2004 AMC 10A 真题与在线限时测验

计时

1:15:00

1.

你和五位朋友需要为慈善机构筹集 $1500$ 美元的捐款，并且平均分担筹款任务。你们每个人需要筹集多少美元？

You and five friends need to raise $ $1500$ in donations for a charity, dividing the fundraising equally. How many dollars will each of you need to raise?

$250$

$300$

$1500$

$7500$

$9000$

答案：A

知识点：钱币分数

难度评级：450

小提示：

包括你在内，先数一共有多少人分担任务。

Including you, count the total number of people sharing the work

大提示：

将 $1500$ 美元平均分给 $6$ 个人。

Divide $ $1500$ evenly among the $6$ people

解答：

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。

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.

对任意三个实数 $a$ 、 $b$ 、 $c$ ，其中 $b \neq c$ ，定义运算 $\diamond$ 为 $\diamond(a, b, c) = \dfrac{a}{b - c}\text{。}$ 求 $\diamond(\diamond(1, 2, 3), \diamond(2, 3, 1), \diamond(3, 1, 2))$ 的值。

For any three real numbers $a,$ $b,$ and $c,$ with $b \neq c,$ the operation $\diamond$ is defined by $\diamond(a, b, c) = \dfrac{a}{b - c}.$ What is $\diamond(\diamond(1, 2, 3), \diamond(2, 3, 1), \diamond(3, 1, 2))?$

$-\dfrac{1}{2}$

$-\dfrac{1}{4}$

$0$

$\dfrac{1}{4}$

$\dfrac{1}{2}$

答案：B

知识点：自定义运算分数

难度评级：980

小提示：

先计算三个内层表达式。

Evaluate the three inner expressions before combining them

大提示：

分别计算 $\diamond(1,2,3) = -1$ 、 $\diamond(2,3,1) = 1$ 、 $\diamond(3,1,2) = -3$ 。

$\diamond(1,2,3) = -1,$ $\diamond(2,3,1) = 1,$ and $\diamond(3,1,2) = -3$

解答：

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。

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.

Alicia 每小时赚 $20$ 美元，其中 $1.45\%$ 会被扣除用于缴纳地方税。她每小时工资中有多少美分用于缴纳地方税？

Alicia earns $ $20$ per hour, of which $1.45\%$ is deducted to pay local taxes. How many cents per hour of Alicia’s wages are used to pay local taxes?

$0.0029$

$0.029$

$0.29$

$2.9$

$29$

答案：E

知识点：百分数单位换算

难度评级：870

小提示：

先将 $20$ 美元转换为美分，再取百分比。

Convert $ $20$ into cents before taking the percentage

大提示：

计算 $2000$ 美分的 $1.45\%$ 。

Compute $1.45\%$ of $2000$ cents

解答：

$20$ 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。

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

Thus, the correct answer is E.

4.

若 $|x - 1| = |x - 2|$ ，则 $x$ 的值是多少？

What is the value of $x$ if $|x - 1| = |x - 2|?$

$-\dfrac{1}{2}$

$\dfrac{1}{2}$

$1$

$\dfrac{3}{2}$

$2$

答案：D

知识点：绝对值中点

难度评级：1030

小提示：

$|x - a|$ 表示 $x$ 到 $a$ 的距离。

$|x - a|$ measures the distance from $x$ to $a$

大提示：

点 $x$ 必须到 $1$ 和 $2$ 的距离相等。

The point $x$ must be equally far from $1$ and $2$

解答：

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。

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.

从图中网格点随机选取三个点，每组三点被选中的概率相同。这三个点在同一直线上的概率是多少？

A set of three points is chosen randomly from the grid shown. Each three-point set has the same probability of being chosen. What is the probability that the points lie on the same straight line?

$\dfrac{1}{21}$

$\dfrac{1}{14}$

$\dfrac{2}{21}$

$\dfrac{1}{7}$

$\dfrac{2}{7}$

答案：C

知识点：基本概率组合格点

难度评级：1240

小提示：

等可能的三点集合共有 $\binom{9}{3}$ 个。

There are $\binom{9}{3}$ equally likely three-point sets

大提示：

数共线三点： $3$ 行、 $3$ 列和 $2$ 条对角线。

Count the collinear triples: $3$ rows, $3$ columns, and $2$ diagonals

解答：

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。

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.

Bertha 有 $6$ 个女儿，没有儿子。她的一些女儿各有 $6$ 个女儿，其余女儿没有女儿。Bertha 的女儿和外孙女总共有 $30$ 人，并且没有曾外孙女。Bertha 的女儿和外孙女中，有多少人没有女儿？

Bertha has $6$ daughters and no sons. Some of her daughters have $6$ daughters, and the rest have none. Bertha has a total of $30$ daughters and granddaughters, and no great-granddaughters. How many of Bertha’s daughters and granddaughters have no daughters?

$22$

$23$

$24$

$25$

$26$

答案：E

知识点：基本计数逻辑推理

难度评级：1170

小提示：

Bertha 有 $30 - 6 = 24$ 个外孙女，而且她们都没有女儿。

Bertha has $30 - 6 = 24$ granddaughters, and none of them have daughters

大提示：

外孙女每 $6$ 人来自一个女儿，求有多少个女儿是母亲。

The granddaughters come in groups of $6,$ so find how many daughters are mothers

解答：

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。

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.

一位杂货商把橙子堆成类似金字塔的形状，长方形底层为 $5$ 个橙子乘 $8$ 个橙子。第一层以上的每个橙子都放在下一层四个橙子形成的凹处。最上层是一排橙子。这个橙子堆共有多少个橙子？

A grocer stacks oranges in a pyramid-like stack whose rectangular base is $5$ oranges by $8$ oranges. Each orange above the first level rests in a pocket formed by four oranges in the level below. The stack is completed by a single row of oranges. How many oranges are in the stack?

$96$

$98$

$100$

$101$

$134$

答案：C

知识点：基本计数求和

难度评级：1100

小提示：

每往上一层，两个方向的橙子数都比下一层少一。

Each layer up has one fewer orange in each dimension than the layer below

大提示：

求和 $5\cdot 8 + 4\cdot 7 + 3\cdot 6 + 2\cdot 5 + 1\cdot 4$ 。

Sum $5\cdot 8 + 4\cdot 7 + 3\cdot 6 + 2\cdot 5 + 1\cdot 4$

解答：

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。

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.

一个游戏按如下规则使用筹码：每轮中，筹码最多的玩家给其他每位玩家各一枚筹码，并且再把一枚筹码放入弃置堆。当某位玩家筹码用完时游戏结束。玩家 $A$ 、 $B$ 、 $C$ 分别以 $15$ 、 $14$ 、 $13$ 枚筹码开始。游戏会进行多少轮？

A game is played with tokens according to the following rule. In each round, the player with the most tokens gives one token to each of the other players and also places one token into a discard pile. The game ends when some player runs out of tokens. Players $A,$ $B,$ and $C$ start with $15,$ $14,$ and $13$ tokens, respectively. How many rounds will there be in the game?

$36$

$37$

$38$

$39$

$40$

答案：B

知识点：过程模拟找规律

难度评级：1390

小提示：

模拟前几轮并寻找循环规律。

Simulate a few rounds and look for a repeating pattern

大提示：

每三轮后，每位玩家的筹码数都正好减少一。

Every three rounds, each player’s total drops by exactly one

解答：

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。

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.

图中， $\angle EAB$ 和 $\angle ABC$ 都是直角， $AB = 4$ 、 $BC = 6$ 、 $AE = 8$ ，且 $\overline{AC}$ 与 $\overline{BE}$ 交于 $D$ 。 $\triangle ADE$ 和 $\triangle BDC$ 的面积之差是多少？

In the figure, $\angle EAB$ and $\angle ABC$ are right angles, $AB = 4,$ $BC = 6,$ $AE = 8,$ and $\overline{AC}$ and $\overline{BE}$ intersect at $D.$ What is the difference between the areas of $\triangle ADE$ and $\triangle BDC?$

$2$

$4$

$5$

$8$

$9$

答案：B

知识点：三角形面积面积分割

难度评级：1330

小提示：

向两个小三角形都加上 $\triangle ABD$ ，会得到两个较大的三角形。

Adding $\triangle ABD$ to each of the two triangles produces two larger triangles

大提示：

减去共有面积后， $[ADE] - [BDC]$ 等于 $[ABE] - [ABC]$ 。

Subtracting the shared area makes $[ADE] - [BDC]$ equal to $[ABE] - [ABC]$

解答：

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。

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.

硬币 $A$ 抛三次，硬币 $B$ 抛四次。两枚公平硬币所得到的正面次数相同的概率是多少？

Coin $A$ is flipped three times and coin $B$ is flipped four times. What is the probability that the number of heads obtained from flipping the two fair coins is the same?

$\dfrac{19}{128}$

$\dfrac{23}{128}$

$\dfrac{1}{4}$

$\dfrac{35}{128}$

$\dfrac{1}{2}$

答案：D

知识点：二项概率分类讨论

难度评级：1470

小提示：

对每个可能的共同正面数 $k$ ，把 $P(A = k)\,P(B = k)$ 相加。

Add up $P(A = k)\,P(B = k)$ over each possible common count $k$

大提示：

$A$ 的计数权重为 $1, 3, 3, 1$ ， $B$ 的计数权重为 $1, 4, 6, 4, 1$ 。

The counts for $A$ are weighted $1, 3, 3, 1$ and for $B$ are $1, 4, 6, 4, 1$

解答：

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。

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.

一家公司用圆柱形罐子销售花生酱。市场研究表明，使用更宽的罐子会增加销量。如果罐子的直径增加 $25\%$ ，而体积不变，那么高度必须减少百分之多少？

A company sells peanut butter in cylindrical jars. Marketing research suggests that using wider jars will increase sales. If the diameter of the jars is increased by $25\%$ without altering the volume, by what percent must the height be decreased?

$10$

$25$

$36$

$50$

$60$

答案：C

知识点：圆柱体积百分数

难度评级：1310

小提示：

体积 $\pi r^2 h$ 不变，所以 $r^2 h$ 保持不变。

The volume $\pi r^2 h$ is unchanged, so $r^2 h$ stays fixed

大提示：

半径乘以 $1.25$ 时， $r^2$ 会乘以 $1.25^2$ 。

Multiplying the radius by $1.25$ multiplies $r^2$ by $1.25^2$

解答：

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。

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.

Henry’s Hamburger Heaven 提供以下汉堡配料：番茄酱、芥末酱、蛋黄酱、番茄、生菜、腌黄瓜、奶酪和洋葱。顾客可以选择一块、两块或三块肉饼，并可选择任意一组配料。共有多少种不同的汉堡可以点？

Henry’s Hamburger Heaven offers its hamburgers with the following condiments: ketchup, mustard, mayonnaise, tomato, lettuce, pickles, cheese, and onions. A customer can choose one, two, or three meat patties, and any collection of condiments. How many different kinds of hamburgers can be ordered?

$24$

$256$

$768$

$40{,}320$

$120{,}960$

答案：C

知识点：乘法原理子集

难度评级：1190

小提示：

$8$ 种配料中每一种都可以选择或不选择。

Each of the $8$ condiments is independently included or left out

大提示：

将 $2^8$ 种配料选择乘以 $3$ 种肉饼数量选择。

Multiply the $2^8$ condiment choices by the $3$ choices of patty count

解答：

$8$ 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。

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.

在一次聚会上，每位男士正好与三位女士跳舞，每位女士正好与两位男士跳舞。共有十二位男士参加聚会。共有多少位女士参加？

At a party, each man danced with exactly three women and each woman danced with exactly two men. Twelve men attended the party. How many women attended the party?

$8$

$12$

$16$

$18$

$24$

答案：D

知识点：双重计数数对计数

难度评级：1190

小提示：

用两种方式数男女跳舞配对数。

Count the man-woman dancing pairs in two ways

大提示：

从男士角度有 $12 \cdot 3$ 对，每位女士贡献 $2$ 对。

There are $12 \cdot 3$ pairs, and each woman accounts for $2$ of them

解答：

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。

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.

Paula 钱包中所有一分、五分、十分和二十五分硬币的平均面值是 $20$ 美分。如果她再多一枚二十五分硬币，平均面值会变为 $21$ 美分。她钱包中有多少枚十分硬币？

The average value of all the pennies, nickels, dimes, and quarters in Paula’s purse is $20$ cents. If she had one more quarter, the average value would be $21$ cents. How many dimes does she have in her purse?

$0$

$1$

$2$

$3$

$4$

答案：A

知识点：平均数钱币一次方程

难度评级：1450

小提示：

若她有 $n$ 枚硬币，总面值为 $20n$ 美分。

If she has $n$ coins, their total value is $20n$ cents

大提示：

加一枚二十五分硬币后， $20n + 25 = 21(n + 1)$ 。

Adding a quarter gives $20n + 25 = 21(n + 1)$

解答：

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。

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.

已知 $-4 \le x \le -2$ ，且 $2 \le y \le 4$ ，求下式的最大可能值： $\dfrac{x + y}{x}\text{？}$

Given that $-4 \le x \le -2$ and $2 \le y \le 4,$ what is the largest possible value of $\dfrac{x + y}{x}?$

$-1$

$-\dfrac{1}{2}$

$0$

$\dfrac{1}{2}$

$1$

答案：D

知识点：不等式最优化代数变形

难度评级：1420

小提示：

改写为 $\dfrac{x + y}{x} = 1 + \dfrac{y}{x}$ 。

Rewrite $\dfrac{x + y}{x} = 1 + \dfrac{y}{x}$

大提示：

因为 $\dfrac{y}{x} \lt 0$ ，要让它的绝对值尽可能小。

Since $\dfrac{y}{x} \lt 0,$ make its absolute value as small as possible

解答：

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。

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.

图中的 $5 \times 5$ 网格包含从 $1 \times 1$ 到 $5 \times 5$ 的各种正方形。有多少个这样的正方形包含阴影中心方格？

The $5 \times 5$ grid shown contains a collection of squares with sizes from $1 \times 1$ to $5 \times 5.$ How many of these squares contain the shaded center square?

$12$

$15$

$17$

$19$

$20$

答案：D

知识点：图形中的形状计数分类讨论

难度评级：1480

小提示：

每个 $3\times3$ 、 $4\times4$ 和 $5\times5$ 的正方形都包含中心方格。

Every $3\times3,$ $4\times4,$ and $5\times5$ square contains the center

大提示：

再数有多少个 $2\times2$ 和 $1\times1$ 的正方形覆盖中心方格。

Then count how many $2\times2$ and $1\times1$ squares cover the center cell

解答：

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。

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.

Brenda 和 Sally 从圆形跑道上一对直径相对的点出发，沿相反方向跑。她们第一次相遇时 Brenda 跑了 $100$ 米。下一次相遇时，Sally 已经从第一次相遇点又跑了 $150$ 米。两人都以恒定速度跑。跑道长多少米？

Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run $100$ meters. They next meet after Sally has run $150$ meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters?

$250$

$300$

$350$

$400$

$500$

答案：C

知识点：相对速度路程、速度与时间比与比例

难度评级：1540

小提示：

从相对点出发，到第一次相遇前两人合计跑了半圈。

Starting from opposite points, together they cover half the track before their first meeting

大提示：

利用恒定的速度比，比较 Brenda 第一次相遇前与两次相遇之间所跑的路程。

Use the constant speed ratio to compare Brenda’s first-meeting distance with her distance between meetings

解答：

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。

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.

三个实数组成一个等差数列，第一项为 $9$ 。若第二项加 $2$ ，第三项加 $20$ ，所得三个数形成等比数列。该等比数列第三项的最小可能值是多少？

A sequence of three real numbers forms an arithmetic progression with a first term of $9.$ If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term of the geometric progression?

$1$

$4$

$36$

$49$

$81$

答案：A

知识点：等差数列等比数列二次方程

难度评级：1630

小提示：

将等差数列写成 $9$ 、 $9 + d$ 、 $9 + 2d$ ，则等比数列为 $9$ 、 $11 + d$ 、 $29 + 2d$ 。

Write the progression as $9,$ $9 + d,$ $9 + 2d,$ so the geometric one is $9,$ $11 + d,$ $29 + 2d$

大提示：

使用 $(11 + d)^2 = 9(29 + 2d)$ ，并选择使第三项较小的 $d$ 。

Use $(11 + d)^2 = 9(29 + 2d)$ and take the value of $d$ giving the smaller third term

解答：

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。

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.

一个白色圆柱形筒仓直径为 $30$ 英尺，高为 $80$ 英尺。如图，筒仓上涂了一条水平宽度为 $3$ 英尺的红色条纹，绕筒仓完整转了两圈。条纹面积是多少平方英尺？

A white cylindrical silo has a diameter of $30$ feet and a height of $80$ feet. A red stripe with a horizontal width of $3$ feet is painted on the silo, as shown, making two complete revolutions around it. What is the area of the stripe in square feet?

$120$

$180$

$240$

$360$

$480$

答案：C

知识点：圆柱展开图（立体几何）平行四边形

难度评级：1600

小提示：

想象把筒仓上的条纹剪下并展开成平面图形。

Imagine cutting the stripe from the silo and unrolling it flat

大提示：

它会变成一个水平宽度为 $3$ 、高为 $80$ 的平行四边形。

It becomes a parallelogram with horizontal width $3$ and height $80$

解答：

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。

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.

点 $E$ 和 $F$ 位于正方形 $ABCD$ 上，使得 $\triangle BEF$ 是等边三角形。 $\triangle DEF$ 的面积与 $\triangle ABE$ 的面积之比是多少？

Points $E$ and $F$ are located on square $ABCD$ so that $\triangle BEF$ is equilateral. What is the ratio of the area of $\triangle DEF$ to that of $\triangle ABE?$

$\dfrac{4}{3}$

$\dfrac{3}{2}$

$\sqrt{3}$

$2$

$1 + \sqrt{3}$

答案：D

知识点：等边三角形面积比勾股定理

难度评级：1790

小提示：

设正方形边长为 $1$ ，并设 $ED = DF = x$ 。

Let the square have side $1$ and set $ED = DF = x$

大提示：

由 $EF^2 = EB^2$ ，得到 $2x^2 = 1 + (1 - x)^2$ 。

From $EF^2 = EB^2,$ get $2x^2 = 1 + (1 - x)^2$

解答：

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。

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.

两条不同直线通过三个同心圆的圆心，三个圆的半径分别为 $3$ 、 $2$ 、 $1$ 。图中阴影区域面积是非阴影区域面积的 $\dfrac{8}{13}$ 。两条直线所成锐角的弧度数是多少？（注： $\pi$ 弧度等于 $180$ 度。）

Two distinct lines pass through the center of three concentric circles of radii $3,$ $2,$ and $1.$ The area of the shaded region in the diagram is $\dfrac{8}{13}$ of the area of the unshaded region. What is the radian measure of the acute angle formed by the two lines? (Note: $\pi$ radians is $180$ degrees.)

$\dfrac{\pi}{8}$

$\dfrac{\pi}{7}$

$\dfrac{\pi}{6}$

$\dfrac{\pi}{5}$

$\dfrac{\pi}{4}$

答案：B

知识点：扇形圆面积圆环一次方程

难度评级：1880

小提示：

设锐角为 $\theta$ ，把每块阴影区域表示为扇形面积。

Let $\theta$ be the acute angle and express each shaded piece as a sector area

大提示：

阴影面积总和为 $3\pi + 3\theta$ ，总面积为 $9\pi$ 。

The shaded pieces total $3\pi + 3\theta,$ out of the whole area $9\pi$

解答：

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。

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.

正方形 $ABCD$ 的边长为 $2$ 。在正方形内部以 $\overline{AB}$ 为直径作一个半圆，从 $C$ 向该半圆作切线，切线与边 $\overline{AD}$ 交于 $E$ 。求 $\overline{CE}$ 的长度。

Square $ABCD$ has side length $2.$ A semicircle with diameter $\overline{AB}$ is constructed inside the square, and the tangent to the semicircle from $C$ intersects side $\overline{AD}$ at $E.$ What is the length of $\overline{CE}?$

$\dfrac{2 + \sqrt{5}}{2}$

$\sqrt{5}$

$\sqrt{6}$

$\dfrac{5}{2}$

$5 - \sqrt{5}$

答案：D

知识点：切线勾股定理

难度评级：1790

小提示：

从同一点作圆的两条切线长度相等： $CF = CB$ ， $EF = EA$ 。

Tangents from a point have equal length: $CF = CB$ and $EF = EA$

大提示：

设 $AE = x$ ，对 $\triangle CDE$ 使用勾股定理： $(2 - x)^2 + 2^2 = (2 + x)^2$ 。

With $AE = x,$ apply the Pythagorean theorem to $\triangle CDE:$ $(2 - x)^2 + 2^2 = (2 + x)^2$

解答：

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。

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.

圆 $A$ 、 $B$ 、 $C$ 两两外切，并且都与圆 $D$ 内切。圆 $B$ 和圆 $C$ 全等。圆 $A$ 半径为 $1$ ，并经过圆 $D$ 的圆心。圆 $B$ 的半径是多少？

Circles $A,$ $B,$ and $C$ are externally tangent to each other and internally tangent to circle $D.$ Circles $B$ and $C$ are congruent. Circle $A$ has radius $1$ and passes through the center of $D.$ What is the radius of circle $B?$

$\dfrac{2}{3}$

$\dfrac{\sqrt{3}}{2}$

$\dfrac{7}{8}$

$\dfrac{8}{9}$

$\dfrac{1 + \sqrt{3}}{3}$

答案：D

知识点：相切圆坐标几何勾股定理

难度评级：1990

小提示：

因为圆 $A$ 经过圆 $D$ 的圆心，并且与该圆内切，所以圆 $D$ 的半径为 $2$ 。

Since $A$ passes through $D$ ’s center and is internally tangent, circle $D$ has radius $2$

大提示：

建立坐标；半径为 $r$ 的圆 $B$ 的圆心到 $A$ 的圆心距离为 $1 + r$ ，到 $D$ 的圆心距离为 $2 - r$ 。

Place the centers on coordinates; circle $B$ of radius $r$ has center at distance $1 + r$ from $A$ and $2 - r$ from $D$

解答：

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。

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.

设 $a_1$ 、 $a_2$ 、 $\ldots$ 是满足以下性质的数列： $a_1 = 1$ ，且对任意正整数 $n$ ，有 $a_{2n} = n \cdot a_n$ 。 $a_{2^{100}}$ 的值是多少？

Let $a_1,$ $a_2,$ $\ldots$ be a sequence with the following properties: $a_1 = 1,$ and $a_{2n} = n \cdot a_n$ for any positive integer $n.$ What is the value of $a_{2^{100}}?$

$1$

$2^{99}$

$2^{100}$

$2^{4950}$

$2^{9999}$

答案：D

知识点：递推三角形数找规律

难度评级：2010

小提示：

计算 $a_{2^1}$ 、 $a_{2^2}$ 、 $a_{2^3}$ 、 $a_{2^4}$ ，并跟踪 $2$ 的指数。

Compute $a_{2^1},$ $a_{2^2},$ $a_{2^3},$ $a_{2^4}$ and track the exponent of $2$

大提示：

指数为 $0$ 、 $1$ 、 $1 + 2$ 、 $1 + 2 + 3$ 、 $\ldots$ 。

The exponents are $0,$ $1,$ $1 + 2,$ $1 + 2 + 3,$ $\ldots$

解答：

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。

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.

三个两两相切、半径为 $1$ 的球放在水平平面上。一个半径为 $2$ 的球放在它们上面。从平面到较大球顶部的距离是多少？

Three mutually tangent spheres of radius $1$ rest on a horizontal plane. A sphere of radius $2$ rests on them. What is the distance from the plane to the top of the larger sphere?

$3 + \dfrac{\sqrt{30}}{2}$

$3 + \dfrac{\sqrt{69}}{3}$

$3 + \dfrac{\sqrt{123}}{4}$

$\dfrac{52}{9}$

$3 + 2\sqrt{2}$

答案：B

知识点：立体几何球重心勾股定理

难度评级：2180

小提示：

三个小球的球心形成边长为 $2$ 的等边三角形，且高度为 $1$ 。

The three small centers form an equilateral triangle of side $2$ at height $1$

大提示：

大球球心位于该三角形重心正上方；重心到每个小球球心的水平距离为 $\dfrac{2\sqrt{3}}{3}$ ，两球心距离为 $1 + 2 = 3$ 。

The big center sits above the triangle’s centroid; the centroid is $\dfrac{2\sqrt{3}}{3}$ from each small center, and the slant distance between centers is $1 + 2 = 3$

解答：

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。

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.