2001 AMC 12 真题（P版）

1 答案 E

给每个数加 \(3\) 后，总和从 \(S\) 变为 \(S + 6\)。再把每个数翻倍，总和也翻倍，得到 \[ 2(S + 6) = 2S + 12\text{。}\]

因此，正确答案是 E。

English

Adding \(3\) to each number raises the sum from \(S\) to \(S + 6.\) Doubling each number doubles the sum, giving \[ 2(S + 6) = 2S + 12. \]

Thus, the correct answer is E.

2 答案 E

设 \(N = 10a + b\)，则 \(P(N) = ab\)，且 \(S(N) = a + b\)，所以 \[ 10a + b = ab + a + b\text{。} \] 化简得 \(9a = ab\)。因为 \(a \neq 0\)，可除以 \(a\)，得到 \(b = 9\)。

\(N\) 的个位数字是 \(9\)。

因此，正确答案是 E。

English

Write \(N = 10a + b.\) Then \(P(N) = ab\) and \(S(N) = a + b,\) so \[ 10a + b = ab + a + b. \] This reduces to \(9a = ab.\) Since \(a \neq 0,\) we can divide by \(a\) to get \(b = 9.\)

The units digit of \(N\) is \(9.\)

Thus, the correct answer is E.

3 答案 B

设她的收入为 \(x \gt 28000\) 美元。用两种描述写出税额，并同乘以 \(100\)，\[ \begin{aligned} &p \cdot 28000 \\ &\quad {}+ (p + 2)(x - 28000) \\ &= (p + 0.25)x\text{。} \end{aligned} \]

展开后，所有含 \(p\) 的项都抵消，剩下 \[ 2x - 56000 = 0.25x\text{，} \] 所以 \(1.75x = 56000\)，且 \(x = 32000\)。

因此，正确答案是 B。

English

Let her income be \(x \gt 28000\) dollars. Writing the tax with both descriptions and multiplying by \(100,\) \[ \begin{aligned} &p \cdot 28000 \\ &\quad {}+ (p + 2)(x - 28000) \\ &= (p + 0.25)x. \end{aligned} \]

Expanding, every term containing \(p\) cancels, leaving \[ 2x - 56000 = 0.25x, \] so \(1.75x = 56000\) and \(x = 32000.\)

Thus, the correct answer is B.

4 答案 D

设平均数为 \(m\)。最小的数是 \(m - 10\)，最大的数是 \(m + 15\)，中间的数是中位数 \(5\)。它们的和是 \(3m\)，因此 \[ (m - 10) + 5 + (m + 15) = 3m\text{。}\]

由此得 \(m = 10\)，所以三个数的和是 \(3m = 30\)。

因此，正确答案是 D。

English

Let \(m\) be the mean. The least number is \(m - 10,\) the greatest is \(m + 15,\) and the middle number is the median \(5.\) Their sum is \(3m,\) so \[ (m - 10) + 5 + (m + 15) = 3m. \]

This gives \(m = 10,\) so the sum of the three numbers is \(3m = 30.\)

Thus, the correct answer is D.

5 答案 D

从 \(1\) 到 \(10000\) 的所有整数的乘积是 \(10000!\)，所以所有奇数的乘积等于 \(10000!\) 除以所有偶数的乘积。

偶数的乘积可分解为 \[ \begin{gathered} 2 \cdot 4 \cdots 10000 \\ = 2^{5000}(1 \cdot 2 \cdots 5000) \\ = 2^{5000} \cdot 5000! \end{gathered}\text{。}\]

因此奇整数的乘积是 \[ \dfrac{10000!}{2^{5000} \cdot 5000!}\text{。}\]

因此，正确答案是 D。

English

The product of every integer from \(1\) to \(10000\) is \(10000!,\) so the product of the odd ones is \(10000!\) divided by the product of the even ones.

The even numbers factor as \[ \begin{gathered} 2 \cdot 4 \cdots 10000 \\ = 2^{5000}(1 \cdot 2 \cdots 5000) \\ = 2^{5000} \cdot 5000!. \end{gathered} \]

Therefore the product of the odd integers is \[ \dfrac{10000!}{2^{5000} \cdot 5000!}. \]

Thus, the correct answer is D.

6 答案 E

四个连续递减的奇数字 \(GHIJ\) 只能是 \(9753\) 或 \(7531\)，因而留给 \(ABC\) 的奇数字是 \(1\) 或 \(9\)。

因为 \(A + B + C = 9\) 且 \(ABC\) 的另外两个数字是偶数，那个奇数字必须是 \(1\)（若为 \(9\)，两个偶数字之和就必须为 \(0\)）。所以两个偶数字之和为 \(8\)。

三个连续递减的偶数字 \(DEF\) 是 \(864\)、\(642\) 或 \(420\)，留给 \(ABC\) 的偶数对分别是 \(\{2, 0\}\)、\(\{8, 0\}\) 或 \(\{8, 6\}\)。只有 \(\{8, 0\}\) 的和为 \(8\)，所以 \(ABC = 810\)，且 \(A = 8\)。

因此，正确答案是 E。

English

The four consecutive decreasing odd digits \(GHIJ\) are either \(9753\) or \(7531,\) leaving one odd digit (\(1\) or \(9\)) for \(ABC.\)

Since \(A + B + C = 9\) and the other two digits of \(ABC\) are even, the odd digit must be \(1\) (a \(9\) would force the two even digits to sum to \(0\)). So the two even digits sum to \(8.\)

The three consecutive decreasing even digits \(DEF\) are \(864,\) \(642,\) or \(420,\) leaving the even pairs \(\{2, 0\},\) \(\{8, 0\},\) or \(\{8, 6\}\) for \(ABC.\) Only \(\{8, 0\}\) sums to \(8,\) so \(ABC = 810\) and \(A = 8.\)

Thus, the correct answer is E.

7 答案 A

设 \(n\) 张票以每张 \(p\) 美元的全价出售。则 \[ np + (140 - n)\dfrac{p}{2} = 2001\text{，} \] 所以 \(p(n + 140)\) \(= 4002\) \(= 2 \cdot 3 \cdot 23 \cdot 29\)。

因为 \(0 \le n \le 140\)，我们需要 \(4002\) 的一个因数满足 \(140 \le n + 140 \le 280\)。唯一这样的因数是 \(174 = 2 \cdot 3 \cdot 29\)，得 \(n = 34\) 且 \(p = 23\)。

全价票收入为 \(34 \cdot 23 = 782\) 美元。

因此，正确答案是 A。

English

Let \(n\) tickets sell at full price \(p\) dollars. Then \[ np + (140 - n)\dfrac{p}{2} = 2001, \] so \(p(n + 140)\) \(= 4002\) \(= 2 \cdot 3 \cdot 23 \cdot 29.\)

Since \(0 \le n \le 140,\) we need a factor of \(4002\) with \(140 \le n + 140 \le 280.\) The only such factor is \(174 = 2 \cdot 3 \cdot 29,\) giving \(n = 34\) and \(p = 23.\)

The full-price tickets raise \(34 \cdot 23 = 782\) dollars.

Thus, the correct answer is A.

8 答案 C

当扇形卷成圆锥时，它的半径 \(10\) 成为母线长，它的弧成为底面圆。

弧长为 \[ \dfrac{252}{360}\cdot 2\pi(10) = \dfrac{7}{10}\cdot 20\pi = 14\pi\text{，} \] 所以底面周长是 \(14\pi\)，底面半径是 \(7\)。

因此该圆锥的底面半径为 \(7\)，母线长为 \(10\)，对应选项 C。

因此，正确答案是 C。

English

When the sector is rolled into a cone, its radius \(10\) becomes the slant height, and its arc becomes the base circle.

The arc length is \[ \dfrac{252}{360}\cdot 2\pi(10) = \dfrac{7}{10}\cdot 20\pi = 14\pi, \] so the base circumference is \(14\pi\) and the base radius is \(7.\)

The cone therefore has base radius \(7\) and slant height \(10,\) which is choice C.

Thus, the correct answer is C.

9 答案 C

取 \(x = 500\)，且令 \(y = \dfrac{6}{5}\)，使得 \(xy = 600\)。则 \[ f(600) = \dfrac{f(500)}{\frac{6}{5}} = \dfrac{3}{\frac{6}{5}} = \dfrac{5}{2}\text{。}\]

因此，正确答案是 C。

English

Choose \(x = 500\) and \(y = \dfrac{6}{5}\) so that \(xy = 600.\) Then \[ f(600) = \dfrac{f(500)}{\frac{6}{5}} = \dfrac{3}{\frac{6}{5}} = \dfrac{5}{2}. \]

Thus, the correct answer is C.

10 答案 D

图案在一个由九个单位正方形组成的 \(3 \times 3\) 方块中重复。其中九个小方格有四个未被五边形覆盖；剩余面积属于五边形。

所以五边形覆盖的比例为 \[ 1 - \dfrac{4}{9} = \dfrac{5}{9} = 55.5\overline{5}\%\text{，} \] 最接近 \(56\)。

因此，正确答案是 D。

English

The pattern repeats over a \(3 \times 3\) block of nine unit squares. Four of these nine squares are not covered by pentagons; the rest of the area belongs to the pentagons.

So the pentagons enclose \[ 1 - \dfrac{4}{9} = \dfrac{5}{9} = 55.5\overline{5}\%, \] which is closest to \(56.\)

Thus, the correct answer is D.

11 答案 D

想象继续取直到五枚筹码全部取出。实际过程停在白筹码上，恰好说明白筹码先于红筹码用完，也就是完整排列中的最后一枚筹码是红色。

五枚筹码中最后一枚等可能是任意一枚，因此它是红色的概率为 \(\dfrac{3}{5}\)。

因此，正确答案是 D。

English

Imagine continuing until all five chips are removed. The process actually stops on a white chip exactly when the whites run out before the reds, i.e. when the last chip in the full ordering is red.

The last of the five chips is equally likely to be any chip, so it is red with probability \(\dfrac{3}{5}.\)

Thus, the correct answer is D.

12 答案 B

不超过 \(2001\) 的 \(3\) 或 \(4\) 的倍数共有 \[ 667 + 500 - 166 = 1001\text{，}\] 这里用到 \(\lfloor \frac{2001}{3} \rfloor = 667\)、\(\lfloor \frac{2001}{4} \rfloor = 500\) 以及 \(\lfloor \frac{2001}{12} \rfloor = 166\)。

其中能被 \(5\) 整除的是 \(15\) 或 \(20\) 的倍数，共有 \[ 133 + 100 - 33 = 200\text{，}\] 这里用到 \(\lfloor \frac{2001}{15} \rfloor = 133\)、\(\lfloor \frac{2001}{20} \rfloor = 100\) 以及 \(\lfloor \frac{2001}{60} \rfloor = 33\)。

所求个数为 \(1001 - 200 = 801\)。

因此，正确答案是 B。

English

Multiples of \(3\) or \(4\) up to \(2001\) number \[ 667 + 500 - 166 = 1001, \] using \(\lfloor \frac{2001}{3} \rfloor = 667,\) \(\lfloor \frac{2001}{4} \rfloor = 500,\) and \(\lfloor \frac{2001}{12} \rfloor = 166.\)

Among these, the ones divisible by \(5\) are multiples of \(15\) or \(20\): \[ 133 + 100 - 33 = 200, \] using \(\lfloor \frac{2001}{15} \rfloor = 133,\) \(\lfloor \frac{2001}{20} \rfloor = 100,\) and \(\lfloor \frac{2001}{60} \rfloor = 33.\)

The count is \(1001 - 200 = 801.\)

Thus, the correct answer is B.

13 答案 E

\(a + b + c\) 是原抛物线在 \(x = 1\) 处的值，\(d + e + f\) 是反射后抛物线在 \(x = 1\) 处的值。

关于 \(y = k\) 反射，会把每个高度 \(y\) 替换成 \(2k - y\)。因此在 \(x = 1\) 处，两条抛物线的高度之和为 \[ (a + b + c) + (d + e + f) = 2k\text{。}\]

因此，正确答案是 E。

English

The value \(a + b + c\) is the first parabola at \(x = 1,\) and \(d + e + f\) is the reflected parabola at \(x = 1.\)

Reflecting the curve about \(y = k\) replaces each height \(y\) by \(2k - y.\) So at \(x = 1\) the two heights sum to \[ (a + b + c) + (d + e + f) = 2k. \]

Thus, the correct answer is E.

14 答案 D

\(\binom{9}{2} = 36\) 对顶点中的每一对，都可以作为恰好两个等边三角形的一条边，因此按重数计共有 \(72\) 个三角形。

三角形 \(A_1 A_4 A_7\)，\(A_2 A_5 A_8\)，和 \(A_3 A_6 A_9\) 的三个顶点都在该集合中，所以每个被数了三次而不是一次，每个多算了 \(2\) 次。

不同三角形的个数为 \(72 - 3 \cdot 2 = 66\)。

因此，正确答案是 D。

English

Each of the \(\binom{9}{2} = 36\) pairs of vertices is a side of exactly two equilateral triangles, giving \(72\) triangles counted with multiplicity.

The triangles \(A_1 A_4 A_7,\) \(A_2 A_5 A_8,\) and \(A_3 A_6 A_9\) have all three vertices in the set, so each is counted three times instead of once, an overcount of \(2\) apiece.

The number of distinct triangles is \(72 - 3 \cdot 2 = 66.\)

Thus, the correct answer is D.

15 答案 B

将昆虫经过的两个面展开到平面上。它们形成一个由两个等边三角形组成、边长为 \(1\) 的菱形。

这两个对棱中点会变成该菱形两条对边的中点，它们之间的直线距离正好是 \(1\)。折回四面体不会改变长度，因此最短路程是 \(1\)。

因此，正确答案是 B。

English

A shortest path leaves the starting edge through one of its two incident faces and reaches the opposite edge through one of its two incident faces. Any such pair of faces shares an edge. Unfolding that pair gives a rhombus of side \(1\) made of two equilateral triangles.

The two opposite-edge midpoints become the midpoints of opposite sides of this rhombus, which are exactly \(1\) unit apart along a straight segment. Folding back preserves the length, so the shortest trip is \(1.\)

Thus, the correct answer is B.

16 答案 D

把 \(16\) 件物品（\(8\) 只袜子和 \(8\) 只鞋）按某个顺序排列：共有 \(16!\) 种排列。

对每条腿，袜子在鞋子之前出现的排列恰好占所有排列的一半。对八条腿同时施加这个限制，需要除以 \(2^8\) 得到 \[ \dfrac{16!}{2^8}\text{。}\]

因此，正确答案是 D。

English

Think of the \(16\) items (\(8\) socks and \(8\) shoes) arranged in some order: there are \(16!\) arrangements.

For each leg, the sock comes before the shoe in exactly half of all arrangements. Imposing this on all eight legs independently divides by \(2^8,\) giving \[ \dfrac{16!}{2^8}. \]

Thus, the correct answer is D.

17 答案 C

当 \(P\) 在以 \(AB\) 为直径的圆上时，\(\angle APB = 90^\circ\)。该圆的圆心为 \((2, 1)\)，半径为 \(\dfrac{|AB|}{2} = \dfrac{\sqrt{20}}{2} = \sqrt{5}\)。当 \(P\) 在这个圆内时，该角为钝角。

相关的半圆完全在五边形内，面积为 \(\dfrac{1}{2}\pi(\sqrt{5})^2 = \dfrac{5\pi}{2}\)。

该五边形等于顶点为 \((0,0)\)，\(C\)，\(D\)，\(E\) 的矩形减去三角形 \(OAB\)，所以面积为 \[ 4(2\pi + 1) - \dfrac{1}{2}(2)(4) = 8\pi\text{。}\]

概率为 \(\dfrac{\frac{5\pi}{2}}{8\pi} = \dfrac{5}{16}\)。

因此，正确答案是 C。

English

\(\angle APB = 90^\circ\) when \(P\) is on the circle with diameter \(AB,\) centered at \((2, 1)\) with radius \(\dfrac{|AB|}{2} = \dfrac{\sqrt{20}}{2} = \sqrt{5}.\) The angle is obtuse when \(P\) is inside this circle.

The relevant half-disk lies wholly within the pentagon, with area \(\dfrac{1}{2}\pi(\sqrt{5})^2 = \dfrac{5\pi}{2}.\)

The pentagon is the rectangle with corners \((0,0),\) \(C,\) \(D,\) \(E\) minus triangle \(OAB,\) so its area is \[ 4(2\pi + 1) - \dfrac{1}{2}(2)(4) = 8\pi. \]

The probability is \(\dfrac{\frac{5\pi}{2}}{8\pi} = \dfrac{5}{16}.\)

Thus, the correct answer is C.

18 答案 D

当两个半径为 \(r\) 和 \(s\) 的相切圆都贴在同一直线上时，它们在直线上的切点距离为 \(2\sqrt{rs}\)。

两个大圆的切点相距 \(2\sqrt{1 \cdot 4} = 4\)。设夹在它们中间的小圆半径为 \(x\)，则它到两边的切点距离相加为 \[ 2\sqrt{1 \cdot x} + 2\sqrt{4 \cdot x} = 4\text{。}\]

于是 \(6\sqrt{x} = 4\)，所以 \(\sqrt{x} = \dfrac{2}{3}\) 且 \(x = \dfrac{4}{9}\)。

因此，正确答案是 D。

English

When two mutually tangent circles of radii \(r\) and \(s\) both rest on a line, the distance between their points of tangency is \(2\sqrt{rs}.\)

The big circles’ contact points are \(2\sqrt{1 \cdot 4} = 4\) apart. Placing the small circle of radius \(x\) between them, its two tangent distances add up: \[ 2\sqrt{1 \cdot x} + 2\sqrt{4 \cdot x} = 4. \]

Then \(6\sqrt{x} = 4,\) so \(\sqrt{x} = \dfrac{2}{3}\) and \(x = \dfrac{4}{9}.\)

Thus, the correct answer is D.

19 答案 A

\(y\)-截距为 \(P(0) = c = 2\)。由韦达定理，零点之积为 \(-c = -2\)，零点的平均数为 \(-\dfrac{a}{3}\)，系数之和为 \(P(1) = 1 + a + b + c\)。

这三者都等于 \(-2\)。由 \(-\dfrac{a}{3} = -2\) 得 \(a = 6\)。

于是 \(1 + a + b + c = -2\) 变为 \(1 + 6 + b + 2 = -2\)，所以 \(b = -11\)。

因此，正确答案是 A。

English

The \(y\)-intercept is \(P(0) = c = 2.\) By Vieta’s formulas the product of the zeros is \(-c = -2,\) the mean of the zeros is \(-\dfrac{a}{3},\) and the sum of the coefficients is \(P(1) = 1 + a + b + c.\)

All three are equal to \(-2.\) From \(-\dfrac{a}{3} = -2\) we get \(a = 6.\)

Then \(1 + a + b + c = -2\) becomes \(1 + 6 + b + 2 = -2,\) so \(b = -11.\)

Thus, the correct answer is A.

20 答案 C

\(\overline{AB}\) 的中点为 \(M = (2, 5)\)，\(\overline{BC}\) 的中点为 \(N = (3, 2)\)。

为了使中点四边形成为正方形，相邻边必须垂直且等长。由于 \(\overrightarrow{NM} = \langle -1, 3 \rangle\)，通向 \(\overline{DA}\) 的中点 \(Q\) 的边向量 \(\overrightarrow{MQ}\) 必须为 \(\langle 3, 1 \rangle\)，所以 \(Q = (5, 6)\)。

因为 \(Q\) 是 \(\overline{DA}\) 的中点，且 \(A = (3, 9)\)，得 \(D = 2Q - A = (7, 3)\)。它的坐标和为 \(7 + 3 = 10\)。

因此，正确答案是 C。

English

The midpoints are \(M = (2, 5)\) of \(\overline{AB}\) and \(N = (3, 2)\) of \(\overline{BC}.\)

For the midpoint quadrilateral to be a square, consecutive sides are perpendicular and equal. With \(\overrightarrow{NM} = \langle -1, 3 \rangle,\) the side \(\overrightarrow{MQ}\) to the midpoint \(Q\) of \(\overline{DA}\) must be \(\langle 3, 1 \rangle,\) so \(Q = (5, 6).\)

Since \(Q\) is the midpoint of \(\overline{DA}\) and \(A = (3, 9),\) we get \(D = 2Q - A = (7, 3).\) The sum of its coordinates is \(7 + 3 = 10.\)

Thus, the correct answer is C.

21 答案 D

给每个方程两边加 \(1\)，左边就可以分解：\[ \begin{aligned} (a + 1)(b + 1) &= 525 = 3 \cdot 5^2 \cdot 7, \\ (b + 1)(c + 1) &= 147 = 3 \cdot 7^2, \\ (c + 1)(d + 1) &= 105 = 3 \cdot 5 \cdot 7 \end{aligned}\text{。}\]

因为 \(525\) 含有因数 \(25\) 而 \(147\) 不能被 \(5\) 整除，所以因数 \(a + 1\) 必须包含这个 \(25\)。在 \(525\) 的因数中，只有 \(a + 1 = 25\) 会使 \(a = 24\) 整除 \(8! = 40320\)。

接着 \(b + 1 = 21\)，\(c + 1 = 7\)，且 \(d + 1 = 15\)，所以 \(b = 20\)，\(c = 6\)，\(d = 14\)。（确实 \(24 \cdot 20 \cdot 6 \cdot 14 = 40320 = 8!\)。）

所以 \(a - d = 24 - 14 = 10\)。

因此，正确答案是 D。

English

Adding \(1\) to each equation factors the left sides: \[ \begin{aligned} (a + 1)(b + 1) &= 525 = 3 \cdot 5^2 \cdot 7, \\ (b + 1)(c + 1) &= 147 = 3 \cdot 7^2, \\ (c + 1)(d + 1) &= 105 = 3 \cdot 5 \cdot 7. \end{aligned} \]

Since \(525\) has a factor of \(25\) while \(147\) is not divisible by \(5,\) the factor \(a + 1\) must carry the \(25.\) Among divisors of \(525,\) only \(a + 1 = 25\) makes \(a = 24\) divide \(8! = 40320.\)

Then \(b + 1 = 21,\) \(c + 1 = 7,\) and \(d + 1 = 15,\) giving \(b = 20,\) \(c = 6,\) \(d = 14.\) (Indeed \(24 \cdot 20 \cdot 6 \cdot 14 = 40320 = 8!.\))

So \(a - d = 24 - 14 = 10.\)

Thus, the correct answer is D.

22 答案 C

三角形 \(EFG\) 的底边 \(FG = \dfrac{1}{3}AB\)，高等于矩形的高，所以它的面积是 \(\dfrac{1}{6}(70) = \dfrac{35}{3}\)。

因为 \(EC \parallel AF\)，三角形 \(AFH\) 与 \(CEH\) 相似，比例为 \(\dfrac{EC}{AF} = \dfrac{3}{2}\)，所以 \(\dfrac{EH}{EF} = \dfrac{3}{5}\)。同理 \(\dfrac{EJ}{EG} = \dfrac{3}{7}\)。

因此 \(\dfrac{[EHJ]}{[EFG]}\) \(= \dfrac{EH}{EF}\cdot\dfrac{EJ}{EG}\) \(= \dfrac{3}{5}\cdot\dfrac{3}{7}\) \(= \dfrac{9}{35}\)，得 \[ [EHJ] = \dfrac{9}{35}\cdot\dfrac{35}{3} = 3\text{。}\]

因此，正确答案是 C。

English

Triangle \(EFG\) has base \(FG = \dfrac{1}{3}AB\) and height equal to the rectangle’s height, so its area is \(\dfrac{1}{6}(70) = \dfrac{35}{3}.\)

Because \(EC \parallel AF,\) triangles \(AFH\) and \(CEH\) are similar with ratio \(\dfrac{EC}{AF} = \dfrac{3}{2},\) so \(\dfrac{EH}{EF} = \dfrac{3}{5}.\) Likewise \(\dfrac{EJ}{EG} = \dfrac{3}{7}.\)

Then \(\dfrac{[EHJ]}{[EFG]}\) \(= \dfrac{EH}{EF}\cdot\dfrac{EJ}{EG}\) \(= \dfrac{3}{5}\cdot\dfrac{3}{7}\) \(= \dfrac{9}{35},\) giving \[ [EHJ] = \dfrac{9}{35}\cdot\dfrac{35}{3} = 3. \]

Thus, the correct answer is C.

23 答案 A

写成 \(P(x)\) \(= (x - r)(x - s)(x^2 + \alpha x + \beta)\)，其中 \(r, s\) 是整数根；比较系数可知 \(\alpha\) 和 \(\beta\) 必须是整数。

另外两个零点为 \[ -\dfrac{\alpha}{2} \pm \dfrac{i\sqrt{4\beta - \alpha^2}}{2}\text{。} \] 若实部为 \(\dfrac{1}{2}\)，则需要 \(\alpha = -1\)，从而虚部为 \(\dfrac{\sqrt{4\beta - 1}}{2}\)。

选项 A 要求 \(\sqrt{4\beta - 1} = \sqrt{11}\)，即 \(\beta = 3\) 是整数，所以可行。其他选项都会迫使 \(\beta\) 不是整数（例如选项 E 需要 \(\beta = 3.5\)，选项 D 需要 \(\alpha = -2\) 且 \(\beta = \tfrac{5}{4}\)）。

因此，正确答案是 A。

English

Writing \(P(x)\) \(= (x - r)(x - s)(x^2 + \alpha x + \beta)\) with integer roots \(r, s,\) matching coefficients forces \(\alpha\) and \(\beta\) to be integers.

The other two zeros are \[ -\dfrac{\alpha}{2} \pm \dfrac{i\sqrt{4\beta - \alpha^2}}{2}. \] A real part of \(\dfrac{1}{2}\) requires \(\alpha = -1,\) making the imaginary part \(\dfrac{\sqrt{4\beta - 1}}{2}.\)

Choice A needs \(\sqrt{4\beta - 1} = \sqrt{11},\) i.e. \(\beta = 3,\) an integer, so it works. The other choices force a non-integer \(\beta\) (for example choice E needs \(\beta = 3.5,\) and choice D needs \(\alpha = -2\) with \(\beta = \tfrac{5}{4}\)).

Thus, the correct answer is A.

24 答案 D

设 \(E\) 为从 \(C\) 到直线 \(AD\) 的垂足。由 \(\triangle ADB\) 的外角可得 \(\angle ADC = 15^\circ + 45^\circ = 60^\circ\)，所以 \(\triangle CDE\) 是一个 \(30\)-\(60\)-\(90\) 三角形，且 \(DE = \tfrac{1}{2}CD = BD\)。

因此 \(\triangle BDE\) 是等腰三角形，\(\angle EBD = \angle BED = 30^\circ\)，又因为 \(\angle ECB = 30^\circ\)，所以 \(\triangle BEC\) 是等腰三角形，\(BE = EC\)。

同时 \(\angle ABE\) \(= 45^\circ - 30^\circ\) \(= 15^\circ\) \(= \angle EAB\)，所以 \(\triangle ABE\) 是等腰三角形，\(AE = BE\)。因此 \(AE = BE = EC\)，使得直角三角形 \(AEC\) 为等腰直角三角形，\(\angle ECA = 45^\circ\)。

所以 \(\angle ACB\) \(= \angle ECA + \angle ECD\) \(= 45^\circ + 30^\circ\) \(= 75^\circ\)。

因此，正确答案是 D。

English

Let \(E\) be the foot of the perpendicular from \(C\) to line \(AD.\) The exterior angle of \(\triangle ADB\) gives \(\angle ADC = 15^\circ + 45^\circ = 60^\circ,\) so \(\triangle CDE\) is a \(30\)-\(60\)-\(90\) triangle with \(DE = \tfrac{1}{2}CD = BD.\)

Then \(\triangle BDE\) is isosceles with \(\angle EBD = \angle BED = 30^\circ,\) and since \(\angle ECB = 30^\circ\) too, \(\triangle BEC\) is isosceles with \(BE = EC.\)

Also \(\angle ABE\) \(= 45^\circ - 30^\circ\) \(= 15^\circ\) \(= \angle EAB,\) so \(\triangle ABE\) is isosceles with \(AE = BE.\) Hence \(AE = BE = EC,\) making right triangle \(AEC\) isosceles with \(\angle ECA = 45^\circ.\)

Therefore \(\angle ACB\) \(= \angle ECA + \angle ECD\) \(= 45^\circ + 30^\circ\) \(= 75^\circ.\)

Thus, the correct answer is D.

25 答案 D

若 \(a, b, c\) 是连续三项，则 \(b = ac - 1\)，所以 \(c = \dfrac{1 + b}{a}\)。反复应用这个关系，前五项依次为 \[ a,\ b,\ \dfrac{1 + b}{a},\ \dfrac{1 + a + b}{ab},\ \dfrac{1 + a}{b}\text{，}\]此后 \(a\) 和 \(b\) 又重新出现，因此数列以 \(5\) 为周期。

此处 \(b = 2000\) 是第二项。令上面另外四项分别等于 \(2001\)，依次解得 \[ \begin{gathered} a=2001,\quad a=1,\\ a=\frac{2001}{4001999},\quad a=4001999 \end{gathered}\text{。}\]这四个值都是正数且互不相同。而第二项本身等于 \(2000\)，再由周期性可知没有其他位置需要考虑。

所以共有 \(4\) 个 \(x\) 值。

因此，正确答案是 D。

English

If \(a, b, c\) are consecutive terms then \(b = ac - 1,\) so \(c = \dfrac{1 + b}{a}.\) Applying this repeatedly, the first five terms are \[ a,\ b,\ \dfrac{1 + b}{a},\ \dfrac{1 + a + b}{ab},\ \dfrac{1 + a}{b}, \] after which \(a\) and \(b\) recur, so the sequence is periodic with period \(5.\)

Here \(b = 2000\) is the second term. Setting each of the other four displayed terms equal to \(2001\) gives, respectively, \[ \begin{gathered} a=2001,\quad a=1,\\ a=\frac{2001}{4001999},\quad a=4001999. \end{gathered} \] These four values are positive and distinct. The second term itself is \(2000,\) and periodicity shows there are no other positions to consider.

So there are \(4\) values of \(x.\)

Thus, the correct answer is D.