) 在以 \((10, 0)\) 为圆心、半径为 \(10\) 的圆上。直线 \(DE\) 上的点可写为 \((5 + t,\, 2t)\)；代入圆方程得到 \(t^2 - 2t - 15 = 0\)，所以 \(t = 5\) 或 \(t = -3\)。

为使 \(E\) 位于 \(B\) 与 \(D\) 之间，取 \(t = -3\)，得到 \(B = (2, -6)\)，它到直线 \(AC\) 的距离为 \(6\)。

于是 \([ACD] = \tfrac12 \cdot 20 \cdot 30 = 300\)，且 \([ABC] = \tfrac12 \cdot 20 \cdot 6 = 60\)，总面积为 \(360\)。

因此，正确答案是 D。

English

Place \(A = (0,0)\) and \(C = (20, 0).\) Since \(\angle ACD = 90^\circ,\) \(D = (20, 30),\) and \(E = (5, 0)\) because \(AE = 5.\)

Since \(\angle ABC = 90^\circ,\) \(B\) lies on the circle of radius \(10\) centered at \((10, 0).\) Line \(DE\) is \((5 + t,\, 2t);\) substituting gives \(t^2 - 2t - 15 = 0,\) so \(t = 5\) or \(t = -3.\)

For \(E\) to lie between \(B\) and \(D,\) take \(t = -3,\) giving \(B = (2, -6),\) a distance \(6\) below line \(AC.\)

Then \([ACD] = \tfrac12 \cdot 20 \cdot 30 = 300\) and \([ABC] = \tfrac12 \cdot 20 \cdot 6 = 60,\) so the total area is \(360.\)

Thus, D is the correct answer.

19 答案 C

令 \(x = 2^{17}\)，则 \(\dfrac{2^{289} + 1}{2^{17} + 1}\) \(= \dfrac{x^{17} + 1}{x + 1}\) \(= x^{16} - x^{15} + \cdots - x + 1\) 是 \(17\) 个幂 \(x^0, x^1, \ldots, x^{16}\) 的交替和。

把每个被减去的幂与它上方刚好被加上的幂配对：\(x^{m+1} - x^m\) \(= 2^{17m}(2^{17} - 1)\) \(= 2^{17m} + 2^{17m+1}\) \(+ \cdots + 2^{17m+16}\)，这是一段 \(17\) 个连续的 \(2\) 的幂。

这样的配对有 \(8\) 组，另有剩下的 \(+2^0\)。这些块占据的指数范围互不重叠，所以幂的总数为 \(8 \cdot 17 + 1 = 137\)。

因此，正确答案是 C。

English

Let \(x = 2^{17}.\) Then \(\dfrac{2^{289} + 1}{2^{17} + 1}\) \(= \dfrac{x^{17} + 1}{x + 1}\) \(= x^{16} - x^{15} + \cdots - x + 1,\) an alternating sum of the \(17\) powers \(x^0, x^1, \ldots, x^{16}.\)

Pair each subtracted power with the added power just above it: \(x^{m+1} - x^m\) \(= 2^{17m}(2^{17} - 1)\) \(= 2^{17m} + 2^{17m+1}\) \(+ \cdots + 2^{17m+16},\) a block of \(17\) consecutive powers of \(2.\)

There are \(8\) such pairs, together with the leftover \(+2^0.\) The blocks occupy disjoint ranges, so the total number of powers is \(8 \cdot 17 + 1 = 137.\)

Thus, C is the correct answer.

20 答案 A

因为 \(T\) 是不等边直角三角形，唯一将 \(T\) 映到自身的等距变换是恒等变换，所以一个序列可行，当且仅当三个变换的复合是恒等变换。

令 \(r\) 表示旋转 \(90^\circ\)，令 \(s\) 表示关于 \(x\) 轴的反射。五个允许的映射是 \(r,r^2,r^3,s,r^2s\)；缺少的非恒等映射是两个对角线反射 \(rs,r^3s\)。在有序三元组中，第三个映射由前两个唯一决定；它被允许，当且仅当前两个的乘积是这五个映射之一。

恰有 \(5\) 个有序对的乘积是恒等变换：每个第一个映射都与它的逆映射配对。得到对角线反射需要一个旋转和一个轴反射；\(r\) 或 \(r^3\) 可以在左侧或右侧与 \(s\) 或 \(r^2s\) 配对，共有 \(2\cdot2\cdot2=8\) 对。剩余 \(25-5-8=12\) 个有序对给出有效序列。

所以 A 是正确答案。

English

Because \(T\) is a scalene right triangle, the only isometry carrying \(T\) to itself is the identity, so a sequence works exactly when the three transformations compose to the identity.

Let \(r\) be the \(90^\circ\) rotation and \(s\) reflection across the \(x\)-axis. The five allowed maps are \(r,r^2,r^3,s,r^2s;\) the missing nonidentity maps are the diagonal reflections \(rs,r^3s.\) In an ordered triple, the third map is forced by the first two and is allowed precisely when their product is one of the five.

Exactly \(5\) ordered pairs have product identity: each first map is paired with its inverse. A diagonal reflection requires one rotation and one axis reflection; \(r\) or \(r^3\) may be paired on either side with \(s\) or \(r^2s,\) giving \(2\cdot2\cdot2=8\) pairs. The remaining \(25-5-8=12\) ordered pairs give valid sequences.

Thus, A is the correct answer.

21 答案 D

写 \(n = 2^a 3^b 5^c 7^d \cdots\)。由于 \(5! = 2^3 \cdot 3 \cdot 5\) 不含其他质数，\(n\) 只能含有 \(2, 3, 5, 7\)。在等式 \(\operatorname{lcm}(5!, n) = 5 \cdot \gcd(10!, n)\) 中逐个匹配质因数的指数。

对于 \(2\)：\(\max(3, a) = \min(8, a)\)，所以 \(3 \le a \le 8\) 有 \(6\) 个值。对于 \(3\)：\(\max(1, b) = \min(4, b)\)，所以 \(1 \le b \le 4\) 有 \(4\) 个值。

对于 \(5\)：\(\max(1, c) = 1 + \min(2, c)\)，且 \(c \ge 1\)，这迫使 \(c = 3\) 有 \(1\) 个值。对于 \(7\)：\(\max(0, d) = \min(1, d)\)，所以 \(d = 0\) 或 \(1\) 有 \(2\) 个值。

总数为 \(6 \cdot 4 \cdot 1 \cdot 2 = 48\)。

因此，正确答案是 D。

English

Write \(n = 2^a 3^b 5^c 7^d \cdots.\) Since \(5! = 2^3 \cdot 3 \cdot 5\) has no other primes, \(n\) can only involve \(2, 3, 5, 7.\) Matching exponents in \(\operatorname{lcm}(5!, n) = 5 \cdot \gcd(10!, n):\)

For \(2:\) \(\max(3, a) = \min(8, a),\) so \(3 \le a \le 8\) gives \(6\) values. For \(3:\) \(\max(1, b) = \min(4, b),\) so \(1 \le b \le 4\) gives \(4\) values.

For \(5:\) \(\max(1, c) = 1 + \min(2, c)\) with \(c \ge 1,\) which forces \(c = 3,\) giving \(1\) value. For \(7:\) \(\max(0, d) = \min(1, d),\) so \(d = 0\) or \(1,\) giving \(2\) values.

The total is \(6 \cdot 4 \cdot 1 \cdot 2 = 48.\)

Thus, D is the correct answer.

22 答案 B

因为 \((a_n + b_n i)^2 = a_n^2 - b_n^2 + 2 a_n b_n i\)，所以 \(a_n b_n\) \(= \tfrac12 \operatorname{Im}\big((2 + i)^{2n}\big)\) \(= \tfrac12 \operatorname{Im}\big((3 + 4i)^n\big)\)。

因此所求和为 \(\tfrac12 \operatorname{Im} \displaystyle\sum_{n=0}^{\infty} \left(\frac{3 + 4i}{7}\right)^n\) \(= \tfrac12 \operatorname{Im}\!\left(\frac{1}{1 - \frac{3 + 4i}{7}}\right)\)。

它等于 \(\tfrac12 \operatorname{Im}\!\left(\dfrac{7}{4 - 4i}\right)\) \(= \tfrac12 \operatorname{Im}\!\left(\dfrac{7(4 + 4i)}{32}\right)\) \(= \tfrac12 \cdot \dfrac{28}{32}\) \(= \dfrac{7}{16}\)。

因此，正确答案是 B。

English

Since \((a_n + b_n i)^2 = a_n^2 - b_n^2 + 2 a_n b_n i,\) we have \(a_n b_n\) \(= \tfrac12 \operatorname{Im}\big((2 + i)^{2n}\big)\) \(= \tfrac12 \operatorname{Im}\big((3 + 4i)^n\big).\)

Therefore the sum is \(\tfrac12 \operatorname{Im} \displaystyle\sum_{n=0}^{\infty} \left(\frac{3 + 4i}{7}\right)^n\) \(= \tfrac12 \operatorname{Im}\!\left(\frac{1}{1 - \frac{3 + 4i}{7}}\right).\)

This equals \(\tfrac12 \operatorname{Im}\!\left(\dfrac{7}{4 - 4i}\right)\) \(= \tfrac12 \operatorname{Im}\!\left(\dfrac{7(4 + 4i)}{32}\right)\) \(= \tfrac12 \cdot \dfrac{28}{32}\) \(= \dfrac{7}{16}.\)

Thus, B is the correct answer.

23 答案 A

重掷一枚骰子、保留点数和为 \(s\) 的两枚骰子时，如果 \(s \le 6\)，获胜概率为 \(\tfrac16\)，否则为 \(0\)。重掷两枚骰子、保留点数为 \(v\) 的一枚骰子时，获胜概率等于两枚骰子点数和为 \(7 - v\) 的方式数除以 \(36\)；当 \(v\) 最小时，这个概率最大。对 \(v=1,2,3\)，这些概率分别是 \(\tfrac5{36},\tfrac4{36},\tfrac3{36}\)，都大于重掷全部骰子的概率 \(\tfrac{15}{216}=\tfrac{5}{72}\)；对 \(v\ge4\)，重掷全部骰子更好。

恰好重掷两枚骰子严格最优，当且仅当最小的两枚骰子点数和至少为 \(7\)（所以重掷一枚不能达到 \(7\)），同时最小骰子的点数是 \(1, 2\)，或 \(3\)（所以保留它优于重掷全部三枚）。

将结果排序为 \(u\le v\le w\)。当 \(u=1\) 时，只有 \((1,6,6)\) 一种，排列数为 \(3\)。当 \(u=2\) 时，\((2,5,5),(2,5,6),(2,6,6)\) 的排列数之和为 \(3+6+3=12\)。当 \(u=3\) 时，从 \(\{4,5,6\}\) 中可重复地选 \(v,w\)，六个三元组的排列数之和为 \(3+6+6+3+6+3=27\)。因此在 \(216\) 个有序结果中，有 \(3+12+27=42\) 个满足条件，概率为 \(\dfrac{42}{216} = \dfrac{7}{36}\)。

所以 A 是正确答案。

English

Rerolling one die, keeping two dice that sum to \(s,\) wins with probability \(\tfrac16\) when \(s \le 6\) and \(0\) otherwise. Rerolling two dice, keeping a die of value \(v,\) wins with probability equal to the number of ways two dice sum to \(7 - v,\) over \(36;\) this is largest when \(v\) is smallest. For \(v=1,2,3,\) these probabilities are \(\tfrac5{36},\tfrac4{36},\tfrac3{36},\) all greater than the reroll-all probability \(\tfrac{15}{216}=\tfrac{5}{72};\) for \(v\ge4,\) rerolling all is better.

Rerolling exactly two dice is strictly best precisely when the two smallest dice sum to at least \(7\) (so rerolling one cannot reach \(7\)) while the smallest die is \(1, 2,\) or \(3\) (so keeping it beats rerolling all three).

Sort the roll as \(u\le v\le w.\) If \(u=1,\) the only possibility is \((1,6,6),\) with \(3\) orderings. If \(u=2,\) the possibilities \((2,5,5),(2,5,6),(2,6,6)\) have \(3+6+3=12\) orderings. If \(u=3,\) choose \(v,w\) with repetition from \(\{4,5,6\};\) the six resulting triples have \(3+6+6+3+6+3=27\) orderings. Thus there are \(3+12+27=42\) qualifying ordered rolls out of \(216,\) a probability of \(\dfrac{42}{216} = \dfrac{7}{36}.\)

Thus, A is the correct answer.

24 答案 B

等边三角形内一点到三个顶点的距离为 \(p, q, r\)，边长为 \(s\) 时，满足 \(3(p^4 + q^4 + r^4 + s^4)\) \(= (p^2 + q^2 + r^2 + s^2)^2\)。

令 \(p^2 = 1\)，\(q^2 = 3\)，\(r^2 = 4\)，并设 \(S = s^2\) 得到 \(3(26 + S^2) = (8 + S)^2\)，所以 \(S^2 - 8S + 7 = 0\)，从而 \(S = 1\) 或 \(S = 7\)。

边长为 \(1\) 的三角形不可能包含一个到某顶点距离为 \(2\) 的点，所以 \(S = 7\)，于是 \(s = \sqrt{7}\)。

因此，正确答案是 B。

English

A point at distances \(p, q, r\) from the vertices of an equilateral triangle of side \(s\) satisfies \(3(p^4 + q^4 + r^4 + s^4)\) \(= (p^2 + q^2 + r^2 + s^2)^2.\)

With \(p^2 = 1,\) \(q^2 = 3,\) \(r^2 = 4,\) letting \(S = s^2\) gives \(3(26 + S^2) = (8 + S)^2,\) so \(S^2 - 8S + 7 = 0\) and \(S = 1\) or \(S = 7.\)

A triangle of side \(1\) cannot contain a point at distance \(2\) from a vertex, so \(S = 7\) and \(s = \sqrt{7}.\)

Thus, B is the correct answer.

25 答案 C

没有负数解，而 \(x=0\) 总是一个解。对 \(n\ge1\) 且 \(x\in[n,n+1)\)，令 \(y=\frac{x}{n}\)。方程变为 \(ay^2-y+1=0\)。它的根必须为实数，所以 \(0\lt a\le\tfrac14\)。

若两个根为 \(\alpha\le\beta\)，它们的和与积都等于 \(\frac{1}{a}\)，所以 \((\alpha-1)(\beta-1)=1\)。写成 \[ \alpha=1+\dfrac1u,\qquad \beta=1+u \] 其中 \(u\ge1\)。此时 \(a=\dfrac{u}{(u+1)^2}\)。根 \(x=n\alpha\) 位于 \([n,n+1)\) 中，当且仅当 \(n\lt u\)，而对正整数 \(n\)，\(n\beta\) 永远不在该区间中。

所要求的正总和保证 \(u\gt1\)。令 \(N\) 为小于 \(u\) 的最大正整数，则 \(N\lt u\le N+1\)。所有解的和为 \[ \dfrac{u+1}{u}\cdot\dfrac{N(N+1)}2=420\text{。}\]因为 \(\dfrac{u+1}{u}\) 随 \(u\) 递减，上述不等式给出 \[ \dfrac{N(N+2)}2\le420\lt\dfrac{(N+1)^2}{2}\text{，}\]从而迫使 \(N=28\)。

代入得到 \(406\cdot\dfrac{u+1}{u}=420\)，所以 \(u=29\)。因此 \(a=\dfrac{29}{30^2}=\dfrac{29}{900}\)。确实，正数解是 \(x=\dfrac{30n}{29}\)，其中 \(1\le n\le28\)，它们的和为 \(420\)。所以 \(p+q=29+900=929\)。

所以 C 是正确答案。

English

There are no negative solutions, while \(x=0\) is always a solution. For \(n\ge1\) and \(x\in[n,n+1),\) put \(y=\frac{x}{n}.\) The equation becomes \(ay^2-y+1=0.\) Its roots must be real, so \(0\lt a\le\tfrac14.\)

If the two roots are \(\alpha\le\beta,\) their sum and product are both \(\frac{1}{a},\) so \((\alpha-1)(\beta-1)=1.\) Write \[ \alpha=1+\dfrac1u,\qquad \beta=1+u \] with \(u\ge1.\) Then \(a=\dfrac{u}{(u+1)^2}.\) The root \(x=n\alpha\) lies in \([n,n+1)\) exactly when \(n\lt u,\) while \(n\beta\) never lies there for a positive integer \(n.\)

The required positive total ensures \(u\gt1.\) Let \(N\) be the largest positive integer less than \(u,\) so \(N\lt u\le N+1.\) The sum of all solutions is therefore \[ \dfrac{u+1}{u}\cdot\dfrac{N(N+1)}2=420. \] Because \(\dfrac{u+1}{u}\) decreases with \(u,\) these inequalities imply \[ \dfrac{N(N+2)}2\le420\lt\dfrac{(N+1)^2}{2}, \] which forces \(N=28.\)

Substitution gives \(406\cdot\dfrac{u+1}{u}=420,\) so \(u=29.\) Hence \(a=\dfrac{29}{30^2}=\dfrac{29}{900}.\) Indeed the positive solutions are \(x=\dfrac{30n}{29}\) for \(1\le n\le28,\) and their sum is \(420.\) Therefore \(p+q=29+900=929.\)

Thus, C is the correct answer.