2013A AMC 12 真题（P版）

1 答案 E

直角三角形 \(ABE\) 的两条直角边为 \(AB = 10\) 和 \(BE\)。由 \(\tfrac12\cdot 10\cdot BE = 40\)，得 \(BE = 8\)。

因此，正确答案是 E。

English

The legs of right triangle \(ABE\) are \(AB = 10\) and \(BE.\) From \(\tfrac12\cdot 10\cdot BE = 40,\) we get \(BE = 8.\)

Thus, the correct answer is E.

2 答案 C

只有当本队得分为偶数时，本队得分才可能是对手的两倍。这些比赛的得分为 \(2, 4, 6, 8, 10\)，因而对手得分为 \(1 + 2 + 3 + 4 + 5 = 15\)。

另外五场比赛的本队得分为 \(1, 3, 5, 7, 9\)，且都是一分惜败，所以对手得分为 \(2 + 4 + 6 + 8 + 10 = 30\)。总分为 \(15 + 30 = 45\)。

因此，正确答案是 C。

English

The team can only score twice as many runs as its opponent when its own score is even. Those games have scores \(2, 4, 6, 8, 10,\) so their opponents scored \(1 + 2 + 3 + 4 + 5 = 15.\)

The other five games had scores \(1, 3, 5, 7, 9\) and were one-run losses, so their opponents scored \(2 + 4 + 6 + 8 + 10 = 30.\) The total is \(15 + 30 = 45.\)

Thus, the correct answer is C.

3 答案 E

所有花中十分之六是粉色，十分之四是红色。因为粉色花中三分之二是康乃馨，所以粉康乃馨占全部花的 \(\tfrac{2}{3}\cdot\tfrac{6}{10} = \tfrac{4}{10}\)。

因为红色花中四分之三是康乃馨，所以红康乃馨占全部花的 \(\tfrac{3}{4}\cdot\tfrac{4}{10} = \tfrac{3}{10}\)。康乃馨合计占 \(\tfrac{4}{10} + \tfrac{3}{10} = \tfrac{7}{10} = 70\%\)。

因此，正确答案是 E。

English

Six tenths of the flowers are pink and four tenths are red. Since two thirds of the pink flowers are carnations, pink carnations make up \(\tfrac{2}{3}\cdot\tfrac{6}{10} = \tfrac{4}{10}\) of the flowers.

Since three fourths of the red flowers are carnations, red carnations make up \(\tfrac{3}{4}\cdot\tfrac{4}{10} = \tfrac{3}{10}\) of the flowers. Together the carnations are \(\tfrac{4}{10} + \tfrac{3}{10} = \tfrac{7}{10} = 70\%.\)

Thus, the correct answer is E.

4 答案 C

从每一项中提出 \(2^{2012}\)，得 \[ \dfrac{2^{2012}(2^2 + 1)}{2^{2012}(2^2 - 1)} = \dfrac{4 + 1}{4 - 1} = \dfrac{5}{3}\text{。}\]

因此，正确答案是 C。

English

Factoring \(2^{2012}\) from each term gives \[ \dfrac{2^{2012}(2^2 + 1)}{2^{2012}(2^2 - 1)} = \dfrac{4 + 1}{4 - 1} = \dfrac{5}{3}. \]

Thus, the correct answer is C.

5 答案 B

总花费为 \(105 + 125 + 175 = 405\)，所以每人的公平份额为 \(\tfrac13\cdot 405 = 135\) 美元。

于是 \(t = 135 - 105 = 30\)，\(d = 135 - 125 = 10\)，所以 \(t - d = 30 - 10 = 20\)。

因此，正确答案是 B。

English

The total spent was \(105 + 125 + 175 = 405,\) so each fair share is \(\tfrac13\cdot 405 = 135\) dollars.

Then \(t = 135 - 105 = 30\) and \(d = 135 - 125 = 10,\) so \(t - d = 30 - 10 = 20.\)

Thus, the correct answer is B.

6 答案 B

如果 Shenille 出手 \(x\) 次三分球、\(30 - x\) 次两分球，那么她的得分为 \[ \begin{gathered} 0.2\cdot 3\cdot x + 0.3\cdot 2\cdot(30 - x) \\ = 0.6x + 0.6(30 - x) \\ = 0.6\cdot 30 = 18 \end{gathered} \] 分。

因此，正确答案是 B。

English

If Shenille attempted \(x\) three-point shots and \(30 - x\) two-point shots, she scored \[ \begin{gathered} 0.2\cdot 3\cdot x + 0.3\cdot 2\cdot(30 - x) \\ = 0.6x + 0.6(30 - x) \\ = 0.6\cdot 30 = 18 \end{gathered} \] points.

Thus, the correct answer is B.

7 答案 C

因为 \(S_9 = S_7 + S_8\)，所以 \(S_8 = 110 - 42 = 68\)。然后 \(S_6 = S_8 - S_7 = 68 - 42 = 26\)，\(S_5 = S_7 - S_6 = 42 - 26 = 16\)，且 \(S_4 = S_6 - S_5 = 26 - 16 = 10\)。

因此，正确答案是 C。

English

Since \(S_9 = S_7 + S_8,\) we get \(S_8 = 110 - 42 = 68.\) Then \(S_6 = S_8 - S_7 = 68 - 42 = 26,\) \(S_5 = S_7 - S_6 = 42 - 26 = 16,\) and \(S_4 = S_6 - S_5 = 26 - 16 = 10.\)

Thus, the correct answer is C.

8 答案 D

同乘 \(xy\) 得 \(x^2 y + 2y = xy^2 + 2x\)，所以 \[ \begin{gathered} x^2 y - xy^2 - 2x + 2y \\ = (x - y)(xy - 2) \\ = 0 \end{gathered}\text{。}\]

因为 \(x \ne y\)，所以 \(xy = 2\)。

因此，正确答案是 D。

English

Multiplying by \(xy\) gives \(x^2 y + 2y = xy^2 + 2x,\) so \[ \begin{gathered} x^2 y - xy^2 - 2x + 2y \\ = (x - y)(xy - 2) \\ = 0. \end{gathered} \]

Since \(x \ne y,\) it follows that \(xy = 2.\)

Thus, the correct answer is D.

9 答案 C

因为 \(EF \parallel AB\)，三角形 \(FEC\) 与三角形 \(ABC\) 相似，而后者是等腰三角形，所以 \(FE = FC\)。

平行四边形 \(ADEF\) 周长的一半为 \(AF + FE\) \(= AF + FC\) \(= AC = 28\)。因此整个周长为 \(56\)。

因此，正确答案是 C。

English

Because \(EF \parallel AB,\) triangle \(FEC\) is similar to triangle \(ABC,\) which is isosceles, so \(FE = FC.\)

Half the perimeter of parallelogram \(ADEF\) is \(AF + FE\) \(= AF + FC\) \(= AC = 28.\) The entire perimeter is \(56.\)

Thus, the correct answer is C.

10 答案 D

若 \(\dfrac{1}{n} = 0.\overline{ab}\)，则 \(\dfrac{99}{n} = \overline{ab}\) 是一个两位数。\(99\) 的正因数为 \(1, 3, 9, 11, 33, 99\)。

只有 \(n = 11, 33, 99\) 时，\(\dfrac{99}{n}\) 分别等于 \(09, 03, 01\)，它们有两个不同的数字。所求和为 \(11 + 33 + 99 = 143\)。

因此，正确答案是 D。

English

If \(\dfrac{1}{n} = 0.\overline{ab},\) then \(\dfrac{99}{n} = \overline{ab},\) a two-digit number. The positive divisors of \(99\) are \(1, 3, 9, 11, 33, 99.\)

Only \(n = 11, 33, 99\) make \(\dfrac{99}{n}\) equal to \(09, 03, 01,\) which have two different digits. The requested sum is \(11 + 33 + 99 = 143.\)

Thus, the correct answer is D.

11 答案 C

设 \(x = DE\)、\(y = FG\)。平行线切出的较小区域是等边三角形或等腰梯形，所以周长为 \[ \begin{gathered} \triangle ADE: 3x, \\ \quad DFGE: 3y - x, \\ \quad FBCG: 3 - y \end{gathered}\text{。}\]

令它们相等，\(3x = 3y - x\) 得 \(4x = 3y\)，且 \(3x = 3 - y\)。解得 \(x = \tfrac{9}{13}\)、\(y = \tfrac{12}{13}\)，所以 \(DE + FG = \tfrac{21}{13}\)。

因此，正确答案是 C。

English

Let \(x = DE\) and \(y = FG.\) The parallel cuts make the small regions equilateral or isosceles trapezoids, so the perimeters are \[ \begin{gathered} \triangle ADE: 3x, \\ \quad DFGE: 3y - x, \\ \quad FBCG: 3 - y. \end{gathered} \]

Setting them equal, \(3x = 3y - x\) gives \(4x = 3y,\) and \(3x = 3 - y.\) Solving yields \(x = \tfrac{9}{13}\) and \(y = \tfrac{12}{13},\) so \(DE + FG = \tfrac{21}{13}.\)

Thus, the correct answer is C.

12 答案 A

若三个角为 \(\alpha - \delta, \alpha, \alpha + \delta\)，它们的和 \(3\alpha = 180^\circ\) 给出 \(\alpha = 60^\circ\)，所以其中一个角是 \(60^\circ\)。

若 \(x\) 对着 \(60^\circ\) 角，由余弦定理得 \[ \begin{gathered} x^2 = 4^2 + 5^2 - 2\cdot 4\cdot 5\cos 60^\circ \\ = 21\text{，} \end{gathered} \] 所以 \(x = \sqrt{21}\)。

若 \(5\) 对着 \(60^\circ\) 角，则 \(25 = x^2 - 4x + 16\)，正解为 \(x = 2 + \sqrt{13}\)。若 \(4\) 对着该角，则 \(16 = x^2 - 5x + 25\)，没有实数解。

所有可能值的和为 \(2 + \sqrt{13} + \sqrt{21}\)，所以 \(a + b + c = 2 + 13 + 21 = 36\)。

因此，正确答案是 A。

English

If the angles are \(\alpha - \delta, \alpha, \alpha + \delta,\) their sum \(3\alpha = 180^\circ\) gives \(\alpha = 60^\circ,\) so one angle is \(60^\circ.\)

If \(x\) is opposite the \(60^\circ\) angle, the Law of Cosines gives \[ \begin{gathered} x^2 = 4^2 + 5^2 - 2\cdot 4\cdot 5\cos 60^\circ \\ = 21, \end{gathered} \] so \(x = \sqrt{21}.\)

If \(5\) is opposite the \(60^\circ\) angle, then \(25 = x^2 - 4x + 16,\) whose positive solution is \(x = 2 + \sqrt{13}.\) If \(4\) is opposite, then \(16 = x^2 - 5x + 25\) has no real solution.

The sum of the possible values is \(2 + \sqrt{13} + \sqrt{21},\) so \(a + b + c = 2 + 13 + 21 = 36.\)

Thus, the correct answer is A.

13 答案 B

由鞋带公式，\(ABCD\) 的面积为 \(\tfrac{15}{2}\)。设直线与 \(\overline{CD}\) 交于 \(G\)。三角形 \(ADG\) 的面积必须为 \(\tfrac{15}{4}\)。

因为 \(AD = 4\) 在 \(x\) 轴上，\(\tfrac12\cdot 4\cdot y_G = \tfrac{15}{4}\) 给出 \(y_G = \tfrac{15}{8}\)。直线 \(CD\) 为 \(y = -3(x - 4)\)，所以 \(x_G = \tfrac{27}{8}\)。

于是 \(p + q + r + s\) \(= 27 + 8 + 15 + 8\) \(= 58\)。

因此，正确答案是 B。

English

By the shoelace formula, the area of \(ABCD\) is \(\tfrac{15}{2}.\) Let the line meet \(\overline{CD}\) at \(G.\) Triangle \(ADG\) must have area \(\tfrac{15}{4}.\)

Since \(AD = 4\) lies on the \(x\)-axis, \(\tfrac12\cdot 4\cdot y_G = \tfrac{15}{4}\) gives \(y_G = \tfrac{15}{8}.\) Line \(CD\) is \(y = -3(x - 4),\) so \(x_G = \tfrac{27}{8}.\)

Then \(p + q + r + s\) \(= 27 + 8 + 15 + 8\) \(= 58.\)

Thus, the correct answer is B.

14 答案 B

因为这些对数成等差数列，\(162, x, y, z, 1250\) 成等比数列。它的公比 \(r\) 满足 \(162 r^4 = 1250\)，所以 \(r^4 = \tfrac{625}{81}\)，且 \(r = \tfrac53\)。

因此 \(x = 162\cdot\tfrac53 = 270\)。

因此，正确答案是 B。

English

Because the logarithms are in arithmetic progression, \(162, x, y, z, 1250\) is a geometric sequence. Its common ratio \(r\) satisfies \(162 r^4 = 1250,\) so \(r^4 = \tfrac{625}{81}\) and \(r = \tfrac53.\)

Therefore \(x = 162\cdot\tfrac53 = 270.\)

Thus, the correct answer is B.

15 答案 D

如果两只父母兔在同一家店，有 \(4\) 种选择，而每只孩子兔必须去另外三家店之一：共有 \(4\cdot 3^3 = 108\) 种。

如果父母兔去不同的店，有 \(4\cdot 3 = 12\) 种选择，而每只孩子兔必须去剩下两家店之一：共有 \(12\cdot 2^3 = 96\) 种。

总数为 \(108 + 96 = 204\)。

因此，正确答案是 D。

English

If the two parents share a store, there are \(4\) choices for it, and each child must go to one of the other three stores: \(4\cdot 3^3 = 108\) ways.

If the parents go to different stores, there are \(4\cdot 3 = 12\) choices, and each child must go to one of the two remaining stores: \(12\cdot 2^3 = 96\) ways.

The total is \(108 + 96 = 204.\)

Thus, the correct answer is D.

16 答案 E

设 \(a, b, c\) 分别为三堆石头的数量。由 \(\dfrac{40a + 50b}{a + b} = 43\)，得 \(7b = 3a\)，所以 \(a = 7k\) 且 \(b = 3k\)。

设 \(\mu_{BC}\) 为 \(B\) 和 \(C\) 的平均重量。利用 \(A, C\) 的平均数 \(44\) 表示出 \(\mu_C = \dfrac{28k + 44c}{c}\)，可得 \(\mu_{BC} = \dfrac{178k + 44c}{3k + c}\)，所以 \((\mu_{BC} - 44)c = (178 - 3\mu_{BC})k\)。

因为 \(B\) 比 \(A\) 重，所以 \(B\) 和 \(C\) 的平均数大于 \(44\)，这迫使 \(178 - 3\mu_{BC} \gt 0\)，即 \(\mu_{BC} \lt \tfrac{178}{3} = 59\tfrac13\)。取 \(k=15c\) 时可以达到 \(59\)：此时堆 \(C\) 的平均重量为 \(464\)，代入上式得 \(\mu_{BC}=59\)。因此最大的整数平均数是 \(59\)。

因此，正确答案是 E。

English

Let \(a, b, c\) be the numbers of rocks in the piles. From \(\dfrac{40a + 50b}{a + b} = 43,\) we get \(7b = 3a,\) so \(a = 7k\) and \(b = 3k.\)

Let \(\mu_{BC}\) be the mean of \(B\) and \(C.\) Using the \(A, C\) mean \(44\) to express \(\mu_C = \dfrac{28k + 44c}{c},\) we find \(\mu_{BC} = \dfrac{178k + 44c}{3k + c},\) so \((\mu_{BC} - 44)c = (178 - 3\mu_{BC})k.\)

Since \(B\) is heavier than \(A,\) the mean of \(B\) and \(C\) exceeds \(44,\) forcing \(178 - 3\mu_{BC} \gt 0,\) i.e. \(\mu_{BC} \lt \tfrac{178}{3} = 59\tfrac13.\) The value \(59\) is attainable by taking \(k=15c;\) then pile \(C\) has mean \(464\) and the displayed formula gives \(\mu_{BC}=59.\) Thus the greatest integer mean is \(59.\)

Thus, the correct answer is E.

17 答案 D

对 \(1 \le k \le 11\)，第 \(k\) 名海盗取走份额之前的金币数，是取走之后金币数的 \(\dfrac{12}{12 - k}\) 倍。所以若留给第 \(12\) 名海盗的是 \(n\) 枚金币，初始金币数为 \[ \dfrac{12^{11}\, n}{11!} = \dfrac{2^{14}\cdot 3^{7}\, n}{5^2\cdot 7\cdot 11}\text{。}\]

使其成为正整数的最小 \(n\) 为 \(5^2\cdot7\cdot11=1925\)。在第 \(k\) 名海盗取金币之前，剩余的金币数等于初始数目乘以 \(\frac{11!}{(12-k)!\,12^{k-1}}\)；代入这个 \(n\) 便可看出它对每个 \(k\) 都是整数。因此每个人分到的都是整数枚金币，而第 \(12\) 名海盗得到 \(1925\) 枚金币。

因此，正确答案是 D。

English

For \(1 \le k \le 11,\) the number of coins before the \(k\)th pirate takes a share is \(\dfrac{12}{12 - k}\) times the number afterward. So if \(n\) coins are left for the \(12\)th pirate, the initial count is \[ \dfrac{12^{11}\, n}{11!} = \dfrac{2^{14}\cdot 3^{7}\, n}{5^2\cdot 7\cdot 11}. \]

The smallest \(n\) making this a positive integer is \(5^2\cdot7\cdot11=1925.\) Before pirate \(k,\) the remaining count is the initial count multiplied by \(\frac{11!}{(12-k)!\,12^{k-1}};\) substituting this \(n\) shows it is an integer for every \(k.\) Hence all shares are integral, and the \(12\)th pirate receives \(1925\) coins.

Thus, the correct answer is D.

18 答案 B

每个小球球心到中心 \(O\) 的距离为 \(2\)，而小球半径为 \(1\)，所以大球半径为 \(3\)。设第八个球半径为 \(r\)，球心 \(G\) 到 \(O\) 的距离为 \(x\)，则 \(x + r = 3\)。

因为 \(G\) 到两个相对的六边形顶点等距，\(GO\) 垂直于通向某个顶点的线，由勾股定理得 \[ \begin{gathered} (r + 1)^2 = 2^2 + x^2 \\ = 4 + (3 - r)^2 \end{gathered}\text{。}\]

化简得 \(2r + 1 = 13 - 6r\)，所以 \(r = \tfrac32\)。

因此，正确答案是 B。

English

Each small center is \(2\) from the center \(O,\) and the small spheres have radius \(1,\) so the large sphere has radius \(3.\) Let the eighth sphere have radius \(r\) and center \(G\) at distance \(x\) from \(O;\) then \(x + r = 3.\)

Since \(G\) is equidistant from two opposite hexagon vertices, \(GO\) is perpendicular to the line to a vertex, and the Pythagorean Theorem gives \[ \begin{gathered} (r + 1)^2 = 2^2 + x^2 \\ = 4 + (3 - r)^2. \end{gathered} \]

This simplifies to \(2r + 1 = 13 - 6r,\) so \(r = \tfrac32.\)

Thus, the correct answer is B.

19 答案 D

由点的幂定理，\(BC\cdot CX = AC^2 - AB^2\)，其中 \(AB\) 是圆的半径。因此 \(BC\cdot CX = 97^2 - 86^2 = 2013\)。

因为 \(BC = BX + CX\) 且 \(CX\) 都是整数，它们是 \(2013 = 3\cdot 11\cdot 61\) 的一对互补因数。由于 \(CX \lt BC \lt AB + AC = 183\)，唯一可能是 \(CX = 33\) 且 \(BC = 61\)。

因此，正确答案是 D。

English

By the Power of a Point Theorem, \(BC\cdot CX = AC^2 - AB^2\) where \(AB\) is the radius. Thus \(BC\cdot CX = 97^2 - 86^2 = 2013.\)

Since \(BC = BX + CX\) and \(CX\) are integers, they are complementary factors of \(2013 = 3\cdot 11\cdot 61.\) As \(CX \lt BC \lt AB + AC = 183,\) the only possibility is \(CX = 33\) and \(BC = 61.\)

Thus, the correct answer is D.

20 答案 B

把元素按模 \(19\) 来看，关系 \(a \succ b\) 恰好在 \(0 \lt (a - b) \bmod 19 \le 9\) 时成立。

\(x\) 有 \(19\) 种选择。固定 \(x\) 后，令 \(y = x + i\)，其中 \(1 \le i \le 9\)。那么 \(z\) 必须满足 \(x + 10 \le z \le x + 9 + i\)，因而有 \(i\) 种选择。

总数为 \(19(1 + 2 + \cdots + 9) = 19\cdot 45\) \(= 855\)。

因此，正确答案是 B。

English

Reading the elements modulo \(19,\) the relation \(a \succ b\) holds exactly when \(0 \lt (a - b) \bmod 19 \le 9.\)

There are \(19\) choices for \(x.\) Once \(x\) is fixed, take \(y = x + i\) for some \(1 \le i \le 9.\) Then \(z\) must satisfy \(x + 10 \le z \le x + 9 + i,\) giving \(i\) choices.

The total is \(19(1 + 2 + \cdots + 9) = 19\cdot 45\) \(= 855.\)

Thus, the correct answer is B.

21 答案 A

设 \(\tiny A_n = \log(n + \log((n-1) + \cdots + \log(3 + \log 2)\cdots))\)。可以检查 \(2 \le n \le 9\) 时 \(0 \lt A_n \lt 1\)，\(10 \le n \le 98\) 时 \(1 \lt A_n \lt 2\)，\(99 \le n \le 997\) 时 \(2 \lt A_n \lt 3\)，且 \(998 \le n \le 9996\) 时 \(3 \lt A_n \lt 4\)。

因此 \(3 \lt A_{2012} \lt 4\)，所以 \(2016 \lt 2013 + A_{2012} \lt 2017\)，从而 \(\log 2016 \lt A \lt \log 2017\)。

因此，正确答案是 A。

English

Let \(\tiny A_n = \log(n + \log((n-1) + \cdots + \log(3 + \log 2)\cdots)).\) One checks \(0 \lt A_n \lt 1\) for \(2 \le n \le 9,\) then \(1 \lt A_n \lt 2\) for \(10 \le n \le 98,\) then \(2 \lt A_n \lt 3\) for \(99 \le n \le 997,\) and \(3 \lt A_n \lt 4\) for \(998 \le n \le 9996.\)

Hence \(3 \lt A_{2012} \lt 4,\) so \(2016 \lt 2013 + A_{2012} \lt 2017\) and therefore \(\log 2016 \lt A \lt \log 2017.\)

Thus, the correct answer is A.

22 答案 E

令 \(m= \frac{n}{11}\)。如果 \(m\) 是四位数，那么 \(n<110000\)，所以六位回文数 \(n\) 的首位和末位都必须是 \(1\)。这迫使回文数 \(m\) 的首位和末位也都是 \(1\)，从而 \(m<2000\) 且 \(n<22000\)，矛盾。因此 \(m\) 是五位回文数 \(\overline{abcba}\)。

写成 \(n=11m=\overline{abcba0}+\overline{abcba}\)，恰好在 \(a+b\le9\) 且 \(b+c\le9\) 时没有进位；所得数字依次为 \(a,a+b,b+c,b+c,a+b,a\)。若 \(a+b\ge10\)，首位与末位不同；若只有 \(b+c\ge10\)，第二位与倒数第二位不同。所以这些条件也是必要的。有效的 \(m\) 的个数是 \[ \sum_{b=0}^{9}(10 - b)(9 - b) = 330\text{。}\]

六位回文数共有 \(9\cdot 10^2 = 900\) 个，所以概率为 \(\dfrac{330}{900} = \dfrac{11}{30}\)。

因此，正确答案是 E。

English

Let \(m= \frac{n}{11}.\) If \(m\) had four digits, then \(n<110000,\) so the first and last digits of the six-digit palindrome \(n\) would both be \(1.\) This forces the first and last digits of the palindromic \(m\) to be \(1,\) hence \(m<2000\) and \(n<22000,\) a contradiction. Therefore \(m\) is a five-digit palindrome \(\overline{abcba}.\)

Writing \(n=11m=\overline{abcba0}+\overline{abcba},\) no carries occur exactly when \(a+b\le9\) and \(b+c\le9;\) the resulting digits are \(a,a+b,b+c,b+c,a+b,a.\) If \(a+b\ge10,\) the leading and trailing digits differ; if only \(b+c\ge10,\) the next pair differs. Thus the conditions are also necessary. The number of valid \(m\) is \[ \sum_{b=0}^{9}(10 - b)(9 - b) = 330. \]

There are \(9\cdot 10^2 = 900\) six-digit palindromes, so the probability is \(\dfrac{330}{900} = \dfrac{11}{30}.\)

Thus, the correct answer is E.

23 答案 C

设 \(A', B', C', D'\) 为各顶点旋转后的像。扫过的区域可分解为四个圆扇形和四个三角形。

因为 \(AP = \sqrt{2}\) 且 \(PC = AC - AP = \sqrt{6}\)，所以 \(A\) 和 \(C\) 处的扇形面积分别为 \(\tfrac{\pi}{2}\) 和 \(\tfrac{3\pi}{2}\)。若 \(H\) 是 \(AA'\) 的中点，则 \(PH=AH=1\) 且 \(HB=\sqrt3\)，所以 \(\triangle BPH\) 是 \(30\)-\(60\)-\(90^\circ\) 三角形，并且 \(PB=2\)。因此沿 \(BC\) 的两个 \(60^\circ\) 扇形面积各为 \(\frac{2\pi}{3}\)。以 \(PH\) 为高的两个三角形贡献 \(\sqrt3-1\)，另一对全等三角形的高为 \(\sqrt3\)，贡献 \(3-\sqrt3\)。所以四个三角形共贡献 \(2\)。

总面积为 \[ \begin{gathered} \dfrac{\pi}{2} + \dfrac{3\pi}{2} + 2\cdot\dfrac{2\pi}{3} + 2 \\ = \dfrac{10\pi + 6}{3} \end{gathered}\text{，}\]因此 \(a + b + c = 10 + 6 + 3 = 19\)。

因此，正确答案是 C。

English

Let \(A', B', C', D'\) be the images of the vertices under the rotation. The swept region decomposes into four circular sectors and four triangles.

Since \(AP = \sqrt{2}\) and \(PC = AC - AP = \sqrt{6},\) the sectors at \(A\) and \(C\) have areas \(\tfrac{\pi}{2}\) and \(\tfrac{3\pi}{2}.\) If \(H\) is the midpoint of \(AA',\) then \(PH=AH=1\) and \(HB=\sqrt3,\) so \(\triangle BPH\) is a \(30\)-\(60\)-\(90^\circ\) triangle and \(PB=2.\) Hence the two \(60^\circ\) sectors along \(BC\) each have area \(\frac{2\pi}{3}.\) The two triangles with altitude \(PH\) contribute \(\sqrt3-1,\) and the other congruent pair has altitude \(\sqrt3\) and contributes \(3-\sqrt3.\) Thus the four triangles contribute \(2.\)

The total area is \[ \begin{gathered} \dfrac{\pi}{2} + \dfrac{3\pi}{2} + 2\cdot\dfrac{2\pi}{3} + 2 \\ = \dfrac{10\pi + 6}{3}, \end{gathered} \] so \(a + b + c = 10 + 6 + 3 = 19.\)

Thus, the correct answer is C.

24 答案 E

将正 \(12\) 边形内接于单位圆。线段长度为 \(d_k = 2\sin(15k^\circ)\)，其中 \(1 \le k \le 6\)，长度 \(d_1, \ldots, d_5\) 各有 \(12\) 条，长度 \(d_6\) 有 \(6\) 条。

比较各和，满足 \(d_a \le d_b \le d_c\) 且 \(d_c \ge d_a + d_b\) 的禁用指标三元组 \((a, b, c)\) 为 \[ \begin{gathered} (1,1,3),(1,1,4),(1,1,5), \\ (1,1,6),(1,2,4),(1,2,5), \\ (1,2,6),(1,3,5),(1,3,6), \\ (2,2,6) \end{gathered}\text{。}\]

前三个以 \(3,4,5\) 结尾的三元组贡献 \(3\binom{12}{2}12\)；两个以 \(6\) 结尾且有重复长度的三元组贡献 \(2\binom{12}{2}6\)；三个长度互异且不含直径的三元组贡献 \(3\cdot12^3\)；剩下两个含直径的三元组贡献 \(2\cdot12^2\cdot6\)。将总和除以 \(\binom{66}{3}\)，得到失败概率 \(\frac{63}{286}\)，所以答案是 \(1-\frac{63}{286}=\frac{223}{286}\)。

因此，正确答案是 E。

English

Inscribe the \(12\)-gon in a unit circle. The segment lengths are \(d_k = 2\sin(15k^\circ)\) for \(1 \le k \le 6,\) with \(12\) segments of each length \(d_1, \ldots, d_5\) and \(6\) of length \(d_6.\)

Comparing sums, the forbidden index triples \((a, b, c)\) with \(d_a \le d_b \le d_c\) and \(d_c \ge d_a + d_b\) are \[ \begin{gathered} (1,1,3),(1,1,4),(1,1,5), \\ (1,1,6),(1,2,4),(1,2,5), \\ (1,2,6),(1,3,5),(1,3,6), \\ (2,2,6). \end{gathered} \]

The first three triples ending in \(3,4,5\) contribute \(3\binom{12}{2}12;\) the two repeated-length triples ending in \(6\) contribute \(2\binom{12}{2}6;\) the three triples of distinct non-diameter lengths contribute \(3\cdot12^3;\) and the two remaining diameter triples contribute \(2\cdot12^2\cdot6.\) Dividing their sum by \(\binom{66}{3}\) gives failure probability \(\frac{63}{286},\) so the answer is \(1-\frac{63}{286}=\frac{223}{286}.\)

Thus, the correct answer is E.

25 答案 A

在上半平面 \(H\) 上，若 \(f(z_1) = f(z_2)\)，则 \((z_1 - z_2)(z_1 + z_2 + i) = 0\)；因为 \(\operatorname{Im}(z_1), \operatorname{Im}(z_2) \gt 0\)，因子 \(z_1 + z_2 + i \ne 0\)，所以 \(f\) 在 \(H\) 上是一一的。

当 \(r\) 取实数时，边界上的值 \(f(r)=r^2+1+ir\) 描出抛物线 \(\operatorname{Re}(w)=(\operatorname{Im}(w))^2+1\)。因为 \(f(i)=-1\) 位于它的左侧，而 \(f\) 在 \(H\) 上连续且一一对应，所以它的像恰好由满足 \(\operatorname{Re}(w)<(\operatorname{Im}(w))^2+1\) 的所有 \(w\) 组成。因此我们计数满足 \(a, b \in \mathbb{Z}\)、\(|a|, |b| \le 10\) 且 \(a \lt b^2 + 1\) 的 \(w = a + ib\)： \[ \begin{gathered} |S| = 21^2 \\ {}- \sum_{b=-3}^{3}(10 - b^2) \\ = 441 - 42 = 399 \end{gathered}\text{。}\]

因此，正确答案是 A。

English

On the upper half-plane \(H,\) if \(f(z_1) = f(z_2)\) then \((z_1 - z_2)(z_1 + z_2 + i) = 0;\) since \(\operatorname{Im}(z_1), \operatorname{Im}(z_2) \gt 0,\) the factor \(z_1 + z_2 + i \ne 0,\) so \(f\) is one-to-one on \(H.\)

For real \(r,\) the boundary values \(f(r)=r^2+1+ir\) trace the parabola \(\operatorname{Re}(w)=(\operatorname{Im}(w))^2+1.\) Since \(f(i)=-1\) lies to its left and \(f\) is continuous and one-to-one on \(H,\) its image consists precisely of the values \(w\) satisfying \(\operatorname{Re}(w)<(\operatorname{Im}(w))^2+1.\) Thus we count \(w = a + ib\) with \(a, b \in \mathbb{Z},\) \(|a|, |b| \le 10,\) and \(a \lt b^2 + 1:\) \[ \begin{gathered} |S| = 21^2 \\ {}- \sum_{b=-3}^{3}(10 - b^2) \\ = 441 - 42 = 399. \end{gathered} \]

Thus, the correct answer is A.