2008A AMC 12 真题（P版）

1 答案 D

从上午 \(8\!:\!30\) 到 \(11\!:\!10\) 是 \(2\) 小时 \(40\) 分，也就是 \(160\) 分钟，完成了三分之一。

整项工作需要 \(3 \cdot 160 = 480\) 分钟，即 \(8\) 小时。从 \(8\!:\!30\) 再过 \(8\) 小时，得到 \(4\!:\!30\)。

所以正确答案是 D。

English

From \(8\!:\!30\) am to \(11\!:\!10\) am is \(2\) hours \(40\) minutes, or \(160\) minutes, to complete one third of the job.

The whole job then takes \(3 \cdot 160 = 480\) minutes, or \(8\) hours. Adding \(8\) hours to \(8\!:\!30\) am gives \(4\!:\!30\) pm.

Thus, D is the correct answer.

2 答案 A

先通分相加，得到 \[ \dfrac{1}{2} + \dfrac{2}{3} = \dfrac{3}{6} + \dfrac{4}{6} = \dfrac{7}{6}\text{。}\]

\(\dfrac{7}{6}\) 的倒数是 \(\dfrac{6}{7}\)。

所以正确答案是 A。

English

Using a common denominator, \[ \dfrac{1}{2} + \dfrac{2}{3} = \dfrac{3}{6} + \dfrac{4}{6} = \dfrac{7}{6}. \]

The reciprocal of \(\dfrac{7}{6}\) is \(\dfrac{6}{7}.\)

Thus, A is the correct answer.

3 答案 C

因为 \(10\) 根香蕉的 \(\tfrac{2}{3}\) 是 \(\tfrac{20}{3}\) 根，价值 \(8\) 个橙子，所以一根香蕉价值 \[ 8 \div \dfrac{20}{3} = \dfrac{24}{20} = \dfrac{6}{5} \] 个橙子。

于是 \(5\) 根香蕉的 \(\tfrac{1}{2}\) 是 \(\tfrac{5}{2}\) 根香蕉，价值 \[ \dfrac{5}{2} \cdot \dfrac{6}{5} = 3 \] 个橙子。

所以正确答案是 C。

English

Since \(\tfrac{2}{3}\) of \(10\) bananas is \(\tfrac{20}{3}\) bananas, worth \(8\) oranges, one banana is worth \[ 8 \div \dfrac{20}{3} = \dfrac{24}{20} = \dfrac{6}{5} \] oranges.

Then \(\tfrac{1}{2}\) of \(5\) bananas is \(\tfrac{5}{2}\) bananas, worth \[ \dfrac{5}{2} \cdot \dfrac{6}{5} = 3 \] oranges.

Thus, C is the correct answer.

4 答案 B

除第一个分母外，每个分母都与前一个分数的分子相消，所以乘积化为 \[ \dfrac{2008}{4} = 502\text{。}\]

所以正确答案是 B。

English

Every denominator except the first cancels with the numerator of the preceding fraction, so the product collapses to \[ \dfrac{2008}{4} = 502. \]

Thus, B is the correct answer.

5 答案 B

通分合并可得 \[ \dfrac{2x}{3} - \dfrac{x}{6} = \dfrac{4x - x}{6} = \dfrac{x}{2}\text{。}\]

该式为整数当且仅当 \(x\) 是偶数。例如 \(x = 4\) 是偶数但不是 \(3\) 的倍数，这就排除了其余每一个说法。

所以正确答案是 B。

English

Combining the fractions, \[ \dfrac{2x}{3} - \dfrac{x}{6} = \dfrac{4x - x}{6} = \dfrac{x}{2}. \]

This is an integer exactly when \(x\) is even. The example \(x = 4\) is even but not a multiple of \(3,\) which rules out every other statement.

Thus, B is the correct answer.

6 答案 A

设标价为 \(x\) 美元。A 店价格为 \(0.85x - 90\)，B 店价格为 \(0.75x\)。

因为 A 店便宜 \(15\) 美元，所以 \[ 0.85x - 90 = 0.75x - 15\text{，} \] 从而 \(0.10x = 75\)，\(x = 750\)。

所以正确答案是 A。

English

Let \(x\) be the sticker price in dollars. Store A charges \(0.85x - 90\) dollars, and store B charges \(0.75x\) dollars.

Since store A is \(15\) dollars cheaper, \[ 0.85x - 90 = 0.75x - 15, \] so \(0.10x = 75\) and \(x = 750.\)

Thus, A is the correct answer.

7 答案 D

以每小时 \(4\) 英里划 \(1\) 英里，需要 \(\tfrac{1}{4}\) 小时，即 \(15\) 分钟。这段时间进入 \(15 \cdot 10 = 150\) 加仑水。

船内最多剩 \(30\) 加仑，所以 LeRoy 必须舀出 \(150 - 30 = 120\) 加仑，用时 \(15\) 分钟，速率为 \[ \dfrac{120}{15} = 8 \] 加仑每分钟。

所以正确答案是 D。

English

Rowing \(1\) mile at \(4\) miles per hour takes \(\tfrac{1}{4}\) hour, or \(15\) minutes. In that time \(15 \cdot 10 = 150\) gallons of water enter the boat.

Since at most \(30\) gallons may remain, LeRoy must bail \(150 - 30 = 120\) gallons in \(15\) minutes, a rate of \[ \dfrac{120}{15} = 8 \] gallons per minute.

Thus, D is the correct answer.

8 答案 C

体积为 \(1\) 的立方体边长为 \(1\)，表面积为 \(6\)。所求立方体表面积为 \(12\)。若边长为 \(s\)，则 \(6s^2 = 12\)，所以 \(s = \sqrt{2}\)。

其体积为 \[ (\sqrt{2})^3 = 2\sqrt{2}\text{。}\]

所以正确答案是 C。

English

The cube with volume \(1\) has side \(1\) and surface area \(6.\) The larger cube has surface area \(12,\) so if its side is \(s,\) then \(6s^2 = 12,\) giving \(s = \sqrt{2}.\)

Its volume is \[ (\sqrt{2})^3 = 2\sqrt{2}. \]

Thus, C is the correct answer.

9 答案 D

屏幕高、宽、对角线成 \(3:4:5\)，所以高度为 \(\tfrac{3}{5} \cdot 27 = 16.2\) 英寸，宽度为 \(\tfrac{4}{5} \cdot 27 = 21.6\) 英寸。

电影宽高比为 \(2:1\)，使用完整宽度时高度为 \(\tfrac{21.6}{2} = 10.8\) 英寸。

因此每条黑条的高度为 \[ \dfrac{16.2 - 10.8}{2} = 2.7 \] 英寸。

所以正确答案是 D。

English

Since the sides and diagonal are in ratio \(3:4:5,\) the height is \(\tfrac{3}{5} \cdot 27 = 16.2\) inches and the width is \(\tfrac{4}{5} \cdot 27 = 21.6\) inches.

The movie has aspect ratio \(2:1,\) so its height is \(\tfrac{21.6}{2} = 10.8\) inches.

Each darkened strip therefore has height \[ \dfrac{16.2 - 10.8}{2} = 2.7 \] inches.

Thus, D is the correct answer.

10 答案 D

Doug 每小时粉刷 \(\tfrac{1}{5}\) 个房间，Dave 每小时粉刷 \(\tfrac{1}{7}\) 个房间，所以合速度为 \(\tfrac{1}{5} + \tfrac{1}{7}\) 个房间每小时。

总时间为 \(t\)，其中 \(1\) 小时用于午饭，所以实际工作 \(t - 1\) 小时。完成一个房间给出 \[ \left(\dfrac{1}{5} + \dfrac{1}{7}\right)(t - 1) = 1\text{。}\]

所以正确答案是 D。

English

In one hour Doug paints \(\tfrac{1}{5}\) of the room and Dave paints \(\tfrac{1}{7},\) so together they paint \(\tfrac{1}{5} + \tfrac{1}{7}\) of the room per hour.

Of the total time \(t,\) one hour is spent at lunch, so they work for \(t - 1\) hours. The fraction painted must equal \(1,\) giving \[ \left(\dfrac{1}{5} + \dfrac{1}{7}\right)(t - 1) = 1. \]

Thus, D is the correct answer.

11 答案 C

每个立方体六面数字和为 \(1 + 2 + 4 + 8 + 16 + 32 = 63\)。由展开图，相对面分别是 \(1\) 与 \(32\)，\(2\) 与 \(16\)，\(4\) 与 \(8\)。

下面两个立方体各隐藏一对上下相对面，应隐藏和最小的 \(4 + 8 = 12\)。最上面立方体只隐藏底面，应隐藏 \(1\)。

最大可见和为 \[ \begin{aligned} &3 \cdot 63 - 2 \cdot 12 - 1 \\ &= 189 - 24 - 1 \\ &= 164 \end{aligned}\text{。}\]

所以正确答案是 C。

English

The six faces of each cube sum to \(1 + 2 + 4 + 8 + 16 + 32 = 63.\) From the pattern, the pairs of opposite faces are \(1\) & \(32,\) \(2\) & \(16,\) and \(4\) & \(8.\)

Each of the two lower cubes hides a pair of opposite faces (top and bottom); hiding the pair \(4 + 8 = 12\) is best. The top cube hides only its bottom face, so hide the \(1.\)

The greatest sum is \[ \begin{aligned} &3 \cdot 63 - 2 \cdot 12 - 1 \\ &= 189 - 24 - 1 \\ &= 164. \end{aligned} \]

Thus, C is the correct answer.

12 答案 B

\(f(x + 1)\) 有定义需要 \(0 \le x + 1 \le 2\)，即 \(-1 \le x \le 1\)，所以 \(g\) 的定义域为 \([-1, 1]\)。

当 \(f(x + 1)\) 取遍 \([0, 1]\) 时，\(1 - f(x + 1)\) 也取遍 \([0, 1]\)，所以 \(g\) 的值域为 \([0, 1]\)。

所以正确答案是 B。

English

The value \(f(x + 1)\) is defined when \(0 \le x + 1 \le 2,\) that is, \(-1 \le x \le 1,\) so the domain of \(g\) is \([-1, 1].\)

As \(f(x + 1)\) ranges over \([0, 1],\) the value \(1 - f(x + 1)\) ranges over \([0, 1]\) as well, so the range of \(g\) is \([0, 1].\)

Thus, B is the correct answer.

13 答案 B

设小圆和大圆半径分别为 \(r\) 和 \(R\)，小圆圆心为 \(E\)。由对称性，\(E\) 在 \(\angle AOB\) 的角平分线上，所以 \(OE\) 与 \(OA\) 成 \(30^\circ\)。

作半径 \(ED\) 垂直于 \(OA\)，得到 \(30\)-\(60\)-\(90\) 三角形，所以 \(OE = 2 \cdot ED = 2r\)。两圆内切还给出 \(OE = R - r\)。

于是 \(R - r = 2r\)，所以 \(R = 3r\)，\(\tfrac{r}{R} = \tfrac{1}{3}\)。于是面积之比为 \[ \left(\dfrac{1}{3}\right)^2 = \dfrac{1}{9}\text{。}\]

所以正确答案是 B。

English

Let \(r\) and \(R\) be the radii of the smaller and larger circles, and let \(E\) be the center of the smaller circle. By symmetry \(E\) lies on the bisector of \(\angle AOB,\) so \(OE\) makes a \(30^\circ\) angle with \(OA.\)

Dropping the radius \(ED\) perpendicular to \(OA\) gives a \(30\)-\(60\)-\(90\) triangle with \(OE = 2 \cdot ED = 2r.\) Since the circles are internally tangent, \(OE = R - r.\)

Then \(R - r = 2r,\) so \(R = 3r\) and \(\tfrac{r}{R} = \tfrac{1}{3}.\) The ratio of areas is \[ \left(\dfrac{1}{3}\right)^2 = \dfrac{1}{9}. \]

Thus, B is the correct answer.

14 答案 A

该区域是以 \(\left(6, -\tfrac{7}{2}\right)\) 为中心的菱形。令 \(2y + 7 = 0\)，得 \(|3x - 18| \le 3\)，所以 \(x \in [5, 7]\)，水平对角线长 \(2\)。

令 \(3x - 18 = 0\)，得 \(|2y + 7| \le 3\)，所以 \(y \in [-5, -2]\)，竖直对角线长 \(3\)。

菱形面积为对角线乘积的一半，即 \[ \dfrac{1}{2} \cdot 2 \cdot 3 = 3\text{。}\]

所以正确答案是 A。

English

The region is a rhombus centered at \(\left(6, -\tfrac{7}{2}\right).\) Setting \(2y + 7 = 0\) gives \(|3x - 18| \le 3,\) so \(x \in [5, 7],\) a horizontal diagonal of length \(2.\)

Setting \(3x - 18 = 0\) gives \(|2y + 7| \le 3,\) so \(y \in [-5, -2],\) a vertical diagonal of length \(3.\)

The area of the rhombus is half the product of its diagonals, \[ \dfrac{1}{2} \cdot 2 \cdot 3 = 3. \]

Thus, A is the correct answer.

15 答案 D

\(2008^2\) 的个位数字为 \(4\)。因为 \(2008\) 是 \(4\) 的倍数，\(2^{2008}\) 的个位数字为 \(6\)。所以 \(k\) 的个位数字为 \(0\)，从而 \(k^2\) 的个位数字也为零。

又 \(2008^2\) 和 \(2^{2008}\) 都是 \(4\) 的倍数，故 \(k\) 是 \(4\) 的倍数，因此 \(2^k\) 的个位数字为 \(6\)。

因此 \(k^2 + 2^k\) 的个位数字为 \(0 + 6 = 6\)。

所以正确答案是 D。

English

The units digit of \(2008^2\) is \(4.\) Since \(2008\) is a multiple of \(4,\) the units digit of \(2^{2008}\) is \(6.\) Thus \(k\) has units digit \(0,\) and so does \(k^2.\)

Both \(2008^2\) and \(2^{2008}\) are multiples of \(4,\) so \(k\) is a multiple of \(4.\) Therefore the units digit of \(2^k\) is \(6.\)

The units digit of \(k^2 + 2^k\) is then \(0 + 6 = 6.\)

Thus, D is the correct answer.

16 答案 D

三项分别为 \(3\log a + 7\log b\)、\(5\log a + 12\log b\) 和 \(8\log a + 15\log b\)。令相邻两项之差相等，得 \[ \begin{aligned} &2\log a + 5\log b \\ &= 3\log a + 3\log b \end{aligned}\text{，}\] 所以 \(\log a = 2\log b\)。

于是第一项为 \((3 \cdot 2 + 7)\log b = 13\log b\)，公差为 \((2 \cdot 2 + 5)\log b = 9\log b\)。

第 \(12\) 项为 \[ \begin{aligned} (13 + 11 \cdot 9)\log b &= 112\log b \\ &= \log(b^{112}) \end{aligned}\text{，}\] 所以 \(n = 112\)。

所以正确答案是 D。

English

The three terms are \(3\log a + 7\log b,\) \(5\log a + 12\log b,\) and \(8\log a + 15\log b.\) Setting the two consecutive differences equal, \[ \begin{aligned} &2\log a + 5\log b \\ &= 3\log a + 3\log b, \end{aligned} \] so \(\log a = 2\log b.\)

The first term is then \((3 \cdot 2 + 7)\log b = 13\log b,\) and the common difference is \((2 \cdot 2 + 5)\log b = 9\log b.\)

The \(12\)th term is \[ \begin{aligned} (13 + 11 \cdot 9)\log b &= 112\log b \\ &= \log(b^{112}), \end{aligned} \] so \(n = 112.\)

Thus, D is the correct answer.

17 答案 D

若 \(a_1\) 为偶数，则 \(a_2 = \frac{a_1}{2} \lt a_1\)，条件失败。

若 \(a_1 \equiv 1 \pmod 4\)，则 \(a_2 = 3a_1 + 1\) 是 \(4\) 的倍数，\(a_3 = \frac{3a_1 + 1}{2}\)，且 \(a_4 = \frac{3a_1 + 1}{4} \le a_1\)，条件也失败。

若 \(a_1 \equiv 3 \pmod 4\)，则 \(a_2\) 为偶数但不是 \(4\) 的倍数，所以 \(a_3 = \frac{3a_1 + 1}{2} \gt a_1\)，且 \(a_3\) 为奇数，\(a_4 = 3a_3 + 1 \gt a_3 \gt a_1\)。此时条件成立。

满足条件的 \(a_1 \le 2008\) 且 \(a_1 \equiv 3 \pmod 4\) 的数恰有 \(\tfrac{2008}{4} = 502\) 个。

所以正确答案是 D。

English

If \(a_1\) is even, then \(a_2 = \frac{a_1}{2} \lt a_1,\) so the condition fails.

If \(a_1 \equiv 1 \pmod 4,\) then \(a_2 = 3a_1 + 1\) is a multiple of \(4,\) so \(a_3 = \frac{3a_1 + 1}{2}\) and \(a_4 = \frac{3a_1 + 1}{4} \le a_1,\) and again the condition fails.

If \(a_1 \equiv 3 \pmod 4,\) then \(a_2\) is even but not a multiple of \(4,\) so \(a_3 = \frac{3a_1 + 1}{2} \gt a_1,\) and \(a_3\) is odd, giving \(a_4 = 3a_3 + 1 \gt a_3 \gt a_1.\) The condition holds.

Exactly \(\tfrac{2008}{4} = 502\) values of \(a_1 \le 2008\) satisfy \(a_1 \equiv 3 \pmod 4.\)

Thus, D is the correct answer.

18 答案 C

设 \(A = (a, 0, 0)\)、\(B = (0, b, 0)\)、\(C = (0, 0, c)\)。于是 \[ \begin{aligned} a^2 + b^2 &= 25, \\ b^2 + c^2 &= 36, \\ a^2 + c^2 &= 49 \end{aligned}\text{。}\]

将三个边长方程相加得 \(a^2 + b^2 + c^2 = 55\)，所以 \(a^2 = 19\)，\(b^2 = 6\)，\(c^2 = 30\)。

体积为 \[ \begin{aligned} \dfrac{1}{6}abc &= \dfrac{1}{6}\sqrt{19 \cdot 6 \cdot 30} \\ &= \dfrac{1}{6}\sqrt{3420} \\ &= \sqrt{95} \end{aligned}\text{。}\]

所以正确答案是 C。

English

Let \(A = (a, 0, 0),\) \(B = (0, b, 0),\) \(C = (0, 0, c).\) Assigning the sides, \[ \begin{aligned} a^2 + b^2 &= 25, \\ b^2 + c^2 &= 36, \\ a^2 + c^2 &= 49. \end{aligned} \]

Adding gives \(a^2 + b^2 + c^2 = 55,\) so \(a^2 = 19,\) \(b^2 = 6,\) and \(c^2 = 30.\)

The volume is \[ \begin{aligned} \dfrac{1}{6}abc &= \dfrac{1}{6}\sqrt{19 \cdot 6 \cdot 30} \\ &= \dfrac{1}{6}\sqrt{3420} \\ &= \sqrt{95}. \end{aligned} \]

Thus, C is the correct answer.

19 答案 C

每项为 \(x^{a + b + c}\)，其中 \(0 \le a \le 27\)，\(0 \le b, c \le 14\)。要得到 \(x^{28}\)，必须有 \(a = 28 - b - c\)。

\((b, c)\) 有 \((14 + 1)^2 = 225\) 种选择。除了 \((b, c) = (0, 0)\) 外，所需的 \(a = 28 - b - c\) 都在 \([0, 27]\) 中。

因此 \(x^{28}\) 的系数为 \(225 - 1 = 224\)。

所以正确答案是 C。

English

Each term is \(x^{a + b + c}\) with \(0 \le a \le 27\) and \(0 \le b, c \le 14.\) To get \(x^{28}\) we need \(a = 28 - b - c.\)

There are \((14 + 1)^2 = 225\) choices for \((b, c).\) For every choice except \((b, c) = (0, 0),\) the required \(a = 28 - b - c\) lies in \([0, 27],\) giving a valid term.

The coefficient of \(x^{28}\) is therefore \(225 - 1 = 224.\)

Thus, C is the correct answer.

20 答案 E

由角平分线定理，\(AD:DB = CA:CB = 3:4\)，所以 \(AD = \tfrac{15}{7}\)，\(BD = \tfrac{20}{7}\)。两个小三角形 \(\triangle ADC\) 和 \(\triangle BCD\) 共用底边 \(CD\)，面积比为 \(3:4\)，面积分别为 \(\tfrac{18}{7}\) 和 \(\tfrac{24}{7}\)。

将 \(\triangle ABC\) 沿 \(CD\) 分割，该线段与两条直角边都成 \(45^\circ\)，于是 \[ \dfrac{3 \cdot CD}{2\sqrt{2}} + \dfrac{4 \cdot CD}{2\sqrt{2}} = 6\text{，}\] 所以 \(CD = \tfrac{12\sqrt{2}}{7}\)。

两个三角形的半周长分别为 \[ \begin{aligned} s_a &= \dfrac{6}{7}(3 + \sqrt{2}), \\ s_b &= \dfrac{6}{7}(4 + \sqrt{2}) \end{aligned}\text{。}\] 利用 \(r = \frac{\text{面积}}{s}\)，可得 \[ \begin{aligned} \dfrac{r_a}{r_b} &= \dfrac{[ADC]}{[BCD]} \cdot \dfrac{s_b}{s_a} \\ &= \dfrac{3}{4} \cdot \dfrac{4 + \sqrt{2}}{3 + \sqrt{2}} \end{aligned}\text{，}\]

有理化得 \(\dfrac{4 + \sqrt{2}}{3 + \sqrt{2}} = \dfrac{10 - \sqrt{2}}{7}\)，因此 \[ \begin{aligned} \dfrac{r_a}{r_b} &= \dfrac{3}{4} \cdot \dfrac{10 - \sqrt{2}}{7} \\ &= \dfrac{3}{28}(10 - \sqrt{2}) \end{aligned}\text{。}\]

所以正确答案是 E。

English

By the Angle Bisector Theorem, \(AD:DB = CA:CB = 3:4,\) so \(AD = \tfrac{15}{7}\) and \(BD = \tfrac{20}{7}.\) The areas of \(\triangle ADC\) and \(\triangle BCD\) share base \(CD,\) so they are in ratio \(3:4,\) namely \(\tfrac{18}{7}\) and \(\tfrac{24}{7}.\)

Splitting \(\triangle ABC\) along \(CD,\) which meets each leg at \(45^\circ,\) gives \[ \dfrac{3 \cdot CD}{2\sqrt{2}} + \dfrac{4 \cdot CD}{2\sqrt{2}} = 6, \] so \(CD = \tfrac{12\sqrt{2}}{7}.\)

The two semiperimeters are \[ \begin{aligned} s_a &= \dfrac{6}{7}(3 + \sqrt{2}), \\ s_b &= \dfrac{6}{7}(4 + \sqrt{2}). \end{aligned} \] Using \(r = \frac{\text{area}}{s},\) \[ \begin{aligned} \dfrac{r_a}{r_b} &= \dfrac{[ADC]}{[BCD]} \cdot \dfrac{s_b}{s_a} \\ &= \dfrac{3}{4} \cdot \dfrac{4 + \sqrt{2}}{3 + \sqrt{2}}, \end{aligned} \]

Rationalizing, \(\dfrac{4 + \sqrt{2}}{3 + \sqrt{2}} = \dfrac{10 - \sqrt{2}}{7},\) so \[ \begin{aligned} \dfrac{r_a}{r_b} &= \dfrac{3}{4} \cdot \dfrac{10 - \sqrt{2}}{7} \\ &= \dfrac{3}{28}(10 - \sqrt{2}). \end{aligned} \]

Thus, E is the correct answer.

21 答案 D

称满足 \(a_1 + a_2 = a_4 + a_5\) 的排列为平衡排列。把排列反向会交换两种严格不等的情形，所以尾重排列与头重排列的个数相同。

总和 \(1 + 2 + 3 + 4 + 5 = 15\) 是奇数，所以在平衡排列中 \(a_3\) 必须是奇数，即 \(1, 3, 5\) 之一。对每种选择，剩下的四个数唯一地分成两组和相等的数对。

这四个数中任何一个都可以作 \(a_1\)（于是 \(a_2\) 随之确定），剩下两个数中任何一个都可以作 \(a_4\)（于是 \(a_5\) 随之确定），共得 \(3 \cdot 4 \cdot 2 = 24\) 个平衡排列。

非平衡排列有 \(120 - 24 = 96\) 个，两种严格不等情形各占一半，所以有 \(\tfrac{96}{2} = 48\) 个尾重排列。

所以正确答案是 D。

English

Call a permutation balanced if \(a_1 + a_2 = a_4 + a_5.\) Reversing the entries swaps the two strict cases, so heavy-tailed and heavy-headed permutations are equally numerous.

The total \(1 + 2 + 3 + 4 + 5 = 15\) is odd, so in a balanced permutation \(a_3\) must be odd, one of \(1, 3, 5.\) For each choice, the remaining four numbers split uniquely into two equal-sum pairs.

Any of the four can be \(a_1\) (fixing \(a_2\)), and either remaining number can be \(a_4\) (fixing \(a_5\)), giving \(3 \cdot 4 \cdot 2 = 24\) balanced permutations.

The other \(120 - 24 = 96\) permutations split evenly, so there are \(\tfrac{96}{2} = 48\) heavy-tailed permutations.

Thus, D is the correct answer.

22 答案 C

取一个餐垫，外侧两角为 \(P\) 和 \(Q\)，令 \(R\) 是圆桌边上与 \(P\) 直径相对的点。则 \(PR = 8\) 为直径，所以 \(\triangle PQR\) 在 \(Q\) 处为直角，且 \(PQ = x\)。

沿 \(QR\) 方向，相邻餐垫的内角形成两边长为 \(x\)、顶角为 \(120^\circ\) 的等腰三角形，其底边为 \(\sqrt{3}\,x\)。因此 \(QR = \sqrt{3}\,x + 2\)。

勾股定理给出 \[ x^2 + \left(\sqrt{3}\,x + 2\right)^2 = 64\text{，} \] 化简为 \(x^2 + \sqrt{3}\,x - 15 = 0\)。

取正根， \[ x = \dfrac{-\sqrt{3} + \sqrt{63}}{2} = \dfrac{3\sqrt{7} - \sqrt{3}}{2}\text{。} \]

所以正确答案是 C。

English

Take one mat with outer corners \(P\) and \(Q,\) and let \(R\) be the point of the table’s edge diametrically opposite \(P.\) Then \(PR = 8\) is a diameter, so \(\triangle PQR\) has a right angle at \(Q,\) with \(PQ = x.\)

Along \(QR,\) the inner corners of neighboring mats meet in an isosceles triangle with two sides of length \(x\) and vertex angle \(120^\circ,\) whose base is \(\sqrt{3}\,x.\) Hence \(QR = \sqrt{3}\,x + 2.\)

The Pythagorean Theorem gives \[ x^2 + \left(\sqrt{3}\,x + 2\right)^2 = 64, \] which simplifies to \(x^2 + \sqrt{3}\,x - 15 = 0.\)

Taking the positive root, \[ x = \dfrac{-\sqrt{3} + \sqrt{63}}{2} = \dfrac{3\sqrt{7} - \sqrt{3}}{2}. \]

Thus, C is the correct answer.

23 答案 D

两边加上 \(1 + i\)，左边变为 \[ \begin{aligned} &z^4 + 4z^3 i - 6z^2 - 4zi + 1 \\ &= (z + i)^4 \end{aligned}\text{，}\] 所以 \((z + i)^4 = 1 + i\)。

令 \(w = z + i\)。四个解等距分布在半径 \(|1 + i|^{\frac{1}{4}} = (2^{\frac{1}{2}})^{\frac{1}{4}} = 2^{\frac{1}{8}}\) 的圆上，形成正方形；减去 \(i\) 只会平移图形。

该正方形的外接圆半径为 \(2^{\frac{1}{8}}\)，所以对角线为 \(2 \cdot 2^{\frac{1}{8}} = 2^{\frac{9}{8}}\)，边长为 \(\tfrac{2^{\frac{9}{8}}}{\sqrt{2}} = 2^{\frac{5}{8}}\)。

面积为 \[ \left(2^{\frac{5}{8}}\right)^2 = 2^{\frac{5}{4}}\text{。}\]

所以正确答案是 D。

English

Adding \(1 + i\) to both sides, the left side becomes \[ \begin{aligned} &z^4 + 4z^3 i - 6z^2 - 4zi + 1 \\ &= (z + i)^4, \end{aligned} \] so \((z + i)^4 = 1 + i.\)

The four solutions for \(w = z + i\) are equally spaced on a circle of radius \(|1 + i|^{\frac{1}{4}} = (2^{\frac{1}{2}})^{\frac{1}{4}} = 2^{\frac{1}{8}},\) and they form a square. Subtracting \(i\) merely translates it.

A square inscribed in a circle of radius \(2^{\frac{1}{8}}\) has diagonal \(2 \cdot 2^{\frac{1}{8}} = 2^{\frac{9}{8}},\) so its side is \(\tfrac{2^{\frac{9}{8}}}{\sqrt{2}} = 2^{\frac{5}{8}}.\)

The area is \[ \left(2^{\frac{5}{8}}\right)^2 = 2^{\frac{5}{4}}. \]

Thus, D is the correct answer.

24 答案 D

取 \(C = (0, 0)\)，\(B = (2, 2\sqrt{3})\)，使 \(\angle C = 60^\circ\) 且 \(BC = 4\)，令 \(A = (x, 0)\)，其中 \(x \gt 0\)。则 \(D = (1, \sqrt{3})\) 是 \(BC\) 的中点。

向量 \(\overrightarrow{AB} = (2-x, 2\sqrt{3})\) 和 \(\overrightarrow{AD} = (1-x, \sqrt{3})\) 的叉积的模为 \(\sqrt{3}\,x\)，点积为 \(x^2 - 3x + 8\)，后者恒为正。因此 \[ \tan(\angle BAD) = \dfrac{\sqrt{3}\,x}{x^2 - 3x + 8}\text{。}\]

导数的符号与 \(8-x^2\) 相同，所以唯一的最大值在 \(x = 2\sqrt{2}\) 处取得。代入得 \[ \begin{aligned} \tan(\angle BAD) &= \dfrac{2\sqrt{6}}{16 - 6\sqrt{2}} \\ &= \dfrac{\sqrt{6}}{8 - 3\sqrt{2}} \\ &= \dfrac{\sqrt{3}}{4\sqrt{2} - 3} \end{aligned}\text{。}\]

所以正确答案是 D。

English

Place \(C = (0, 0),\) \(B = (2, 2\sqrt{3})\) so that \(\angle C = 60^\circ\) and \(BC = 4,\) and let \(A = (x, 0)\) with \(x \gt 0.\) Then \(D = (1, \sqrt{3})\) is the midpoint of \(BC.\)

The vectors \(\overrightarrow{AB} = (2-x, 2\sqrt{3})\) and \(\overrightarrow{AD} = (1-x, \sqrt{3})\) have cross-product magnitude \(\sqrt{3}\,x\) and dot product \(x^2 - 3x + 8,\) which is always positive. Hence \[ \tan(\angle BAD) = \dfrac{\sqrt{3}\,x}{x^2 - 3x + 8}. \]

The derivative has the sign of \(8-x^2,\) so the unique maximum occurs at \(x = 2\sqrt{2}.\) Substituting, \[ \begin{aligned} \tan(\angle BAD) &= \dfrac{2\sqrt{6}}{16 - 6\sqrt{2}} \\ &= \dfrac{\sqrt{6}}{8 - 3\sqrt{2}} \\ &= \dfrac{\sqrt{3}}{4\sqrt{2} - 3}. \end{aligned} \]

Thus, D is the correct answer.

25 答案 D

令 \(z_n = a_n + b_n i\)，则 \[ \begin{aligned} z_{n+1} &= (\sqrt{3}\,a_n - b_n) \\ &\quad {}+ (\sqrt{3}\,b_n + a_n)i \\ &= (a_n + b_n i)(\sqrt{3} + i)\text{，} \end{aligned} \] 所以 \(z_{n+1} = z_n(\sqrt{3} + i)\)，且 \(z_{100} = z_1(\sqrt{3} + i)^{99}\)。

因为 \(\sqrt{3} + i = 2(\cos 30^\circ + i\sin 30^\circ)\)，棣莫弗定理给出 \((\sqrt{3} + i)^{99}\) \(= 2^{99}(\cos 2970^\circ + i\sin 2970^\circ)\)。而 \(2970^\circ\) 与 \(90^\circ\) 同终边，所以它等于 \(2^{99} i\)。

于是 \(2 + 4i = z_1 \cdot 2^{99} i\)，所以 \[ z_1 = \dfrac{2 + 4i}{2^{99} i} = \dfrac{4 - 2i}{2^{99}}\text{。} \]

因此 \(a_1 = \tfrac{4}{2^{99}}\)，\(b_1 = -\tfrac{2}{2^{99}}\)，所以 \[ a_1 + b_1 = \dfrac{2}{2^{99}} = \dfrac{1}{2^{98}}\text{。} \]

所以正确答案是 D。

English

Let \(z_n = a_n + b_n i.\) Then \[ \begin{aligned} z_{n+1} &= (\sqrt{3}\,a_n - b_n) \\ &\quad {}+ (\sqrt{3}\,b_n + a_n)i \\ &= (a_n + b_n i)(\sqrt{3} + i), \end{aligned} \] so \(z_{n+1} = z_n(\sqrt{3} + i)\) and \(z_{100} = z_1(\sqrt{3} + i)^{99}.\)

Since \(\sqrt{3} + i = 2(\cos 30^\circ + i\sin 30^\circ),\) De Moivre’s theorem gives \((\sqrt{3} + i)^{99}\) \(= 2^{99}(\cos 2970^\circ + i\sin 2970^\circ).\) As \(2970^\circ\) is coterminal with \(90^\circ,\) this equals \(2^{99} i.\)

Thus \(2 + 4i = z_1 \cdot 2^{99} i,\) so \[ z_1 = \dfrac{2 + 4i}{2^{99} i} = \dfrac{4 - 2i}{2^{99}}. \]

Then \(a_1 = \tfrac{4}{2^{99}}\) and \(b_1 = -\tfrac{2}{2^{99}},\) so \[ a_1 + b_1 = \dfrac{2}{2^{99}} = \dfrac{1}{2^{98}}. \]

Thus, D is the correct answer.