2002A AMC 10 真题（P版）

1 答案 D

提出公因式得 \[\dfrac{10^{2000}(1+100)}{2\cdot 10^{2001}}=\dfrac{101}{20}=5.05\text{，}\]它最接近 \(5\)。

所以正确答案是 D。

English

Factoring gives \[\dfrac{10^{2000}(1+100)}{2\cdot 10^{2001}}=\dfrac{101}{20}=5.05,\] which is closest to \(5.\)

Thus, the correct answer is D.

2 答案 C

\[ \begin{aligned} (2,12,9) &= \dfrac{2}{12}+\dfrac{12}{9}+\dfrac{9}{2} \\ &= \dfrac{1}{6}+\dfrac{4}{3}+\dfrac{9}{2} \end{aligned}\text{。}\]通分到分母 \(6\)，得到 \(\dfrac{1+8+27}{6}=\dfrac{36}{6}=6\)。

所以正确答案是 C。

English

\[ \begin{aligned} (2,12,9) &= \dfrac{2}{12}+\dfrac{12}{9}+\dfrac{9}{2} \\ &= \dfrac{1}{6}+\dfrac{4}{3}+\dfrac{9}{2}. \end{aligned} \] Over a denominator of \(6,\) this is \(\dfrac{1+8+27}{6}=\dfrac{36}{6}=6.\)

Thus, the correct answer is C.

3 答案 B

这个四层指数塔有五种加括号方式。其中 \((2^2)^{\left(2^2\right)}\)、\(\left(2^{\left(2^2\right)}\right)^2\)、\(\left(\left(2^2\right)^2\right)^2\) 都等于 \(2^{8}=256\)。另外两种都给出标准值 \(2^{16}=65{,}536\)。

所以只可能得到一个其他值，即 \(256\)。

所以正确答案是 B。

English

There are five ways to parenthesize the tower. Three of them, \((2^2)^{\left(2^2\right)},\) \(\left(2^{\left(2^2\right)}\right)^2,\) and \(\left(\left(2^2\right)^2\right)^2,\) all equal \(2^{8}=256.\) The other two both give the standard value \(2^{16}=65{,}536.\)

So exactly one other value, \(256,\) is possible.

Thus, the correct answer is B.

4 答案 E

取 \(n=1\)。此时 \(m\cdot 1\le m+1\)，也就是 \(m\le m+1\)，对每个正整数 \(m\) 都成立。

所以每个正整数 \(m\) 都符合条件，共有无穷多个。

所以正确答案是 E。

English

Take \(n=1.\) Then \(m\cdot 1\le m+1\) becomes \(m\le m+1,\) which holds for every positive integer \(m.\)

So every positive integer \(m\) works, giving infinitely many.

Thus, the correct answer is E.

5 答案 C

外圈小圆的圆心到中心的距离为 \(2\)，再加上它自己的半径 \(1\)，得到大圆半径为 \(3\)。

大圆面积为 \(9\pi\)，七个单位圆总面积为 \(7\pi\)，所以阴影面积为 \(9\pi-7\pi=2\pi\)。

所以正确答案是 C。

English

The center of a surrounding circle is \(2\) from the center (two radii), and adding its own radius \(1\) gives a large radius of \(3.\)

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

Thus, the correct answer is C.

6 答案 A

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

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

所以正确答案是 A。

English

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

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

Thus, the correct answer is A.

7 答案 A

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

面积之比是 \(\left(\dfrac{r_A}{r_B}\right)^2=\dfrac{4}{9}\)。

所以正确答案是 A。

English

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

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

画出网格线和对角线，把旗帜分成全等的直角三角形。数一数可得蓝色区域有 \(24\) 个，白色区域有 \(24\) 个，红色区域有 \(16\) 个。

因此 \(B=W\)。

所以正确答案是 A。

English

Divide the flag into congruent right triangles by drawing the grid lines and diagonals. Counting gives \(24\) triangles in the blue region, \(24\) in the white region, and \(16\) in the red region.

Hence \(B=W.\)

Thus, the correct answer is A.

9 答案 B

两个方程相加，得到 \[ \begin{gathered} 1001C-2002A+1001B \\ {}+3003A \\ = 1001A+1001B+1001C \\ = 9009 \end{gathered}\text{。}\]

所以 \(A+B+C=9\)，平均数为 \(\dfrac{9}{3}=3\)。

所以正确答案是 B。

English

Adding the equations, \[ \begin{gathered} 1001C-2002A+1001B \\ {}+3003A \\ = 1001A+1001B+1001C \\ = 9009. \end{gathered} \]

So \(A+B+C=9\) and the average is \(\dfrac{9}{3}=3.\)

Thus, the correct answer is B.

10 答案 A

因式分解得 \[ \begin{gathered} (2x+3)\left[(x-4)+(x-6)\right] \\ = (2x+3)(2x-10) \\ = 0 \end{gathered}\text{。}\]

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

所以正确答案是 A。

English

Factoring, \[ \begin{gathered} (2x+3)\left[(x-4)+(x-6)\right] \\ = (2x+3)(2x-10) \\ = 0. \end{gathered} \]

The roots are \(-\dfrac{3}{2}\) and \(5,\) which sum to \(\dfrac{7}{2}.\)

Thus, the correct answer is A.

11 答案 B

文件总大小为 \(3(0.8)+12(0.7)+15(0.4)\) \(=16.8\) 兆字节。任何装有 \(0.8\) 兆字节文件的软盘，只能再放一个 \(0.4\) 兆字节文件，因为 \(0.8+0.7\gt 1.44\)。因此每张这样的软盘至少浪费 \(0.24\) 兆字节，三张共至少浪费 \(0.72\) 兆字节，有效需求至少为 \(16.8+0.72=17.52\) 兆字节，所以至少需要 \(\left\lceil\dfrac{17.52}{1.44}\right\rceil=13\) 张软盘。

这个数可以达到：\(3\) 张各放一个 \(0.8\) 兆字节文件和一个 \(0.4\) 兆字节文件；\(6\) 张各放两个 \(0.7\) 兆字节文件；\(4\) 张各放三个 \(0.4\) 兆字节文件。

所以正确答案是 B。

English

The files need \(3(0.8)+12(0.7)+15(0.4)\) \(=16.8\) mb. On any disk holding a \(0.8\) mb file, only one \(0.4\) mb file fits alongside it (since \(0.8+0.7\gt 1.44\)), leaving at least \(0.24\) mb wasted. Across the three such disks that is at least \(0.72\) mb, so the effective demand is at least \(16.8+0.72=17.52\) mb, requiring at least \(\left\lceil\dfrac{17.52}{1.44}\right\rceil=13\) disks.

This is achievable: \(3\) disks each hold one \(0.8\) file and one \(0.4\) file, \(6\) disks each hold two \(0.7\) files, and \(4\) disks each hold three \(0.4\) files.

Thus, the correct answer is B.

12 答案 B

设准时所需时间为 \(t\) 小时。题中的 \(3\) 分钟等于 \(0.05\) 小时，所以 \(40(t+0.05)=60(t-0.05)\)。化简得 \(40t+2=60t-3\)，从而 \(t=0.25\)。

路程为 \(40(0.30)=12\) 英里，因此准时速度为 \(\dfrac{12}{0.25}=48\) 英里每小时。

所以正确答案是 B。

English

Let \(t\) hours be the on-time travel time. Since \(3\) minutes is \(0.05\) hours, \(40(t+0.05)=60(t-0.05).\) Then \(40t+2=60t-3,\) so \(t=0.25.\)

The distance is \(40(0.30)=12\) miles, so the required speed is \(\dfrac{12}{0.25}=48\) mph.

Thus, the correct answer is B.

13 答案 B

因为 \(15^2+20^2=225+400\) \(=625=25^2\)，这是一个直角三角形，两条直角边为 \(15\) 和 \(20\)，面积为 \(\dfrac{1}{2}(15)(20)=150\)。

最短高对应最长边 \(25\)，长度为 \(\dfrac{2\cdot 150}{25}=12\)。

所以正确答案是 B。

English

Since \(15^2+20^2=225+400\) \(=625=25^2,\) the triangle is right with legs \(15\) and \(20,\) and area \(\dfrac{1}{2}(15)(20)=150.\)

The shortest altitude falls to the longest side \(25,\) and equals \(\dfrac{2\cdot 150}{25}=12.\)

Thus, the correct answer is B.

14 答案 B

设两个质数根为 \(p\)、\(q\)，则 \(p+q=63\)，\(pq=k\)。因为 \(63\) 是奇数，其中一个质数必须是 \(2\)，从而另一个是 \(61\)，它也是质数。

因此 \(k=2\cdot 61=122\)，所以只有这一个可能取值。

所以正确答案是 B。

English

If the roots are primes \(p\) and \(q,\) then \(p+q=63\) and \(pq=k.\) Because \(63\) is odd, one prime must be \(2,\) making the other \(61,\) which is prime.

So \(k=2\cdot 61=122\) is the only possible value.

Thus, the correct answer is B.

15 答案 E

两位质数不能以 \(2\)、\(4\)、\(5\)、\(6\) 结尾，所以这些数字必须是十位数字，而 \(1\)、\(3\)、\(7\)、\(9\) 是个位数字。

四个数的和为 \(10(2+4+5+6)\) \(+(1+3+7+9)\) \(=170+20=190\)。例如 \(\{23,47,59,61\}\) 是一组可行构造。

所以正确答案是 E。

English

A two-digit prime cannot end in \(2,\) \(4,\) \(5,\) or \(6,\) so these four are the tens digits and \(1,\) \(3,\) \(7,\) and \(9\) are the units digits.

The sum is \(10(2+4+5+6)\) \(+(1+3+7+9)\) \(=170+20=190.\) One valid set is \(\{23,47,59,61\}.\)

Thus, the correct answer is E.

16 答案 B

设公共值为 \(k\)。则 \(a=k-1\)，\(b=k-2\)，\(c=k-3\)，\(d=k-4\)，所以 \(a+b+c+d=4k-10\)。

因为 \(a+b+c+d+5=k\)，可得 \(4k-10+5=k\)，所以 \(3k=5\)，\(k=\dfrac{5}{3}\)。于是 \(a+b+c+d=k-5\) \(=\dfrac{5}{3}-5=-\dfrac{10}{3}\)，这就是所求值。

所以正确答案是 B。

English

Let the common value be \(k.\) Then \(a=k-1,\) \(b=k-2,\) \(c=k-3,\) \(d=k-4,\) so \(a+b+c+d=4k-10.\)

Since \(a+b+c+d+5=k,\) we get \(4k-10+5=k,\) so \(3k=5\) and \(k=\dfrac{5}{3}.\) Then \(a+b+c+d=k-5\) \(=\dfrac{5}{3}-5=-\dfrac{10}{3}.\)

Thus, the correct answer is B.

17 答案 D

第一次转移 \(2\) 盎司咖啡后，第 \(1\) 杯有 \(2\) 盎司咖啡，第 \(2\) 杯有 \(2\) 盎司咖啡和 \(4\) 盎司奶油，共 \(6\) 盎司。

从第 \(2\) 杯倒回一半，即 \(3\) 盎司，其中含 \(1\) 盎司咖啡和 \(2\) 盎司奶油，留下第 \(1\) 杯有 \(3\) 盎司咖啡和 \(2\) 盎司奶油。奶油所占比例为 \(\dfrac{2}{2+3}=\dfrac{2}{5}\)。

所以正确答案是 D。

English

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

Transferring back half of cup \(2\) (that is \(3\) oz, consisting of \(1\) oz coffee and \(2\) oz cream) leaves cup \(1\) with \(3\) 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.

18 答案 D

\(8\) 个角上的骰子各露 \(3\) 个面，最小和为 \(1+2+3=6\)，贡献 \(8\cdot 6=48\)。另外 \(12\) 个棱上骰子各露 \(2\) 个面，最小和为 \(1+2=3\)，贡献 \(12\cdot 3=36\)。

\(6\) 个面中心骰子各露 \(1\) 个面，最小是 \(1\)，贡献 \(6\)；内部骰子贡献 \(0\)。总和为 \(48+36+6=90\)。

所以正确答案是 D。

English

The \(8\) corner dice show \(3\) faces each, minimized at \(1+2+3=6,\) contributing \(8\cdot 6=48.\) The \(12\) edge dice show \(2\) faces, minimized at \(1+2=3,\) contributing \(12\cdot 3=36.\)

The \(6\) face-center dice show \(1\) face, minimized at \(1,\) contributing \(6,\) and the hidden interior die contributes \(0.\) The total is \(48+36+6=90.\)

Thus, the correct answer is D.

19 答案 E

在拴点处，狗屋挡住了 \(120^\circ\) 的内角，剩下半径为 \(2\) 的 \(240^\circ\) 扇形，面积为 \(\dfrac{240}{360}\pi(2)^2=\dfrac{8\pi}{3}\)。

绕过两个相邻顶点时，各剩 \(1\) 码绳长，每处扫过 \(60^\circ\) 扇形，总面积为 \(2\cdot\dfrac{60}{360}\pi(1)^2=\dfrac{\pi}{3}\)。总面积为 \(\dfrac{8\pi}{3}+\dfrac{\pi}{3}=3\pi\)。

所以正确答案是 E。

English

At the tether vertex the hexagon blocks its \(120^\circ\) interior angle, leaving a \(240^\circ\) sector of radius \(2:\) area \(\dfrac{240}{360}\pi(2)^2=\dfrac{8\pi}{3}.\)

Wrapping around each of the two adjacent vertices, \(1\) yard of rope remains and sweeps a \(60^\circ\) sector: \(2\cdot\dfrac{60}{360}\pi(1)^2=\dfrac{\pi}{3}.\) The total is \(\dfrac{8\pi}{3}+\dfrac{\pi}{3}=3\pi.\)

Thus, the correct answer is E.

20 答案 D

因为 \(HC\parallel AG\)，所以 \(\triangle DHC\sim\triangle DGA\)，于是 \(\dfrac{HC}{AG}=\dfrac{DC}{DA}=\dfrac{1}{3}\)，即 \(HC=\dfrac{AG}{3}\)。

因为 \(JE\parallel AG\)，所以 \(\triangle FJE\sim\triangle FGA\)，于是 \(\dfrac{JE}{AG}=\dfrac{FE}{FA}=\dfrac{1}{5}\)，即 \(JE=\dfrac{AG}{5}\)。

因此 \(\dfrac{HC}{JE}=\dfrac{\frac{AG}{3}}{\frac{AG}{5}}=\dfrac{5}{3}\)。

所以正确答案是 D。

English

Since \(HC\parallel AG,\) \(\triangle DHC\sim\triangle DGA,\) so \(\dfrac{HC}{AG}=\dfrac{DC}{DA}=\dfrac{1}{3},\) giving \(HC=\dfrac{AG}{3}.\)

Since \(JE\parallel AG,\) \(\triangle FJE\sim\triangle FGA,\) so \(\dfrac{JE}{AG}=\dfrac{FE}{FA}=\dfrac{1}{5},\) giving \(JE=\dfrac{AG}{5}.\)

Therefore \(\dfrac{HC}{JE}=\dfrac{\frac{AG}{3}}{\frac{AG}{5}}=\dfrac{5}{3}.\)

Thus, the correct answer is D.

21 答案 D

八个整数的和是 \(8\cdot 8=64\)。数列 \(6\)、\(6\)、\(6\)、\(8\)、\(8\)、\(8\)、\(8\)、\(14\) 的平均数、中位数、唯一众数和极差都等于 \(8\)，所以最大值 \(14\) 可以达到。

若最大值至少为 \(16\)，极差条件会使最小值至少为 \(8\)。平均数为 \(8\) 就会迫使八个整数全都等于 \(8\)，与极差条件矛盾。

若最大值为 \(15\)，极差 \(8\) 会使最小值为 \(7\)，所以八个整数都至少为 \(7\)。其余七个数的和是 \(64-15=49=7\cdot 7\)，迫使它们全都等于 \(7\)。但这样中位数和众数会是 \(7\)，而不是 \(8\)，矛盾。

所以正确答案是 D。

English

The sum is \(8\cdot 8=64.\) 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.

If the largest were at least \(16,\) the range condition would make the smallest at least \(8.\) A mean of \(8\) would then force all eight integers to equal \(8,\) contradicting the range.

If the largest were \(15,\) the range \(8\) forces the smallest to be \(7,\) so all eight integers are at least \(7.\) The other seven then sum to \(64-15=49=7\cdot 7,\) forcing every one of them to equal \(7.\) But then the median and mode would be \(7,\) not \(8,\) a contradiction.

Thus, the correct answer is D.

22 答案 C

从 \(n^2\) 块瓷砖开始，一次操作会移除 \(n\) 个完全平方编号，剩 \(n^2-n\) 块。下一次操作会移除 \(n-1\) 个完全平方编号，剩 \(n^2-n-(n-1)=(n-1)^2\) 块。

因此每两次操作会把 \(n^2\) 减到 \((n-1)^2\)。从 \(10^2=100\) 减到 \(1^2=1\)，需要 \(2(10-1)=18\) 次操作。

所以正确答案是 C。

English

Starting from \(n^2\) tiles, one operation removes the \(n\) perfect squares, leaving \(n^2-n.\) The next operation removes \(n-1\) perfect squares, leaving \(n^2-n-(n-1)=(n-1)^2.\)

So every two operations reduce \(n^2\) to \((n-1)^2.\) Going from \(10^2=100\) down to \(1^2=1\) takes \(2(10-1)=18\) operations.

Thus, the correct answer is C.

23 答案 D

设 \(M\) 是 \(BC\) 的中点。因为 \(BE=CE\)，所以 \(EM\perp BC\)，且 \(EM=\sqrt{10^2-6^2}=8\)。由对称性 \(AE=ED\)；设 \(AB=CD=x\)、\(AE=ED=y\)。

由周长条件，\(2y+(2x+12)\) \(=2(10+10+12)=64\)，所以 \(x+y=26\)。勾股定理又给出 \(y^2=EM^2+(x+6)^2\) \(=64+(x+6)^2\)。

代入 \(y=26-x\)，得到 \((26-x)^2=64+(x+6)^2\)，即 \(676-52x=100+12x\)，所以 \(64x=576\)，从而 \(x=9\)。

所以正确答案是 D。

English

Let \(M\) be the midpoint of \(BC.\) Since \(BE=CE,\) \(EM\perp BC\) and \(EM=\sqrt{10^2-6^2}=8.\) By symmetry \(AE=ED;\) write \(AB=CD=x\) and \(AE=ED=y.\)

The perimeter condition gives \(2y+(2x+12)\) \(=2(10+10+12)=64,\) so \(x+y=26.\) Also \(y^2=EM^2+(x+6)^2\) \(=64+(x+6)^2.\)

Substituting \(y=26-x,\) \((26-x)^2=64+(x+6)^2,\) which simplifies to \(676-52x=100+12x,\) so \(64x=576\) and \(x=9.\)

Thus, the correct answer is D.

24 答案 A

Tina 的 \(10\) 个等可能数对给出的和为 \(3\)、\(4\)、\(5\)、\(5\)、\(6\)、\(6\)、\(7\)、\(7\)、\(8\)、\(9\)。若和为 \(s\)，Sergio 的数大于它的概率是 \(\dfrac{10-s}{10}\)。

对十个数对取平均，成功概率为 \(\small\dfrac{7+6+5+5+4+4+3+3+2+1}{100}\) \(=\dfrac{40}{100}=\dfrac{2}{5}\)。

所以正确答案是 A。

English

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

Averaging the winning probability over the ten pairs, the total is \(\small\dfrac{7+6+5+5+4+4+3+3+2+1}{100}\) \(=\dfrac{40}{100}=\dfrac{2}{5}.\)

Thus, the correct answer is A.

25 答案 C

延长 \(DA\) 和 \(CB\)，交于 \(P\)。因为 \(DC\parallel AB\)，所以 \(\triangle PDC\sim\triangle PAB\)，相似比为 \(\dfrac{39}{52}=\dfrac{3}{4}\)。于是 \(\dfrac{PD}{PD+5}=\dfrac{3}{4}\)，得 \(PD=15\)，同理 \(PC=36\)。

于是 \(PD:PC:DC=15:36:39\) \(=3\cdot(5:12:13)\)，所以 \(\angle P\) 是直角。梯形 \(ABCD\) 的面积为 \[ \begin{gathered} \dfrac{1}{2}(PA)(PB) \\ {}-\dfrac{1}{2}(PD)(PC) \\ = \dfrac{1}{2}(20)(48) \\ {}-\dfrac{1}{2}(15)(36) \\ = 480-270=210 \end{gathered}\text{。}\]

所以正确答案是 C。

English

Extend \(DA\) and \(CB\) to meet at \(P.\) Since \(DC\parallel AB,\) \(\triangle PDC\sim\triangle PAB\) with ratio \(\dfrac{39}{52}=\dfrac{3}{4}.\) From \(\dfrac{PD}{PD+5}=\dfrac{3}{4}\) we get \(PD=15,\) and similarly \(PC=36.\)

Then \(PD:PC:DC=15:36:39\) \(=3\cdot(5:12:13),\) so \(\angle P\) is a right angle. The area of \(ABCD\) is \[ \begin{gathered} \dfrac{1}{2}(PA)(PB) \\ {}-\dfrac{1}{2}(PD)(PC) \\ = \dfrac{1}{2}(20)(48) \\ {}-\dfrac{1}{2}(15)(36) \\ = 480-270=210. \end{gathered} \]

Thus, the correct answer is C.