2004B AMC 12 真题（P版）

1 答案 A

每次训练罚中球数都是前一次的两倍，所以倒着推要除以二。第五次是 \(48\)，前几次分别是 \(24\)、\(12\)、\(6\)、\(3\) 球。

因此正确答案是 A。

English

Each practice she made twice the previous, so going backward we halve. From the fifth practice at \(48,\) the earlier practices had \(24,\) \(12,\) \(6,\) and \(3\) free throws.

Thus, the correct answer is A.

2 答案 D

为了使结果最大，取 \(d = 0\)。此时 \(a, b, c\) 分别取 \(1, 2, 3\) 中的不同值；当 \(c = 1\) 且 \(a^b = 3^2 = 9\) 时，\(c \cdot a^b\) 最大。这给出 \(1 \cdot 9 - 0 = 9\)，比 \(2^3 = 8\) 以及其他分配都大。

因此正确答案是 D。

English

To maximize, set \(d = 0.\) With \(a, b, c\) taking \(1, 2, 3,\) the term \(c \cdot a^b\) is largest when \(c = 1\) and \(a^b = 3^2 = 9.\) This gives \(1 \cdot 9 - 0 = 9,\) which beats \(2^3 = 8\) and the other assignments.

Thus, the correct answer is D.

3 答案 A

分解得 \(1296 = 6^4 = 2^4 \cdot 3^4\)。比较指数可得 \(x = 4\)、\(y = 4\)，所以 \(x + y = 8\)。

因此正确答案是 A。

English

Factoring, \(1296 = 6^4 = 2^4 \cdot 3^4.\) Matching exponents gives \(x = 4\) and \(y = 4,\) so \(x + y = 8.\)

Thus, the correct answer is A.

4 答案 B

从 \(10\) 到 \(99\) 共有 \(90\) 个整数。个位是 \(7\) 的有十个，十位是 \(7\) 的有九个。由于 \(77\) 被重复计算，所以至少有一个 \(7\) 的数共有 \(10 + 9 - 1 = 18\) 个。概率为 \(\dfrac{18}{90} = \dfrac{1}{5}\)。

因此正确答案是 B。

English

There are \(90\) integers from \(10\) to \(99.\) Ten have a units digit \(7,\) and nine have a tens digit \(7.\) Since \(77\) is counted twice, there are \(10 + 9 - 1 = 18\) with at least one \(7.\) The probability is \(\dfrac{18}{90} = \dfrac{1}{5}.\)

Thus, the correct answer is B.

5 答案 A

兑换后得到 \(\dfrac{10d}{7}\) 加元。花掉 \(60\) 后，她有 \(\dfrac{10d}{7} - 60 = d\)。因此 \(\dfrac{3d}{7} = 60\)，所以 \(d = 140\)，数字和为 \(1 + 4 + 0 = 5\)。

因此正确答案是 A。

English

Exchanging gives \(\dfrac{10d}{7}\) Canadian dollars. After spending \(60,\) she has \(\dfrac{10d}{7} - 60 = d.\) Then \(\dfrac{3d}{7} = 60,\) so \(d = 140.\) The sum of its digits is \(1 + 4 + 0 = 5.\)

Thus, the correct answer is A.

6 答案 A

西南和东南方向互相垂直，所以机场位于一个直角三角形的直角顶点，两条直角边长为 \(8\) 和 \(10\)。两个市中心之间的距离为 \(\sqrt{10^2 + 8^2} = \sqrt{164} \approx 12.8\)，最接近 \(13\)。

因此正确答案是 A。

English

Southwest and southeast are perpendicular, so the airport sits at the right angle of a right triangle with legs \(8\) and \(10.\) The distance between downtowns is \(\sqrt{10^2 + 8^2} = \sqrt{164} \approx 12.8,\) closest to \(13.\)

Thus, the correct answer is A.

7 答案 B

正方形面积为 \(100\)，圆面积为 \(100\pi\)。二者重叠部分是位于正方形内的四分之一圆，面积为 \(25\pi\)。并集面积为 \(100 + 100\pi - 25\pi = 100 + 75\pi\)。

因此正确答案是 B。

English

The square has area \(100\) and the circle has area \(100\pi.\) Their overlap is the quarter of the circle lying inside the square, with area \(25\pi.\) The union is \(100 + 100\pi - 25\pi = 100 + 75\pi.\)

Thus, the correct answer is B.

8 答案 D

各排罐头数为 \(1, 3, 5, \ldots, (2n - 1)\)，前 \(n\) 个奇数之和为 \(n^2\)。令 \(n^2 = 100\)，得 \(n = 10\)。

因此正确答案是 D。

English

The rows contain \(1, 3, 5, \ldots, (2n - 1)\) cans, and the sum of the first \(n\) odd numbers is \(n^2.\) Setting \(n^2 = 100\) gives \(n = 10.\)

Thus, the correct answer is D.

9 答案 E

将 \((-3, 2)\) 顺时针旋转 \(90^\circ\)，按 \((x, y) \to (y, -x)\)，得到 \(B = (2, 3)\)。关于 \(y = x\) 反射会交换坐标，所以 \(C = (3, 2)\)。

因此正确答案是 E。

English

Rotating \((-3, 2)\) by \(90^\circ\) clockwise sends \((x, y) \to (y, -x),\) giving \(B = (2, 3).\) Reflecting in \(y = x\) swaps coordinates, giving \(C = (3, 2).\)

Thus, the correct answer is E.

10 答案 A

圆环面积为 \(\pi b^2 - \pi c^2\)。因为 \(\overline{XZ}\) 在 \(Z\) 处与小圆相切，所以它垂直于半径 \(\overline{OZ}\)，于是 \(\triangle OZX\) 在 \(Z\) 处为直角。由此 \(b^2 = c^2 + a^2\)，所以 \(b^2 - c^2 = a^2\)，面积为 \(\pi a^2\)。

因此正确答案是 A。

English

The annulus area is \(\pi b^2 - \pi c^2.\) Because \(\overline{XZ}\) is tangent to the smaller circle at \(Z,\) it is perpendicular to radius \(\overline{OZ},\) so \(\triangle OZX\) is right-angled at \(Z.\) Then \(b^2 = c^2 + a^2,\) giving \(b^2 - c^2 = a^2.\) The area is \(\pi a^2.\)

Thus, the correct answer is A.

11 答案 D

每个 \(100\) 分比平均分 \(76\) 高 \(24\) 分，所以五个学生共高出 \(120\) 分。这些必须由低于平均分的分数抵消，而每个剩余学生最多低 \(76 - 60 = 16\) 分。因此至少需要 \(\dfrac{120}{16} = 7.5\)，即 \(8\) 名额外学生，总数为 \(13\)。五个 \(100\) 分和八个 \(61\) 分可以达到这一情况。

因此正确答案是 D。

English

Each score of \(100\) is \(24\) above the mean, so the five contribute \(120\) points above \(76.\) These must be balanced by points below the mean, and each remaining student is at most \(76 - 60 = 16\) below. So at least \(\dfrac{120}{16} = 7.5,\) hence \(8\) more students are needed, for a total of \(13.\) Five \(100\)s and eight \(61\)s achieve this.

Thus, the correct answer is D.

12 答案 C

递推给出 \(2001\)、\(2002\)、\(2003\)、\(2000\)、\(2005\)、\(1998, \ldots\)。偶数项为 \(2002, 2000, 1998, \ldots\)，每次减少 \(2\)。

更确切地说，用归纳法可以验证 \(a_{2k} = 2004 - 2k\) 且 \(a_{2k+1} = 2001 + 2k\)。因此 \[ \begin{aligned} a_{2004} &= a_{2\cdot1002} \\ &= 2004 - 2(1002) = 0 \end{aligned}\text{。}\]

因此正确答案是 C。

English

The rule gives \(2001,\) \(2002,\) \(2003,\) \(2000,\) \(2005,\) \(1998, \ldots\) The even-indexed terms are \(2002, 2000, 1998, \ldots,\) decreasing by \(2.\)

More precisely, the recurrence verifies inductively that \(a_{2k} = 2004 - 2k\) and \(a_{2k+1} = 2001 + 2k.\) Therefore \[ \begin{aligned} a_{2004} &= a_{2\cdot1002} \\ &= 2004 - 2(1002) = 0. \end{aligned} \]

Thus, the correct answer is C.

13 答案 A

因为 \(f(f^{-1}(x)) = x\)，所以 \(a(bx + a) + b = x\)。比较系数得 \(ab = 1\) 和 \(a^2 + b = 0\)。于是 \(b = \frac{1}{a}\)，且 \(a^2 + \frac{1}{a} = 0\)，所以 \(a^3 = -1\) 得到 \(a = -1\)、\(b = -1\)。因此 \(a + b = -2\)。

因此正确答案是 A。

English

Since \(f(f^{-1}(x)) = x,\) we have \(a(bx + a) + b = x.\) Matching terms gives \(ab = 1\) and \(a^2 + b = 0.\) Then \(b = \frac{1}{a}\) and \(a^2 + \frac{1}{a} = 0,\) so \(a^3 = -1,\) giving \(a = -1\) and \(b = -1.\) Thus \(a + b = -2.\)

Thus, the correct answer is A.

14 答案 D

因为 \(5^2 + 12^2 = 13^2\)，\(\triangle ABC\) 在 \(C\) 处为直角，面积为 \(\tfrac12 (5)(12) = 30\)。小直角三角形 \(\triangle AMJ\) 和 \(\triangle NBK\) 都与 \(\triangle ABC\) 相似，斜边分别为 \(AM = 5 - 4 = 1\) 和 \(BN = 12 - 4 = 8\)。它们的面积为 \(\left(\dfrac{1}{13}\right)^2 (30)\) 和 \(\left(\dfrac{8}{13}\right)^2 (30)\)。

五边形是剩余部分：\[ \begin{gathered} \left(1 - \dfrac{1}{169} - \dfrac{64}{169}\right)(30) \\ {}= \dfrac{104}{169}(30) = \dfrac{240}{13} \end{gathered}\text{。}\]

因此正确答案是 D。

English

Since \(5^2 + 12^2 = 13^2,\) \(\triangle ABC\) is right-angled at \(C\) with area \(\tfrac12 (5)(12) = 30.\) The small right triangles \(\triangle AMJ\) and \(\triangle NBK\) are each similar to \(\triangle ABC,\) with hypotenuses \(AM = 5 - 4 = 1\) and \(BN = 12 - 4 = 8.\) Their areas are \(\left(\dfrac{1}{13}\right)^2 (30)\) and \(\left(\dfrac{8}{13}\right)^2 (30).\)

The pentagon is what remains: \[ \begin{gathered} \left(1 - \dfrac{1}{169} - \dfrac{64}{169}\right)(30) \\ {}= \dfrac{104}{169}(30) = \dfrac{240}{13}. \end{gathered} \]

Thus, the correct answer is D.

15 答案 B

设 Jack 为 \(10x + y\)，Bill 为 \(10y + x\)，则 \(10x + y + 5 = 2(10y + x + 5)\)，所以 \(8x = 19y + 5\)。检验数字，只有 \(y = 1, x = 3\) 可行，因此 Jack 是 \(31\) 岁，Bill 是 \(13\) 岁。差为 \(31 - 13 = 18\)。

因此正确答案是 B。

English

Let Jack be \(10x + y\) and Bill be \(10y + x.\) Then \(10x + y + 5 = 2(10y + x + 5),\) so \(8x = 19y + 5.\) Testing digits, only \(y = 1, x = 3\) works, so Jack is \(31\) and Bill is \(13.\) The difference is \(31 - 13 = 18.\)

Thus, the correct answer is B.

16 答案 C

写 \(z = x + iy\) 则 \(f(z) = i(x - iy) = y + ix\)。令 \(f(z) = z\) 得 \(y = x\)，这是一条过原点的直线。条件 \(|z| = 5\) 是一个圆，过圆心的直线与圆相交于 \(2\) 个点。

因此正确答案是 C。

English

Writing \(z = x + iy,\) we get \(f(z) = i(x - iy) = y + ix.\) Setting \(f(z) = z\) gives \(y = x,\) which is a line through the origin. The condition \(|z| = 5\) is a circle, and a line through the center meets the circle in \(2\) points.

Thus, the correct answer is C.

17 答案 A

以 \(2\) 为底的对数之和为 \(\log_2(r_1 r_2 r_3) = 5\)，所以 \(r_1 r_2 r_3 = 2^5 = 32\)。由 Vieta 公式，方程 \(8x^3 + 4ax^2 + 2bx + a\) 的根的乘积为 \(-\dfrac{a}{8}\)。因此 \(-\dfrac{a}{8} = 32\) 得到 \(a = -256\)。

因此正确答案是 A。

English

The sum of the base-\(2\) logarithms is \(\log_2(r_1 r_2 r_3) = 5,\) so \(r_1 r_2 r_3 = 2^5 = 32.\) By Vieta’s formulas on \(8x^3 + 4ax^2 + 2bx + a,\) the product of the roots is \(-\dfrac{a}{8}.\) Thus \(-\dfrac{a}{8} = 32,\) giving \(a = -256.\)

Thus, the correct answer is A.

18 答案 E

设 \(B = (a, b)\)、\(A = (-a, -b)\)。则 \(4a^2 + 7a - 1 = b\)，且 \(4a^2 - 7a - 1 = -b\)。相减得 \(14a = 2b\)，所以 \(b = 7a\)。再由 \(4a^2 + 7a - 1 = 7a\) 得 \(a^2 = \dfrac14\)，且 \(b^2 = 49a^2 = \dfrac{49}{4}\)。因此 \(AB = 2\sqrt{a^2 + b^2}\) \(= 2\sqrt{\dfrac{50}{4}} = 5\sqrt{2}\)。

因此正确答案是 E。

English

Let \(B = (a, b)\) and \(A = (-a, -b).\) Then \(4a^2 + 7a - 1 = b\) and \(4a^2 - 7a - 1 = -b.\) Subtracting gives \(14a = 2b,\) so \(b = 7a.\) Then \(4a^2 + 7a - 1 = 7a\) gives \(a^2 = \dfrac14,\) and \(b^2 = 49a^2 = \dfrac{49}{4}.\) So \(AB = 2\sqrt{a^2 + b^2}\) \(= 2\sqrt{\dfrac{50}{4}} = 5\sqrt{2}.\)

Thus, the correct answer is E.

19 答案 A

轴截面是梯形 \(ABCD\)，平行边长为 \(2\) 和 \(18\)，且有一个内切圆（球的大圆）。从 \(B\) 和 \(C\) 引出的切线段长度相等，所以斜边 \(BC = 18 + 2 = 20\)。从 \(C\) 向下底作垂线，得到一个水平直角边为 \(18 - 2 = 16\) 的直角三角形，所以高为 \(\sqrt{20^2 - 16^2} = 12\)。球的半径是高的一半，即 \(6\)。

因此正确答案是 A。

English

The axial cross-section is a trapezoid \(ABCD\) with parallel sides \(2\) and \(18\) and an inscribed circle (a great circle of the sphere). By equal tangent lengths from \(B\) and \(C,\) the slant side \(BC = 18 + 2 = 20.\) Dropping a perpendicular from \(C\) to the bottom base gives a right triangle with horizontal leg \(18 - 2 = 16,\) so the height is \(\sqrt{20^2 - 16^2} = 12.\) The sphere’s radius is half the height, \(6.\)

Thus, the correct answer is A.

20 答案 B

共有 \(2^6 = 64\) 种涂色。存在合适摆放方式的情况包括：六个面全同色（\(2\) 种），恰好五个面同色（ \(2 \cdot 6 = 12\) 种），或四个面为一种颜色且另外两个面为另一种颜色并位于相对面（ \(2 \cdot 3 = 6\) 种）。共有 \(2 + 12 + 6 = 20\) 种有利涂色，所以概率为 \(\dfrac{20}{64} = \dfrac{5}{16}\)。

因此正确答案是 B。

English

There are \(2^6 = 64\) colorings. A suitable orientation exists when all six faces are one color (\(2\) ways), exactly five faces are one color (\(2 \cdot 6 = 12\) ways), or four faces are one color with the other color on a pair of opposite faces (\(2 \cdot 3 = 6\) ways). That is \(2 + 12 + 6 = 20\) favorable colorings, so the probability is \(\dfrac{20}{64} = \dfrac{5}{16}.\)

Thus, the correct answer is B.

21 答案 C

斜率 \(a\) 和 \(b\) 是使直线 \(y = mx\) 与椭圆恰好交于一点的 \(m\) 值。代入得 \[ \begin{aligned} &(3m^2 + m + 2)x^2 \\ &\quad {}- (20m + 11)x + 40 = 0\text{。} \end{aligned} \] 令判别式为零，得到 \(-80m^2 + 280m - 199 = 0\)。由 Vieta 公式，\(a + b = \dfrac{280}{80} = \dfrac{7}{2}\)。

因此正确答案是 C。

English

The slopes \(a\) and \(b\) are the values of \(m\) for which \(y = mx\) meets the ellipse in exactly one point. Substituting gives \[ \begin{aligned} &(3m^2 + m + 2)x^2 \\ &\quad {}- (20m + 11)x + 40 = 0. \end{aligned} \] Setting its discriminant to zero yields \(-80m^2 + 280m - 199 = 0.\) By Vieta’s formulas, \(a + b = \dfrac{280}{80} = \dfrac{7}{2}.\)

Thus, the correct answer is C.

22 答案 C

由各行、列、对角线乘积相等，可把每一项都写成关于 \(b\) 的式子：\(h = \dfrac{100}{b}\)、\(g = \dfrac{100}{c}\)、\(f = \dfrac{100}{d}\)。比较行和列得 \(c = \dfrac{20}{b}\)、\(d = \dfrac{4}{b}\)，因此 \(g = 5b\)，且 \(e = 10\)。

所有项都是正整数当且仅当 \(b = 1, 2\)，或 \(4\)，对应 \(g = 5, 10, 20\)。它们的和为 \(35\)。

因此正确答案是 C。

English

From the equal row, column, and diagonal products, every entry can be written in terms of \(b:\) \(h = \dfrac{100}{b},\) \(g = \dfrac{100}{c},\) \(f = \dfrac{100}{d}.\) Comparing rows and columns gives \(c = \dfrac{20}{b}\) and \(d = \dfrac{4}{b},\) hence \(g = 5b\) and \(e = 10.\)

All entries are positive integers exactly when \(b = 1, 2,\) or \(4,\) giving \(g = 5, 10, 20.\) Their sum is \(35.\)

Thus, the correct answer is C.

23 答案 C

设整数零点为 \(a\)。另外两个零点是无理共轭数 \(\dfrac{a}{2} \pm r\)，它们的和 \(a\) 等于整数零点。由 \(x^2\) 项系数的 Vieta 公式，\(a + a = 2004\)，所以 \(a = 1002\)，共轭对为 \(501 \pm r\)。

系数为整数当且仅当 \(r^2\) 是正整数；零点为正且互不相同当 \(1 \le r^2 \le 501^2 - 1 = 251{,}000\)。由于 \(r\) 不能是整数，排除 \(500\) 个平方数 \(r^2 = 1^2, \ldots, 500^2\)，剩下 \(251{,}000 - 500 = 250{,}500\) 个 \(n\) 的值。

因此正确答案是 C。

English

Let the integer zero be \(a.\) The other two zeros are irrational conjugates \(\dfrac{a}{2} \pm r,\) whose sum \(a\) equals the integer zero. Vieta’s formula on the \(x^2\) coefficient gives \(a + a = 2004,\) so \(a = 1002\) and the conjugate pair is \(501 \pm r.\)

The coefficients are integers exactly when \(r^2\) is a positive integer, and the zeros are positive and distinct when \(1 \le r^2 \le 501^2 - 1 = 251{,}000.\) Since \(r\) cannot be an integer, we exclude the \(500\) perfect-square values \(r^2 = 1^2, \ldots, 500^2,\) leaving \(251{,}000 - 500 = 250{,}500\) values of \(n.\)

Thus, the correct answer is C.

24 答案 B

令 \(\angle DBE = \alpha\)、\(\angle DBC = \beta\)。由于 \(\overline{BD}\) 是等腰三角形的高，\(\angle CBE = \alpha - \beta\)，且 \(\angle ABE = \alpha + \beta\)。等比数列给出 \(\tan(\alpha - \beta)\tan(\alpha + \beta) = \tan^2\alpha\)，化简为 \(\tan^2\beta(\tan^4\alpha - 1) = 0\)，所以 \(\tan\alpha = 1\)，\(\alpha = 45^\circ\)。

设 \(DC = a\)、\(BD = b\)，等差数列 \(\cot\angle DBE\)、\(\cot\angle CBE\)、\(\cot\angle DBC\) 变为 \(1, \dfrac{b + a}{b - a}, \dfrac{b}{a}\)，从而推出 \(b = 3a\)。又 \(BE = 10\)，且 \(\angle DBE = 45^\circ\)，所以 \(b = \dfrac{BE}{\sqrt2} = 5\sqrt2\)，于是 \(a = \dfrac{5\sqrt2}{3}\)。

\(\triangle ABC\) 的面积为 \(\tfrac12 (AC)(BD) = ab\) \(= 5\sqrt2 \cdot \dfrac{5\sqrt2}{3}\) \(= \dfrac{50}{3}\)。

因此正确答案是 B。

English

Let \(\angle DBE = \alpha\) and \(\angle DBC = \beta.\) Since \(\overline{BD}\) is the altitude of the isosceles triangle, \(\angle CBE = \alpha - \beta\) and \(\angle ABE = \alpha + \beta.\) The geometric progression gives \(\tan(\alpha - \beta)\tan(\alpha + \beta) = \tan^2\alpha,\) which simplifies to \(\tan^2\beta(\tan^4\alpha - 1) = 0,\) so \(\tan\alpha = 1\) and \(\alpha = 45^\circ.\)

Writing \(DC = a\) and \(BD = b,\) the arithmetic progression \(\cot\angle DBE,\) \(\cot\angle CBE,\) \(\cot\angle DBC\) becomes \(1, \dfrac{b + a}{b - a}, \dfrac{b}{a},\) forcing \(b = 3a.\) With \(BE = 10\) and \(\angle DBE = 45^\circ,\) we get \(b = \dfrac{BE}{\sqrt2} = 5\sqrt2,\) so \(a = \dfrac{5\sqrt2}{3}.\)

The area of \(\triangle ABC\) is \(\tfrac12 (AC)(BD) = ab\) \(= 5\sqrt2 \cdot \dfrac{5\sqrt2}{3}\) \(= \dfrac{50}{3}.\)

Thus, the correct answer is B.

25 答案 B

具有任意固定位数的最小 \(2\) 的幂位于 \(10^k\) 和 \(2\cdot10^k\) 之间，所以首位数字为 \(1\)。因为 \(2^{2004}\) 是以 \(1\) 开头的 \(604\) 位数，它是位数为 \(604\) 的第一个幂。因此集合 \(S\) 对从 \(1\) 到 \(603\) 的每一种位数恰含一个首位为 \(1\) 的数，共 \(603\) 个。

在每个首位为 \(1\) 的幂之后，下一个幂的首位为 \(2\) 或 \(3\)，再下一个幂的首位为 \(4, 5, 6\)，或 \(7\)。所以有 \(603\) 个元素首位为 \(2\) 或 \(3\)，另有 \(603\) 个元素首位从 \(4\) 到 \(7\)，剩余 \(2004 - 3(603) = 195\) 个元素首位为 \(8\) 或 \(9\)。

最后，把首位为 \(8\) 或 \(9\) 的幂除以二，会得到前一个首位为 \(4\) 的幂；把任意首位为 \(4\) 的幂乘以二正好是逆操作。这是一个双射，所以首位数字为 \(4\) 的元素有 \(195\) 个。

所以正确答案是 B。

English

The smallest power of \(2\) with any given digit-count lies between \(10^k\) and \(2\cdot10^k,\) so it has leading digit \(1.\) Because \(2^{2004}\) is a \(604\)-digit number beginning with \(1,\) it is the first \(604\)-digit power. Thus the powers in \(S\) contain exactly one leading-\(1\) number for each digit-count from \(1\) through \(603,\) or \(603\) in all.

After each leading-\(1\) power, the next power leads with \(2\) or \(3,\) and the following power leads with \(4, 5, 6,\) or \(7.\) Hence \(603\) elements lead with \(2\) or \(3,\) another \(603\) lead with \(4\) through \(7,\) and the remaining \(2004 - 3(603) = 195\) lead with \(8\) or \(9.\)

Finally, halving a power that leads with \(8\) or \(9\) produces the preceding power with leading digit \(4,\) and doubling any leading-\(4\) power reverses this. This is a bijection, so there are \(195\) elements with first digit \(4.\)

Thus, the correct answer is B.