2010 AMC 10B 真题与在线限时测验

计时

1:15:00

1.

求 $100(100-3)-(100 \cdot 100-3)$ 的值。

What is $100(100-3)-(100 \cdot 100-3)?$

$-20,000$

$-10,000$

$-297$

$-6$

$0$

答案：C

知识点：运算顺序整数运算

难度评级：560

解答：

第一项为 $\begin{align*}100(100 - 3) &= 100 \cdot 97 \\&= 9700\end{align*}$ 而第二项为 $\begin{align*}100 \cdot 100 - 3 &= 10000 - 3\\ &= 9997.\end{align*}$

因此差为 $9700 - 9997 = - 297.$

所以正确答案是 C。

We have that $\begin{align*}100(100 - 3) &= 100 \cdot 97 \\&= 9700\end{align*}$ and $\begin{align*}100 \cdot 100 - 3 &= 10000 - 3\\ &= 9997.\end{align*}$

As such, their difference is: $9700 - 9997 = - 297.$

Thus, C is the correct answer.

2.

Makayla 在 $9$ 小时工作日中参加了两场会议。第一场会议用了 $45$ 分钟，第二场会议用时是第一场的两倍。她工作日的百分之多少用于参加会议？

Makayla attended two meetings during her $9$ -hour work day. The first meeting took $45$ minutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?

$15$

$20$

$25$

$30$

$35$

答案：C

知识点：百分数单位换算

难度评级：870

解答：

$45$ 分钟等于 $\dfrac{45}{60} = \dfrac{3}{4}$ 小时。第二场会议用时为 $2 \cdot \dfrac{3}{4} = \dfrac{3}{2}$ 小时，所以两场会议共用 $\dfrac{3}{4} + \dfrac{3}{2} = \dfrac{9}{4}$ 小时。这占整个工作日的 $100 \% \cdot \dfrac{\frac{9}{4}}{9} = 100 \% \cdot \dfrac{1}{4} = 25 \%.$

所以正确答案是 C。

Note that $45$ minutes is $\dfrac{45}{60} = \dfrac{3}{4}$ hours. The second meeting is then $2 \cdot \dfrac{3}{4} = \dfrac{3}{2}$ hours. Both meetings take a total of $\dfrac{3}{4} + \dfrac{3}{2} = \dfrac{9}{4}$ hours then. The percent of Makayla's work day spent attending meetings is $100 \% \cdot \dfrac{\frac{9}{4}}{9} = 100 \% \cdot \dfrac{1}{4} = 25 \%.$

Thus, C is the correct answer.

3.

一个抽屉里有红、绿、蓝、白四种颜色的袜子，每种颜色至少 $2$ 只。至少要从抽屉里取出多少只袜子，才能保证有一双同色袜子？

A drawer contains red, green, blue, and white socks with at least $2$ of each color. What is the minimum number of socks that must be pulled from the drawer to guarantee a matching pair?

$3$

$4$

$5$

$8$

$9$

答案：C

知识点：抽屉原理

难度评级：720

解答：

为了尽量延迟出现同色一双，最多可以先取出每种颜色各一只袜子。

一共有 $4$ 种颜色，因此取出四只时仍可能没有同色一双。

再取一只就必然与其中一种颜色配成一双，所以需要 $4 + 1 = 5$ 只。

所以正确答案是 C。

To maximize the number of socks, we want to grab as many single socks as possible before getting a pair.

There are $4$ colors, which means that we can draw one sock of each color before drawing a pair.

This means that it takes at least $4 + 1 = 5$ socks to be drawn before a pair is guaranteed.

Thus, C is the correct answer.

4.

对实数 $x$ ，定义 $\heartsuit(x)$ 为 $x$ 和 $x^2$ 的平均数。求 $\heartsuit(1)+\heartsuit(2)+\heartsuit(3)$ 的值。

For a real number $x,$ define $\heartsuit(x)$ to be the average of $x$ and $x^2.$ What is $\heartsuit(1)+\heartsuit(2)+\heartsuit(3)?$

$3$

$6$

$10$

$12$

$20$

答案：C

知识点：自定义运算平均数

难度评级：870

解答：

依次计算得 $\heartsuit(1) = \dfrac{1 + 1^2}{2} = 1,$ $\heartsuit(2) = \dfrac{2 + 2^2}{2} = 3,$ 以及 $\heartsuit(3) = \dfrac{3 + 3^2}{2} = 6.$

因此总和为 $1 + 3 + 6 = 10.$

所以正确答案是 C。

We have $\heartsuit(1) = \dfrac{1 + 1^2}{2} = 1,$ $\heartsuit(2) = \dfrac{2 + 2^2}{2} = 3,$ and $\heartsuit(3) = \dfrac{3 + 3^2}{2} = 6.$

Then $1 + 3 + 6 = 10.$

Thus, C is the correct answer.

5.

某个有 $31$ 天的月份中，星期一和星期三的天数相同。这个月的第一天可能是七天中的多少种？

A month with $31$ days has the same number of Mondays and Wednesdays. How many of the seven days of the week could be the first day of this month?

$2$

$3$

$4$

$5$

$6$

答案：B

知识点：日期与时间模运算

难度评级：960

解答：

$31$ 除以 $7$ 余 $3$ ，所以从该月第一天开始的连续三个星期几会出现五次，其余出现四次。

要使星期一和星期三次数相同，它们不能恰好只有一个在这三个多出的星期几中。

检查可知，该月只能从星期一、星期四或星期五开始。

所以正确答案是 B。

Note that $31$ days leaves a remainder of $3$ when divided by $7.$

This means that if the month starts on a Saturday, Sunday, Tuesday, or Wednesday, there will be an uneven number of Mondays and Wednesdays.

Then the month can only start on a Monday, Thursday, or Friday.

Thus, B is the correct answer.

6.

圆心为 $O$ ， $\overline{AB}$ 是直径，点 $C$ 在圆上，且 $\angle COB = 50^\circ$ 。 $\angle CAB$ 的度数是多少？

A circle is centered at $O,$ $\overline{AB}$ is a diameter and $C$ is a point on the circle with $\angle COB = 50^\circ.$ What is the degree measure of $\angle CAB?$

$20$

$25$

$45$

$50$

$65$

答案：B

知识点：圆等腰三角形导角

难度评级：960

解答：

参考下图：

由于直径的两端与圆心在同一直线上， $\angle AOC = 180^{\circ} - 50^{\circ} = 130^{\circ}.$

$\triangle AOC$ 是等腰三角形，所以 $\angle CAO = \dfrac{180^{\circ} - 130^{\circ}}{2} = 25^{\circ}.$

因为 $\angle CAB = \angle CAO$ ，所以 $\angle CAB = 25^{\circ}$ 。

所以正确答案是 B。

Consider the following diagram:

We have that $\angle AOC = 180^{\circ} - 50^{\circ} = 130^{\circ}.$

Since $\triangle AOC$ is isosceles, we have that $\angle CAO = \dfrac{180^{\circ} - 130^{\circ}}{2} = 25^{\circ}.$

Since $\angle CAB = \angle CAO,$ we have that $\angle CAB = 25^{\circ}.$

Thus, B is the correct answer.

7.

一个三角形的边长为 $10$ 、 $10$ 和 $12$ 。一个长方形宽为 $4$ ，面积等于该三角形的面积。这个长方形的周长是多少？

A triangle has side lengths $10,$ $10,$ and $12.$ A rectangle has width $4$ and area equal to the area of the triangle. What is the perimeter of this rectangle?

$16$

$24$

$28$

$32$

$36$

答案：D

知识点：三角形面积勾股定理周长

难度评级：1220

解答：

向长为 $12$ 的边作高。

由于另外两边相等，高把底边平分，每半段为 $12 \div 2 = 6$ ，斜边为 $10$ 。

所得直角三角形的斜边为十，一条直角边为六，所以高为 $\sqrt{10^2 - 6^2} = \sqrt{64} = 8.$

三角形面积为 $\dfrac{8 \cdot 12}{2} = 48.$

长方形宽为四，所以长为 $48 \div 4 = 12$ ，周长为 $2(4 + 12) = 2 \cdot 16 = 32.$

所以正确答案是 D。

To find the area of the triangle, we can drop the altitude to the side of length $12.$

Then we have a right triangle with one leg $12 \div 2 = 6$ and hypotenuse $10.$

The other leg has length $\sqrt{10^2 - 6^2} = \sqrt{64} = 8.$

The area of the triangle is then $\dfrac{8 \cdot 12}{2} = 48.$

The length of the rectangle is then $48 \div 4 = 12.$ Its perimeter is $2(4 + 12) = 2 \cdot 16 = 32.$

Thus, D is the correct answer.

8.

一张学校戏剧票价为 $x$ 美元，其中 $x$ 是整数。一个 $9$ 年级小组买票共花 $\$48$ ，一个 $10$ 年级小组买票共花 $\$64$ 。 $x$ 可能有多少个值？

A ticket to a school play costs $x$ dollars, where $x$ is a whole number. A group of $9$ th graders buys tickets costing a total of $\$48,$ and a group of $10$ th graders buys tickets costing a total of $\$64.$ How many values for $x$ are possible?

$1$

$2$

$3$

$4$

$5$

答案：E

知识点：最大公约数因数个数

难度评级：1020

解答：

$x$ 必须同时整除 $48$ 和 $64$ ，因此必须整除它们的最大公因数。

$48$ 和 $64$ 的最大公因数是 $16$ ，所以 $x$ 必须整除 $16$ 。

$16$ 的正因数都是不超过 $16$ 的 $2$ 的幂，共有 $5$ 个。

所以正确答案是 E。

Note that $x$ must divide both $48$ and $64,$ which means that means it must divide their greatest common divisor.

The greatest common divisor of $48$ and $64$ is $16,$ which means $x$ divides $16.$

$16$ has $5$ factors, namely all the powers of $2$ up to $16.$

Thus, E is the correct answer.

9.

Larry 的老师让他把数字代入 $a$ 、 $b$ 、 $c$ 、 $d$ 和 $e$ ，再计算表达式 $a-(b-(c-(d+e)))$ 并求值。Larry 忽略了括号，但加减号本身算对了，并且碰巧得到了正确结果。他给 $a$ 、 $b$ 、 $c$ 、 $d$ 代入的数分别是 $1$ 、 $2$ 、 $3$ 、 $4$ 。他给 $e$ 代入了什么数？

Lucky Larry's teacher asked him to substitute numbers for $a,$ $b,$ $c,$ $d,$ and $e$ in the expression $a-(b-(c-(d+e)))$ and evaluate the result. Larry ignored the parentheses but added and subtracted correctly and obtained the correct result by coincidence. The number Larry substituted for $a,$ $b,$ $c,$ and $d$ were $1,$ $2,$ $3,$ and $4,$ respectively. What number did Larry substitute for $e?$

$-5$

$-3$

$0$

$3$

$5$

答案：D

知识点：运算顺序一次方程

难度评级：1280

解答：

忽略括号时，Larry 会得到 $1 - 2 - 3 - 4 + e = e - 8.$

按括号正确计算， $\begin{align*} &1 - (2 -(3 - (4 + e)))\\ & = 1 - (2 - (-1 - e))\\ &= 1 - (3 + e) \\ &=-2 - e.\end{align*}$

两者相等，所以 $e - 8 = -2 - e$ $e = 3.$

所以正确答案是 D。

Ignoring the parentheses, Larry would get $1 - 2 - 3 - 4 + e = e - 8.$

Evaluating with the parentheses, one would get $\begin{align*} &1 - (2 -(3 - (4 + e)))\\ & = 1 - (2 - (-1 - e))\\ &= 1 - (3 + e) \\ &=-2 - e.\end{align*}$

Both of these values are the same, so $e - 8 = -2 - e$ $e = 3.$

Thus, D is the correct answer.

10.

如果不下雨，Shelby 骑踏板车的速度为每小时 $30$ 英里；如果下雨，速度为每小时 $20$ 英里。今天上午她在晴天中骑行，傍晚在雨中骑行，共用 $40$ 分钟行驶了 $16$ 英里。她在雨中骑了多少分钟？

Shelby drives her scooter at a speed of $30$ miles per hour if it is not raining, and $20$ miles per hour if it is raining. Today she drove in the sun in the morning and in the rain in the evening, for a total of $16$ miles in $40$ minutes. How many minutes did she drive in the rain?

$18$

$21$

$24$

$27$

$30$

答案：C

知识点：路程、速度与时间一次方程

难度评级：1370

解答：

$40$ 分钟是 $\frac{2}{3}$ 小时。设 Shelby 在晴天中骑了 $h$ 小时。

则雨中骑了 $\frac{2}{3} - h$ 小时，所以总路程为 $30h + 20\left(\dfrac{2}{3} - h\right) = 10h + \dfrac{40}{3}$ 英里。这个值等于 $16$ ，因此 $10h + \dfrac{40}{3} = 16$ $h = \dfrac{4}{15}.$

因此雨中骑行时间为 $60\left(\dfrac{2}{3} - \dfrac{4}{15}\right) = 60 \cdot \dfrac{2}{5} = 24$

所以正确答案是 C。

Note that $40$ minutes is $\frac{2}{3}$ hours. Let Shelby drive $h$ hours in the sun.

Then she drives $\frac{2}{3} - h$ hours in the rain, which means she travels a total of $30h + 20\left(\dfrac{2}{3} - h\right) = 10h + \dfrac{40}{3}$ miles. We have this equals $16,$ so $10h + \dfrac{40}{3} = 16$ $h = \dfrac{4}{15}.$

Then Shelby drives $60\left(\dfrac{2}{3} - \dfrac{4}{15}\right) = 60 \cdot \dfrac{2}{5} = 24$ minutes in the rain.

Thus, C is the correct answer.

11.

一位购物者计划购买一件标价大于 $\$100$ 的商品，并且可以使用三张优惠券中的任意一张。优惠券 A 可按标价打 $15\%$ 折扣，优惠券 B 可直接减 $\$30$ ，优惠券 C 可对标价超过 $\$ 100$ 的部分打 $25\%$ 折扣。

设 $x$ 和 $y$ 分别为使优惠券 A 节省的钱不少于优惠券 B 或 C 的最小和最大标价。求 $y - x$ 。

A shopper plans to purchase an item that has a listed price greater than $\$100$ and can use any one of the three coupons. Coupon A gives $15\%$ off the listed price, Coupon B gives $\$30$ off the listed price, and Coupon C gives $25\%$ off the amount by which the listed price exceeds $\$ 100.$

Let $x$ and $y$ be the smallest and largest prices, respectively, for which Coupon A saves at least as many dollars as Coupon B or C. What is $y - x?$

$50$

$60$

$75$

$80$

$100$

答案：A

知识点：百分数不等式

难度评级：1540

解答：

设商品标价为 $p$ 。优惠券 A 节省 $.15p$ ，优惠券 B 节省 $\$ 30$ 。

优惠券 C 节省 $.25(p - 100) = .25p - 25.$

因此必须满足 $.15p \ge 30$ $p \geq 200$ 以及 $.15p \ge .25p - 25$ $250 \geq p.$

这给出 $x = 200$ 、 $y = 250$ ，所以 $y - x = 50$ 。

所以正确答案是 A。

Let $p$ be the price of the item. Then coupon A saves $.15p.$ Coupon B saves $\$ 30.$

Coupon C will save $.25(p - 100) = .25p - 25.$

We must have that $.15p \ge 30$ $p \geq 200$ and $.15p \ge .25p - 25$ $250 \geq p.$

This shows that $x = 200$ and $y = 250.$ Therefore $y - x = 50.$

Thus, A is the correct answer.

12.

学年初，Wells 老师数学课上 $50\%$ 的学生对“你热爱数学吗？”回答“是”， $50\%$ 回答“否”。学年末， $70\%$ 回答“是”， $30\%$ 回答“否”。总共有 $x\%$ 的学生在学年初和学年末给出了不同回答。 $x$ 的最大可能值与最小可能值之差是多少？

At the beginning of the school year, $50\%$ of all students in Mr. Wells' math class answered "Yes" to the question "Do you love math", and $50\%$ answered "No." At the end of the school year, $70\%$ answered "Yes" and $30\%$ answered "No." Altogether, $x\%$ of the students gave a different answer at the beginning and end of the school year. What is the difference between the maximum and the minimum possible values of $x?$

$0$

$20$

$40$

$60$

$80$

答案：D

知识点：百分数极限情形界定

难度评级：1420

解答：

为了最小化 $x$ ，尽量让学生保持原答案。

由于回答“是”的比例从百分之五十增到百分之七十，至少有 $70 - 50 = 20$ 个百分点所对应的学生必须改变答案。

为了最大化 $x$ ，可以让所有原来回答“否”的学生都改成“是”，但最终仍需要百分之七十回答“是”。

这要求原来回答“是”的学生中有 $70 - 50 = 20$ 个百分点仍回答“是”，所以原来回答“是”的百分之五十中，最多有 $50 - 20 = 30$ 个百分点可以改变。

因此最多有 $50 + 30 = 80$ 个百分点的学生改变答案。最大值与最小值之差为 $80 - 20 = 60$ 。

所以正确答案是 D。

To minimize $x,$ we want to have as many kids as possible maintain their answer.

We then need at least $70 - 50 = 20$ percent of the students to change their answer.

To maximize $x,$ we can have everybody that answered no change their answer, but some who answered yes must stay yes.

We need $70 - 50 = 20$ percent of the people who said yes to stay yes, which means only $50 - 20 = 30$ percent can switch.

This makes the maximum percent of students that can switch $50 + 30 = 80$ percent. The difference is then $80 - 20 = 60.$

Thus, D is the correct answer.

13.

方程 $x = |2x-|60-2x||$ 的所有解之和是多少？

What is the sum of all the solutions of $x = |2x-|60-2x||?$

$32$

$60$

$92$

$120$

$124$

答案：C

知识点：绝对值分类讨论

难度评级：1660

解答：

先处理外层绝对值。有两种可能： $x = 2x - |60 - 2x|$ 或 $x = -2x + |60 - 2x|.$

这两个方程分别化为 $x = |60 - 2x|$ 和 $3x = |60 - 2x|.$

对第一个方程，有 $2$ 个线性方程要解： $x = 60 - 2x \text{ and } x = 2x - 60.$

分别求解得 $3x = 60$ $x = 20$ 以及 $-x = -60$ $x = 60.$

对第二个方程，有 $3x = 60 - 2x \text{ and } 3x = 2x - 60.$

再次分别求解，得 $5x = 60$ $x = 12$ 和 $x = -60$ 。由于原方程右边非负， $x$ 不能为负数。

所有解之和为 $20 + 60 + 12 = 92.$

所以正确答案是 C。

We first take care of the outer absolute value. We have either $x = 2x - |60 - 2x|$ or $x = -2x + |60 - 2x|.$

These simplify to $x = |60 - 2x|$ and $3x = |60 - 2x|.$

We have $2$ cases for each equation. For the first one, we have $x = 60 - 2x \text{ and } x = 2x - 60.$

Solving both gives us $3x = 60$ $x = 20$ and $-x = -60$ $x = 60.$

For the other equation, we have $3x = 60 - 2x \text{ and } 3x = 2x - 60.$

Again solving both, we have $5x = 60$ $x = 12$ and $x = -60.$ Note that $x$ cannot be negative since the right side of the original equation is nonnegative.

Adding up all the solutions gives us $20 + 60 + 12 = 92.$

Thus, C is the correct answer.

14.

数 $1, 2, 3,\cdots, 98, 99$ 和 $x$ 的平均数为 $100x$ 。求 $x$ 。

The average of the numbers $1, 2, 3,\cdots, 98, 99,$ and $x$ is $100x.$ What is $x?$

$\dfrac{49}{101}$

$\dfrac{50}{101}$

$\dfrac{1}{2}$

$\dfrac{51}{101}$

$\dfrac{50}{99}$

答案：B

知识点：平均数等差数列一次方程

难度评级：1370

解答：

前 $n$ 个正整数的和为 $\dfrac{n(n + 1)}{2}$ 。

因此有 $\dfrac{\frac{99 \cdot 100}{2} + x}{100} = 100x,$ 化简得 $99 \cdot 50 = (100^2 - 1)x$ $= 101 \cdot 99x,$ 所以 $x = \dfrac{50}{101}$ 。

所以正确答案是 B。

Recall that the sum of the first $n$ integers is $\dfrac{n(n + 1)}{2}.$

Then, we have that $\dfrac{\frac{99 \cdot 100}{2} + x}{100} = 100x,$ which simplifies to $99 \cdot 50 = (100^2 - 1)x$ $= 101 \cdot 99x,$ by difference of squares. Dividing gives us $x = \dfrac{50}{101}.$

Thus, B is the correct answer.

15.

在一场 $50$ 题的选择题数学竞赛中，学生每答对一题得 $4$ 分，空题得 $0$ 分，答错一题得 $-1$ 分。Jesse 的总分为 $99$ 。Jesse 最多可能答对多少题？

On a $50$ -question multiple choice math contest, students receive $4$ points for a correct answer, $0$ points for an answer left blank, and $-1$ point for an incorrect answer. Jesse’s total score on the contest was $99.$ What is the maximum number of questions that Jesse could have answered correctly?

$25$

$27$

$29$

$31$

$33$

答案：C

知识点：一次方程不等式最优化

难度评级：1420

解答：

设 Jesse 答对 $x$ 题，答错 $y$ 题。

因此必须同时满足 $4x - y = 99 \text{ and } x + y \leq 50.$

由 $y = 4x - 99$ 代入得到 $5x - 99 \leq 50.$

因为 $x$ 是整数，而 $x \leq 29.8$ ，所以最大值为 $29$ 。

所以正确答案是 C。

Let $x$ be the number of questions Jesse answered correctly and $y$ be the number he answered incorrectly.

Then $4x - y = 99 \text{ and } x + y \leq 50.$

We have that $y = 4x - 99$ $5x - 99 \leq 50.$

Rearranging and simplifying tells us that $x \leq 29.8.$ Since $x$ is an integer, its maximum value is $29.$

Thus, C is the correct answer.

16.

一个边长为 $1$ 的正方形和一个半径为 $\dfrac{\sqrt{3}}{3}$ 的圆有相同圆心。圆内但正方形外的面积是多少？

A square of side length $1$ and a circle of radius $\dfrac{\sqrt{3}}{3}$ share the same center. What is the area inside the circle, but outside the square?

$\dfrac{\pi}{3}-1$

$\dfrac{2\pi}{9}-\dfrac{\sqrt{3}}{3}$

$\dfrac{\pi}{18}$

$\dfrac{1}{4}$

$\dfrac{2\pi}{9}$

答案：B

知识点：圆面积扇形特殊直角三角形面积分割

难度评级：1790

解答：

从 $O$ 向 $\overline{AB}$ 作高，垂足为 $X$ 。

由边长关系， $\triangle OBX$ 是 $30-60-90$ 特殊直角三角形。

扇形 $AOB$ 的面积为 $\dfrac{1}{6} \cdot \pi \cdot \left(\dfrac{\sqrt3}{3}\right)^2 = \dfrac{\pi}{18}.$

$\triangle AOB$ 的面积为 $\dfrac{1}{2} \cdot 2 \cdot \dfrac{\sqrt3}{6} \cdot \dfrac{1}{2} = \dfrac{\sqrt3}{12}.$

因此该边外、圆内的一个圆弓形面积为 $\dfrac{\pi}{18} - \dfrac{\sqrt3}{12}.$

这样的区域有 $4$ 个，总面积为 $4\left(\dfrac{\pi}{18} - \dfrac{\sqrt3}{12}\right) = \dfrac{2\pi}{9} - \dfrac{\sqrt3}{3}.$

所以正确答案是 B。

Drop the altitude from $O$ to $\overline{AB}$ at $X.$

Looking at the side lengths, we see that $\triangle OBX$ is a $30-60-90$ triangle.

This means that the area of sector $AOB$ is $\dfrac{1}{6} \cdot \pi \cdot \left(\dfrac{\sqrt3}{3}\right)^2 = \dfrac{\pi}{18}.$

We also have that the area of $\triangle AOB$ is $\dfrac{1}{2} \cdot 2 \cdot \dfrac{\sqrt3}{6} \cdot \dfrac{1}{2} = \dfrac{\sqrt3}{12}.$

The area of the sector outside the square and inside the circle is then $\dfrac{\pi}{18} - \dfrac{\sqrt3}{12}.$

There are $4$ of these regions, which gives us a total area of $4\left(\dfrac{\pi}{18} - \dfrac{\sqrt3}{12}\right) = \dfrac{2\pi}{9} - \dfrac{\sqrt3}{3}.$

Thus, B is the correct answer.

17.

Euclid 市的每所高中都派出一支 $3$ 人队伍参加数学竞赛。每位参赛者的分数都不同。Andrea 的分数是所有学生中的中位数，并且她是自己队伍中得分最高的。Andrea 的队友 Beth 和 Carla 分别排第 $37^\text{th}$ 和第 $64^\text{th}$ ，这个城市有多少所学校？

Every high school in the city of Euclid sent a team of $3$ students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed $37^\text{th}$ and $64^\text{th},$ respectively. How many schools are in the city?

$22$

$23$

$24$

$25$

$26$

答案：B

知识点：中位数（数据）不等式逻辑推理

难度评级：1600

解答：

设有 $x$ 所学校，则共有 $3x$ 名参赛者。

因为每个分数不同且存在唯一中位数，参赛者总数必须为奇数，所以 $x$ 为奇数。

至少有 $23$ 支队伍，否则不会有第 $64^\text{th}$ 名。

另一方面，如果超过 $23$ 支队伍，Andrea 的中位名次会在 Beth 之后，这与 Andrea 是队内最高分矛盾。

因此有 $23$ 所学校。

所以正确答案是 B。

Let there be $x$ schools. Then there are $3x$ participants in the contest.

Since everyone has a unique score, there is also a unique median. This means $x$ is odd.

Note there are at least $23$ teams, since otherwise a $64^\text{th}$ placed wouldn't exist.

We also have at most $23$ teams, since otherwise we have Andrea's place as being greater than Beth's.

This means that there are $23$ teams.

Thus, B is the correct answer.

18.

正整数 $a$ 、 $b$ 、 $c$ 从集合 $\{1, 2, 3,\dots, 2010\}$ 中随机且独立地有放回选取。

$abc + ab + a$ 能被 $3$ 整除的概率是多少？

Positive integers $a,$ $b,$ and $c$ are randomly and independently selected with replacement from the set $\{1, 2, 3,\dots, 2010\}.$

What is the probability that $abc + ab + a$ is divisible by $3?$

$\dfrac{1}{3}$

$\dfrac{29}{81}$

$\dfrac{31}{81}$

$\dfrac{11}{27}$

$\dfrac{13}{27}$

答案：E

知识点：模运算基本概率分类讨论

难度评级：1660

解答：

注意如果 $a$ 能被 $3$ 整除，则整个表达式也能被三整除。 $abc + ab + a = a(bc + b + 1).$

因为 $2010$ 能被 $3$ 整除，所以 $a$ 能被 $3$ 整除的概率为 $\frac{1}{3}$ 。

若 $a$ 不能被 $3$ 整除，要使整个表达式能被 $3$ 整除，就需要 $bc + b + 1$ 能被 $3$ 整除。

这可写成 $bc + b = b(c + 1) \equiv 2 \pmod{3}.$

只有一个因子模 $3$ 为 $2$ 、另一个模 $3$ 为 $1$ 时才成立。

对于每一种次序，两个因子各有 $\frac{1}{3}$ 的概率取到所需的余数，所以该次序的概率为 $\dfrac{1}{3} \cdot \dfrac{1}{3} = \dfrac{1}{9}.$

共有两种次序，所以条件概率为 $\dfrac{2}{9}$ 。

$\dfrac{1}{3} \cdot 1 + \dfrac{2}{3} \cdot \dfrac{2}{9} = \dfrac{13}{27}.$

所以正确答案是 E。

Note that $abc + ab + a = a(bc + b + 1).$ This means that if $a$ is divisible by $3,$ the whole expression is as well.

Since $2010$ is divisible by $3,$ we have that $a$ is divisible by $3$ with probability $\frac{1}{3}.$

Now consider $a$ not divisible by $3.$ For the expression to be divisible by $3,$ we must have that $bc + b + 1$ is divisible by $3.$

This means that $bc + b = b(c + 1) \equiv 2 \pmod{3}.$

The only possibility for this is that one of the factors is $2$ mod $3$ and the other is $1$ mod $3.$

For each of the two cases, there is a $\frac{1}{3}$ chance that each of the factors is the desired modulus, for a probability of $\dfrac{1}{3} \cdot \dfrac{1}{3} = \dfrac{1}{9}.$

There are two cases, which means that this happens with a $\dfrac{2}{9}$ probability.

The total probability is then $\dfrac{1}{3} \cdot 1 + \dfrac{2}{3} \cdot \dfrac{2}{9} = \dfrac{13}{27}.$

Thus, E is the correct answer.

19.

圆心为 $O$ 的圆面积为 $156\pi$ 。 $\triangle ABC$ 是等边三角形， $\overline{BC}$ 是圆的一条弦， $OA = 4\sqrt{3}$ ，且点 $O$ 在 $\triangle ABC$ 外。求三角形 $ABC$ 的边长。

A circle with center $O$ has area $156\pi.$ Triangle $ABC$ is equilateral, $\overline{BC}$ is a chord on the circle, $OA = 4\sqrt{3},$ and point $O$ is outside $\triangle ABC.$ What is the side length of $\triangle ABC?$

$2\sqrt{3}$

$6$

$4\sqrt{3}$

$12$

$18$

答案：B

知识点：弦等边三角形勾股定理

难度评级：1860

解答：

参考下图：

圆半径满足 $BO = \sqrt{156}$ 。

延长 $\overline{AO}$ 与 $\overline{BC}$ 交于 $X$ 。设 $s$ 为 $\triangle ABC$ 的边长。

于是 $BX = \dfrac{s}{2}$ 且 $AX = \dfrac{s\sqrt3}{2}.$

$\triangle OXB$ 是直角三角形，所以 $(\sqrt{156})^2 = \left(\dfrac{s}{2}\right)^2$ $+ \left(\dfrac{s\sqrt3}{2} + 4\sqrt3\right)^2.$

化简得 $156 = s^2 + 12s + 48$ $s^2 + 12s - 108 = 0$ $(s - 6)(s + 18) = 0.$

由于 $s$ 为正， $s = 6$ 。

所以正确答案是 B。

Consider the following diagram:

Using the formula for the area of a circle, we have that $BO = \sqrt{156}$ since it is a radius.

Extend $\overline{AO}$ to intersect $\overline{BC}$ at $X.$ Let $s$ be the side length of $\triangle ABC.$

Then we have that $BX = \dfrac{s}{2}$ and $AX = \dfrac{s\sqrt3}{2}.$

We have that $\triangle OXB$ is right, which means that we can apply the Pythagorean Theorem. This gives us $(\sqrt{156})^2 = \left(\dfrac{s}{2}\right)^2$ $+ \left(\dfrac{s\sqrt3}{2} + 4\sqrt3\right)^2.$

Simplifying, we get $156 = s^2 + 12s + 48$ $s^2 + 12s - 108 = 0$ $(s - 6)(s + 18) = 0.$

Since $s$ is positive, we must have that $s = 6.$

Thus, B is the correct answer.

20.

两个圆位于正六边形 $ABCDEF$ 外。第一个圆与 $\overline{AB}$ 相切，第二个圆与 $\overline{DE}$ 相切。两个圆都与直线 $BC$ 和 $FA$ 相切。第二个圆面积与第一个圆面积的比是多少？

Two circles lie outside regular hexagon $ABCDEF.$ The first is tangent to $\overline{AB},$ and the second is tangent to $\overline{DE}.$ Both are tangent to lines $BC$ and $FA.$ What is the ratio of the area of the second circle to that of the first circle?

$18$

$27$

$36$

$81$

$108$

答案：D

知识点：正多边形切线特殊直角三角形面积比

难度评级：1960

解答：

参考下图：

设正六边形边长为 $1$ 。较小圆内切于一个边长为 $1$ 的等边三角形。

这个等边三角形的内切圆半径为 $\dfrac{\sqrt3}{6}$ 。面积为 $\pi \cdot \left(\dfrac{\sqrt3}{6}\right)^2 = \dfrac{\pi}{12}.$

设较大圆圆心为 $O$ ，从 $O$ 向 $\overline{GH}$ 作垂线，垂足为 $J$ 。并连接 $\overline{OG}$ 。

$\triangle OJG$ 是直角三角形，且由 $\angle HGI = 60^{\circ}$ 可知， $\triangle OJG$ 是 $30-60-90$ 三角形。

设 $OJ = r$ 。由三十、六十、九十度三角形关系， $OG = 2r$ 。接下来用 $OG$ 表示较大圆的位置。

于是 $OG = \dfrac{\sqrt3}{2} + \sqrt3 + r.$

代入 $OG$ ，得 $2r = \dfrac{\sqrt3}{2} + \sqrt3 + r.$ 化简得 $r = \dfrac{3\sqrt3}{2}.$

较大圆面积为 $\pi \cdot \left(\dfrac{3\sqrt3}{2}\right)^2 = \dfrac{27}{4}\pi.$

所求比值为 $\dfrac{\frac{27\pi}{4}}{\frac{\pi}{12}} = 81.$

所以正确答案是 D。

Consider the following diagram:

Assume the regular hexagon has side length $1.$ The smaller circle is inscribed in an equilateral triangle of side length $1.$

The inradius of this equilateral triangle is $\dfrac{\sqrt3}{6}.$ The area of the circle is then $\pi \cdot \left(\dfrac{\sqrt3}{6}\right)^2 = \dfrac{\pi}{12}.$

Let $O$ be the center of the larger circle. Drop the perpendicular from $O$ to $\overline{GH}$ at $J.$ Draw $\overline{OG}.$

We have that $\triangle OJG$ is right. Since $\angle HGI = 60^{\circ},$ we also have that $\triangle OJG$ is a $30-60-90$ triangle.

Let $OJ = r.$ Then $OG = 2r.$ We also have that $OG$ is the sum of the height of the hexagon, equilateral triangle, and radius of the circle.

Then $OG = \dfrac{\sqrt3}{2} + \sqrt3 + r.$

Substituting in $OG,$ we get $2r = \dfrac{\sqrt3}{2} + \sqrt3 + r.$ Simplifying gives us $r = \dfrac{3\sqrt3}{2}.$

The area of the larger circle is then $\pi \cdot \left(\dfrac{3\sqrt3}{2}\right)^2 = \dfrac{27}{4}\pi.$

The desired ratio is then $\dfrac{\frac{27\pi}{4}}{\frac{\pi}{12}} = 81.$

Thus, D is the correct answer.

21.

从 $1000$ 到 $10,000$ 之间随机选择一个回文数。它能被 $7$ 整除的概率是多少？

A palindrome between $1000$ and $10,000$ is chosen at random. What is the probability that it is divisible by $7?$

$\dfrac{1}{10}$

$\dfrac{1}{9}$

$\dfrac{1}{7}$

$\dfrac{1}{6}$

$\dfrac{1}{5}$

答案：E

知识点：回文数整除性基本概率

难度评级：1540

解答：

任意一个 $4$ 位数都可写成 $abcd$ ，展开后为 $10^3a + 10^2b + 10c + d.$ 对回文数， $a = d$ 且 $b = c$ ，所以可化为 $1001a + 110b.$ 因为 $1001$ 能被 $7$ 整除，所以还需要 $110b$ 能被 $7$ 整除。

$110$ 不能被 $7$ 整除，因此 $b$ 必须为 $0$ 或 $7$ 。

有 $9$ 种 $a$ 和 $2$ 种 $b$ ，符合条件的回文数共有 $9 \cdot 2 = 18$ 个。

四位回文数总数为 $9 \cdot 10$ ，因为千位有 $9$ 种选择、百位有 $10$ 种选择。

所求概率为 $\dfrac{18}{9 \cdot 10} = \dfrac{1}{5}.$

所以正确答案是 E。

Note that we can express any $4$ digit number as $abcd.$ This can be expressed in long form as $10^3a + 10^2b + 10c + d.$ Since in a palindrome, we have that $a = d$ and $b = c.$ We can simplify this to get $1001a + 110b.$ Note that $1001$ is divisible by $7.$ This means that $110b$ must also be divisible by $7.$

The only way for this to happen is if $b$ is $0$ or $7$ since $110$ is not divisible by $7.$

There are $9$ options for $a$ and $2$ options for $b,$ for a total of $9 \cdot 2 = 18$ palindromes.

The total number of palindromes is $9 \cdot 10$ since there are $9$ options for the thousands digit and $10$ options for the hundreds digit.

The desired probability is then $\dfrac{18}{9 \cdot 10} = \dfrac{1}{5}.$

Thus, E is the correct answer.

22.

七块不同的糖果要分到三个袋子中。红袋和蓝袋必须各至少得到一块糖，白袋可以为空。有多少种分法？

Seven distinct pieces of candy are to be distributed among three bags. The red bag and the blue bag must each receive at least one piece of candy; the white bag may remain empty. How many arrangements are possible?

$1930$

$1931$

$1932$

$1933$

$1934$

答案：C

知识点：补集计数容斥原理乘法原理

难度评级：1790

解答：

无限制时，每块糖有三个袋子可选，共 $3^7 = 2187.$

要数无效安排，需要考虑红袋或蓝袋为空的情况。

若红袋为空，则每块糖只有 $2$ 个袋子可选，共 $2^7 = 128$ 种安排。蓝袋为空也同样有 $128$ 种。

这两类有一个重叠情况：红袋和蓝袋都为空。因此有效分法的数量为 $2187 - (128 + 128 - 1) = 1932.$

所以正确答案是 C。

We can count this with complementary counting. The total number of ways to distribute the candies with no restrictions is $3^7 = 2187.$

To find the number of invalid arrangements, we have to count the number of ways where either the red or blue bag is empty.

For the case where the red bag is empty, each candy has $2$ options for the bag that goes into. There are then $2^7 = 128$ arrangements for this case. Similarly, there are $128$ arrangements for the case where the blue bag is empty.

There is an overlap of one case where both bags are empty. The final answer is then $2187 - (128 + 128 - 1) = 1932.$

Thus, C is the correct answer.

23.

一个 $3 \times 3$ 数组中填入数字 $1$ 到 $9$ ，每个数字恰好使用一次，并且每一行和每一列中的数字都按递增顺序排列。这样的数组有多少个？

The entries in a $3 \times 3$ array include all the digits from $1$ through $9,$ arranged so that the entries in every row and column are in increasing order. How many such arrays are there?

$18$

$24$

$36$

$42$

$60$

答案：D

知识点：有限制的排列分类讨论

难度评级：2030

解答：

$1$ 和 $9$ 必须分别在左上角和右下角。 $2$ 和 $8$ 必须分别与它们相邻。

中心格只能为 $4$ 、 $5$ 或 $6$ 。

情况一：中心格为 $4$ 。

此时 $3$ 必须与 $1$ 相邻，因为没有其他小于 $4$ 的位置可放。

与 $8$ 相邻的那个格有三种选择，其余两格随之确定。因此共有 $2 \cdot 2 \cdot 3 = 12$ 种： $2$ 有两个位置， $8$ 有两个位置，而与 $8$ 相邻的格有三种选择。

情况二：中心格为 $5$ 。

按 $3$ 的位置分类。若 $3$ 在右上角， $4$ 必须与 $1$ 相邻。

若 $8$ 在 $9$ 上方，其余两格被唯一确定；若它在 $9$ 左边，其余两格可以任意对调。

现在考虑 $3$ 在 $1$ 下方的情况。 $8$ 有两个可能位置，与 $8$ 相邻的格可选任意剩余数字。

其余两格随之确定。因此这种情况共有 $2(1 + 2 + 2 \cdot 3) = 18.$ 种排法。乘以二是因为 $2$ 可以在 $1$ 的右边或下方。

情况三：中心为 $6$ 。

这与情况 $1$ 类似，只是固定的数字是 $7$ 而不是 $3$ ，因此也有十二种。

总安排数为 $12 + 18 + 12 = 42.$

所以正确答案是 D。

Note that $1$ and $9$ must be in the top left and bottom right corners respectively. We must also have that $2$ and $8$ are next to these squares.

We can then case on the center square. Note that the only possible values are $4,$ $5,$ or $6.$

Case 1: the center is $4$

The $3$ is necessarily next to the $1,$ since there is no other option that is less than $4.$

Any number can be in the square next to the $8,$ but the other two squares are then fixed. There are $2 \cdot 2 \cdot 3 = 12$ cases (two places for the $2,$ two places for the $8,$ and three choices for the square adjacent to $8$ ).

Case 2: the center is $5$

We can case on the position of the $3.$ If the $3$ is in the top right square, the $4$ is necessarily next to the $1.$

If the $8$ is above the $9,$ then the other two squares are fixed. If it is to the left of the $9,$ the other two squares can be filled arbitrarily.

Now consider when the $3$ is below the $1.$ There are two spots for the $8,$ and the square next to the $8$ can be any number.

The other two squares are then fixed. This means that this case has a total of $2(1 + 2 + 2 \cdot 3) = 18.$ We multiply by two since the $2$ can be either to the right of or below the $1.$

Case 3: the center is $6$

This is similar to case $1$ since the $7$ is fixed instead of the $3.$

The total number of arrangements is then $12 + 18 + 12 = 42.$

Thus, D is the correct answer.

24.

Raiders 和 Wildcats 的一场高中篮球赛在第一节结束时打平。Raiders 四节每节得分构成一个递增等比数列，Wildcats 四节每节得分构成一个递增等差数列。第四节结束时 Raiders 以一分获胜。两队得分都不超过 $100$ 。两队上半场总共得了多少分？

A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than $100$ points. What was the total number of points scored by the two teams in the first half?

$30$

$31$

$32$

$33$

$34$

答案：E

知识点：等比数列等差数列丢番图方程

难度评级：2180

解答：

设两队第一节得分均为 $a$ 。设 Raiders 的公比为 $r$ ，Wildcats 的公差为 $d$ 。

将 $r=m/n$ 写成最简分数。由于 Raiders 四节得分都是整数， $a=n^3A$ ，其中 $A$ 为正整数，且总分为 $A(n^3+n^2m+nm^2+m^3)\le100$ ，所以只需检查 $\dfrac{3}{2},2,3,4$ 。

对 $r=\dfrac{3}{2}$ 或 $r=4$ ，Raiders 的可行总分无法使 Wildcats 总分成为 $4a+6d$ 的形式。对 $r=3$ ，方程 $40a=4a+6d+1$ 模 $6$ 不可能。

对 $r=2$ ，方程为 $15a=4a+6d+1$ ，所以 $11a=6d+1$ 。总分不超过 $100$ 的唯一正整数解是 $a=5,d=9$ 。

上半场总分为 $5+10+5+14=34$ 。

所以正确答案是 E。

Let the first-quarter score for each team be $a.$ Let the Raiders have common ratio $r$ and let the Wildcats have common difference $d.$

Write $r=m/n$ in lowest terms. Since the Raiders' four quarter scores are integers, $a=n^3A$ for some positive integer $A$ , and their total is $A(n^3+n^2m+nm^2+m^3)\le100$ . Thus the only possible ratios are $\dfrac{3}{2},2,3,4$ .

For $r=\dfrac{3}{2}$ or $r=4$ , the only possible Raiders totals give Wildcats totals that are not of the form $4a+6d$ . For $r=3$ , the equation $40a=4a+6d+1$ is impossible modulo $6$ .

For $r=2$ , the equation is $15a=4a+6d+1$ , so $11a=6d+1$ . The only positive solution with total at most $100$ is $a=5,d=9$ .

The first-half total is $5+10+5+14=34.$

Thus, E is the correct answer.

25.

设 $a \gt 0$ ，且 $P(x)$ 是一个整系数多项式，满足 $\begin{aligned} P(1) &= P(3) \\ &= P(5) = P(7) = a, \end{aligned}$ 并且 $P(2) = P(4) = P(6) = P(8)$ $= -a.$ $a$ 的最小可能值是多少？

Let $a \gt 0,$ and let $P(x)$ be a polynomial with integer coefficients such that $\begin{aligned} P(1) &= P(3) \\ &= P(5) = P(7) = a, \end{aligned}$ and $P(2) = P(4) = P(6) = P(8)$ $= -a.$ What is the smallest possible value of $a?$

$105$

$315$

$945$

$7!$

$8!$

答案：B

知识点：多项式整除性最小公倍数

难度评级：2350

解答：

因为 $1,3,5,7$ 是 $P(x)-a$ 的根，可写成 $P(x)-a=(x-1)(x-3)$ $\cdot(x-5)(x-7)Q(x)$ ，其中 $Q(x)$ 也有整数系数。

代入 $x=2,4,6,8$ ，得到 $-2a=-15Q(2)$ $=9Q(4)$ $=-15Q(6)$ $=105Q(8)$ 。因此 $a$ 必须是 $\operatorname{lcm}(15,9,105)=315$ 的倍数。

这个下界可以达到：取 $Q(x)=42$ $+(x-2)(x-6)(60-8x)$ ，并定义 $P(x)=315$ $+(x-1)(x-3)$ $\cdot(x-5)(x-7)Q(x)$ 。该多项式有整数系数并满足所有要求。

所以正确答案是 B。

Because $1,3,5,7$ are roots of $P(x)-a$ , write $P(x)-a=(x-1)(x-3)$ $\cdot(x-5)(x-7)Q(x)$ , where $Q(x)$ has integer coefficients.

Substituting $x=2,4,6,8$ gives $-2a=-15Q(2)$ $=9Q(4)$ $=-15Q(6)$ $=105Q(8)$ . Hence $a$ must be a multiple of $\operatorname{lcm}(15,9,105)=315$ .

This lower bound is attainable: take $Q(x)=42$ $+(x-2)(x-6)(60-8x)$ and define $P(x)=315$ $+(x-1)(x-3)$ $\cdot(x-5)(x-7)Q(x)$ . This polynomial has integer coefficients and satisfies the required values.

Thus, B is the correct answer.