2002A AMC 12 真题（P版）

1 答案 A

提出 \(2x+3\) 得到 \((2x+3)\big[(x-4)+(x-6)\big]\) \(= (2x+3)(2x-10)\) \(= 0\)。

两个根为 \(-\dfrac{3}{2}\) 和 \(5\)，它们的和是 \(-\dfrac{3}{2} + 5 = \dfrac{7}{2}\)。

因此，正确答案是 A。

English

Factoring out \(2x+3\) gives \((2x+3)\big[(x-4)+(x-6)\big]\) \(= (2x+3)(2x-10)\) \(= 0.\)

The roots are \(-\dfrac{3}{2}\) and \(5,\) whose sum is \(-\dfrac{3}{2} + 5 = \dfrac{7}{2}.\)

Thus, the correct answer is A.

2 答案 A

设原数为 \(x\)。Cindy 计算了 \(\dfrac{x-9}{3} = 43\)，所以 \(x - 9 = 129\)，\(x = 138\)。

正确计算为 \(\dfrac{138-3}{9} = \dfrac{135}{9} = 15\)。

因此，正确答案是 A。

English

Let \(x\) be the number. Cindy computed \(\dfrac{x-9}{3} = 43,\) so \(x - 9 = 129\) and \(x = 138.\)

The correct computation gives \(\dfrac{138-3}{9} = \dfrac{135}{9} = 15.\)

Thus, the correct answer is A.

3 答案 B

\(2^{2^{2^2}}\) 的五种加括号方式给出 \[ \begin{aligned} &\left(\left(2^2\right)^2\right)^2 = 2^8, \\ &\left(2^{2^2}\right)^2 = 2^8, \\ &\left(2^2\right)^{2^2} = 2^8 \end{aligned}\text{，}\]\[2^{\left(2^2\right)^2} = 2^{16},\quad 2^{2^{2^2}} = 2^{16}\text{。}\]

因此只可能得到 \(2^{16} = 65{,}536\) 和 \(2^8 = 256\)。除标准值外，恰好还有 \(1\) 个其他值。

因此，正确答案是 B。

English

The five parenthesizations of \(2^{2^{2^2}}\) give \[ \begin{aligned} &\left(\left(2^2\right)^2\right)^2 = 2^8, \\ &\left(2^{2^2}\right)^2 = 2^8, \\ &\left(2^2\right)^{2^2} = 2^8, \end{aligned} \] \[2^{\left(2^2\right)^2} = 2^{16},\quad 2^{2^{2^2}} = 2^{16}.\]

So the only values are \(2^{16} = 65{,}536\) and \(2^8 = 256.\) Besides the standard value there is exactly \(1\) other.

Thus, the correct answer is B.

4 答案 B

设这个角为 \(x\)，则 \(90 - x = \dfrac14(180 - x)\)，所以 \(360 - 4x = 180 - x\)。

由此 \(3x = 180\)，所以 \(x = 60\)。

因此，正确答案是 B。

English

Let the angle be \(x.\) Then \(90 - x = \dfrac14(180 - x),\) so \(360 - 4x = 180 - x.\)

This gives \(3x = 180,\) so \(x = 60.\)

Thus, the correct answer is B.

5 答案 C

六个外圈单位圆都与中心单位圆相切，所以它们的圆心到中心距离为 \(2\)。再加上一个半径，大圆半径为 \(3\)，面积为 \(9\pi\)。

七个单位圆的总面积为 \(7\pi\)，所以阴影区域面积为 \(9\pi - 7\pi = 2\pi\)。

因此，正确答案是 C。

English

Each of the six outer unit circles is tangent to the central unit circle, so its center is \(2\) units from the center. Adding one more radius, the large circle has radius \(3\) and area \(9\pi.\)

The seven unit circles have total area \(7\pi,\) so the shaded region has area \(9\pi - 7\pi = 2\pi.\)

Thus, the correct answer is C.

6 答案 E

取 \(n = 1\)，不等式变为 \(m \le m + 1\)，这对每个正整数 \(m\) 都成立。

所以每个正整数 \(m\) 都满足条件，共有无穷多个。

因此，正确答案是 E。

English

Taking \(n = 1,\) the inequality becomes \(m \le m + 1,\) which holds for every positive integer \(m.\)

So every positive integer \(m\) works, and there are infinitely many.

Thus, the correct answer is E.

7 答案 A

弧长相等给出 \(\dfrac{45}{360}\cdot 2\pi R_A = \dfrac{30}{360}\cdot 2\pi R_B\)，所以 \(\dfrac{R_A}{R_B} = \dfrac{30}{45} = \dfrac{2}{3}\)。

面积比为 \(\left(\dfrac{R_A}{R_B}\right)^2 = \dfrac{4}{9}\)。

因此，正确答案是 A。

English

Equal arc lengths give \(\dfrac{45}{360}\cdot 2\pi R_A = \dfrac{30}{360}\cdot 2\pi R_B,\) so \(\dfrac{R_A}{R_B} = \dfrac{30}{45} = \dfrac{2}{3}.\)

The ratio of areas is \(\left(\dfrac{R_A}{R_B}\right)^2 = \dfrac{4}{9}.\)

Thus, the correct answer is A.

8 答案 A

画出所示的对角线后，整面旗帜被分成全等三角形。数一数可得，蓝色区域有 \(24\) 个三角形，白色区域有 \(24\) 个，红色正方形有 \(16\) 个。

由于蓝色和白色区域含有相同数量的三角形，所以 \(B = W\)。

因此，正确答案是 A。

English

Drawing the diagonals shown, the entire flag is tiled by congruent triangles. Counting, there are \(24\) triangles in the blue region, \(24\) in the white region, and \(16\) in the red square.

Since the blue and white regions contain the same number of triangles, \(B = W.\)

Thus, the correct answer is A.

9 答案 B

文件总大小为 \(3(0.8)+12(0.7)\) \(+15(0.4) = 16.8\) mb，所以仅按容量至少需要 \(\dfrac{16.8}{1.44} = 11\tfrac{2}{3}\) 张软盘。

含有 \(0.8\)-mb 文件的软盘只能再放一个 \(0.4\)-mb 文件，至少留下 \(0.24\) mb 空间。三个 \(0.8\)-mb 文件总共至少浪费 \(3(0.24) = 0.72\) mb，超过半张软盘容量，因此至少需要 \(13\) 张。

十三张也足够：六张各放两个 \(0.7\)-mb 文件，三张各放一个 \(0.8\)-mb 文件和一个 \(0.4\)-mb 文件，四张各放三个 \(0.4\)-mb 文件。

因此，正确答案是 B。

English

The files need \(3(0.8)+12(0.7)\) \(+15(0.4) = 16.8\) mb, so at least \(\dfrac{16.8}{1.44} = 11\tfrac{2}{3}\) disks by volume alone.

A disk containing a \(0.8\)-mb file has room for only one more \(0.4\)-mb file, leaving at least \(0.24\) mb unused. Across the three \(0.8\)-mb files this wastes at least \(3(0.24) = 0.72\) mb, over half a disk, forcing at least \(13\) disks.

Thirteen suffice: six disks each hold two \(0.7\)-mb files, three disks each hold one \(0.8\)-mb file plus one \(0.4\)-mb file, and four disks each hold three \(0.4\)-mb files.

Thus, the correct answer is B.

10 答案 D

倒出一半咖啡后，杯 \(1\) 有 \(2\) oz 咖啡；杯 \(2\) 有 \(2\) oz 咖啡和 \(4\) oz 奶油，共 \(6\) oz。

杯 \(2\) 倒回一半，会转移 \(1\) oz 咖啡和 \(2\) oz 奶油。于是杯 \(1\) 有 \(2 + 1 = 3\) oz 咖啡和 \(2\) oz 奶油。

奶油所占比例为 \(\dfrac{2}{2+3} = \dfrac{2}{5}\)。

因此，正确答案是 D。

English

After transferring half the coffee, cup \(1\) has \(2\) oz coffee and cup \(2\) has \(2\) oz coffee and \(4\) oz cream, a total of \(6\) oz.

Transferring half of cup \(2\) back moves \(1\) oz coffee and \(2\) oz cream. Cup \(1\) then holds \(2 + 1 = 3\) oz coffee and \(2\) oz cream.

The fraction that is cream is \(\dfrac{2}{2+3} = \dfrac{2}{5}.\)

Thus, the correct answer is D.

11 答案 B

设准时所需时间为 \(t\) 小时。三分钟是 \(0.05\) 小时，所以 \(40(t+0.05) = 60(t-0.05)\)。

得 \(40t + 2 = 60t - 3\)，所以 \(t = 0.25\)。路程为 \(40(0.25+0.05) = 12\) 英里，所需速度为 \(\dfrac{12}{0.25} = 48\) 英里每小时。

因此，正确答案是 B。

English

Let \(t\) be the time in hours to arrive on time. Since three minutes is \(0.05\) hours, \(40(t+0.05) = 60(t-0.05).\)

This gives \(40t + 2 = 60t - 3,\) so \(t = 0.25.\) The distance is \(40(0.25+0.05) = 12\) miles, and the required speed is \(\dfrac{12}{0.25} = 48\) miles per hour.

Thus, the correct answer is B.

12 答案 B

如果两个根是质数 \(p\) 和 \(q\)，则由韦达定理 \(p + q = 63\)，且 \(pq = k\)。

因为 \(63\) 是奇数，所以一个质数必须是偶数，即 \(2\)，另一个是 \(61\)。两者都是质数，所以 \(k = 2\cdot 61 = 122\) 是唯一可能值。

因此，正确答案是 B。

English

If the roots are primes \(p\) and \(q,\) then by Vieta’s formulas \(p + q = 63\) and \(pq = k.\)

Since \(63\) is odd, one prime must be even, namely \(2,\) and the other is \(61.\) Both are prime, so \(k = 2\cdot 61 = 122\) is the only possible value.

Thus, the correct answer is B.

13 答案 C

正数 \(x\) 与其倒数相差 \(1\) 时，\(x - \dfrac1x = 1\) 或 \(x - \dfrac1x = -1\)，即 \(x^2 - x - 1 = 0\) 或 \(x^2 + x - 1 = 0\)。

正根为 \(\dfrac{1+\sqrt5}{2}\) 和 \(\dfrac{-1+\sqrt5}{2}\)，它们互为倒数。它们的和是 \(a + b = \sqrt5\)。

因此，正确答案是 C。

English

A positive number \(x\) differs from its reciprocal by \(1\) when \(x - \dfrac1x = 1\) or \(x - \dfrac1x = -1,\) i.e. \(x^2 - x - 1 = 0\) or \(x^2 + x - 1 = 0.\)

The positive roots are \(\dfrac{1+\sqrt5}{2}\) and \(\dfrac{-1+\sqrt5}{2},\) which are reciprocals of each other. Their sum is \(a + b = \sqrt5.\)

Thus, the correct answer is C.

14 答案 D

利用 \(\log a^2 = 2\log a\) 并合并对数，\[ \begin{aligned} N &= \log_{2002} 11^2 + \log_{2002} 13^2 \\ &\quad {}+ \log_{2002} 14^2 \\ &= \log_{2002}(11\cdot 13\cdot 14)^2 \end{aligned}\text{。}\]

因为 \(11\cdot 13\cdot 14 = 2002\)，所以这是 \(\log_{2002} 2002^2 = 2\)。

因此，正确答案是 D。

English

Using \(\log a^2 = 2\log a\) and adding logs, \[ \begin{aligned} N &= \log_{2002} 11^2 + \log_{2002} 13^2 \\ &\quad {}+ \log_{2002} 14^2 \\ &= \log_{2002}(11\cdot 13\cdot 14)^2. \end{aligned} \]

Since \(11\cdot 13\cdot 14 = 2002,\) this is \(\log_{2002} 2002^2 = 2.\)

Thus, the correct answer is D.

15 答案 D

集合 \(6, 6, 6, 8, 8, 8, 8, 14\) 的平均数、中位数、唯一众数和极差都等于 \(8\)，所以 \(14\) 可以达到。

假设最大值为 \(15\)。极差 \(8\) 迫使最小值为 \(7\)。因为 \(8\) 是众数，它在排好序的八个数中重复出现；再结合中位数 \(8\)，中间两个数只能是 \(8, 8\)。于是 \(7 + 8 + 8 + 15 = 38\)，剩下四个数的和为 \(64 - 38 = 26\)，平均为 \(6.5\)。这样至少有一个数会小于 \(7\)，与最小值矛盾。因此 \(15\) 不可能。

因此，正确答案是 D。

English

The collection \(6, 6, 6, 8, 8, 8, 8, 14\) has mean, median, unique mode, and range all equal to \(8,\) so \(14\) is attainable.

Suppose the largest were \(15.\) The range \(8\) forces the smallest to be \(7.\) Because \(8\) is the mode, it occurs in the sorted list; together with median \(8,\) this forces the two middle values to be \(8, 8.\) Then \(7 + 8 + 8 + 15 = 38,\) so the remaining four values sum to \(64 - 38 = 26,\) averaging \(6.5.\) At least one would be below \(7,\) contradicting the minimum. So \(15\) is impossible.

Thus, the correct answer is D.

16 答案 A

Tina 的十个数对的和为 \(3, 4, 5, 5, 6, 6, 7, 7, 8, 9\)。对于和 \(s\)，Sergio 的数大于它的概率为 \(\dfrac{10 - s}{10}\)。

相应的 \(10 - s\) 值为 \(7, 6, 5, 5, 4, 4, 3, 3, 2, 1\)，总和为 \(40\)。总概率为 \(\dfrac{40}{10\cdot 10} = \dfrac{2}{5}\)。

因此，正确答案是 A。

English

Tina’s ten pairs have sums \(3, 4, 5, 5, 6, 6, 7, 7, 8, 9.\) For a sum \(s,\) Sergio’s number exceeds it with probability \(\dfrac{10 - s}{10}.\)

The corresponding values of \(10 - s\) are \(7, 6, 5, 5, 4, 4, 3, 3, 2, 1,\) totaling \(40.\) The overall probability is \(\dfrac{40}{10\cdot 10} = \dfrac{2}{5}.\)

Thus, the correct answer is A.

17 答案 B

偶数字 \(4, 6, 8\) 不能作为多位质数的个位，所以每个都必须出现在十位或更高位，至少贡献 \(40 + 60 + 80 = 180\)。另外六个数字至少贡献 \(1 + 2 + 3 + 5 + 7 + 9 = 27\)，因此总和至少为 \(207\)。

这个下界可以达到，例如 \(\{2, 3, 5, 41, 67, 89\}\)，它们的和为 \(207\)。

因此，正确答案是 B。

English

The even digits \(4, 6, 8\) cannot be the units digit of a multi-digit prime, so each must appear in a tens place or higher, contributing at least \(40 + 60 + 80 = 180.\) The other six digits contribute at least \(1 + 2 + 3 + 5 + 7 + 9 = 27,\) so the sum is at least \(207.\)

This bound is achieved, for example by \(\{2, 3, 5, 41, 67, 89\},\) whose sum is \(207.\)

Thus, the correct answer is B.

18 答案 C

圆心为 \(A = (10, 0)\) 和 \(B = (-15, 0)\)，半径分别为 \(6\) 和 \(9\)，所以 \(AB = 25\)。最短切线是内公切线，它在点 \(D\) 与 \(\overline{AB}\) 相交，并按 \(6 : 9\) 分割该线段，因此 \(D = (0, 0)\)。

直角三角形 \(APD\) 和 \(BQD\) 相似，比例为 \(2 : 3\)。因而 \(PD = \sqrt{10^2 - 6^2} = 8\)，\(QD = \sqrt{15^2 - 9^2} = 12\)，所以 \(PQ = 8 + 12 = 20\)。

因此，正确答案是 C。

English

The centers are \(A = (10, 0)\) and \(B = (-15, 0),\) with radii \(6\) and \(9,\) so \(AB = 25.\) The shortest tangent is the internal one, meeting \(\overline{AB}\) at a point \(D\) that splits it in the ratio \(6 : 9,\) giving \(D = (0, 0).\)

The right triangles \(APD\) and \(BQD\) are similar with ratio \(2 : 3.\) Then \(PD = \sqrt{10^2 - 6^2} = 8\) and \(QD = \sqrt{15^2 - 9^2} = 12,\) so \(PQ = 8 + 12 = 20.\)

Thus, the correct answer is C.

19 答案 D

图像在 \(x = -2\) 和 \(x = 1\) 处达到 \(6\)，所以 \(f(f(x)) = 6\) 要求 \(f(x) = -2\) 或 \(f(x) = 1\)。

水平线 \(y = -2\) 与图像相交两次，\(y = 1\) 与图像相交四次，共有 \(2 + 4 = 6\) 个解。

因此，正确答案是 D。

English

The graph reaches \(6\) at \(x = -2\) and \(x = 1,\) so \(f(f(x)) = 6\) requires \(f(x) = -2\) or \(f(x) = 1.\)

The horizontal line \(y = -2\) meets the graph twice, and \(y = 1\) meets it four times, giving \(2 + 4 = 6\) solutions.

Thus, the correct answer is D.

20 答案 C

因为 \(0.\overline{ab} = \dfrac{\overline{ab}}{99}\)，最简分母必须整除 \(99 = 3^2\cdot 11\)。它的因数为 \(1, 3, 9, 11, 33, 99\)。

分母为 \(1\) 会要求 \(\overline{ab} = 99\)，即 \(a = b = 9\)，但这种情况被排除。其余各个分母都能达到：分子 \(33, 11, 9, 3\) 和 \(1\) 分别约分后得到分母 \(3, 9, 11, 33\) 和 \(99\)，所以共有 \(5\) 种可能的分母。

所以正确答案是 C。

English

Since \(0.\overline{ab} = \dfrac{\overline{ab}}{99},\) the reduced denominator divides \(99 = 3^2\cdot 11.\) The divisors are \(1, 3, 9, 11, 33, 99.\)

The denominator \(1\) would require \(\overline{ab} = 99,\) i.e. \(a = b = 9,\) which is excluded. Each is achievable: numerators \(33, 11, 9, 3,\) and \(1\) reduce to denominators \(3, 9, 11, 33,\) and \(99,\) respectively. Thus there are \(5\) possible denominators.

Thus, the correct answer is C.

21 答案 B

继续写数列得到 \(4, 7, 1, 8, 9, 7, 6\)，\(3, 9, 2, 1, 3, 4, 7, 1, \ldots\)，它以 \(12\) 为周期重复。每个 \(12\) 项循环的和为 \(60\)。

满足 \(60k \le 10{,}000\) 的最大 \(k\) 是 \(k = 166\)，因此 \(S_{12\cdot 166} = 9960\)。再加上下一轮的 \(4, 7, 1, 8, 9, 7, 6\)，增加 \(42\)，总和超过 \(10{,}000\)。所以 \(n = 12\cdot 166 + 7 = 1999\)。

因此，正确答案是 B。

English

Continuing the sequence gives \(4, 7, 1, 8, 9, 7, 6,\) \(3, 9, 2, 1, 3, 4, 7, 1, \ldots,\) which repeats with period \(12.\) Each block of \(12\) terms sums to \(60.\)

The largest \(k\) with \(60k \le 10{,}000\) is \(k = 166,\) giving \(S_{12\cdot 166} = 9960.\) Adding the next terms \(4, 7, 1, 8, 9, 7, 6\) contributes \(42,\) pushing the total past \(10{,}000.\) So \(n = 12\cdot 166 + 7 = 1999.\)

Thus, the correct answer is B.

22 答案 C

因为 \(AB = 10\) 且 \(\angle ABC = 60^\circ\)，这个 \(30\)-\(60\)-\(90\) 三角形满足 \(BC = 5\)，\(AC = 5\sqrt3\)。

在 \(\overline{AC}\) 上取点 \(E\)，使 \(CE = 5\)，则 \(BE = \sqrt{5^2 + 5^2} = 5\sqrt2\)。当 \(D\) 沿 \(\overline{AC}\) 移动时，\(BD = \sqrt{25 + CD^2}\)。它超过 \(5\sqrt2\) 正好等价于 \(CD \gt 5\)，也就是 \(D\) 位于 \(E\) 的外侧；这又当且仅当 \(P\) 在 \(\triangle ABE\) 内。

所求概率为 \[ \begin{aligned} \dfrac{[ABE]}{[ABC]} &= \dfrac{EA}{CA} \\ &= \dfrac{5\sqrt3 - 5}{5\sqrt3} \\ &= \dfrac{3 - \sqrt3}{3} \end{aligned}\text{。}\]

因此，正确答案是 C。

English

Since \(AB = 10\) and \(\angle ABC = 60^\circ,\) the \(30\)-\(60\)-\(90\) triangle has \(BC = 5\) and \(AC = 5\sqrt3.\)

Place \(E\) on \(\overline{AC}\) with \(CE = 5;\) then \(BE = \sqrt{5^2 + 5^2} = 5\sqrt2.\) As \(D\) moves along \(\overline{AC},\) \(BD = \sqrt{25 + CD^2}\) exceeds \(5\sqrt2\) exactly when \(CD \gt 5,\) i.e. when \(D\) lies beyond \(E,\) which happens iff \(P\) is inside \(\triangle ABE.\)

The probability is \[ \begin{aligned} \dfrac{[ABE]}{[ABC]} &= \dfrac{EA}{CA} \\ &= \dfrac{5\sqrt3 - 5}{5\sqrt3} \\ &= \dfrac{3 - \sqrt3}{3}. \end{aligned} \]

Thus, the correct answer is C.

23 答案 D

因为 \(D\) 在 \(\overline{BC}\) 的垂直平分线上，所以 \(DB = DC = 7\)。角平分线 \(\overline{BD}\) 给出 \(\dfrac{AB}{BC} = \dfrac{AD}{DC} = \dfrac{9}{7}\)，设 \(AB = 9x\)，\(BC = 7x\)。

令 \(\theta = \angle ABD = \angle DBC\)。在等腰三角形 \(\triangle BDC\) 中，垂足是 \(\overline{BC}\) 的中点 \(M\)，所以 \(\cos\theta = \dfrac{BM}{BD} = \dfrac{\frac{7x}{2}}{7} = \dfrac{x}{2}\)。

在 \(\triangle ABD\) 中应用余弦定理：\[9^2 = (9x)^2 + 7^2 - 2(9x)(7)\cdot\dfrac{x}{2}\text{，}\]化简得 \(81 = 18x^2 + 49\)，所以 \(x = \dfrac43\)，\(AB = 12\)。

现在 \(\triangle ABD\) 的三边为 \(9, 7, 12\)。由海伦公式，半周长 \(s = 14\)，面积为 \(\sqrt{14\cdot 5\cdot 7\cdot 2} = \sqrt{980} = 14\sqrt5\)。

因此，正确答案是 D。

English

Since \(D\) lies on the perpendicular bisector of \(\overline{BC},\) \(DB = DC = 7.\) The angle bisector \(\overline{BD}\) gives \(\dfrac{AB}{BC} = \dfrac{AD}{DC} = \dfrac{9}{7},\) so write \(AB = 9x\) and \(BC = 7x.\)

Let \(\theta = \angle ABD = \angle DBC.\) In isosceles \(\triangle BDC,\) the foot of the perpendicular is the midpoint \(M\) of \(\overline{BC},\) so \(\cos\theta = \dfrac{BM}{BD} = \dfrac{\frac{7x}{2}}{7} = \dfrac{x}{2}.\)

Applying the Law of Cosines in \(\triangle ABD:\) \[9^2 = (9x)^2 + 7^2 - 2(9x)(7)\cdot\dfrac{x}{2},\] which simplifies to \(81 = 18x^2 + 49,\) so \(x = \dfrac43\) and \(AB = 12.\)

Now \(\triangle ABD\) has sides \(9, 7, 12.\) By Heron’s formula with \(s = 14,\) the area is \(\sqrt{14\cdot 5\cdot 7\cdot 2} = \sqrt{980} = 14\sqrt5.\)

Thus, the correct answer is D.

24 答案 E

设 \(z = a + bi\)。方程为 \(z^{2002} = \overline{z}\)。取模长，得 \(|z|^{2002} = |z|\)，所以 \(|z|\big(|z|^{2001} - 1\big) = 0\)，因此 \(|z| = 0\) 或 \(|z| = 1\)。

若 \(|z| = 0\)，则 \((a, b) = (0, 0)\)，得到一个解。若 \(|z| = 1\)，则 \(\overline{z} = \dfrac1z\)，所以 \(z^{2002} = \dfrac1z\)，即 \(z^{2003} = 1\)，它有 \(2003\) 个不同的根。

总共有 \(1 + 2003 = 2004\) 个有序对。

因此，正确答案是 E。

English

Let \(z = a + bi.\) The equation is \(z^{2002} = \overline{z}.\) Taking magnitudes, \(|z|^{2002} = |z|,\) so \(|z|\big(|z|^{2001} - 1\big) = 0,\) giving \(|z| = 0\) or \(|z| = 1.\)

If \(|z| = 0,\) then \((a, b) = (0, 0),\) one solution. If \(|z| = 1,\) then \(\overline{z} = \dfrac1z,\) so \(z^{2002} = \dfrac1z,\) i.e. \(z^{2003} = 1,\) which has \(2003\) distinct roots.

Altogether there are \(1 + 2003 = 2004\) ordered pairs.

Thus, the correct answer is E.

25 答案 B

将非零系数替换为它们的平均数，会保持系数总和不变，所以 \(P\) 和 \(Q\) 的系数和相同。由于 \(P(1)\) 和 \(Q(1)\) 都等于各自的系数和，\(P(1) = Q(1)\)。

因此 \(y = P(x)\) 和 \(y = Q(x)\) 的图像必须在 \(x = 1\) 处相交。唯一显示在 \(x = 1\) 处相交的是图 B。（其中 \(P(x) = 2x^4 - 3x^2 - 3x - 4\)，\(Q(x) = -2x^4 - 2x^2 - 2x - 2\)。）

因此，正确答案是 B。

English

Replacing the nonzero coefficients by their mean keeps the total of the coefficients unchanged, so \(P\) and \(Q\) have the same coefficient sum. Since \(P(1)\) and \(Q(1)\) each equal that sum, \(P(1) = Q(1).\)

Therefore the graphs of \(y = P(x)\) and \(y = Q(x)\) must cross at \(x = 1.\) The only choice showing an intersection at \(x = 1\) is graph B. (There, \(P(x) = 2x^4 - 3x^2 - 3x - 4\) and \(Q(x) = -2x^4 - 2x^2 - 2x - 2.\))

Thus, the correct answer is B.