2015B AMC 12 真题（P版）

1 答案 C

因为 \((-2)^{-2} = \dfrac{1}{(-2)^2} = \dfrac14\)，所以 \(2 - \dfrac14 = \dfrac74\)。

因此，正确选项是 C。

English

Since \((-2)^{-2} = \dfrac{1}{(-2)^2} = \dfrac14,\) we get \(2 - \dfrac14 = \dfrac74.\)

Thus, the correct answer is C.

2 答案 B

前两项任务一共用 \(100\) 分钟，所以每项任务用 \(50\) 分钟。

第三项任务在下午 \(2{:}40\) 后 \(50\) 分钟完成，即下午 \(3{:}30\)。

因此，正确选项是 B。

English

The first two tasks together take \(100\) minutes, so each task takes \(50\) minutes.

The third task finishes \(50\) minutes after \(2{:}40\) PM, at \(3{:}30\) PM.

Thus, the correct answer is B.

3 答案 A

写作 \(2x + 3y = 100\)。如果 \(28\) 被写了两次，则 \(3y = 100 - 56 = 44\)，这不是 \(3\) 的倍数。

所以 \(28\) 被写了三次：\(2x = 100 - 84 = 16\)，得 \(x = 8\)。

因此，正确选项是 A。

English

Write \(2x + 3y = 100.\) If \(28\) were written twice, then \(3y = 100 - 56 = 44,\) which is not a multiple of \(3.\)

So \(28\) is written three times: \(2x = 100 - 84 = 16,\) giving \(x = 8.\)

Thus, the correct answer is A.

4 答案 B

Marta 第 \(6\) 名，所以 Jack 第 \(5\) 名。Jack 比 Todd 落后 \(2\) 名，所以 Todd 第 \(3\) 名。Todd 比 Rand 落后 \(1\) 名，所以 Rand 第 \(2\) 名。

Rand 比 Hikmet 领先 \(6\) 名，所以 Hikmet 第 \(8\) 名。（David 第 \(10\) 名。）

因此，正确选项是 B。

English

Marta is \(6\)th, so Jack is \(5\)th. Jack is \(2\) behind Todd, so Todd is \(3\)rd. Todd is \(1\) behind Rand, so Rand is \(2\)nd.

Rand is \(6\) ahead of Hikmet, so Hikmet is \(8\)th. (David is \(10\)th.)

Thus, the correct answer is B.

5 答案 B

Sharks 队在前 \(3\) 场中赢了 \(1\) 场。要用最少的额外比赛达到 \(95\%\)，他们应赢下全部 \(N\) 场额外比赛，此时胜率为 \(\dfrac{1 + N}{3 + N}\)。

要求 \(\dfrac{1 + N}{3 + N} \ge \dfrac{19}{20}\)，得 \(20 + 20N \ge 57 + 19N\)，所以 \(N \ge 37\)。

因此，正确选项是 B。

English

The Sharks won \(1\) of the first \(3\) games. To reach \(95\%\) with the fewest extra games, they should win all \(N\) additional games, giving a win fraction \(\dfrac{1 + N}{3 + N}.\)

Requiring \(\dfrac{1 + N}{3 + N} \ge \dfrac{19}{20}\) gives \(20 + 20N \ge 57 + 19N,\) so \(N \ge 37.\)

Thus, the correct answer is B.

6 答案 A

表格主体有 \(13 \times 13 = 169\) 个表项。乘积为奇数，当且仅当两个因数都是奇数。

在 \(0, 1, \ldots, 12\) 中有 \(6\) 个奇数，因此有 \(6 \times 6 = 36\) 个奇数表项。所求比例是 \(\dfrac{36}{169} = 0.213\ldots \approx 0.21\)。

因此，正确选项是 A。

English

The body has \(13 \times 13 = 169\) entries. A product is odd exactly when both factors are odd.

There are \(6\) odd numbers among \(0, 1, \ldots, 12,\) giving \(6 \times 6 = 36\) odd entries. The fraction is \(\dfrac{36}{169} = 0.213\ldots \approx 0.21.\)

Thus, the correct answer is A.

7 答案 D

一个正 \(15\) 边形有 \(L = 15\) 条对称轴，它的最小旋转对称角是 \(R = \dfrac{360}{15} = 24\) 度。

于是 \(L + R = 15 + 24 = 39\)。

因此，正确选项是 D。

English

A regular \(15\)-gon has \(L = 15\) lines of symmetry, and its smallest angle of rotational symmetry is \(R = \dfrac{360}{15} = 24\) degrees.

Then \(L + R = 15 + 24 = 39.\)

Thus, the correct answer is D.

8 答案 D

因为 \(625 = 5^4\)，所以 \[ \begin{gathered} 625^{\log_5 2015} = 5^{4\log_5 2015} \\ = \left(5^{\log_5 2015}\right)^4 = 2015^4 \end{gathered}\text{。}\]

取四次方根得 \(\left(2015^4\right)^{\frac{1}{4}} = 2015\)。

因此，正确选项是 D。

English

Since \(625 = 5^4,\) we have \[ \begin{gathered} 625^{\log_5 2015} = 5^{4\log_5 2015} \\ = \left(5^{\log_5 2015}\right)^4 = 2015^4. \end{gathered} \]

Taking the fourth root gives \(\left(2015^4\right)^{\frac{1}{4}} = 2015.\)

Thus, the correct answer is D.

9 答案 C

设 \(x\) 为 Larry 获胜的概率。他以概率 \(\dfrac12\) 立即获胜；或者两名玩家都没击中（概率 \(\dfrac14\)），游戏重新开始。

所以 \(x = \dfrac12 + \dfrac14 x\)，得 \(\dfrac34 x = \dfrac12\)，因此 \(x = \dfrac23\)。

因此，正确选项是 C。

English

Let \(x\) be the probability Larry wins. He wins right away with probability \(\dfrac12,\) or both players miss (probability \(\dfrac14\)) and the game restarts.

So \(x = \dfrac12 + \dfrac14 x,\) giving \(\dfrac34 x = \dfrac12\) and \(x = \dfrac23.\)

Thus, the correct answer is C.

10 答案 C

设三条不同边为 \(a \lt b \lt c\)。因为 \(a + b \gt c\)，周长大于 \(2c\)，所以 \(2c \lt 15\)，从而 \(c \le 6\)。

周长小于 \(15\) 的不等边三元组为 \((6,5,3)\)，\((6,5,2)\)，\((6,4,3)\)，\((5,4,3)\)，\((5,4,2)\)，和 \((4,3,2)\)。其中只有 \((5,4,3)\) 是直角三角形，剩下 \(5\) 个。

因此，正确选项是 C。

English

Let the distinct sides be \(a \lt b \lt c.\) Since \(a + b \gt c,\) the perimeter exceeds \(2c,\) so \(2c \lt 15\) and \(c \le 6.\)

The scalene triples with perimeter less than \(15\) are \((6,5,3),\) \((6,5,2),\) \((6,4,3),\) \((5,4,3),\) \((5,4,2),\) and \((4,3,2).\) Of these, only \((5,4,3)\) is a right triangle, leaving \(5.\)

Thus, the correct answer is C.

11 答案 E

这条直线与坐标轴交于 \((5, 0)\) 和 \((0, 12)\)，所以该三角形是直角三角形，两条直角边为 \(5\) 和 \(12\)，斜边为 \(13\)，面积为 \(30\)。

两条高是直角边 \(5\) 和 \(12\)；到斜边的高为 \(\dfrac{2 \cdot 30}{13} = \dfrac{60}{13}\)。三条高的和为 \(17 + \dfrac{60}{13} = \dfrac{281}{13}\)。

因此，正确选项是 E。

English

The line meets the axes at \((5, 0)\) and \((0, 12),\) so the triangle is right with legs \(5\) and \(12\) and hypotenuse \(13.\) Its area is \(30.\)

Two altitudes are the legs \(5\) and \(12;\) the altitude to the hypotenuse is \(\dfrac{2 \cdot 30}{13} = \dfrac{60}{13}.\) The sum is \(17 + \dfrac{60}{13} = \dfrac{281}{13}.\)

Thus, the correct answer is E.

12 答案 D

因式分解得 \((x - b)\bigl(2x - (a + c)\bigr) = 0\)，所以两个根是 \(b\) 和 \(\dfrac{a + c}{2}\)。它们的和是 \(b + \dfrac{a + c}{2}\)。

使用互不相同的数字，取 \(b = 9\) 且 \(a + c = 8 + 7 = 15\)，得 \(9 + 7.5 = 16.5\)。

因此，正确选项是 D。

English

Factoring gives \((x - b)\bigl(2x - (a + c)\bigr) = 0,\) so the roots are \(b\) and \(\dfrac{a + c}{2}.\) Their sum is \(b + \dfrac{a + c}{2}.\)

Using distinct digits, take \(b = 9\) and \(a + c = 8 + 7 = 15,\) giving \(9 + 7.5 = 16.5.\)

Thus, the correct answer is D.

13 答案 B

角 \(BAC\) 和 \(BDC\) 截弧 \(BC\)，所以 \(\angle BDC = 70^\circ\)。于是 \(\angle ADC = \angle ADB\) \(+ \angle BDC = 110^\circ\)。

因为 \(ABCD\) 是圆内接四边形，\(\angle ABC = 180^\circ - 110^\circ = 70^\circ\) \(= \angle BAC\)。因此 \(\triangle ABC\) 是等腰三角形，\(AC = BC = 6\)。

因此，正确选项是 B。

English

Angles \(BAC\) and \(BDC\) subtend arc \(BC,\) so \(\angle BDC = 70^\circ.\) Then \(\angle ADC = \angle ADB\) \(+ \angle BDC = 110^\circ.\)

Since \(ABCD\) is cyclic, \(\angle ABC = 180^\circ - 110^\circ = 70^\circ\) \(= \angle BAC.\) Thus \(\triangle ABC\) is isosceles with \(AC = BC = 6.\)

Thus, the correct answer is B.

14 答案 D

设 \(z\) 为圆和三角形的公共面积。所求差为 \[ \begin{aligned} &(\text{圆} - z) - (\text{三角形} - z) \\ &= \text{圆} - \text{三角形} \end{aligned}\text{。}\]

圆的面积为 \(\pi \cdot 2^2 = 4\pi\)，等边三角形的面积为 \(\dfrac{\sqrt3}{4}\cdot 4^2 = 4\sqrt3\)。差为 \(4\pi - 4\sqrt3 = 4(\pi - \sqrt3)\)。

因此，正确选项是 D。

English

Let \(z\) be the area shared by the circle and triangle. The requested difference is \[ \begin{aligned} &(\text{circle} - z) - (\text{triangle} - z) \\ &= \text{circle} - \text{triangle}. \end{aligned} \]

The circle has area \(\pi \cdot 2^2 = 4\pi,\) and the equilateral triangle has area \(\dfrac{\sqrt3}{4}\cdot 4^2 = 4\sqrt3.\) The difference is \(4\pi - 4\sqrt3 = 4(\pi - \sqrt3).\)

Thus, the correct answer is D.

15 答案 D

数学和科学给 \(8\) 分，所以 Rachelle 还需要英语和历史至少给 \(6\) 分。英语得 C 的概率是 \(1 - \dfrac16 - \dfrac14 = \dfrac{7}{12}\)，历史得 C 的概率是 \(1 - \dfrac14 - \dfrac13 = \dfrac{5}{12}\)。

统一到分母 \(144\)：\(8\) 分的概率为 \(\dfrac16\cdot\dfrac14 = \dfrac{6}{144}\)；\(7\) 分的概率为 \(\dfrac16\cdot\dfrac13 + \dfrac14\cdot\dfrac14 = \dfrac{17}{144}\)；\(6\) 分的概率为 \(\dfrac16\cdot\dfrac{5}{12} + \dfrac14\cdot\dfrac13\) \(+ \dfrac{7}{12}\cdot\dfrac14 = \dfrac{43}{144}\)。

总概率为 \(\dfrac{6 + 17 + 43}{144} = \dfrac{66}{144} = \dfrac{11}{24}\)。

因此，正确选项是 D。

English

Math and Science give \(8\) points, so Rachelle needs at least \(6\) more from English and History. The chance of a C is \(1 - \dfrac16 - \dfrac14 = \dfrac{7}{12}\) in English and \(1 - \dfrac14 - \dfrac13 = \dfrac{5}{12}\) in History.

Working over a denominator of \(144:\) \(8\) points has probability \(\dfrac16\cdot\dfrac14 = \dfrac{6}{144};\) \(7\) points has \(\dfrac16\cdot\dfrac13 + \dfrac14\cdot\dfrac14 = \dfrac{17}{144};\) and \(6\) points has \(\dfrac16\cdot\dfrac{5}{12} + \dfrac14\cdot\dfrac13\) \(+ \dfrac{7}{12}\cdot\dfrac14 = \dfrac{43}{144}.\)

The total is \(\dfrac{6 + 17 + 43}{144} = \dfrac{66}{144} = \dfrac{11}{24}.\)

Thus, the correct answer is D.

16 答案 C

六边形中心到顶点的距离为 \(6\)。侧棱长为 \(8\)，所以棱锥高为 \(\sqrt{8^2 - 6^2} = \sqrt{28} = 2\sqrt7\)。

六边形面积为 \(\dfrac{3\sqrt3}{2}\cdot 6^2 = 54\sqrt3\)。因此体积为 \(\dfrac13 \cdot 54\sqrt3 \cdot 2\sqrt7 = 36\sqrt{21}\)。

因此，正确选项是 C。

English

The distance from the hexagon’s center to a vertex is \(6.\) A lateral edge has length \(8,\) so the pyramid’s height is \(\sqrt{8^2 - 6^2} = \sqrt{28} = 2\sqrt7.\)

The hexagon’s area is \(\dfrac{3\sqrt3}{2}\cdot 6^2 = 54\sqrt3.\) Thus the volume is \(\dfrac13 \cdot 54\sqrt3 \cdot 2\sqrt7 = 36\sqrt{21}.\)

Thus, the correct answer is C.

17 答案 D

令两个概率相等，并消去 \(\tfrac14\) 和 \(\tfrac34\) 的公共幂，得 \(\binom{n}{2}\cdot\dfrac34 = \binom{n}{3}\cdot\dfrac14\)。

这化为 \(\dfrac{n(n-1)}{2}\cdot 3\) \(= \dfrac{n(n-1)(n-2)}{6}\)，所以 \(\dfrac32 = \dfrac{n-2}{6}\)，得 \(n - 2 = 9\)，即 \(n = 11\)。

因此，正确选项是 D。

English

Setting the two probabilities equal and cancelling the common powers of \(\tfrac14\) and \(\tfrac34\) gives \(\binom{n}{2}\cdot\dfrac34 = \binom{n}{3}\cdot\dfrac14.\)

This becomes \(\dfrac{n(n-1)}{2}\cdot 3\) \(= \dfrac{n(n-1)(n-2)}{6},\) so \(\dfrac32 = \dfrac{n-2}{6},\) giving \(n - 2 = 9\) and \(n = 11.\)

Thus, the correct answer is D.

18 答案 D

一个合数至少有两个质因子（按重数计），而最小质数是 \(2\)，所以最小可能值是 \(2 + 2 = 4\)。

每个大于 \(3\) 的整数都能取到：\(r(2^k) = 2k\) 覆盖所有 \(\ge 4\) 的偶数值，\(r(2^k\cdot 3) = 2k + 3\) 覆盖所有 \(\ge 5\) 的奇数值。所以值域是大于 \(3\) 的整数。

因此，正确选项是 D。

English

A composite number has at least two prime factors (with multiplicity), and the smallest prime is \(2,\) so the least possible value is \(2 + 2 = 4.\)

Every integer greater than \(3\) is attained: \(r(2^k) = 2k\) covers the even values \(\ge 4,\) and \(r(2^k\cdot 3) = 2k + 3\) covers the odd values \(\ge 5.\) So the range is the integers greater than \(3.\)

Thus, the correct answer is D.

19 答案 C

圆心 \(O\) 位于 \(XY\) 和 \(ZW\) 的垂直平分线上，而它们分别就是 \(AB\) 和 \(AC\) 的垂直平分线。因此 \(O\) 是 \(\triangle ABC\) 的外心；又因为 \(\angle C = 90^\circ\)，所以 \(O\) 是 \(AB\) 的中点。

令 \(a = \tfrac12 BC\) 且 \(b = \tfrac12 CA\)。则 \(a^2 + b^2 = 6^2\)，由 \(OX^2 = OW^2\) 得 \(12^2 + 6^2 = b^2 + (a + 2b)^2\)。因为左边是 \(5(a^2+b^2)\)，相减得到 \(4a(b-a)=0\)。所以 \(a=b\)，再由 \(2a^2=36\) 得 \(a=b=3\sqrt2\)。因此 \(BC=CA=6\sqrt2\)，周长为 \(12+12\sqrt2\)。

因此，正确答案是 C。

English

The center \(O\) of the circle lies on the perpendicular bisectors of \(XY\) and \(ZW,\) which are the same as those of \(AB\) and \(AC.\) So \(O\) is the circumcenter of \(\triangle ABC,\) and since \(\angle C = 90^\circ,\) \(O\) is the midpoint of \(AB.\)

Let \(a = \tfrac12 BC\) and \(b = \tfrac12 CA.\) Then \(a^2 + b^2 = 6^2,\) and computing \(OX^2 = OW^2\) gives \(12^2 + 6^2 = b^2 + (a + 2b)^2.\) Because the left side is \(5(a^2+b^2),\) subtracting gives \(4a(b-a)=0.\) Thus \(a=b,\) and \(2a^2=36\) gives \(a=b=3\sqrt2.\) Therefore \(BC=CA=6\sqrt2,\) and the perimeter is \(12+12\sqrt2.\)

Thus, the correct answer is C.

20 答案 B

在每一行中从左到右应用递推，得到 \[ \begin{array}{c|ccccc} i\backslash j&0&1&2&3&4\\ \hline 0&1&2&3&4&0\\ 1&2&3&4&0&1\\ 2&3&0&2&4&1\\ 3&0&3&4&1&0\\ 4&3&1&3&1&3\\ 5&1&1&1&1&1 \end{array} \] 如果某一行全是 \(1\)，递推就会使下一行也全是 \(1\)。因此由归纳法可知，对每个 \(i\ge5\) 都有 \(f(i,2)=1\)。

因为 \(2015 \ge 5\)，所以 \(f(2015, 2) = 1\)。

因此，正确选项是 B。

English

Applying the recursion from left to right in each row gives \[ \begin{array}{c|ccccc} i\backslash j&0&1&2&3&4\\ \hline 0&1&2&3&4&0\\ 1&2&3&4&0&1\\ 2&3&0&2&4&1\\ 3&0&3&4&1&0\\ 4&3&1&3&1&3\\ 5&1&1&1&1&1 \end{array} \] If a row consists entirely of \(1\)s, the recursion makes the next row entirely \(1\)s as well. Hence, by induction, \(f(i,2)=1\) for every \(i\ge5.\)

Since \(2015 \ge 5,\) we get \(f(2015, 2) = 1.\)

Thus, the correct answer is B.

21 答案 D

一段 \(t\) 级的楼梯需要 Cozy 跳 \(\left\lceil \tfrac{t}{2} \right\rceil\) 次，Dash 跳 \(\left\lceil \tfrac{t}{5} \right\rceil\) 次，我们需要二者差为 \(19\)。

设 Dash 跳了 \(d+1\) 次。那么 \(t\) 是 \(5d+1\)、\(5d+2\)、\(5d+3\)、\(5d+4\)、\(5d+5\) 之一。Cozy 跳了 \(d+20\) 次，所以 \(t\) 是 \(2d+39\) 或 \(2d+40\)。令这两组值相等，只有在以下三种情形中 \(d\) 才是整数：\[ \begin{gathered} 5d+3=2d+39,\\ 5d+1=2d+40,\\ 5d+4=2d+40 \end{gathered}\text{。}\] 它们分别给出 \(t=63,66,64\)。因此有效值为 \(t=63\)、\(64\) 和 \(66\)，所以 \(s = 63 + 64 + 66 = 193\)。它的各位数字之和是 \(1 + 9 + 3 = 13\)。

因此，正确选项是 D。

English

A staircase of \(t\) steps takes Cozy \(\left\lceil \tfrac{t}{2} \right\rceil\) jumps and Dash \(\left\lceil \tfrac{t}{5} \right\rceil\) jumps, and we need the difference to equal \(19.\)

Suppose Dash makes \(d+1\) jumps. Then \(t\) is one of \(5d+1,\) \(5d+2,\) \(5d+3,\) \(5d+4,\) \(5d+5.\) Cozy makes \(d+20\) jumps, so \(t\) is either \(2d+39\) or \(2d+40.\) Equating these two lists gives an integer \(d\) only in the three cases \[ \begin{gathered} 5d+3=2d+39,\\ 5d+1=2d+40,\\ 5d+4=2d+40. \end{gathered} \] These yield respectively \(t=63,66,64.\) Thus the valid values are \(t=63,\) \(64,\) and \(66,\) so \(s = 63 + 64 + 66 = 193.\) Its digit sum is \(1 + 9 + 3 = 13.\)

Thus, the correct answer is D.

22 答案 D

先想象每个人都移动到正对面的椅子。条件变为：每个人必须坐在同一把椅子或相邻椅子上。保持原座位的人数必须为偶数（否则一个奇数长度的空段无法填满）。

若 \(0\) 人保持原座位，所有人一齐向左移、一齐向右移，或与相邻者两两交换：共 \(4\) 种。若 \(2\) 人保持原座位，这两人必须相对或相邻，给出 \(3+6=9\) 种选择，其余的人只能与相邻者两两交换。若 \(4\) 人保持原座位，另外两人必须占据相邻座位并互相交换，给出 \(6\) 种选择。若全部 \(6\) 人都不动，有 \(1\) 种。总数是 \(4+9+6+1=20\)。

因此，正确选项是 D。

English

First imagine everyone moves to the chair directly opposite. The condition becomes: each person must sit in the same chair or an adjacent one. The number of people who keep their seat must be even (otherwise an odd-length gap cannot be filled).

If \(0\) keep their seat, everyone shifts left, shifts right, or swaps with a neighbor: \(4\) ways. If \(2\) keep their seats, those two must be opposite or adjacent, giving \(3+6=9\) choices, and the remaining people are forced to swap in adjacent pairs. If \(4\) keep their seats, the other two must occupy adjacent seats and swap, giving \(6\) choices. If all \(6\) stay, there is \(1\) way. The total is \(4+9+6+1=20.\)

Thus, the correct answer is D.

23 答案 B

体积和表面积在数值上相等，意味着 \(abc=2(ab+bc+ca)\)。两边除以 \(abc\) 得 \(1=\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\le\frac{6}{a}\)，所以 \(a\le6\)。情形 \(a=1\) 和 \(a=2\) 都没有正数解。当 \(a\ne2\) 时，令 \[ \begin{gathered} u=(a-2)b-2a,\\ v=(a-2)c-2a \end{gathered}\text{。}\]原方程可因式分解为 \(uv=4a^2\)。

当 \(a=3\) 时，\((b-6)(c-6)=36\) 给出 \((b,c)=(7,42)\)，\((8,24)\)，\((9,18)\)，\((10,15)\)，\((12,12)\)。当 \(a=4\) 时，\((b-4)(c-4)=16\) 给出 \((5,20),(6,12),(8,8)\)。当 \(a=5\) 时，两个因数模 \(3\) 的同余条件只留下有效数对 \((b,c)=(5,10)\)，当 \(a=6\) 时，\((b-3)(c-3)=9\) 只留下 \((b,c)=(6,6)\)。因此共有 \(5+3+1+1=10\) 个三元组。

因此，正确答案是 B。

English

Numerically equal volume and surface area means \(abc=2(ab+bc+ca).\) Dividing by \(abc\) gives \(1=\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\le\frac{6}{a},\) so \(a\le6.\) The cases \(a=1\) and \(a=2\) give no positive solutions. For \(a\ne2,\) set \[ \begin{gathered} u=(a-2)b-2a,\\ v=(a-2)c-2a. \end{gathered} \] The equation then factors as \(uv=4a^2.\)

For \(a=3,\) \((b-6)(c-6)=36\) gives \((b,c)=(7,42),\) \((8,24),\) \((9,18),\) \((10,15),\) \((12,12).\) For \(a=4,\) \((b-4)(c-4)=16\) gives \((5,20),(6,12),(8,8).\) For \(a=5,\) the congruence of the two factors modulo \(3\) leaves only the valid pair \((b,c)=(5,10),\) and for \(a=6,\) \((b-3)(c-3)=9\) leaves only \((b,c)=(6,6).\) Thus there are \(5+3+1+1=10\) triples.

Thus, the correct answer is B.

24 答案 D

因为每个圆心到 \(P\) 和 \(Q\) 等距，所以四个圆心和 \(R\) 都位于 \(PQ\) 的垂直平分线上，并且 \(PR=24\)。先考虑半径比为 \(5{:}8\)、彼此距离为 \(39\) 的两个圆心。如果 \(R\) 位于它们之间，令 \(y = AR\)，并令 \(x = \tfrac15\) 乘以圆 \(A\) 的半径。则 \(y^2 + 24^2 = 25x^2\) 且 \((39 - y)^2 + 24^2 = 64x^2\)。相减得到 \(x^2 = 39 - 2y\)，所以 \(y^2 + 50y - 399 = 0\)，从而 \(y=7\)。此时 \(x=5\)，两个圆心到 \(R\) 的距离为 \(7\) 和 \(32\)，半径为 \(25\) 和 \(40\)。

如果两个圆心位于 \(R\) 的同一侧，它们到该点的距离为 \(w\) 和 \(w+39\)。类似的方程给出 \(w^2-50w-399=0\)，所以 \(w=57\)，两个距离为 \(57\) 和 \(96\)。此时半径为 \(5\sqrt{153}\) 和 \(8\sqrt{153}\)。如果两对 \((A,B)\) 和 \((C,D)\) 都采用同一种位置关系，就会出现两对同半径的全等圆，与题设矛盾。因此两对必须分别采用一种位置关系。它们的距离和分别为 \(7+32=39\) 和 \(57+96=153\)，所以所求总和为 \(39+153=192\)。

因此，正确答案是 D。

English

Since every center is equidistant from \(P\) and \(Q,\) all four centers and \(R\) lie on the perpendicular bisector of \(PQ,\) with \(PR=24.\) First consider two centers whose radii are in the ratio \(5{:}8\) and whose distance apart is \(39.\) If \(R\) lies between them, let \(y = AR\) and \(x = \tfrac15\) of circle \(A\)’s radius. Then \(y^2 + 24^2 = 25x^2\) and \((39 - y)^2 + 24^2 = 64x^2.\) Subtracting gives \(x^2 = 39 - 2y,\) so \(y^2 + 50y - 399 = 0\) and \(y=7.\) Here \(x=5,\) so the two center distances from \(R\) are \(7\) and \(32,\) and the radii are \(25\) and \(40.\)

If instead the two centers lie on the same side of \(R,\) their distances are \(w\) and \(w+39.\) The analogous equations give \(w^2-50w-399=0,\) hence \(w=57,\) and the distances are \(57\) and \(96.\) In this case the radii are \(5\sqrt{153}\) and \(8\sqrt{153}.\) Using the same placement for both pairs \((A,B)\) and \((C,D)\) would give two congruent circles of each radius, contrary to the hypothesis. Thus one pair uses each placement. Their distance sums are \(7+32=39\) and \(57+96=153,\) so the requested total is \(39+153=192.\)

Thus, the correct answer is D.

25 答案 B

令 \(P_0 = 0\)，并设 \(z = e^{\frac{\pi i}{6}}\)，则长度为 \(k\)、方向为 \(z^{k-1}\) 的每一步给出 \(P_{2015} = \sum_{k=1}^{2015} k z^{k-1}\)。对这个级数求和（对几何级数求导），得到 \[ \begin{aligned} &P_{2015} = \dfrac{1}{(z - 1)^2} \\ &\quad {}\cdot \bigl(2015 z^{2016} - 2016 z^{2015} + 1\bigr) \end{aligned}\text{。}\]

因为 \(z^{12} = 1\)，所以 \(z^{2016} = 1\) 且 \(z^{2015} = \tfrac1z\)，因此 \(P_{2015} = \dfrac{2016}{z(z - 1)}\)。使用 \(|z - 1|^2 = 2 - \sqrt3 = \dfrac{(\sqrt3 - 1)^2}{2}\) 以及 \(|z| = 1\)，距离为 \(\dfrac{2016}{|z - 1|} = 1008\sqrt6 + 1008\sqrt2\)。

因此 \(a + b + c + d = 1008 + 6 + 1008\) \(+ 2 = 2024\)。

因此，正确选项是 B。

English

Place \(P_0 = 0\) and let \(z = e^{\frac{\pi i}{6}},\) so each step of length \(k\) in direction \(z^{k-1}\) gives \(P_{2015} = \sum_{k=1}^{2015} k z^{k-1}.\) Summing this (a differentiated geometric series) leads to \[ \begin{aligned} &P_{2015} = \dfrac{1}{(z - 1)^2} \\ &\quad {}\cdot \bigl(2015 z^{2016} - 2016 z^{2015} + 1\bigr). \end{aligned} \]

Since \(z^{12} = 1,\) we have \(z^{2016} = 1\) and \(z^{2015} = \tfrac1z,\) so \(P_{2015} = \dfrac{2016}{z(z - 1)}.\) Using \(|z - 1|^2 = 2 - \sqrt3 = \dfrac{(\sqrt3 - 1)^2}{2}\) and \(|z| = 1,\) the distance is \(\dfrac{2016}{|z - 1|} = 1008\sqrt6 + 1008\sqrt2.\)

Hence \(a + b + c + d = 1008 + 6 + 1008\) \(+ 2 = 2024.\)

Thus, the correct answer is B.