2013 AMC 10A 详解 | LIVE by Po-Shen Loh

1.

乘出租车的费用为起步价 $\$1.50$ ，加上每英里 $\$0.25$ 。乘坐 $5$ 英里出租车要花多少钱？

A taxi ride costs $\$1.50$ plus $\$0.25$ per mile traveled. How much does a $5$ -mile taxi ride cost?

$\$2.25$

$\$2.50$

$\$2.75$

$\$3.00$

$\$3.25$

知识点：速率钱币

难度评级：450

小提示：

先计算里程费用。

Compute the mileage charge first

大提示：

再加上固定起步价。

Add the fixed starting charge after the mileage charge

解答：

里程费用为 $\$0.25\cdot 5=\$1.25$ 。

加上起步价，费用为 $\$1.50+\$1.25=\$2.75$ 。

所以正确答案是 C。

The mileage charge is $\$0.25\cdot 5=\$1.25$ .

Adding the fixed charge gives $\$1.50+\$1.25=\$2.75$ .

Thus, C is the correct answer.

2.

Alice 正在做一批饼干，需要 $2\frac{1}{2}$ 杯糖。可惜她的量杯每次只能装 $\frac{1}{4}$ 杯糖。她必须装满这个量杯多少次，才能得到所需的糖量？

Alice is making a batch of cookies and needs $2\frac{1}{2}$ cups of sugar. Unfortunately, her measuring cup holds only $\frac{1}{4}$ cup of sugar. How many times must she fill that cup to get the correct amount of sugar?

$8$

$10$

$12$

$16$

$20$

知识点：分数

难度评级：560

小提示：

先把带分数化成假分数。

Convert the mixed number of cups to an improper fraction

大提示：

用需要的糖量除以量杯容量。

Divide the needed sugar by the cup size

解答：

Alice 需要 $2\frac12=\frac52$ 杯糖。

每次量杯装 $\frac14$ 杯，所以需要 $\frac52\div\frac14=\frac52\cdot4=10$ 次。

所以正确答案是 B。

Alice needs $2\frac12=\frac52$ cups of sugar.

Each full measuring cup gives $\frac14$ cup, so the number of fillings is $\frac52\div\frac14=\frac52\cdot4=10$ .

Thus, B is the correct answer.

3.

正方形 $ABCD$ 的边长为 $10$ 。点 $E$ 在 $\overline{BC}$ 上，且 $\triangle ABE$ 的面积为 $40$ 。 $BE$ 是多少？

Square $ABCD$ has side length $10.$ Point $E$ is on $\overline{BC},$ and the area of $\triangle ABE$ is $40.$ What is $BE?$

$4$

$5$

$6$

$7$

$8$

知识点：三角形面积

难度评级：770

小提示：

把 $AB$ 当作 $\triangle ABE$ 的高。

Use $AB$ as the height of $\triangle ABE$

大提示：

令 $\frac12\cdot AB\cdot BE$ 等于给定面积。

Set $\frac12\cdot AB\cdot BE$ equal to the given area

解答：

由三角形面积公式， $40 = \dfrac{1}{2} \cdot 10 \cdot BE\text{。}$ 因此 $40 = 5BE$ $BE = 8\text{。}$ 所以正确答案是 E。

We have by the formula for the area of a triangle that $40 = \dfrac{1}{2} \cdot 10 \cdot BE.$ This gives us $40 = 5BE$ $BE = 8.$ Thus, E is the correct answer.

4.

一支垒球队打了十场比赛，得分分别为 $1$ 、 $2$ 、 $3$ 、 $4$ 、 $5$ 、 $6$ 、 $7$ 、 $8$ 、 $9$ 和 $10$ 分。他们恰好有五场输一分。在其余每场比赛中，他们的得分是对手的两倍。对手总共得了多少分？

A softball team played ten games, scoring $1,$ $2,$ $3,$ $4,$ $5,$ $6,$ $7,$ $8,$ $9,$ and $10$ runs. They lost by one run in exactly five games. In each of their other games, they scored twice as many runs as their opponent. How many total runs did their opponents score?

$35$

$40$

$45$

$50$

$55$

知识点：奇偶性分类讨论

难度评级：1070

小提示：

得分是对手两倍的比赛中，本队得分必须是偶数。

The doubled-score games must be the even scoring games

大提示：

偶数得分对应对手得一半；奇数得分对应输一分。

Pair odd scores with one-run losses and even scores with halved opponent scores

解答：

若本队得分是对手两倍，则本队得分必须为偶数。

因此得 $2, 4, 6, 8,$ 和 $10$ 分的比赛中，对手分别得 $1, 2, 3, 4,$ 和 $5$ 分。

这些分数的和为 $1 + 2 + 3 + 4 + 5 = 15\text{。}$

其余比赛中，对手分别得 $2, 4, 6, 8,$ 和 $10$ 分。

这些分数的和为 $2 + 4 + 6 + 8 + 10 = 30\text{。}$

对手总得分为 $15 + 30 = 45\text{。}$ 所以正确答案是 C。

Note that if they scored twice as many runs as their opponents, then they scored an even number of runs.

This means in the games where they scored $2, 4, 6, 8,$ and $10$ runs, their opponents scored $1, 2, 3, 4,$ and $5$ runs respectively.

This sums to $1 + 2 + 3 + 4 + 5 = 15.$

In the other games, their opponents scored $2, 4, 6, 8,$ and $10$ runs.

This sums to $2 + 4 + 6 + 8 + 10 = 30.$

The total number of runs is then $15 + 30 = 45.$ Thus, C is the correct answer.

5.

Tom、Dorothy 和 Sammy 一起度假，并同意平均分摊费用。旅途中 Tom 支付了 $\$105$ ，Dorothy 支付了 $\$125$ ，Sammy 支付了 $\$175$ 。为了平均分摊，Tom 给 Sammy $t$ 美元，Dorothy 给 Sammy $d$ 美元。 $t-d$ 是多少？

Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid $\$105,$ Dorothy paid $\$125,$ and Sammy paid $\$175.$ In order to share costs equally, Tom gave Sammy $t$ dollars, and Dorothy gave Sammy $d$ dollars. What is $t-d?$

$15$

$20$

$25$

$30$

$35$

知识点：平均数钱币

难度评级：900

小提示：

先求总费用和每个人应承担的份额。

First find the total trip cost and each person’s fair share

大提示：

Tom 和 Dorothy 各把低于应付份额的差额给 Sammy。

Tom and Dorothy each pay Sammy the amount they are short of a fair share

解答：

总费用为 $\$105+\$125+\$175=\$405$ ，所以每人应承担 $\$405\div3=\$135$ 。

Tom 少付 $\$30$ ，Dorothy 少付 $\$10$ ，因此 $t=30$ 且 $d=10$ 。

所以 $t-d=30-10=20$ 。

所以正确答案是 B。

The total cost was $\$105+\$125+\$175=\$405$ , so each person’s share was $\$405\div3=\$135$ .

Tom paid $\$30$ less than his share, and Dorothy paid $\$10$ less than her share, so $t=30$ and $d=10$ .

Therefore $t-d=30-10=20$ .

Thus, B is the correct answer.

6.

Joey 和他的五个兄弟年龄分别为 $3$ 、 $5$ 、 $7$ 、 $9$ 、 $11$ 和 $13$ 。一天下午，两个年龄和为 $16$ 的兄弟去看电影，两个小于 $10$ 岁的兄弟去打棒球，Joey 和 $5$ 岁的兄弟留在家里。Joey 几岁？

Joey and his five brothers are ages $3,$ $5,$ $7,$ $9,$ $11,$ and $13.$ One afternoon two of his brothers whose ages sum to $16$ went to the movies, two brothers younger than $10$ went to play baseball, and Joey and the $5$ -year-old stayed home. How old is Joey?

$3$

$7$

$9$

$11$

$13$

知识点：逻辑推理分类讨论

难度评级：1140

小提示：

列出年龄和为 $16$ 的配对，并记得 $5$ 岁的人留在家里。

List the age pairs whose sum is $16$ , remembering the $5$ -year-old stayed home

大提示：

选定看电影的两人后，打棒球的必须是剩下两个小于 $10$ 岁的兄弟。

After choosing the movie pair, the baseball pair must be two remaining brothers younger than $10$

解答：

年龄和为 $16$ 的配对有 $(3,13)$ 、 $(5,11)$ 、 $(7,9)$ 。

$5$ 岁的人留在家里，所以看电影的不能是 $(5,11)$ 。若 $7$ 岁和 $9$ 岁的兄弟去看电影，则剩下小于 $10$ 岁的只有 $3$ 岁的兄弟，不够两人去打棒球。

因此看电影的是 $(3,13)$ ，打棒球的是 $(7,9)$ ，剩下的 Joey 是 $11$ 岁。

所以正确答案是 D。

The age pairs that sum to $16$ are $(3,13)$ , $(5,11)$ , and $(7,9)$ .

The $5$ -year-old stayed home, so the movie pair cannot be $(5,11)$ . If $7$ and $9$ went to the movies, then the only remaining brother younger than $10$ would be the $3$ -year-old, not enough for baseball.

Thus the movie pair was $(3,13)$ , the baseball pair was $(7,9)$ , and Joey is the remaining brother, age $11$ .

Thus, D is the correct answer.

7.

一名学生必须从英语、代数、几何、历史、美术和拉丁语中选择四门课。课程组合必须包含英语，并且至少包含一门数学课。有多少种选择方式？

A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?

$6$

$8$

$9$

$12$

$16$

知识点：组合补集计数

难度评级：1140

小提示：

英语固定，再从剩下五门课中选三门。

English is fixed, so choose three of the five remaining courses

大提示：

先数所有选法，再去掉没有数学课的选法。

Count all three-course choices from the remaining courses, then remove the choice with no math

解答：

英语必须选，所以还要从代数、几何、历史、美术、拉丁语这 $5$ 门中选 $3$ 门。

这样的选法共有 $\binom53=10$ 种，但其中只有历史、美术、拉丁语这一种不含数学课。

因此有效选法为 $10-1=9$ 种。

所以正确答案是 C。

English is required, so choose the other $3$ courses from the $5$ courses Algebra, Geometry, History, Art, and Latin.

There are $\binom53=10$ such choices, but one of them, History-Art-Latin, contains no mathematics course.

Therefore the number of valid programs is $10-1=9$ .

Thus, C is the correct answer.

8.

求下列式子的值： $\dfrac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}}\text{？}$

What is the value of $\dfrac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}} ?$

$-1$

$1$

$\dfrac{5}{3}$

$2013$

$2^{4024}$

知识点：指数因式分解

难度评级：870

小提示：

提取较小的幂 $2^{2012}$ 。

Factor out the smaller power $2^{2012}$

大提示：

提取后括号中只剩 $2^2$ 。

After factoring, only $2^2$ remains in each parenthesis

解答：

提取 $2^{2012}$ ， $\dfrac{2^{2012}(2^2 + 1)}{2^{2012}(2^2 - 1)} = \dfrac{5}{3}\text{。}$

所以正确答案是 C。

Factoring out a $2^{2012},$ we get: $\dfrac{2^{2012}(2^2 + 1)}{2^{2012}(2^2 - 1)} = \dfrac{5}{3}.$

Thus, C is the correct answer.

9.

在最近一场篮球比赛中，Shenille 只投三分球和两分球。她三分球命中率为 $20\%$ ，两分球命中率为 $30\%$ 。Shenille 共出手 $30$ 次。她得了多少分？

In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on $20\%$ of her three-point shots and $30\%$ of her two-point shots. Shenille attempted $30$ shots. How many points did she score?

$12$

$18$

$24$

$30$

$36$

知识点：百分数一次方程

难度评级：1020

小提示：

设两分球和三分球出手数分别为 $x$ 和 $y$ 。

Let $x$ and $y$ be the numbers of two-point and three-point attempts

大提示：

计算每种出手平均贡献多少分。

Compare the points earned per attempted shot type after applying the success percentages

解答：

设两分球出手 $x$ 次，三分球出手 $y$ 次。

那么 Shenille 命中 $0.3x$ 个两分球和 $0.2y$ 个三分球，总得分为 $2 \cdot 0.3x + 3 \cdot 0.2y = 0.6x + 0.6y\text{。}$

我们知道 $x + y = 30$ $0.6(x + y) = 18\text{。}$

所以正确答案是 B。

Let $x$ be the number of two-point shots and $y$ be the number of three-point shots.

Then, Shenille makes $0.3x$ two-point shots and $0.2y$ three-point shots, for a total score of $2 \cdot 0.3x + 3 \cdot 0.2y = 0.6x + 0.6y.$

We know that $x + y = 30$ $0.6(x + y) = 18.$

Thus, B is the correct answer.

10.

一束花包含粉玫瑰、红玫瑰、粉康乃馨和红康乃馨。粉色花中有三分之一是玫瑰，红色花中有四分之三是康乃馨，全部花中十分之六是粉色的。康乃馨占全部花的百分之几？

A flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations?

$15$

$30$

$40$

$60$

$70$

知识点：分数百分数

难度评级：1280

小提示：

分别计算粉康乃馨和红康乃馨。

Break the bouquet into pink carnations and red carnations

大提示：

对粉花和红花分别使用给出的比例。

Use the given fractions of pink and red flowers separately

解答：

设总花数为 $x$ 。粉色花有 $0.6x$ 朵，红色花有 $0.4x$ 朵。

于是粉玫瑰有 $\dfrac{1}{3} \cdot 0.6x = 0.2x$ 朵，所以粉康乃馨有 $0.4x$ 朵。

另外红康乃馨有 $\dfrac{3}{4} \cdot 0.4x = 0.3x$ 朵。因此康乃馨共有 $0.3x + 0.4x = 0.7x$ 朵，占全部花的 $70\%$ 。

所以正确答案是 E。

Let the total number of flowers be $x.$ There are $0.6x$ pink flowers and $0.4x$ red flowers.

Then there are $\dfrac{1}{3} \cdot 0.6x = 0.2x$ pink roses, which means there are $0.4x$ pink carnations.

There are also $\dfrac{3}{4} \cdot 0.4x = 0.3x$ red carnations. This means there are $0.3x + 0.4x = 0.7x$ carnations. This is $70\%$ of the total flowers.

Thus, E is the correct answer.

11.

学生会要从成员中选出一个两人迎宾委员会和一个三人策划委员会。选出两人迎宾委员会恰好有 $10$ 种方法。学生可以同时在两个委员会中任职。三人策划委员会有多少种不同选法？

A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly $10$ ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person planning committee be selected?

$10$

$12$

$15$

$18$

$25$

知识点：组合二次方程

难度评级：1020

小提示：

设成员数为 $n$ ，并使用 $\binom n2=10$ 。

Let $n$ be the number of council members and use $\binom n2=10$

大提示：

求出 $n$ 后，计算 $\binom n3$ 。

After finding $n$ , count $\binom n3$ planning committees

解答：

设学生会有 $x$ 名成员。选二人委员会的方法数为 $\binom{x}{2} = \dfrac{x(x - 1)}{2}\text{。}$ 这个值等于 $10$ ，所以 $x^2 - x = 20$ $x^2 - x - 20 = 0\text{。}$ 因式分解得 $(x - 5)(x + 4) = 0$ $x = 5\text{，}$ 因为学生人数不能为负数。

于是选 $3$ 人策划委员会的方法数为 $\binom{5}{3} = \binom{5}{2} = 10\text{。}$

所以正确答案是 A。

Let $x$ be the number of students. Then the number of ways to pick a two-person committee is $\binom{x}{2} = \dfrac{x(x - 1)}{2}.$ We know that this equals $10,$ so $x^2 - x = 20$ $x^2 - x - 20 = 0.$ Factoring yields $(x - 5)(x + 4) = 0$ $x = 5,$ since there cannot be a negative number of students.

Then, the number of ways to pick a $3$ -person committee is $\binom{5}{3} = \binom{5}{2} = 10.$

Thus, A is the correct answer.

12.

在 $\triangle ABC$ 中， $AB=AC=28$ ， $BC=20$ 。点 $D$ ， $E$ 和 $F$ 分别在边 $\overline{AB}$ 、 $\overline{BC}$ 和 $\overline{AC}$ 上，且 $\overline{DE}$ 与 $\overline{AC}$ 平行， $\overline{EF}$ 与 $\overline{AB}$ 平行。平行四边形 $ADEF$ 的周长是多少？

In $\triangle ABC,$ $AB=AC=28$ and $BC=20.$ Points $D,$ $E,$ and $F$ are on sides $\overline{AB},$ $\overline{BC},$ and $\overline{AC},$ respectively, such that $\overline{DE}$ and $\overline{EF}$ are parallel to $\overline{AC}$ and $\overline{AB},$ respectively. What is the perimeter of parallelogram $ADEF?$

$48$

$52$

$56$

$60$

$72$

知识点：相似平行四边形周长

难度评级：1420

小提示：

利用平行线得到靠近 $B$ 和 $C$ 的相似三角形。

Use the parallels to get similar triangles near $B$ and $C$

大提示：

把平行四边形的对边改写成 $AB$ 和 $AC$ 的部分。

Rewrite opposite sides of the parallelogram as pieces of $AB$ and $AC$

解答：

由平行线可知 $\triangle DBE \sim \triangle ABC$ ，且 $\triangle FEC \sim \triangle ABC$ 。

因为原三角形两腰相等，所以在这些相似三角形中 $DB = DE$ ， $FE = FC$ 。平行四边形 $ADEF$ 的周长为 $AD + DE + EF + AF$ $= AD + DB + FC + AF$ $= AB + AC$ $= 56\text{。}$

所以正确答案是 C。

Note that $\triangle DBE \sim \triangle ABC$ and $\triangle FEC \sim \triangle ABC$ due to the parallel lines.

This tells us that $DB = DE$ and $FE = FC.$ We have that the perimeter of $ADEF$ is $AD + DE + EF + AF$ $= AD + DB + FC + AF$ $= AB + AC$ $= 56.$

Thus, C is the correct answer.

13.

有多少个三位数不能被 $5$ 整除、各位数字和小于 $20$ ，并且百位数字等于个位数字？

How many three-digit numbers are not divisible by $5,$ have digits that sum to less than $20,$ and have the first digit equal to the third digit?

$52$

$60$

$66$

$68$

$70$

知识点：数字分类讨论

难度评级：1370

小提示：

这样的数形如 $aba$ 。

The number has form $aba$

大提示：

按重复数字 $a$ 分类，并排除使数能被 $5$ 整除的情况。

Case on the repeated digit $a$ , excluding values that make the number divisible by $5$

解答：

要使这个数不能被 $5$ 整除，个位数字不能是 $0$ 或 $5$ 。

设百位和个位数字为 $x$ ，十位数字为 $y$ 。需要 $2x + y \lt 20\text{。}$ 按 $x$ 的 $8$ 种可能取值分类：

若 $x$ 为 $1, 2, 3$ 或 $4$ ，则任意数字 $y$ 都满足，因为 $y \lt 10$ 。

若 $x = 6$ ，则 $y \lt 8$ ，有 $8$ 种选择。

若 $x = 7$ ，则 $y \lt 6$ ，有 $6$ 种选择。

若 $x = 8$ ，则 $y \lt 4$ ，有 $4$ 种选择。

若 $x = 9$ ，则 $y \lt 2$ ，有 $2$ 种选择。

这样总共有 $4 \cdot 10 + 8 + 6 + 4 + 2 = 60$ 组解。

所以正确答案是 B。

Note that for the number to not be divisible by $5,$ the units digits cannot be either $0$ or $5.$

Let $x$ be the hundreds and units digit and $y$ be the tens digit. Then we want $2x + y \lt 20.$ Casing on the $8$ possible values of $x,$ we get:

If $x$ is $1, 2, 3,$ or $4,$ then $y$ can be anything since $y \lt 10.$

If $x = 6,$ then $y \lt 8,$ which gives us $8$ solutions.

If $x = 7,$ then $y \lt 6,$ which gives us $6$ solutions.

If $x = 8,$ then $y \lt 4,$ which gives us $4$ solutions.

If $x = 9,$ then $y \lt 2,$ which gives us $2$ solutions.

This gives us a total of $4 \cdot 10 + 8 + 6 + 4 + 2 = 60$ solutions.

Thus, B is the correct answer.

14.

从边长为 $3$ 的实心立方体的每个角上各去掉一个边长为 $1$ 的实心小立方体。剩下的立体有多少条棱？

A solid cube of side length $1$ is removed from each corner of a solid cube of side length $3.$ How many edges does the remaining solid have?

$36$

$60$

$72$

$84$

$108$

知识点：多面体正方体

难度评级：1540

小提示：

每去掉一个角，都会产生新的外露棱。

Each corner removal creates new exposed edges

大提示：

数原立方体的棱，再加上每个被去掉角产生的新棱。

Count the original cube edges plus the new edges from each removed corner

解答：

去掉角上的小立方体不会消除原来大立方体的十二条外边界棱，只会增加新的棱。

每去掉一个角，会新增 $9$ 条棱。

共有 $8$ 个角，所以新增 $8 \cdot 9 = 72$ 条棱。再加上原来的 $12$ 条棱，总数为 $72 + 12 = 84$ 。

所以正确答案是 D。

Removing the cubes does not remove any edges from the original cube. It only adds edges.

After removing each cube, we can see that $9$ extra edges are added to the solid.

$8$ cubes are removed, which means $8 \cdot 9 = 72$ edges are added to the original $12$ edges, for a total of $72 + 12 = 84$ edges.

Thus, D is the correct answer.

15.

一个三角形的两条边长为 $10$ 和 $15$ 。到第三边的高的长度等于到这两条已知边的高的长度的平均数。第三边长是多少？

Two sides of a triangle have lengths $10$ and $15.$ The length of the altitude to the third side is the average of the lengths of the altitudes to the two given sides. How long is the third side?

$6$

$8$

$9$

$12$

$18$

知识点：三角形面积高线比与比例

难度评级：1660

小提示：

用每条边与对应高表示同一个两倍面积。

Express the same doubled area using each side and its altitude

大提示：

用平均高条件把未知第三边和已知高联系起来。

Use the average-altitude condition to relate the unknown third side to a known altitude

解答：

设边长 $10$ 的边对应的高为 $h_1$ ，另一条已知边对应的高为 $h_2$ 。

同一三角形面积相同，所以 $10h_1 = 15h_2$ $h_1 = \dfrac{3}{2}h_2\text{。}$

第三边对应的高是另外两个高的平均数，因此为 $\dfrac{h_2 + \frac{3}{2}h_2}{2} = \dfrac{5}{4}h_2\text{。}$

设第三边长为 $x$ 。由面积相等， $\dfrac{5}{4}h_2x = 15h_2$ $x = 12\text{。}$

所以正确答案是 D。

Let $h_1$ be the length of the altitude to the side of length $10$ and similarly define $h_2$ for the other given side.

We have that $10h_1 = 15h_2$ $h_1 = \dfrac{3}{2}h_2.$

The third altitude is the average of the other two, which makes its length $\dfrac{h_2 + \frac{3}{2}h_2}{2} = \dfrac{5}{4}h_2.$

Let the third side have length $x.$ Then $\dfrac{5}{4}h_2x = 15h_2$ $x = 12.$

Thus, D is the correct answer.

16.

顶点为 $(6, 5)$ 、 $(8, -3)$ 和 $(9, 1)$ 的三角形关于直线 $x = 8$ 反射，得到第二个三角形。两个三角形并集的面积是多少？

A triangle with vertices $(6, 5),$ $(8, -3),$ and $(9, 1)$ is reflected about the line $x = 8$ to create a second triangle. What is the area of the union of the two triangles?

$9$

$\dfrac{28}{3}$

$10$

$\dfrac{31}{3}$

$\dfrac{32}{3}$

知识点：坐标几何反射（几何）三角形面积对称性

难度评级：2060

小提示：

把不在 $x=8$ 上的两个顶点反射到另一侧。

Reflect the two non-fixed vertices across $x=8$

大提示：

找出两个重叠三角形形成的公共竖直线段。

Find the shared vertical segment created by the two overlapping triangles

解答：

反射后三角形顶点为 $(7,1)$ 、 $(8,-3)$ 、 $(10,5)$ 。

直线经过 $(6,5)$ 和 $(9,1)$ ，方程为 $y=-\frac43x+13$ ，它与 $x=8$ 相交于 $y=\frac73$ 。由对称性，并集可看作两个全等三角形，每个三角形的竖直底边从 $(8,-3)$ 到 $(8,\frac73)$ 。

这条底边长度为 $\frac73+3=\frac{16}{3}$ ，每个三角形的水平高为 $2$ 。因此并集面积为 $2\left(\frac12\cdot\frac{16}{3}\cdot2\right)=\frac{32}{3}$ 。

所以正确答案是 E。

The reflected triangle has vertices $(7,1)$ , $(8,-3)$ , and $(10,5)$ .

The line through $(6,5)$ and $(9,1)$ is $y=-\frac43x+13$ , so it meets $x=8$ at $y=\frac73$ . By symmetry, the union is two congruent triangles with vertical base from $(8,-3)$ to $(8,\frac73)$ .

That base has length $\frac73+3=\frac{16}{3}$ , and each triangle has horizontal height $2$ . Hence the union area is $2\left(\frac12\cdot\frac{16}{3}\cdot2\right)=\frac{32}{3}$ .

Thus, E is the correct answer.

17.

Daphne 的三位好友 Alice、Beatrix 和 Claire 会定期来拜访她。Alice 每三天来一次，Beatrix 每四天来一次，Claire 每五天来一次。

昨天三人都拜访了 Daphne。接下来 $365$ 天中，恰好有两位好友来访的天数是多少？

Daphne is visited periodically by her three best friends: Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day.

All three friends visited Daphne yesterday. How many days of the next $365$ -day period will exactly two friends visit her?

$48$

$54$

$60$

$66$

$72$

知识点：最小公倍数容斥原理区间内整数计数

难度评级：1420

小提示：

用最小公倍数数每一对同时来访的天数。

Count pairwise meeting days using least common multiples

大提示：

从每一对的计数中减去三人都来的天数。

Subtract the triple-meeting days from each pair count

解答：

两两同时来访的周期分别为 $\operatorname{lcm}(3,4)=12$ 、 $\operatorname{lcm}(3,5)=15$ 、 $\operatorname{lcm}(4,5)=20$ 天。

在接下来 $365$ 天中，这三种两两同时来访的天数分别为 $\left\lfloor\frac{365}{12}\right\rfloor=30$ 、 $\left\lfloor\frac{365}{15}\right\rfloor=24$ 、 $\left\lfloor\frac{365}{20}\right\rfloor=18$ 。

三人都来访的周期为 $60$ 天，共 $\left\lfloor\frac{365}{60}\right\rfloor=6$ 天。恰好两人来访的天数为 $(30-6)+(24-6)$ $+(18-6)=54$ 。

所以正确答案是 B。

Pairwise visits occur every $\operatorname{lcm}(3,4)=12$ , $\operatorname{lcm}(3,5)=15$ , and $\operatorname{lcm}(4,5)=20$ days.

In the next $365$ days, these pair counts are $\left\lfloor\frac{365}{12}\right\rfloor=30$ , $\left\lfloor\frac{365}{15}\right\rfloor=24$ , and $\left\lfloor\frac{365}{20}\right\rfloor=18$ .

All three visit every $60$ days, which happens $\left\lfloor\frac{365}{60}\right\rfloor=6$ times. Subtracting those days from each pair count gives $(30-6)+(24-6)$ $+(18-6)=54$ .

Thus, B is the correct answer.

18.

设点 $A=(0,0)$ ， $B=(1,2)$ ， $C=(3,3)$ ， $D=(4,0)$ 。一条经过 $A$ 的直线把四边形 $ABCD$ 分成面积相等的两部分。该直线与 $\overline{CD}$ 交于点 $\left(\dfrac{p}{q}, \dfrac{r}{s}\right)$ ，其中两个分数均为最简形式。 $p+q+r+s$ 是多少？

Let points $A=(0,0),$ $B=(1,2),$ $C=(3,3),$ and $D=(4,0).$ Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A.$ This line intersects $\overline{CD}$ at point $\left(\dfrac{p}{q}, \dfrac{r}{s}\right),$ where these fractions are in lowest terms. What is $p+q+r+s?$

$54$

$58$

$62$

$70$

$75$

知识点：坐标几何面积分割一次方程

难度评级：1660

小提示：

先求四边形 $ABCD$ 的总面积。

First compute the total area of $ABCD$

大提示：

经过 $A$ 的分割线使 $\triangle ADG$ 的面积等于四边形面积的一半。

The cutting line through $A$ makes $\triangle ADG$ half the quadrilateral area

解答：

设分割线与 $\overline{CD}$ 交于 $G$ 。如图，从 $B$ 、 $C$ 和 $G$ 向 $x$ -轴作垂线。

由图中分割可得 $\triangle ABF$ 、梯形 $BCEF$ 、 $\triangle CDE$ 的面积分别为 $1$ 、 $5$ 和 $\frac32$ ，所以 $[ABCD]=\frac{15}{2}$ 。

因此 $\triangle ADG$ 面积为 $\frac{15}{4}$ 。因为 $AD=4$ ，点 $G$ 的高度为 $\frac{15}{8}$ 。

直线 $CD$ 的方程为 $y=-3x+12$ ，所以 $\frac{15}{8}=-3x+12$ ，得到 $x=\frac{27}{8}$ 。因此 $p+q+r+s$ $=27+8+15+8$ $=58$ 。

所以正确答案是 B。

Let the cutting line meet $\overline{CD}$ at $G$ . Drop perpendiculars from $B$ , $C$ , and $G$ to the $x$ -axis as in the diagram.

The areas of $\triangle ABF$ , trapezoid $BCEF$ , and $\triangle CDE$ are $1$ , $5$ , and $\frac32$ , respectively, so $[ABCD]=\frac{15}{2}$ .

Thus $\triangle ADG$ has area $\frac{15}{4}$ . Since $AD=4$ , the height of $G$ is $\frac{15}{8}$ .

The line $CD$ has equation $y=-3x+12$ , so $\frac{15}{8}=-3x+12$ , giving $x=\frac{27}{8}$ . Therefore $p+q+r+s$ $=27+8+15+8$ $=58$ .

Thus, B is the correct answer.

19.

在 $10$ 进制中，数 $2013$ 的末位数字是 $3$ 。另一方面，在 $9$ 进制中，同一个数写作 $(2676)_9$ ，末位数字是 $6$ 。有多少个正整数 $b$ ，使得 $2013$ 的 $b$ 进制表示以数字 $3$ 结尾？

In base $10,$ the number $2013$ ends in the digit $3.$ In base $9,$ on the other hand, the same number is written as $(2676)_9$ and ends in the digit $6.$ For how many positive integers $b$ does the base- $b$ -representation of $2013$ end in the digit $3?$

$6$

$9$

$13$

$16$

$18$

知识点：进制模运算因数个数

难度评级：1420

小提示：

进制表示的末位数字就是除以底数的余数。

A final base digit is a remainder

大提示：

底数必须大于 $3$ ，并且整除 $2013-3$ 。

The base must be greater than $3$ and divide $2013-3$

解答：

进制表示的末位数字就是除以底数后的余数。

因此问题等价于寻找所有正整数 $b$ ，使 $2013$ 除以 $b$ 余 $3$ 。

这意味着 $b$ 必须整除 $2010$ 。此外 $b \geq 4$ ，否则余数不可能是 $3$ 。

$2010$ 的质因数分解为 $2010 = 2 \cdot 3 \cdot 5 \cdot 67\text{。}$ 因此 $2010$ 有 $(1 + 1)^4 = 2^4 = 16$ 个正因数。它有 $3$ 个小于 $4$ 的因数，即 $1, 2,$ 和 $3$ ，都不能作为底数。因此有效的 $b$ 有 $16 - 3 = 13$ 个。

所以正确答案是 C。

Note that the units digit represents the remainder when the number is divided by the base.

The question then boils down to finding all numbers, $b,$ such that $2013$ leaves a remainder of $3$ when divided by $b.$

This means that $b$ must divide $2010.$ Also note that $b \geq 4,$ since otherwise the remainder cannot be $3.$

The prime factorization of $2010$ is $2010 = 2 \cdot 3 \cdot 5 \cdot 67.$ Then, $2010$ has $(1 + 1)^4 = 2^4 = 16$ factors. It has $3$ factors less than $4,$ namely $1, 2,$ and $3.$ This means there are $16 - 3 = 13$ valid values for $b.$

Thus, C is the correct answer.

20.

一个单位正方形绕其中心旋转 $45^\circ$ 。正方形内部扫过区域的面积是多少？

A unit square is rotated $45^\circ$ about its center. What is the area of the region swept out by the interior of the square?

$1 - \dfrac{\sqrt2}{2} + \dfrac{\pi}{4}$

$\dfrac{1}{2} + \dfrac{\pi}{4}$

$2 - \sqrt2 + \dfrac{\pi}{4}$

$\dfrac{\sqrt2}{2} + \dfrac{\pi}{4}$

$1 + \dfrac{\sqrt2}{4} + \dfrac{\pi}{8}$

知识点：变换扇形面积分割

难度评级：2060

小提示：

只看扫过区域的一个象限。

Work with one quadrant of the swept region

大提示：

把该象限分解成一个扇形和两个直角三角形部分。

Decompose that quadrant into a sector and two right-triangle pieces

解答：

考虑扫过区域的四分之一，最后乘以 $4$ 。

其中扇形圆心角为 $45^\circ$ ，半径为 $\frac{\sqrt2}{2}$ ，面积为 $\frac18\pi\left(\frac{\sqrt2}{2}\right)^2=\frac{\pi}{16}$ 。

该四分之一区域中的两个直角三角形面积分别为 $\frac12\cdot\frac12\cdot\frac12=\frac18$ 和 $\frac12\left(\frac{\sqrt2-1}{2}\right)^2$ 。

把四分之一区域的面积和乘以 $4$ ，得到 $2-\sqrt2+\frac{\pi}{4}$ 。

所以正确答案是 C。

Consider one quarter of the swept region and multiply its area by $4$ .

The sector has angle $45^\circ$ and radius $\frac{\sqrt2}{2}$ , so its area is $\frac18\pi\left(\frac{\sqrt2}{2}\right)^2=\frac{\pi}{16}$ .

The two right-triangle pieces in that quarter have areas $\frac12\cdot\frac12\cdot\frac12=\frac18$ and $\frac12\left(\frac{\sqrt2-1}{2}\right)^2$ .

Multiplying the quarter-area sum by $4$ gives $2-\sqrt2+\frac{\pi}{4}$ .

Thus, C is the correct answer.

21.

$12$ 名海盗同意按如下方式分一箱金币。第 $k^{\text{th}}$ 个拿份额的海盗取走箱中剩余金币的 $\dfrac{k}{12}$ 。箱中最初的金币数是使得每名海盗都能得到正整数枚金币的最小数。第 $12^{\text{th}}$ 个海盗得到多少枚金币？

A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^{\text{th}}$ pirate to take a share takes $\dfrac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the $12^{\text{th}}$ pirate receive?

$720$

$1296$

$1728$

$1925$

$3850$

知识点：整除性质因数分解逆推法

难度评级：2300

小提示：

从第十二个海盗得到的金币数反向推。

Work backward from the number left for the last pirate

大提示：

反推第 $k$ 个海盗之前的数量时，要乘以 $\frac{12}{12-k}$ 。

The remaining amount is multiplied by $\frac{12}{12-k}$ when reversing pirate $k$

解答：

反向计算。若第 $12$ 个海盗分到时还剩 $n$ 枚金币，那么在第 $k$ 个海盗拿走份额之前，箱中的金币数是他拿完之后的 $\frac{12}{12-k}$ 倍。

因此最初的金币数为 $n\cdot\frac{12^{11}}{11!}$ 。

由于 $\frac{12^{11}}{11!}=\frac{2^{14}3^7}{5\cdot7\cdot11}$ ，使最初金币数为整数的最小 $n$ 是 $5^2\cdot7\cdot11=1925$ 。

这个值确实可以达到。在第 $k$ 个海盗拿走份额之前，金币数为 $2^{14}3^7\cdot \frac{11!}{(12-k)!\,12^{k-1}}\text{。}$ 对 $k=1,2,\ldots,12$ ，这些数全是整数。每个海盗拿到的份额是相邻两个剩余量之差，所以每份也都是整数。

因此第 $12$ 个海盗得到 $1925$ 枚金币，正确答案是 D。

Work backward. If $n$ coins remain for the $12$ th pirate, then before pirate $k$ took a share, the chest had $\frac{12}{12-k}$ times as many coins as it had afterward.

Therefore the initial number of coins is $n\cdot\frac{12^{11}}{11!}$ .

Since $\frac{12^{11}}{11!}=\frac{2^{14}3^7}{5\cdot7\cdot11}$ , the smallest $n$ that makes the initial number an integer is $5^2\cdot7\cdot11=1925$ .

This value is attainable. The number of coins present just before pirate $k$ takes a share is $2^{14}3^7\cdot \frac{11!}{(12-k)!\,12^{k-1}}.$ For $k=1,2,\ldots,12$ , all these amounts are integers. Each pirate’s share is the difference between two consecutive remaining amounts, so every share is an integer as well.

Thus, the $12$ th pirate receives $1925$ coins, and D is the correct answer.

22.

六个半径为 $1$ 的球摆放成它们的球心位于边长为 $2$ 的正六边形的六个顶点。这六个球都内切于一个大球，大球球心为该正六边形的中心。第八个球外切于这六个小球，并内切于大球。第八个球的半径是多少？

Six spheres of radius $1$ are positioned so that their centers are at the vertices of a regular hexagon of side length $2.$ The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?

$\sqrt2$

$\dfrac{3}{2}$

$\dfrac{5}{3}$

$\sqrt3$

$2$

知识点：球立体几何勾股定理

难度评级：1970

小提示：

大球半径为 $3$ 。

The large sphere has radius $3$

大提示：

连接大球球心、小球球心和第八个球的球心，使用直角三角形。

Use a right triangle joining the large center, a small-sphere center, and the eighth-sphere center

解答：

六个半径为 $1$ 的小球球心构成边长为 $2$ 的正六边形，所以每个小球球心到大球球心的距离为 $2$ 。因此大球半径为 $3$ 。

设第八个球半径为 $r$ ，球心到大球球心的距离为 $x$ 。内切给出 $x+r=3$ ，所以 $x=3-r$ 。

由图中的直角三角形， $(r+1)^2=2^2+(3-r)^2$ 。化简得 $2r+1=13-6r$ ，所以 $r=\frac32$ 。

所以正确答案是 B。

The centers of the six radius- $1$ spheres form a regular hexagon of side length $2$ , so each is $2$ units from the large sphere’s center. Hence the large sphere has radius $3$ .

Let the eighth sphere have radius $r$ , and let its center be distance $x$ from the large sphere’s center. Internal tangency gives $x+r=3$ , so $x=3-r$ .

Using the right triangle between the large center, a small-sphere center, and the eighth-sphere center, $(r+1)^2=2^2+(3-r)^2$ . Thus $2r+1=13-6r$ , so $r=\frac32$ .

Thus, B is the correct answer.

23.

在 $\triangle ABC$ 中， $AB = 86$ ，且 $AC=97$ 。以 $A$ 为圆心、 $AB$ 为半径的圆与 $\overline{BC}$ 交于点 $B$ 和 $X$ 。此外， $\overline{BX}$ 和 $\overline{CX}$ 的长度都是整数。 $BC$ 是多少？

In $\triangle ABC,$ $AB = 86,$ and $AC=97.$ A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X.$ Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC?$

$11$

$28$

$33$

$61$

$72$

知识点：圆幂质因数分解三角不等式

难度评级：2010

小提示：

从点 $C$ 使用圆的幂。

Use power of a point from $C$

大提示：

分解 $AC^2-AB^2$ ，再用三角形不等式选择有效因数对。

Factor $AC^2-AB^2$ and use the triangle inequality to choose the valid factor pair

解答：

由点 $C$ 的圆幂， $CB\cdot CX$ $=AC^2-AB^2$ $=97^2-86^2$ 。

右边等于 $(97-86)(97+86)$ $=11\cdot183$ $=2013$ $=3\cdot11\cdot61$ 。

$CX$ 和 $BX$ 都是整数，所以 $BC=BX+CX$ 是与 $CX$ 配对的整数因数。还需满足 $CX<BC<86+97=183$ ，唯一可行因数对为 $CX=33$ 、 $BC=61$ 。

所以正确答案是 D。

By power of a point from $C$ , $CB\cdot CX$ $=AC^2-AB^2$ $=97^2-86^2$ .

This equals $(97-86)(97+86)$ $=11\cdot183$ $=2013$ $=3\cdot11\cdot61$ .

Both $CX$ and $BX$ are integers, so $BC=BX+CX$ is an integer factor paired with $CX$ . Also $CX<BC<86+97=183$ , so the only possible pair is $CX=33$ , $BC=61$ .

Thus, D is the correct answer.

24.

Central 高中与 Northern 高中进行十五子棋比赛。每所学校有三名选手，规则要求每名选手都与对方学校的每名选手比赛两局。比赛分六轮进行，每轮同时进行三局。共有多少种不同的赛程安排？

Central High School is competing against Northern High School in a backgammon match. Each school has three players, and the contest rules require that each player play two games against each of the other school’s players. The match takes place in six rounds, with three games played simultaneously in each round. In how many different ways can the match be scheduled?

$540$

$600$

$720$

$810$

$900$

知识点：有限制的排列多重集排列分类讨论

难度评级：2390

小提示：

先固定一名选手六轮的对手安排。

Fix one player’s six-round schedule first

大提示：

对第二名选手，数每轮都避开第一名选手对手的字符串。

For the second player, count strings avoiding the first player’s opponent in every round

解答：

记 Central 的选手为 $A,B,C$ ，Northern 的选手为 $X,Y,Z$ 。

选手 $A$ 的六轮对手字符串中 $X,Y,Z$ 各出现两次，所以有 $\frac{6!}{2!2!2!}=90$ 种。

固定 $A$ 的字符串，例如 $XXYYZZ$ 。 $B$ 的字符串也必须含两个 $X,Y,Z$ ，且每一轮不能与 $A$ 的对手相同。

若 $B$ 的前两位为 $Y,Z$ 的某种顺序，中间两位必须为 $X,Z$ 的某种顺序，后两位必须为 $X,Y$ 的某种顺序，共 $2^3=8$ 种。另有 $YYZZXX$ 和 $ZZXXYY$ 两种，共 $10$ 种 $B$ 的字符串。

一旦 $A$ 和 $B$ 的赛程确定， $C$ 的赛程被迫确定。因此总数为 $90\cdot10=900$ 。

所以正确答案是 E。

Label Central’s players $A,B,C$ and Northern’s players $X,Y,Z$ .

Player $A$ ’s six-round opponent string contains two each of $X,Y,Z$ , so it can be chosen in $\frac{6!}{2!2!2!}=90$ ways.

For a fixed $A$ -string, say $XXYYZZ$ , player $B$ ’s string must also contain two each of $X,Y,Z$ and cannot match $A$ ’s opponent in any position.

If $B$ ’s first two entries are $Y,Z$ in either order, then the middle two entries must be $X,Z$ in either order and the last two must be $X,Y$ in either order, giving $2^3=8$ strings. The two remaining possibilities are $YYZZXX$ and $ZZXXYY$ , for $10$ total $B$ -strings.

Once $A$ and $B$ are scheduled, $C$ ’s schedule is forced. Hence there are $90\cdot10=900$ schedules.

Thus, E is the correct answer.

25.

在一个正八边形中画出全部 $20$ 条对角线。在八边形内部，不在边界上的位置，有多少个不同的点是两条或更多对角线的交点？

All $20$ diagonals are drawn in a regular octagon. At how many distinct points in the interior of the octagon (not on the boundary) do two or more diagonals intersect?

$49$

$65$

$70$

$96$

$128$

知识点：交点计数正多边形组合

难度评级：2180

小提示：

先从由四个顶点确定的 $\binom84$ 个交点开始。

Start with $\binom84$ intersections from four vertices

大提示：

再修正中心点和另外八个三线共点的位置。

Correct for the center and the eight other triple-concurrence points

解答：

若没有三条对角线共点，每选 $4$ 个顶点会确定一个内部交点，共 $\binom84=70$ 个。

通过对顶点的 $4$ 条长对角线都在中心相交。中心原本被数了 $\binom42=6$ 次，但应只数一次，所以减去 $5$ 。

另外还有 $8$ 个对称位置各有 $3$ 条对角线共点。每个位置原本被数 $\binom32=3$ 次，但应只数一次，所以再减去 $8(3-1)=16$ 。

不同内部交点数为 $70-5-16=49$ 。

所以正确答案是 A。

If no three diagonals were concurrent, each choice of $4$ vertices would determine one interior intersection, giving $\binom84=70$ .

The $4$ long diagonals through opposite vertices all meet at the center, so the center was counted $\binom42=6$ times and should be counted once. Subtract $5$ .

There are also $8$ symmetric points where $3$ diagonals meet. Each was counted $\binom32=3$ times and should be counted once, so subtract $8(3-1)=16$ .

The number of distinct interior intersection points is $70-5-16=49$ .

Thus, A is the correct answer.