1999 AMC 12 真题（P版）

1 答案 E

两两分组可得 \[ \begin{aligned} &(1-2) + (3-4) + \cdots \\ &\quad {}+ (97-98) + 99\text{。} \end{aligned} \] 共有 \(49\) 组，每组等于 \(-1\)，所以总和为 \(-49 + 99 = 50\)。

所以正确答案是 E。

English

Pairing consecutive terms gives \[ \begin{aligned} &(1-2) + (3-4) + \cdots \\ &\quad {}+ (97-98) + 99. \end{aligned} \] There are \(49\) pairs, each equal to \(-1,\) so the sum is \(-49 + 99 = 50.\)

Thus, the correct answer is E.

2 答案 A

边长为 \(1\) 和边长为 \(2\) 的等边三角形形状相同但大小不同，所以它们相似但不全等。每个等边三角形都是凸的、等角的（每个角都是 \(60^\circ\)），也是正多边形，因此唯一错误的命题是所有等边三角形都彼此全等。

所以正确答案是 A。

English

Equilateral triangles with side lengths \(1\) and \(2\) have the same shape but different sizes, so they are similar but not congruent. Every equilateral triangle is convex, equiangular (all angles \(60^\circ\)), and a regular polygon, so the only false statement is that they are all congruent.

Thus, the correct answer is A.

3 答案 E

正中间的数为 \[ \dfrac12\left(\dfrac18 + \dfrac{1}{10}\right) = \dfrac12 \cdot \dfrac{18}{80} = \dfrac{9}{80}\text{。}\]

所以正确答案是 E。

English

The halfway point is the average \[ \dfrac12\left(\dfrac18 + \dfrac{1}{10}\right) = \dfrac12 \cdot \dfrac{18}{80} = \dfrac{9}{80}. \]

Thus, the correct answer is E.

4 答案 A

比 \(5\) 的倍数小 \(1\) 的数个位为 \(4\) 或 \(9\)，而比 \(4\) 的倍数大 \(1\) 的数是奇数，所以这些数必须满足 \(\equiv 9 \pmod{20}\)。符合条件的数是 \(9, 29, 49, 69, 89\)。

其中质数只有 \(29\) 和 \(89\)，和为 \(29 + 89 = 118\)。

所以正确答案是 A。

English

A number that is \(1\) less than a multiple of \(5\) ends in \(4\) or \(9,\) and one that is \(1\) greater than a multiple of \(4\) is odd. Together these give numbers \(\equiv 9 \pmod{20},\) namely \(9, 29, 49, 69, 89.\)

Among these, only \(29\) and \(89\) are prime, and their sum is \(29 + 89 = 118.\)

Thus, the correct answer is A.

5 答案 C

若建议零售价为 \(P\)，则标价为 \(0.7P\)。Alice 支付其中一半，即 \(0.35P\)，也就是建议零售价的 \(35\%\)。

所以正确答案是 C。

English

If the suggested retail price is \(P,\) then the marked price is \(0.7P.\) Alice pays half of this, \(0.35P,\) which is \(35\%\) of the suggested retail price.

Thus, the correct answer is C.

6 答案 D

有 \[ \begin{aligned} 2^{1999} \cdot 5^{2001} &= 2^{1999} \cdot 5^{1999} \cdot 5^2 \\ &= 25 \cdot 10^{1999}\text{，} \end{aligned} \] 这就是 \(25\) 后面跟 \(1999\) 个零，所以数字和为 \(2 + 5 = 7\)。

所以正确答案是 D。

English

Write \[ \begin{aligned} 2^{1999} \cdot 5^{2001} &= 2^{1999} \cdot 5^{1999} \cdot 5^2 \\ &= 25 \cdot 10^{1999}, \end{aligned} \] which is \(25\) followed by \(1999\) zeros. The sum of the digits is \(2 + 5 = 7.\)

Thus, the correct answer is D.

7 答案 B

每个锐内角对应一个大于 \(90^\circ\) 的外角。因为凸多边形的外角和为 \(360^\circ\)，其中最多有三个外角能大于 \(90^\circ\)。这个上界可以达到：取一个等边六边形，其外角交替为 \(100^\circ\) 和 \(20^\circ\)。它的边方向分成两组三条，每组内相差 \(120^\circ\)，所以图形闭合；内角交替为 \(80^\circ\) 与 \(160^\circ\)。因此最大可能数为三。

所以正确答案是 B。

English

Each acute interior angle corresponds to an exterior angle greater than \(90^\circ.\) Since the exterior angles of a convex polygon sum to \(360^\circ,\) at most three of them can exceed \(90^\circ.\) This bound is attainable: take a hexagon with equal side lengths and exterior angles alternating \(100^\circ\) and \(20^\circ.\) Its edge directions split into two triples \(120^\circ\) apart, so it closes, and its interior angles alternate between \(80^\circ\) and \(160^\circ.\) Hence the largest possible number is three.

Thus, the correct answer is B.

8 答案 D

设 Walter 在 \(1994\) 年底为 \(w\) 岁，则祖母为 \(2w\) 岁。两人的出生年份为 \(1994 - w\) 和 \(1994 - 2w\)，且 \[ \begin{aligned} &(1994 - w) \\ &\quad {}+ (1994 - 2w) = 3838\text{。} \end{aligned} \] 因此 \(3988 - 3w = 3838\)，得 \(w = 50\)。

到 \(1999\) 年底，Walter 年龄为 \(50 + 5 = 55\)。

所以正确答案是 D。

English

Let Walter be \(w\) years old at the end of \(1994,\) so his grandmother is \(2w.\) Their birth years are \(1994 - w\) and \(1994 - 2w,\) and \[ \begin{aligned} &(1994 - w) \\ &\quad {}+ (1994 - 2w) = 3838. \end{aligned} \] This gives \(3988 - 3w = 3838,\) so \(w = 50.\)

At the end of \(1999,\) Walter will be \(50 + 5 = 55.\)

Thus, the correct answer is D.

9 答案 D

\(29792\) 之后的回文数依次为 \(29892, 29992, 30003\) 和 \(30103\)。三小时内最多行驶 \(3 \cdot 75 = 225\) 英里。

到 \(30103\) 需要行驶 \(30103 - 29792 = 311\) 英里，太远。到 \(30003\) 需要行驶 \(30003 - 29792 = 211\) 英里，平均速度为 \(\dfrac{211}{3} = 70\tfrac13\) 英里每小时。

所以正确答案是 D。

English

The palindromes after \(29792\) are \(29892, 29992, 30003,\) and \(30103.\) In three hours Ashley can drive at most \(3 \cdot 75 = 225\) miles.

Reaching \(30103\) would require \(30103 - 29792 = 311\) miles, which is too far. Reaching \(30003\) requires \(30003 - 29792 = 211\) miles, giving average speed \(\dfrac{211}{3} = 70\tfrac13\) miles per hour.

Thus, the correct answer is D.

10 答案 C

命题 I 和 III 不可能同时为真，因此唯一的假命题必定是其中之一。

所以命题 II 和 IV 都为真，故“II 为真”必定正确。实际上数字只能是 \(1\) 或 \(3\)。若是 \(1\)，选项 B 和 D 为假；若是 \(3\)，选项 A 为假；选项 E 总不正确。只有 C 必定成立。

所以正确答案是 C。

English

Statements I and III cannot both be true, so the single false statement is one of them. Therefore statements II and IV are both true, which makes “II is true” necessarily correct.

The digit is thus \(1\) or \(3.\) If it were \(1,\) then (B) and (D) are false; if it were \(3,\) then (A) is false; and (E) is always incorrect. Only (C) is guaranteed.

Thus, the correct answer is C.

11 答案 A

总共使用 \(\frac{\$137.94}{\$0.02} = 6897\) 个数字。编号 \(1\) 到 \(9\) 使用 \(9\) 个数字，\(10\) 到 \(99\) 使用 \(2 \cdot 90 = 180\) 个数字，\(100\) 到 \(999\) 使用 \(3 \cdot 900 = 2700\) 个数字。

剩余数字数为 \(6897 - 2700 - 180 - 9 = 4008\)，可编号 \(\frac{4008}{4} = 1002\) 个四位数储物柜。所以总数为 \(999 + 1002 = 2001\)。

所以正确答案是 A。

English

Labeling costs \(\frac{\$137.94}{\$0.02} = 6897\) digits. Lockers \(1\)-\(9\) use \(9\) digits, lockers \(10\)-\(99\) use \(2 \cdot 90 = 180\) digits, and lockers \(100\)-\(999\) use \(3 \cdot 900 = 2700\) digits.

The remaining digits number \(6897 - 2700 - 180 - 9 = 4008,\) which label \(\frac{4008}{4} = 1002\) four-digit lockers. In all there are \(999 + 1002 = 2001\) lockers.

Thus, the correct answer is A.

12 答案 C

交点的 \(x\) 坐标是 \(p(x) - q(x)\) 的根。由于两个四次多项式首项系数都为 \(1\)，相减时 \(x^4\) 项抵消，所以 \(p(x) - q(x)\) 次数至多为 \(3\)，最多有 \(3\) 个实根。这个上界可以达到，例如取 \(p(x) = x^4\) 和 \(q(x) = x^4 - x(x-1)(x+1)\)，它们的图像在 \(x = -1, 0, 1\) 处相交。

所以正确答案是 C。

English

The \(x\)-coordinates of the intersection points are the roots of \(p(x) - q(x).\) Because both leading coefficients are \(1,\) the \(x^4\) terms cancel, so \(p(x) - q(x)\) has degree at most \(3\) and therefore at most \(3\) roots. The bound is attainable, for example by taking \(p(x) = x^4\) and \(q(x) = x^4 - x(x-1)(x+1),\) whose graphs intersect at \(x = -1, 0, 1.\)

Thus, the correct answer is C.

13 答案 C

取立方根得 \(a_{n+1} = \sqrt[3]{99}\, a_n\)，因此该数列首项为 \(1\)，公比为 \(\sqrt[3]{99}\)。\[ a_{100} = \left(\sqrt[3]{99}\right)^{99} = 99^{33}\text{。}\]

所以正确答案是 C。

English

Taking cube roots, \(a_{n+1} = \sqrt[3]{99}\, a_n,\) so the sequence is geometric with first term \(1\) and ratio \(\sqrt[3]{99}.\) Then \[ a_{100} = \left(\sqrt[3]{99}\right)^{99} = 99^{33}. \]

Thus, the correct answer is C.

14 答案 A

若共唱 \(N\) 首歌，则总演唱人次为 \(3N\)。Alina 和 Tina 的演唱次数都严格介于 \(4\) 和 \(7\) 之间，所以各为 \(5\) 或 \(6\)。

因此 \(3N = 7 + 4 + (\text{Alina}) + (\text{Tina})\)，可能为 \(21, 22\)，或 \(23\)。只有 \(21\) 是 \(3\) 的倍数，所以 \(N = 7\)。

所以正确答案是 A。

English

If \(N\) songs are sung, the total number of girl-appearances is \(3N.\) Alina and Tina each sang strictly between \(4\) and \(7,\) so each sang \(5\) or \(6.\)

Then \(3N = 7 + 4 + (\text{Alina}) + (\text{Tina}),\) which is \(21, 22,\) or \(23.\) Only \(21\) is a multiple of \(3,\) so \(N = 7.\)

Thus, the correct answer is A.

15 答案 E

因为 \(\sec^2 x - \tan^2 x = 1\)，所以 \[ \begin{aligned} &(\sec x - \tan x)(\sec x + \tan x) \\ &\quad = 1\text{。} \end{aligned} \] 已知 \(\sec x - \tan x = 2\)，故 \(\sec x + \tan x = \tfrac12 = 0.5\)。

所以正确答案是 E。

English

Since \(\sec^2 x - \tan^2 x = 1,\) we have \[ \begin{aligned} &(\sec x - \tan x)(\sec x + \tan x) \\ &\quad = 1. \end{aligned} \] With \(\sec x - \tan x = 2,\) it follows that \(\sec x + \tan x = \tfrac12 = 0.5.\)

Thus, the correct answer is E.

16 答案 C

半对角线为 \(5\) 和 \(12\)，所以菱形边长为 \(\sqrt{5^2 + 12^2} = 13\)。对角线形成的四个直角三角形之一的直角边为 \(5\) 和 \(12\)，面积为 \(30\)。

从中心到边长 \(13\) 的边的高为 \(\dfrac{2 \cdot 30}{13} = \dfrac{60}{13}\)，这就是内切圆半径。

所以正确答案是 C。

English

The half-diagonals are \(5\) and \(12,\) so each side of the rhombus is \(\sqrt{5^2 + 12^2} = 13.\) One of the four right triangles formed by the diagonals has legs \(5\) and \(12\) and area \(30.\)

The altitude from the center to the side of length \(13\) is \(\dfrac{2 \cdot 30}{13} = \dfrac{60}{13},\) which is the inscribed circle’s radius.

Thus, the correct answer is C.

17 答案 C

由余数定理，\(P(19) = 99\)，\(P(99) = 19\)。设 \[ \begin{aligned} &P(x) = (x - 19)(x - 99)Q(x) \\ &\quad {}+ ax + b\text{。} \end{aligned} \] 则 \[ 19a + b = 99, \qquad 99a + b = 19\text{。} \]

相减得 \(80a = -80\)，所以 \(a = -1\)，进而 \(b = 118\)。余数为 \(-x + 118\)。

所以正确答案是 C。

English

By the Remainder Theorem, \(P(19) = 99\) and \(P(99) = 19.\) Write \[ \begin{aligned} &P(x) = (x - 19)(x - 99)Q(x) \\ &\quad {}+ ax + b. \end{aligned} \] Then \[ 19a + b = 99, \qquad 99a + b = 19. \]

Subtracting gives \(80a = -80,\) so \(a = -1\) and \(b = 118.\) The remainder is \(-x + 118.\)

Thus, the correct answer is C.

18 答案 E

当 \(x\) 在 \((0, 1)\) 中变化时，\(\log x\) 覆盖所有负实数。余弦函数在 \(\tfrac{\pi}{2} - n\pi\) 处为零，其中 \(n\) 取遍每个正整数，而这些值全都是负数，因此 \(f\) 有无限多个零点。

所以正确答案是 E。

English

As \(x\) ranges over \((0, 1),\) \(\log x\) ranges over all negative real numbers. The cosine function is zero at \(\tfrac{\pi}{2} - n\pi\) for every positive integer \(n,\) all of which are negative, so \(f\) has infinitely many zeros.

Thus, the correct answer is E.

19 答案 C

设 \(AD = n\)，\(CD = m\)。因为 \(\triangle ADB\) 在 \(D\) 处为直角，\(AB^2 = n^2 + 57\)。又 \(AB = AC = m + n\)，所以 \[ (m + n)^2 = n^2 + 57\text{，} \] 化简为 \(m(m + 2n) = 57\)。

正整数解为 \(m = 1, n = 28\)，此时 \(AC = 29\)；以及 \(m = 3, n = 8\)，此时 \(AC = 11\)。因此 \(AC\) 的最小可能值为 \(11\)。

所以正确答案是 C。

English

Let \(AD = n\) and \(CD = m.\) Since \(\triangle ADB\) is right-angled at \(D,\) \(AB^2 = n^2 + 57.\) Also \(AB = AC = m + n,\) so \[ (m + n)^2 = n^2 + 57, \] which simplifies to \(m(m + 2n) = 57.\)

The positive integer solutions are \(m = 1, n = 28\) (giving \(AC = 29\)) and \(m = 3, n = 8\) (giving \(AC = 11\)). The smallest possible value of \(AC\) is \(11.\)

Thus, the correct answer is C.

20 答案 E

对 \(n \ge 3\) 有 \((n - 1)a_n = a_1 + \cdots + a_{n-1}\)。因此 \[ a_{n+1} = \dfrac{(n-1)a_n + a_n}{n} = a_n\text{，} \] 所以从 \(a_3\) 开始数列为常数。故 \(a_3 = a_9 = 99\)。

又 \(a_3 = \dfrac{a_1 + a_2}{2} = \dfrac{19 + a_2}{2} = 99\)，得 \(a_2 = 179\)。

所以正确答案是 E。

English

For \(n \ge 3,\) \((n - 1)a_n = a_1 + \cdots + a_{n-1}.\) Then \[ a_{n+1} = \dfrac{(n-1)a_n + a_n}{n} = a_n, \] so the sequence is constant from \(a_3\) onward. Hence \(a_3 = a_9 = 99.\)

Since \(a_3 = \dfrac{a_1 + a_2}{2} = \dfrac{19 + a_2}{2} = 99,\) we get \(a_2 = 179.\)

Thus, the correct answer is E.

21 答案 B

因为 \(20^2 + 21^2 = 841 = 29^2\)，该三角形为直角三角形，斜边 \(29\) 是外接圆直径。因此最大区域 \(C\) 是该直径一侧的半圆。

另一半圆由三角形以及区域 \(A\)、\(B\) 组成。两个半圆面积相等，而三角形面积为 \(\tfrac12 \cdot 20 \cdot 21 = 210\)，所以 \[ A + B + 210 = C\text{。}\]

所以正确答案是 B。

English

Since \(20^2 + 21^2 = 841 = 29^2,\) the triangle is right-angled, and its hypotenuse of length \(29\) is a diameter of the circle. Thus the largest region \(C\) is the semicircle on one side of that diameter.

The other semicircle consists of the triangle together with regions \(A\) and \(B.\) Since the two semicircles are congruent and the triangle has area \(\tfrac12 \cdot 20 \cdot 21 = 210,\) we get \[ A + B + 210 = C. \]

Thus, the correct answer is B.

22 答案 C

在交点处有 \(-|x-a|+b=|x-c|+d\)，即 \[|x-a|+|x-c|=b-d\text{。}\]因为两个交点彼此分离，它们分别位于以 \(a\) 和 \(c\) 为端点的区间的两侧。这两个解关于 \(x=\tfrac{a+c}{2}\) 对称。它们的 \(x\) 坐标为 \(2\) 和 \(8\)，所以 \(a+c=2+8=10\)。

所以正确答案是 C。

English

At an intersection, \(-|x-a|+b=|x-c|+d,\) or \[|x-a|+|x-c|=b-d.\] Because there are two isolated intersections, they lie on opposite sides of the interval with endpoints \(a\) and \(c.\) The two solutions are symmetric about \(x=\tfrac{a+c}{2}.\) Their \(x\)-coordinates are \(2\) and \(8,\) so \(a+c=2+8=10.\)

Thus, the correct answer is C.

23 答案 E

等角六边形的每个内角为 \(120^\circ\)。延长边 \(FA\) 与 \(BC\)、\(BC\) 与 \(DE\)、\(DE\) 与 \(FA\)，会形成一个大等边三角形，并切去三个等边小角三角形。

设 \(EF=e\)，\(FA=f\)。把六条边按彼此相差 \(60^\circ\) 的方向分解，可得 \(f-e=4\) 和 \(e+f=6\)，所以 \(e=1\)，\(f=5\)。于是建立在 \(AB, CD\) 和 \(EF\) 上的角三角形都是等边三角形。大三角形的边长为 \(1+4+2=7\)，而被切去的三角形边长分别为 \(1,2\) 和 \(1\)。所以面积为 \[ \begin{aligned} &\frac{\sqrt3}{4}\left(7^2 - 1^2 - 2^2 - 1^2\right) \\ &\quad = \frac{43\sqrt3}{4} \end{aligned}\text{。}\]

所以正确答案是 E。

English

Each interior angle is \(120^\circ,\) so extending sides \(FA\) and \(BC,\) \(BC\) and \(DE,\) and \(DE\) and \(FA\) cuts off three equilateral corner triangles and forms a large equilateral triangle.

Let \(EF=e\) and \(FA=f.\) Resolving the six sides in directions separated by \(60^\circ\) gives \(f-e=4\) and \(e+f=6,\) so \(e=1\) and \(f=5.\) The corner triangles built on \(AB, CD,\) and \(EF\) are therefore equilateral. The large triangle has side \(1+4+2=7,\) while the removed triangles have sides \(1,2,\) and \(1.\) The area is \[ \begin{aligned} &\frac{\sqrt3}{4}\left(7^2 - 1^2 - 2^2 - 1^2\right) \\ &\quad = \frac{43\sqrt3}{4}. \end{aligned} \]

Thus, the correct answer is E.

24 答案 B

共有 \(\binom{6}{2} = 15\) 条弦，因此选择四条弦的方法数为 \(\binom{15}{4} = 1365\)。四条弦形成凸四边形，当且仅当它们是某 \(4\) 个圆上点所形成四边形的四条边。

任选四个点恰好给出一个这样的四边形，所以有利情况为 \(\binom{6}{4} = 15\)，概率为 \(\dfrac{15}{1365} = \dfrac{1}{91}\)。

所以正确答案是 B。

English

There are \(\binom{6}{2} = 15\) chords, so \(\binom{15}{4} = 1365\) ways to select four of them. A convex quadrilateral arises exactly when the four chords are the sides of a quadrilateral on four of the six points, and each choice of \(4\) points gives exactly one such quadrilateral.

Hence there are \(\binom{6}{4} = 15\) favorable outcomes, and the probability is \(\dfrac{15}{1365} = \dfrac{1}{91}.\)

Thus, the correct answer is B.

25 答案 B

两边乘以 \(7! = 5040\)，得 \[ \begin{aligned} &3600 = 2520a_2 + 840a_3 \\ &\quad {}+ 210a_4 + 42a_5 \\ &\quad {}+ 7a_6 + a_7\text{。} \end{aligned} \] 模 \(7\) 得 \(a_7 = 2\)。

再计算 \(\dfrac{3600 - 2}{7} = 514\) \(= 360a_2 + 120a_3\) \(+ 30a_4 + 6a_5\) \(+ a_6\)。模 \(6\) 得 \(a_6 = 4\)，继续同样步骤得到 \(a_5 = 0, a_4 = 1, a_3 = 1, a_2 = 1\)。

所求和为 \(1 + 1 + 1 + 0 + 4 + 2 = 9\)。

所以正确答案是 B。

English

Multiplying by \(7! = 5040\) gives \[ \begin{aligned} &3600 = 2520a_2 + 840a_3 \\ &\quad {}+ 210a_4 + 42a_5 \\ &\quad {}+ 7a_6 + a_7. \end{aligned} \] Reducing modulo \(7,\) \(a_7 = 2.\)

Then \(\dfrac{3600 - 2}{7} = 514\) \(= 360a_2 + 120a_3\) \(+ 30a_4 + 6a_5\) \(+ a_6.\) Reducing modulo \(6\) gives \(a_6 = 4,\) and continuing this way yields \(a_5 = 0, a_4 = 1, a_3 = 1, a_2 = 1.\)

The sum is \(1 + 1 + 1 + 0 + 4 + 2 = 9.\)

Thus, the correct answer is B.

26 答案 D

设两个全等的多边形为正 \(a\) 边形，另一个为正 \(b\) 边形，在点 \(A\) 处相接。内角条件为 \[ \begin{aligned} &2 \cdot 180\left(1 - \tfrac2a\right) \\ &\quad {}+ 180\left(1 - \tfrac2b\right) = 360\text{，} \end{aligned} \] 化简得 \((a - 4)(b - 2) = 8\)。

解为 \((a, b)\) 等于 \((5, 10), (6, 6), (8, 4)\) 或 \((12, 3)\)。新多边形的周长为 \(2a + b - 6\)，分别等于 \(14, 12, 14\) 和 \(21\)。最大值为 \(21\)。

所以正确答案是 D。

English

Let two congruent \(a\)-gons and one \(b\)-gon meet at \(A.\) Their interior angles satisfy \[ \begin{aligned} &2 \cdot 180\left(1 - \tfrac2a\right) \\ &\quad {}+ 180\left(1 - \tfrac2b\right) = 360, \end{aligned} \] which reduces to \((a - 4)(b - 2) = 8.\)

The solutions \((a, b)\) are \((5, 10), (6, 6), (8, 4),\) and \((12, 3).\) The new polygon’s perimeter is \(2a + b - 6,\) giving \(14, 12, 14,\) and \(21.\) The largest is \(21.\)

Thus, the correct answer is D.

27 答案 A

将两个方程平方后相加：\[ \begin{aligned} &9 + 16 \\ &\quad {}+ 24\small(\sin A \cos B + \cos A \sin B) \\ &\quad = 37\text{，} \end{aligned} \] 因此 \(24\sin(A + B) = 12\)，所以 \(\sin(A + B) = \tfrac12\)。

因为第三个内角与前两个内角之和互补，所以 \(\sin C = \sin(A + B) = \tfrac12\)，即 \(\angle C = 30^\circ\) 或 \(150^\circ\)。若 \(\angle C = 150^\circ\)，则 \(A \lt 30^\circ\)，会使 \(3\sin A + 4\cos B \lt 6\)，产生矛盾。因此 \(\angle C = 30^\circ\)。

所以正确答案是 A。

English

Squaring both equations and adding gives \[ \begin{aligned} &9 + 16 \\ &\quad {}+ 24\small(\sin A \cos B + \cos A \sin B) \\ &\quad = 37, \end{aligned} \] so \(24\sin(A + B) = 12\) and \(\sin(A + B) = \tfrac12.\)

Then \(\sin C = \sin(A + B) = \tfrac12,\) so \(\angle C = 30^\circ\) or \(150^\circ.\) If \(\angle C = 150^\circ,\) then \(A \lt 30^\circ,\) making \(3\sin A + 4\cos B \lt 6,\) a contradiction. Hence \(\angle C = 30^\circ.\)

Thus, the correct answer is A.

28 答案 E

设 \(a, b, c\) 分别为 \(-1\)、\(1\)、\(2\) 的个数。则 \(-a + b + 2c = 19\)，且 \(a + b + 4c = 99\)，解得 \(a = 40 - c\)、\(b = 59 - 3c\)，并且 \(0 \le c \le 19\)。

立方和为 \(-a + b + 8c = 19 + 6c\)。最小值在 \(c = 0\) 时为 \(19\)，最大值在 \(c = 19\) 时为 \(133\)。因此 \(\dfrac{M}{m} = \dfrac{133}{19} = 7\)。

所以正确答案是 E。

English

Let \(a, b, c\) be the numbers of \(-1\)s, \(1\)s, and \(2\)s. Then \(-a + b + 2c = 19\) and \(a + b + 4c = 99,\) giving \(a = 40 - c\) and \(b = 59 - 3c\) with \(0 \le c \le 19.\)

The sum of cubes is \(-a + b + 8c = 19 + 6c.\) The minimum is at \(c = 0\) (value \(19\)) and the maximum at \(c = 19\) (value \(133\)), so \(\dfrac{M}{m} = \dfrac{133}{19} = 7.\)

Thus, the correct answer is E.

29 答案 C

设 \(O\) 为内切球和外接球的共同球心。从 \(O\) 把正四面体分成四个全等部分可知，外接球半径是内切球半径的 \(3\) 倍，所以外接球的体积是内切球体积 \(V\) 的 \(27\) 倍。

若其余四个小球之一的半径为 \(s\)，则它的球心到 \(O\) 的距离为 \(r+s\)；从 \(O\) 量起，这段距离同时又等于 \(3r-s\)，所以 \(s=r\)。这五个球的内部互不相交（中心球与其余四球分别相切），所以并集体积为 \(5V\)。概率为 \(\dfrac{5V}{27V} = \dfrac{5}{27} \approx 0.185\)，最接近 \(0.2\)。

所以正确答案是 C。

English

Let \(O\) be the common center of the inscribed and circumscribed spheres. Splitting the tetrahedron into four congruent pieces from \(O\) shows the circumradius is \(3\) times the inradius, so the circumscribed sphere has \(27\) times the inscribed sphere’s volume \(V.\)

If one of the four other small spheres has radius \(s,\) its center is \(r+s\) from \(O\) and also \(3r-s\) from \(O,\) so \(s=r.\) These five spheres have disjoint interiors (the central sphere is tangent to each of the other four), so their union has volume \(5V.\) The probability is \(\dfrac{5V}{27V} = \dfrac{5}{27} \approx 0.185,\) closest to \(0.2.\)

Thus, the correct answer is C.

30 答案 D

令 \(z = -33\)，方程变为 \(m^3 + n^3 + z^3 - 3mnz = 0\)，因式分解得 \[ \begin{aligned} &(m + n - 33) \\ &\quad {}\cdot \scriptsize\left(m^2 + n^2 + 33^2 - mn + 33m + 33n\right) \\ &= 0 \end{aligned}\text{。}\]

第二个因子等于 \(\scriptsize\tfrac12\big[(m - n)^2 + (m + 33)^2 + (n + 33)^2\big]\)，只在 \((m, n) = (-33, -33)\) 时为 \(0\)，且满足 \(mn \ge 0\)。

否则 \(m + n = 33\)。由 \(mn \ge 0\)，二者同为非负，得到 \((0, 33), (1, 32), \ldots, (33, 0)\)，共 \(34\) 对。总计 \(35\) 对。

所以正确答案是 D。

English

Writing \(z = -33,\) the equation becomes \(m^3 + n^3 + z^3 - 3mnz = 0,\) which factors as \[ \begin{aligned} &(m + n - 33) \\ &\quad {}\cdot \scriptsize\left(m^2 + n^2 + 33^2 - mn + 33m + 33n\right) \\ &= 0. \end{aligned} \]

The second factor equals \(\scriptsize\tfrac12\big[(m - n)^2 + (m + 33)^2 + (n + 33)^2\big],\) which is \(0\) only at \((m, n) = (-33, -33);\) this satisfies \(mn \ge 0.\)

Otherwise \(m + n = 33.\) With \(mn \ge 0\) both are nonnegative, giving \((0, 33), (1, 32), \ldots, (33, 0),\) which is \(34\) pairs. Together there are \(35\) solutions.

Thus, the correct answer is D.