2004A AMC 10 真题（P版）

1 答案 A

包括你在内共有 \(6\) 个人平均分担筹款。每人需要筹集 \[ \dfrac{1500}{6} = 250 \] 美元。

所以正确答案是 A。

English

Including you, there are \(6\) people sharing the fundraising equally. Each must raise \[ \dfrac{1500}{6} = 250 \] dollars.

Thus, the correct answer is A.

2 答案 B

三个内层表达式的值为 \[ \begin{aligned} \diamond(1,2,3) &= \dfrac{1}{2-3} = -1, \\ \diamond(2,3,1) &= \dfrac{2}{3-1} = 1, \\ \diamond(3,1,2) &= \dfrac{3}{1-2} = -3 \end{aligned}\text{。}\]

因此 \[ \begin{aligned} \diamond(-1, 1, -3) &= \dfrac{-1}{1 - (-3)} \\ &= -\dfrac{1}{4} \end{aligned}\text{。}\]

所以正确答案是 B。

English

The inner values are \[ \begin{aligned} \diamond(1,2,3) &= \dfrac{1}{2-3} = -1, \\ \diamond(2,3,1) &= \dfrac{2}{3-1} = 1, \\ \diamond(3,1,2) &= \dfrac{3}{1-2} = -3. \end{aligned} \]

Therefore \[ \begin{aligned} \diamond(-1, 1, -3) &= \dfrac{-1}{1 - (-3)} \\ &= -\dfrac{1}{4}. \end{aligned} \]

Thus, the correct answer is B.

3 答案 E

\(20\) 美元等于 \(2000\) 美分，地方税为每小时 \[ 0.0145 \times 2000 = 29 \] 美分。

所以正确答案是 E。

English

Since $\(20\) equals \(2000\) cents, the local tax is \[ 0.0145 \times 2000 = 29 \] cents per hour.

Thus, the correct answer is E.

4 答案 D

由于 \(|x - 1|\) 和 \(|x - 2|\) 分别表示 \(x\) 到 \(1\) 和 \(2\) 的距离，所以点 \(x\) 到 \(1\) 和 \(2\) 的距离相等。

到一和二距离相等的点是它们的中点：\[ x = \dfrac{1 + 2}{2} = \dfrac{3}{2}\text{。}\]

所以正确答案是 D。

English

Since \(|x - 1|\) and \(|x - 2|\) are the distances from \(x\) to \(1\) and \(2,\) the point \(x\) is equidistant from \(1\) and \(2.\)

That midpoint is \[ x = \dfrac{1 + 2}{2} = \dfrac{3}{2}. \]

Thus, the correct answer is D.

5 答案 C

三点集合总数为 \[ \binom{9}{3} = 84\text{。}\]

共线三点包括 \(3\) 行、\(3\) 列和 \(2\) 条主对角线，共 \(8\) 组。

因此所求概率为 \[ \dfrac{8}{84} = \dfrac{2}{21}\text{。}\]

所以正确答案是 C。

English

The number of three-point sets is \[ \binom{9}{3} = 84. \]

The collinear triples are the \(3\) rows, the \(3\) columns, and the \(2\) main diagonals, for a total of \(8.\)

The probability is therefore \[ \dfrac{8}{84} = \dfrac{2}{21}. \]

Thus, the correct answer is C.

6 答案 E

Bertha 有 \(30 - 6 = 24\) 个外孙女，这些外孙女都没有女儿。

她们来自 \(\frac{24}{6} = 4\) 个 Bertha 的女儿，所以正好 \(4\) 人有女儿，没有女儿的人数为 \[ 30 - 4 = 26\text{。}\]

所以正确答案是 E。

English

Bertha has \(30 - 6 = 24\) granddaughters, none of whom have daughters.

These granddaughters belong to \(\frac{24}{6} = 4\) of Bertha’s daughters. So exactly \(4\) women have daughters, and the number with no daughters is \[ 30 - 4 = 26. \]

Thus, the correct answer is E.

7 答案 C

共有五层，每层都比下面一层少一行且少一列。橙子总数为 \[ \begin{aligned} &5\cdot 8 + 4\cdot 7 + 3\cdot 6 \\ &\quad {}+ 2\cdot 5 + 1\cdot 4 \\ &= 40 + 28 + 18 + 10 + 4 = 100 \end{aligned}\text{。}\]

所以正确答案是 C。

English

There are five layers, each one shorter and narrower than the one below. The total number of oranges is \[ \begin{aligned} &5\cdot 8 + 4\cdot 7 + 3\cdot 6 \\ &\quad {}+ 2\cdot 5 + 1\cdot 4 \\ &= 40 + 28 + 18 + 10 + 4 = 100. \end{aligned} \]

Thus, the correct answer is C.

8 答案 B

前三轮后，筹码数从 \((15, 14, 13)\) 变为 \((14, 13, 12)\)。

一般地，每三轮后每位玩家都减少一枚筹码。\(36\) 轮后筹码数为 \((3, 2, 1)\)。第 \(37\) 轮中，领先者给出三枚筹码后降到 \(0\)，游戏结束。

所以正确答案是 B。

English

After the first three rounds the counts go from \((15, 14, 13)\) to \((14, 13, 12).\) In general, every three rounds each player loses exactly one token.

After \(36\) rounds the counts are \((3, 2, 1).\) On the \(37\)th round the leader gives away three tokens and drops to \(0,\) ending the game.

Thus, the correct answer is B.

9 答案 B

设 \([ABD]\) 是两个大三角形共有的面积。则 \([ABE] = [ADE] + [ABD]\)，且 \([ABC] = [BDC] + [ABD]\)。

相减得 \[ \begin{aligned} &[ADE] - [BDC] \\ &= [ABE] - [ABC] \end{aligned}\text{。}\]由于 \(\angle EAB\) 和 \(\angle ABC\) 都是直角，\[ \begin{aligned} [ABE] &= \tfrac12(4)(8) = 16, \\ [ABC] &= \tfrac12(4)(6) = 12 \end{aligned}\text{。}\]

因此差为 \(16 - 12 = 4\)。

所以正确答案是 B。

English

Let \([ABD]\) be the area shared by both large triangles. Then \([ABE] = [ADE] + [ABD]\) and \([ABC] = [BDC] + [ABD].\)

Subtracting, \[ \begin{aligned} &[ADE] - [BDC] \\ &= [ABE] - [ABC]. \end{aligned} \] Since \(\angle EAB\) and \(\angle ABC\) are right angles, \[ \begin{aligned} [ABE] &= \tfrac12(4)(8) = 16, \\ [ABC] &= \tfrac12(4)(6) = 12. \end{aligned} \]

The difference is \(16 - 12 = 4.\)

Thus, the correct answer is B.

10 答案 D

两者正面数相同可能为 \(0\)、\(1\)、\(2\) 或 \(3\)。硬币 \(A\) 的权重为 \(1, 3, 3, 1\)，总数为 \(8\)；硬币 \(B\) 的权重为 \(1, 4, 6, 4, 1\)，总数为 \(16\)。

所以概率为 \[ \begin{aligned} &\dfrac{1\cdot 1 + 3\cdot 4 + 3\cdot 6 + 1\cdot 4}{8 \cdot 16} \\ &= \dfrac{35}{128} \end{aligned}\text{。}\]

所以正确答案是 D。

English

The two coins match when both show \(0,\) \(1,\) \(2,\) or \(3\) heads. Coin \(A\) has weights \(1, 3, 3, 1\) out of \(8\) and coin \(B\) has weights \(1, 4, 6, 4, 1\) out of \(16.\)

The probability is \[ \begin{aligned} &\dfrac{1\cdot 1 + 3\cdot 4 + 3\cdot 6 + 1\cdot 4}{8 \cdot 16} \\ &= \dfrac{35}{128}. \end{aligned} \]

Thus, the correct answer is D.

11 答案 C

保持 \(\pi r^2 h\) 不变，而半径乘以 \(1.25\)，则高度必须乘以 \[ \dfrac{1}{1.25^2} = \dfrac{1}{1.5625} = 0.64\text{。}\]

因此高度变为原来的 \(64\%\)，减少了 \(36\%\)。

所以正确答案是 C。

English

Keeping \(\pi r^2 h\) constant while multiplying the radius by \(1.25\) requires the height to be multiplied by \[ \dfrac{1}{1.25^2} = \dfrac{1}{1.5625} = 0.64. \]

So the height becomes \(64\%\) of the original, a decrease of \(36\%.\)

Thus, the correct answer is C.

12 答案 C

\(8\) 种配料各自独立选择是否加入，共有 \(2^8 = 256\) 种配料组合。

每种组合还有 \(3\) 种肉饼数量选择，所以汉堡种类为 \[ 3 \times 256 = 768\text{。}\]

所以正确答案是 C。

English

Each of the \(8\) condiments is independently in or out, giving \(2^8 = 256\) condiment combinations.

For each of these there are \(3\) choices of patty count, so the number of hamburgers is \[ 3 \times 256 = 768. \]

Thus, the correct answer is C.

13 答案 D

从男士角度数，跳舞配对数为 \(12 \cdot 3 = 36\)。每位女士恰好在 \(2\) 对中，所以女士人数为 \[ \dfrac{36}{2} = 18\text{。}\]

所以正确答案是 D。

English

The number of dancing pairs is \(12 \cdot 3 = 36,\) counting from the men’s side. Each woman was in exactly \(2\) pairs, so the number of women is \[ \dfrac{36}{2} = 18. \]

Thus, the correct answer is D.

14 答案 A

若有 \(n\) 枚硬币，总面值为 \(20n\) 美分。加一枚二十五分硬币后，方程为 \[ 20n + 25 = 21(n + 1)\text{，}\]解得 \(n = 4\)。

如果二十五分硬币至多有两枚，其余硬币每枚最多值 \(10\) 美分，那么四枚硬币的总值至多为 \(2(25)+2(10)=70\) 美分。因此必须有三枚二十五分硬币，第四枚硬币的面值为 \(5\) 美分。钱包中有三枚二十五分硬币和一枚五分硬币，所以十分硬币有 \(0\) 枚。

所以正确答案是 A。

English

With \(n\) coins the total value is \(20n\) cents. Adding a quarter gives \[ 20n + 25 = 21(n + 1), \] so \(n = 4.\)

If there were at most two quarters, the other coins would be worth at most \(10\) cents each, so four coins would total at most \(2(25)+2(10)=70\) cents. Thus there must be three quarters, leaving \(5\) cents for the fourth coin. The purse contains three quarters and one nickel, so it has \(0\) dimes.

Thus, the correct answer is A.

15 答案 D

写成 \(\dfrac{x + y}{x} = 1 + \dfrac{y}{x}\)。这里 \(\dfrac{y}{x} \lt 0\)，所以表达式最大时 \(\left|\dfrac{y}{x}\right|\) 最小。

这在 \(y = 2\)、\(x = -4\) 时发生，\[ 1 + \dfrac{2}{-4} = 1 - \dfrac{1}{2} = \dfrac{1}{2}\text{。}\]

所以正确答案是 D。

English

Write \(\dfrac{x + y}{x} = 1 + \dfrac{y}{x}.\) Here \(\dfrac{y}{x} \lt 0,\) so the expression is largest when \(\left|\dfrac{y}{x}\right|\) is smallest.

That happens with \(y = 2\) and \(x = -4,\) giving \[ 1 + \dfrac{2}{-4} = 1 - \dfrac{1}{2} = \dfrac{1}{2}. \]

Thus, the correct answer is D.

16 答案 D

所有 \(5\times5\)、\(4\times4\)、\(3\times3\) 的正方形都包含中心方格，它们共有 \[ 1^2 + 2^2 + 3^2 = 14 \] 个。

较小的正方形中，有 \(4\) 个 \(2\times2\) 正方形和 \(1\) 个 \(1\times1\) 正方形覆盖中心方格，于是共有 \[ 14 + 4 + 1 = 19\text{。} \]

所以正确答案是 D。

English

Every \(5\times5,\) \(4\times4,\) and \(3\times3\) square contains the center cell, and there are \[ 1^2 + 2^2 + 3^2 = 14 \] of them.

Among the smaller squares, \(4\) of the \(2\times2\) squares and \(1\) of the \(1\times1\) squares cover the center, giving \[ 14 + 4 + 1 = 19. \]

Thus, the correct answer is D.

17 答案 C

第一次相遇前，两人合计跑了半圈。两次相遇之间，两人合计跑了一整圈，是前一段总路程的两倍，所以 Brenda 在这段中跑了 \(2 \cdot 100 = 200\) 米。

Sally 在同一段中跑了 \(150\) 米，所以跑道全长为 \[ 200 + 150 = 350\text{。}\]

所以正确答案是 C。

English

Before the first meeting the two together cover half the track. Between the first and second meetings they together cover a full track, which is twice as far, so Brenda runs \(2 \cdot 100 = 200\) meters in that stretch.

Sally runs \(150\) meters in the same stretch, so the full track length is \[ 200 + 150 = 350. \]

Thus, the correct answer is C.

18 答案 A

等差数列为 \(9\)、\(9 + d\)、\(9 + 2d\)，所以新的三个数为 \(9\)、\(11 + d\)、\(29 + 2d\)。

等比条件给出 \[ (11 + d)^2 = 9(29 + 2d)\text{，}\]化简得 \(d^2 + 4d - 140 = 0\)，所以 \(d = 10\) 或 \(d = -14\)。

第三项 \(29 + 2d\) 分别为 \(49\) 和 \(1\)。最小是 \(1\)。

所以正确答案是 A。

English

The arithmetic progression is \(9,\) \(9 + d,\) \(9 + 2d,\) so the geometric progression is \(9,\) \(11 + d,\) \(29 + 2d.\)

The geometric condition gives \[ (11 + d)^2 = 9(29 + 2d), \] which simplifies to \(d^2 + 4d - 140 = 0,\) so \(d = 10\) or \(d = -14.\)

The third terms \(29 + 2d\) are \(49\) and \(1.\) The smallest is \(1.\)

Thus, the correct answer is A.

19 答案 C

将条纹展开后，它成为一个平行四边形。其底边（水平宽度）为 \(3\) 英尺，高跨过整个筒仓，为 \(80\) 英尺。

因此面积为 \[ 3 \times 80 = 240 \] 平方英尺。

所以正确答案是 C。

English

Unrolling the stripe flattens it into a parallelogram. Its base (the horizontal width) is \(3\) feet and its height spans the full \(80\) feet of the silo.

The area is therefore \[ 3 \times 80 = 240 \] square feet.

Thus, the correct answer is C.

20 答案 D

设正方形边长为 \(1\)，\(ED = DF = x\)，\(AE = 1 - x\)。

由于 \(\triangle BEF\) 是等边三角形，\(EF^2 = EB^2\)，所以 \[ 2x^2 = 1 + (1 - x)^2\text{，}\]化简得 \(x^2 = 2(1 - x)\)。

此外 \([DEF] = \tfrac12 x^2\)，\([ABE] = \tfrac12(1 - x)\)，所以 \[ \begin{aligned} \dfrac{[DEF]}{[ABE]} &= \dfrac{x^2}{1 - x} \\ &= \dfrac{2(1 - x)}{1 - x} = 2 \end{aligned}\text{。}\]

所以正确答案是 D。

English

Let the square have side \(1,\) and by symmetry let \(ED = DF = x,\) so \(AE = 1 - x.\)

Since \(\triangle BEF\) is equilateral, \(EF^2 = EB^2,\) giving \[ 2x^2 = 1 + (1 - x)^2, \] which simplifies to \(x^2 = 2(1 - x).\)

The right triangles have areas \([DEF] = \tfrac12 x^2\) and \([ABE] = \tfrac12(1 - x),\) so \[ \begin{aligned} \dfrac{[DEF]}{[ABE]} &= \dfrac{x^2}{1 - x} \\ &= \dfrac{2(1 - x)}{1 - x} = 2. \end{aligned} \]

Thus, the correct answer is D.

21 答案 B

设锐角为 \(\theta\)。阴影区域分三部分：单位圆中的两个锐角扇形总面积为 \(\theta\)；半径 \(1\) 到 \(2\) 的圆环中两个钝角扇形总面积为 \(3(\pi - \theta)\)；半径 \(2\) 到 \(3\) 的圆环中两个锐角扇形总面积为 \(5\theta\)。

相加得到阴影面积 \[ \theta + 3(\pi - \theta) + 5\theta = 3\pi + 3\theta\text{。}\]

阴影面积是非阴影面积的 \(\dfrac{8}{13}\)，因此是总面积 \(9\pi\) 的 \(\dfrac{8}{21}\)。于是 \[ 3\pi + 3\theta = \dfrac{8}{21}(9\pi) = \dfrac{24\pi}{7}\text{，}\]解得 \(\theta = \dfrac{\pi}{7}\)。

所以正确答案是 B。

English

Let \(\theta\) be the acute angle. The shaded region has three parts: two acute sectors of the unit disk with total area \(\theta,\) two obtuse sectors of the ring between radii \(1\) and \(2\) with total area \(3(\pi - \theta),\) and two acute sectors of the ring between radii \(2\) and \(3\) with total area \(5\theta.\)

Adding these gives a shaded area of \[ \theta + 3(\pi - \theta) + 5\theta = 3\pi + 3\theta. \]

The shaded region is \(\dfrac{8}{13}\) of the unshaded region, so it is \(\dfrac{8}{21}\) of the total area \(9\pi.\) Then \[ 3\pi + 3\theta = \dfrac{8}{21}(9\pi) = \dfrac{24\pi}{7}, \] which gives \(\theta = \dfrac{\pi}{7}.\)

Thus, the correct answer is B.

22 答案 D

设 \(F\) 是 \(CE\) 与半圆的切点，令 \(x = AE\)。从同一点作圆的切线长度相等，所以 \(CF = CB = 2\)，且 \(EF = EA = x\)，于是 \(CE = 2 + x\)。

在直角三角形 \(CDE\) 中，\(DE = 2 - x\)、\(DC = 2\)，所以 \[ (2 - x)^2 + 2^2 = (2 + x)^2\text{。}\]解得 \(x = \dfrac{1}{2}\)，于是 \(CE = 2 + \dfrac{1}{2} = \dfrac{5}{2}\)。

所以正确答案是 D。

English

Let \(F\) be the point where \(CE\) touches the semicircle and let \(x = AE.\) Since tangents from a point are equal, \(CF = CB = 2\) and \(EF = EA = x,\) so \(CE = 2 + x.\)

In right triangle \(CDE,\) we have \(DE = 2 - x\) and \(DC = 2,\) so \[ (2 - x)^2 + 2^2 = (2 + x)^2. \] This gives \(x = \dfrac{1}{2},\) hence \(CE = 2 + \dfrac{1}{2} = \dfrac{5}{2}.\)

Thus, the correct answer is D.

23 答案 D

因为圆 \(A\) 经过圆 \(D\) 的圆心，并且与圆 \(D\) 内切，所以圆 \(D\) 的半径为 \(2\)。把圆 \(D\) 的圆心放在原点，圆 \(A\) 的圆心放在 \((-1, 0)\)。

设圆 \(B\) 的半径为 \(r\)，圆心为 \((x, r)\)；由圆 \(B\) 和圆 \(C\) 关于水平轴对称，可得切线关系 \[ \begin{aligned} (x + 1)^2 + r^2 &= (1 + r)^2, \\ x^2 + r^2 &= (2 - r)^2 \end{aligned}\text{。}\]

两式相减得 \(x = 3r - 2\)。代入第二式得 \(9r^2 - 8r = 0\)，所以 \(r = \dfrac{8}{9}\)。

所以正确答案是 D。

English

Because circle \(A\) passes through \(D\)’s center and is internally tangent to \(D,\) circle \(D\) has radius \(2.\) Place \(D\)’s center at the origin and \(A\)’s center at \((-1, 0).\)

Let circle \(B\) have radius \(r\) and center \((x, r),\) using the symmetry of \(B\) and \(C\) about the horizontal axis. Tangency gives \[ \begin{aligned} (x + 1)^2 + r^2 &= (1 + r)^2, \\ x^2 + r^2 &= (2 - r)^2. \end{aligned} \]

Subtracting yields \(x = 3r - 2.\) Substituting into the second equation gives \(9r^2 - 8r = 0,\) so \(r = \dfrac{8}{9}.\)

Thus, the correct answer is D.

24 答案 D

反复使用递推式，得到 \[ \begin{aligned} a_{2^1} &= 2^0, \\ a_{2^2} &= 2^1, \\ a_{2^3} &= 2^{1+2}, \\ a_{2^4} &= 2^{1+2+3}, \ldots \end{aligned} \] 因此一般地，\(a_{2^n} = 2^{1 + 2 + \cdots + (n - 1)} = 2^{\frac{n(n-1)}{2}}\)。

当 \(n = 100\) 时，指数为 \(\dfrac{100 \cdot 99}{2} = 4950\)，所以 \(a_{2^{100}} = 2^{4950}\)。

所以正确答案是 D。

English

Applying the rule repeatedly, \[ \begin{aligned} a_{2^1} &= 2^0, \\ a_{2^2} &= 2^1, \\ a_{2^3} &= 2^{1+2}, \\ a_{2^4} &= 2^{1+2+3}, \ldots \end{aligned} \] so in general \(a_{2^n} = 2^{1 + 2 + \cdots + (n - 1)} = 2^{\frac{n(n-1)}{2}}.\)

For \(n = 100,\) the exponent is \(\dfrac{100 \cdot 99}{2} = 4950,\) so \(a_{2^{100}} = 2^{4950}.\)

Thus, the correct answer is D.

25 答案 B

三个小球球心形成边长为 \(2\) 的等边三角形，每个球心距平面 \(1\)。设其重心为 \(D\)，则它到每个顶点的距离为 \(\dfrac{2\sqrt{3}}{3}\)。

大球球心 \(E\) 位于 \(D\) 正上方，且 \(E\) 到每个小球球心的距离为 \(1 + 2 = 3\)，所以 \[ \begin{aligned} DE &= \sqrt{3^2 - \left(\dfrac{2\sqrt{3}}{3}\right)^2} \\ &= \sqrt{9 - \dfrac{4}{3}} = \dfrac{\sqrt{69}}{3} \end{aligned}\text{。}\]

再加上平面到 \(D\) 的 \(1\) 个单位，以及从 \(E\) 到大球顶部的 \(2\) 个单位，得到 \[ 1 + \dfrac{\sqrt{69}}{3} + 2 = 3 + \dfrac{\sqrt{69}}{3}\text{。}\]

所以正确答案是 B。

English

The three small centers form an equilateral triangle of side \(2,\) each \(1\) unit above the plane. Its centroid \(D\) is at distance \(\dfrac{2\sqrt{3}}{3}\) from each vertex.

The large sphere’s center \(E\) sits directly above \(D,\) and the distance between \(E\) and a small center is \(1 + 2 = 3.\) Thus \[ \begin{aligned} DE &= \sqrt{3^2 - \left(\dfrac{2\sqrt{3}}{3}\right)^2} \\ &= \sqrt{9 - \dfrac{4}{3}} = \dfrac{\sqrt{69}}{3}. \end{aligned} \]

Adding the \(1\) unit from the plane to \(D\) and the \(2\) units from \(E\) to the top of the large sphere gives \[ 1 + \dfrac{\sqrt{69}}{3} + 2 = 3 + \dfrac{\sqrt{69}}{3}. \]

Thus, the correct answer is B.