2010 AMC 10A 详解 | LIVE by Po-Shen Loh

1.

Mary 书架最上层有五本书，它们的宽度分别为 $6$ 、 $\dfrac{1}{2}$ 、 $1$ 、 $2.5$ 和 $10$ 厘米。

这些书的平均宽度是多少厘米？

Mary’s top book shelf holds five books with the following widths, in centimeters: $6,$ $\dfrac{1}{2},$ $1,$ $2.5,$ and $10.$

What is the average book width, in centimeters?

$1$

$2$

$3$

$4$

$5$

知识点：平均数小数

难度评级：560

小提示：

先把宽度相加，再除以书的本数。

Add the widths before dividing by the number of books

大提示：

计算 $\left(6+\dfrac12+1+2.5+10\right)\div5$ 。

Compute $\left(6+\dfrac12+1+2.5+10\right)\div5$

解答：

五个宽度之和为 $6+\dfrac12+1+2.5+10=20$ 。平均宽度为 $20\div5=4$ 。

所以正确答案是 D。

Adding the five widths gives $6+\dfrac12+1+2.5+10=20.$ The average is $20\div5=4.$

Thus, D is the correct answer.

2.

四个相同的正方形和一个长方形拼在一起，形成如图所示的一个大正方形。这个长方形的长是宽的多少倍？

Four identical squares and one rectangle are placed together to form one large square as shown. The length of the rectangle is how many times as large as its width?

$\dfrac{5}{4}$

$\dfrac{4}{3}$

$\dfrac{3}{2}$

$2$

$3$

知识点：比与比例正方形（几何）矩形

难度评级：870

小提示：

设每个小正方形边长为 $s$ 。

Let each small square have side length $s$

大提示：

长方形的长为 $4s$ ，宽比大正方形边长少一个小正方形边长。

The rectangle is $4s$ long and one small square shorter than the large square

解答：

不妨设每个小正方形的边长为 $1$ 。

这样长方形的长为 $4$ ，宽为 $4 - 1 = 3$ 。

所求长宽比为 $\dfrac{4}{3}$ 。

所以正确答案是 B。

WLOG, let the side lengths of the squares be $1.$

This means that the length of the rectangle is $4.$ We also have that the width must be $4 - 1 = 3.$

The desired ratio is then $\dfrac{4}{3}.$

Thus, B is the correct answer.

3.

Tyrone 有 $97$ 颗弹珠，Eric 有 $11$ 颗弹珠。之后 Tyrone 给了 Eric 一些弹珠，使得最后 Tyrone 的弹珠数是 Eric 的两倍。Tyrone 给了 Eric 多少颗弹珠？

Tyrone had $97$ marbles and Eric had $11$ marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?

$3$

$13$

$18$

$25$

$29$

知识点：一次方程比与比例

难度评级：960

小提示：

弹珠总数保持不变。

The total number of marbles stays fixed

大提示：

最后的数量比为 $2:1$ 。

The final amounts are in the ratio $2:1$

解答：

设 Eric 最后有 $x$ 颗弹珠，则 Tyrone 最后有 $2x$ 颗。

总弹珠数为 $97 + 11 = 108$ ，所以 $3x = 108$ $x = 36\text{。}$

因此 Tyrone 最后有 $36 \cdot 2 = 72$ 颗弹珠，他给出了 $97 - 72 = 25$ 颗。

所以正确答案是 D。

Let $x$ be the number of marbles that Eric ends up with. Then Tyrone ends up with $2x.$

The total number of marbles is $97 + 11 = 108,$ so $3x = 108$ $x = 36.$

Then, Tyrone ends up with $36 \cdot 2 = 72$ marbles. This means he has to give away $97 - 72 = 25$ marbles.

Thus, D is the correct answer.

4.

一本要录成光盘的书，朗读需要 $412$ 分钟。每张光盘最多能容纳 $56$ 分钟的朗读。假设使用尽可能少的光盘，并且每张光盘包含相同长度的朗读内容。每张光盘包含多少分钟朗读？

A book that is to be recorded onto compact discs takes $412$ minutes to read aloud. Each disc can hold up to $56$ minutes of reading. Assume that the smallest possible number of discs is used and that each disc contains the same length of reading. How many minutes of reading will each disc contain?

$50.2$

$51.5$

$52.4$

$53.8$

$55.2$

知识点：整除性估算

难度评级：1030

小提示：

先确定所需光盘的最少数量，再求平均。

Determine the minimum number of discs before averaging

大提示：

$7$ 张光盘容量不够，而 $8$ 张足够。

$7$ discs hold too little, while $8$ discs suffice

解答：

注意 $7 \cdot 56 = 392$ ，而 $8 \cdot 56 = 448$ ，所以最少需要 $8$ 张光盘。

每张光盘包含的朗读时间为 $412 \div 8 = 51.5\text{。}$

所以正确答案是 B。

Note that $7 \cdot 56 = 392$ and $8 \cdot 56 = 448,$ which means that the minimum number of discs needed is $8.$

Then the minutes of reading that each disc contains is $412 \div 8 = 51.5.$

Thus, B is the correct answer.

5.

一个圆的周长为 $24\pi$ ，其面积为 $k\pi$ 。求 $k$ 的值。

The area of a circle whose circumference is $24\pi$ is $k\pi.$ What is the value of $k?$

$6$

$12$

$24$

$36$

$144$

知识点：圆周长圆面积

难度评级：960

小提示：

先用周长求半径。

Use the circumference to find the radius first

大提示：

由 $2\pi r=24\pi$ ，所求 $k$ 为 $r^2$ 。

From $2\pi r=24\pi$ , the desired $k$ is $r^2$

解答：

圆的周长公式为 $2 \pi r$ ，所以 $24\pi = 2\pi r$ $r = 12\text{。}$

圆面积为 $\pi r^2$ ，因此 $k \pi = \pi 12^2$ $k = 144\text{。}$

所以正确答案是 E。

Recall that the formula for the circumference of a circle is $2 \pi r.$ We then have that $24\pi = 2\pi r$ $r = 12.$

The area of a circle is $\pi r^2,$ so we have that $k \pi = \pi 12^2$ $k = 144.$

Thus, E is the correct answer.

6.

对正数 $x$ 和 $y$ ，运算 $\spadesuit (x,y)$ 定义为 $\spadesuit (x,y) = x-\dfrac{1}{y}$ 求 $\spadesuit (2,\spadesuit (2,2))$ 。

For positive numbers $x$ and $y$ the operation $\spadesuit (x,y)$ is defined as $\spadesuit (x,y) = x-\dfrac{1}{y}$ What is $\spadesuit (2,\spadesuit (2,2))?$

$\dfrac{2}{3}$

$1$

$\dfrac{4}{3}$

$\dfrac{5}{3}$

$2$

知识点：自定义运算分数

难度评级：1020

小提示：

先计算内层的 $\spadesuit$ 表达式。

Evaluate the inner $\spadesuit$ expression first

大提示：

$\spadesuit(2,2)=2-\dfrac12$ ，然后把它作为第二个输入。

$\spadesuit(2,2)=2-\dfrac12$ , then use that as the second input

解答：

先计算内层，得到 $\spadesuit (2, 2) = 2 - \dfrac{1}{2} = \dfrac{3}{2}\text{。}$ 于是 $\spadesuit \left(2, \dfrac{3}{2}\right) = 2 - \dfrac{1}{\frac{3}{2}} = \dfrac{4}{3}\text{。}$

所以正确答案是 C。

Evaluating the inner expression, we get $\spadesuit (2, 2) = 2 - \dfrac{1}{2} = \dfrac{3}{2}.$ Then we have $\spadesuit \left(2, \dfrac{3}{2}\right) = 2 - \dfrac{1}{\frac{3}{2}} = \dfrac{4}{3}.$

Thus, C is the correct answer.

7.

Crystal 为每日跑步规划了一条路线。她先向正北跑一英里，然后向东北跑一英里，再向东南跑一英里。最后一段她沿直线跑回起点。最后一段长多少英里？

Crystal has a running course marked out for her daily run. She starts this run by heading due north for one mile. She then runs northeast for one mile, then southeast for one mile. The last portion of her run takes her on a straight line back to where she started. How far, in miles, is this last portion of her run?

$1$

$\sqrt{2}$

$\sqrt{3}$

$2$

$2\sqrt{2}$

知识点：勾股定理向量

难度评级：1220

小提示：

东北和东南两段的竖直分量相互抵消。

The northeast and southeast vertical components cancel

大提示：

总位移的北向分量为 $1$ ，东向分量为 $\sqrt2$ 。

The net displacement has north component $1$ and east component $\sqrt2$

解答：

从图中可以看出，最后一段是一个直角三角形的斜边。

一条直角边来自正北方向的 $1$ 英里，另一条直角边为 $\sqrt{1^2 + 1^2} = \sqrt2\text{。}$

因此最后一段距离为 $\sqrt{\sqrt2^2 + 1^2} = \sqrt3\text{。}$

所以正确答案是 C。

From the diagram, we see that the distance traveled is the hypotenuse of a right triangle.

One of the legs is just $1$ from running due north. The other leg is $\sqrt{1^2 + 1^2} = \sqrt2.$

The final distance is then $\sqrt{\sqrt2^2 + 1^2} = \sqrt3.$

Thus, C is the correct answer.

8.

Tony 每天工作 $2$ 小时，并且每满一岁，每小时工资为 $\$0.50$ 。在六个月期间，Tony 工作了 $50$ 天，赚了 $\$630$ 。这六个月结束时 Tony 几岁？

Tony works $2$ hours a day and is paid $\$0.50$ per hour for each full year of his age. During a six month period Tony worked $50$ days and earned $\$630.$ How old was Tony at the end of the six month period?

$9$

$11$

$12$

$13$

$14$

知识点：年龄问题速率极限情形界定

难度评级：1370

小提示：

计算总工作小时数和平均小时工资。

Compute the total hours worked and average hourly pay

大提示：

用平均小时工资除以 $\$0.50$ ，得到平均年龄。

Divide the average hourly pay by $\$0.50$ to get the average age

解答：

Tony 共工作了 $2\cdot50=100$ 小时，所以平均时薪是 $\$630\div100=\$6.30$ 。因为他的时薪等于其整岁年龄乘以 $\$0.50$ ，所以工作日的平均年龄是 $6.30\div0.50=12.6$ 岁。在六个月内，他的年龄最多变化一岁，因此他必定有些工作日在 $12$ 岁，有些在 $13$ 岁。所以这段时间结束时他是 $13$ 岁。

所以正确答案是 D。

Tony worked $2\cdot50=100$ hours, so his average hourly pay was $\$630\div100=\$6.30.$ Because his hourly pay is $\$0.50$ times his age in full years, his average age on the days he worked was $6.30\div0.50=12.6$ years. During a six-month period his age can change by at most one, so he must have worked some days at age $12$ and some at age $13.$ Thus he was $13$ at the end of the period.

Thus, D is the correct answer.

9.

回文数，例如 $83438$ ，是一个数字反过来读仍相同的数。数 $x$ 和 $x + 32$ 分别是三位数和四位数回文数。 $x$ 的各位数字之和是多少？

A palindrome, such as $83438,$ is a number that remains the same when its digits are reversed. The numbers $x$ and $x + 32$ are three-digit and four-digit palindromes, respectively. What is the sum of the digits of $x?$

$20$

$21$

$22$

$23$

$24$

知识点：回文数数字极限情形界定

难度评级：1280

小提示：

将四位回文数限制在 $1000$ 到 $1031$ 之间。

Bound the four-digit palindrome between $1000$ and $1031$

大提示：

找出该区间内唯一的回文数，再减去 $32$ 。

Find the only palindrome in that interval, then subtract $32$

解答：

因为 $x$ 至多为 $999$ ，所以 $x + 32$ 至多为 $1031$ 。

同时 $x + 32$ 至少为 $1000$ 。

这个范围内唯一的回文数是 $1001$ ，所以 $x + 32$ 必须等于它。

于是可求出该数： $x + 32 = 1001$ $x = 969\text{。}$

各位数字之和为 $9 + 6 + 9 = 24\text{。}$

所以正确答案是 E。

Note that $x$ is at most $999.$ This means that $x + 32$ has a maximum of $1031.$

Similarly, we have that the minimum value of $x + 32$ is $1000.$

The only palindrome in this range is $1001,$ so this is what $x + 32$ equals.

Then $x + 32 = 1001$ $x = 969.$

The sum of the digits is then $9 + 6 + 9 = 24.$

Thus, E is the correct answer.

10.

Marvin 在闰年 $2008$ 年五月 $27$ 日星期二过生日。他的生日下一次落在星期六是哪一年？

Marvin had a birthday on Tuesday, May $27$ in the leap year $2008.$ In what year will his birthday next fall on a Saturday?

$2011$

$2012$

$2013$

$2015$

$2017$

知识点：日期与时间模运算

难度评级：1480

小提示：

每个非闰年会使同一日期的星期向后移 $1$ 天。

Each non-leap year shifts the weekday forward by $1$

大提示：

因为五月 $27$ 日在闰日之后，闰年会使它向后移 $2$ 天。

Since May $27$ is after leap day, leap years shift it forward by $2$

解答：

普通年有 $365 = 52 \cdot 7 + 1\text{，}$ 所以同一日期在下一年会向后移一天。

对于闰年中位于二月二十九日之后的日期，星期会向后移两天。

因此 $2009$ 年这一天是星期三， $2010$ 年是星期四。

$2011$ 年是星期五；由于 $2012$ 年是闰年，这一天变为星期日。

接下来三年每年向后移一天。到 $2016$ 年又因闰年向后移两天，落在星期五。

最后， $2017$ 年这一天是星期六。

所以正确答案是 E。

Note that on a normal year, we have that $365 = 52 \cdot 7 + 1,$ which means that for a specific day, it moves to the day after the next year.

On a leap year, the day of the week moves forward two since there is an extra day.

Then in $2009,$ this day falls on a Wednesday. In $2010,$ it falls on a Thursday.

Similarly, in $2011,$ it falls on a Friday. In $2012,$ however, since it is a leap year, it falls on a Sunday.

Now, for the next three years, the day moves forward one. Then in $2016,$ it moves forward two, landing on a Friday.

Finally, in $2017,$ the day of the week is a Saturday.

Thus, E is the correct answer.

11.

不等式 $a \le 2x + 3 \le b$ 的解区间长度为 $10$ 。求 $b - a$ 。

The length of the interval of solutions of the inequality $a \le 2x + 3 \le b$ is $10.$ What is $b - a?$

$6$

$10$

$15$

$20$

$30$

知识点：不等式一次方程

难度评级：1280

小提示：

解出 $x$ 的两个端点。

Solve the inequality for the two endpoints of $x$

大提示：

区间长度为 $\dfrac{b-3}{2}-\dfrac{a-3}{2}$ 。

The interval length is $\dfrac{b-3}{2}-\dfrac{a-3}{2}$

解答：

分别解两边的不等式，得到 $a \le 2x + 3$ $x \ge \dfrac{a - 3}{2}$ 以及 $2x + 3 \le b$ $x \le \dfrac{b - 3}{2}\text{。}$

因此解区间的长度满足 $\dfrac{b - 3}{2} - \dfrac{a - 3}{2} = 10\text{，}$ 化简得 $(b - 3) - (a - 3) = 20$ $b - a = 20\text{。}$

所以正确答案是 D。

Splitting the inequality into two of them and solving gives us $a \le 2x + 3$ $x \ge \dfrac{a - 3}{2}$ and $2x + 3 \le b$ $x \le \dfrac{b - 3}{2}.$

The range of the solutions is then $\dfrac{b - 3}{2} - \dfrac{a - 3}{2} = 10,$ which then simplifying gives us $(b - 3) - (a - 3) = 20$ $b - a = 20.$

Thus, D is the correct answer.

12.

Logan 正在制作他的城镇的比例模型。该城市的水塔高 $40$ 米，顶部是一个可容纳 $100{,}000$ 升水的球体。Logan 的迷你水塔可容纳 $0.1$ 升水。他应该把水塔做成多少米高？

Logan is constructing a scaled model of his town. The city’s water tower stands $40$ meters high, and the top portion is a sphere that holds $100{,}000$ liters of water. Logan’s miniature water tower holds $0.1$ liters. How tall, in meters, should Logan make his tower?

$0.04$

$\dfrac{0.4}{\pi}$

$0.4$

$\dfrac{4}{\pi}$

$4$

知识点：长度、面积与体积的缩放关系比与比例

难度评级：1420

小提示：

体积比例是长度比例的立方。

Volume scale is the cube of length scale

大提示：

比较 $0.1$ 升和 $100000$ 升，再取立方根。

Compare $0.1$ liters to $100000$ liters, then take a cube root

解答：

实际水塔的容量是迷你水塔的 $\dfrac{100,000}{0.1} = 1,000,000$ 倍。这是体积比，所以高度比为 $(1,000,000)^{\frac{1}{3}} = 100\text{。}$ 因此迷你水塔的高度为 $\dfrac{40}{100} = 0.4\text{。}$

所以正确答案是 C。

The miniature tower holds $\dfrac{100,000}{0.1} = 1,000,000$ times less water than the actual tower. Since this is the ratio for volumes, the ratio of heights is $(1,000,000)^{\frac{1}{3}} = 100.$ This means that the height of the miniature tower is $\dfrac{40}{100} = 0.4.$

Thus, C is the correct answer.

13.

Angelina 先以平均 $80$ 千米/小时的速度驾驶，然后停车加油 $20$ 分钟。停车后，她以平均 $100$ 千米/小时的速度驾驶。包括停车时间在内，她总共用 $3$ 小时行驶了 $250$ 千米。下列哪个方程可用来求她停车前驾驶的时间 $t$ ，单位为小时？

Angelina drove at an average rate of $80$ kph and then stopped $20$ minutes for gas. After the stop, she drove at an average rate of $100$ kph. Altogether she drove $250$ km in a total trip time of $3$ hours including the stop. Which equation could be used to solve for the time $t$ in hours that she drove before her stop?

$80t + 100\left(\dfrac83 - t\right) = 250$

$80t = 250$

$100t = 250$

$90t = 250$

$80\left(\dfrac83 - t\right) + 100t = 250$

知识点：路程、速度与时间一次方程

难度评级：1370

小提示：

从总时间中减去 $20$ 分钟的停车时间。

Subtract the $20$ -minute stop from the total time

大提示：

停车后，她驾驶了 $3-\dfrac13-t$ 小时。

After the stop, she drives for $3-\dfrac13-t$ hours

解答：

停车前，Angelina 行驶了 $80t$ 千米。

停车用时 $\frac{1}{3}$ 小时，所以总驾驶时间为 $3-\frac13=\frac83$ 小时；停车后她驾驶 $\frac83-t$ 小时，行驶 $100\left(\frac83-t\right)$ 千米。

因此方程为 $80t+100\left(\frac83-t\right)=250$ 。

所以正确答案是 A。

Before the stop, Angelina drove $80t$ km.

The stop takes $\frac{1}{3}$ of an hour, so her total driving time is $3-\frac13=\frac83$ hours. After the stop, she drives for $\frac83-t$ hours, covering $100\left(\frac83-t\right)$ km.

The total distance equation is $80t+100\left(\frac83-t\right)=250.$

Thus, A is the correct answer.

14.

三角形 $ABC$ 满足 $AB=2 \cdot AC$ 。点 $D$ 和 $E$ 分别在 $\overline{AB}$ 和 $\overline{BC}$ 上，且 $\angle BAE = \angle ACD$ 。设 $F$ 为线段 $AE$ 和 $CD$ 的交点，并且 $\triangle CFE$ 是等边三角形。 $\angle ACB$ 是多少？

Triangle $ABC$ has $AB=2 \cdot AC.$ Let $D$ and $E$ be on $\overline{AB}$ and $\overline{BC},$ respectively, such that $\angle BAE = \angle ACD.$ Let $F$ be the intersection of segments $AE$ and $CD,$ and suppose that $\triangle CFE$ is equilateral. What is $\angle ACB?$

$60^\circ$

$75^\circ$

$90^\circ$

$105^\circ$

$120^\circ$

知识点：导角等边三角形特殊直角三角形

难度评级：1660

小提示：

设 $\angle BAE=\angle ACD=x$ 。

Let $\angle BAE=\angle ACD=x$

大提示：

用 $\triangle CFE$ 求 $\angle AFC$ ，再追角求 $\angle BAC$ 。

Use $\triangle CFE$ to find $\angle AFC$ , then angle-chase $\angle BAC$

解答：

设 $\angle BAE = \angle ACD = x$ 。因为 $\triangle CFE$ 是等边三角形，所以 $\angle CFE = 60^{\circ}$ 。

因此 $\angle AFC = 180^{\circ} - \angle CFE = 120^{\circ}\text{。}$

因此 $\begin{aligned} \angle FAC &= 180^{\circ} - 120^{\circ} - x\\ &=60^{\circ} - x \\ &= \angle EAC\end{aligned}\text{。}$

于是 $\begin{aligned} \angle BAC &= \angle BAE + \angle EAC\\ &= x + 60^{\circ} - x \\&= 60^{\circ} \end{aligned}\text{。}$

由于 $AB = 2 \cdot AC$ 且 $\angle BAC = 60^{\circ}$ ， $\triangle ABC$ 是 $30-60-90$ 三角形。

所以正确答案是 C。

Let $\angle BAE = \angle ACD = x.$ Note that $\angle CFE = 60^{\circ}$ since $\triangle CFE$ is equilateral.

We then have that $\angle AFC = 180^{\circ} - \angle CFE = 120^{\circ}.$

Then: $\begin{aligned} \angle FAC &= 180^{\circ} - 120^{\circ} - x\\ &=60^{\circ} - x \\ &= \angle EAC.\end{aligned}$

We then get that $\begin{aligned} \angle BAC &= \angle BAE + \angle EAC\\ &= x + 60^{\circ} - x \\&= 60^{\circ}. \end{aligned}$

Since $AB = 2 \cdot AC$ and $\angle BAC = 60^{\circ},$ we have that $\triangle ABC$ is a $30-60-90$ triangle.

Thus, C is the correct answer.

15.

在一个神奇沼泽中，有两种会说话的两栖动物：蟾蜍说的话总是真的，青蛙说的话总是假的。Brian、Chris、LeRoy 和 Mike 四只两栖动物住在这个沼泽里，并作出如下陈述。

Brian：“Mike 和我是不同物种。”

Chris：“LeRoy 是青蛙。”

LeRoy：“Chris 是青蛙。”

Mike：“我们四个中至少有两个是蟾蜍。”

这四只两栖动物中有多少只是青蛙？

In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.

Brian: “Mike and I are different species.”

Chris: “LeRoy is a frog.”

LeRoy: “Chris is a frog.”

Mike: “Of the four of us, at least two are toads.”

How many of these four amphibians are frogs?

$0$

$1$

$2$

$3$

$4$

知识点：说真话者与说谎者逻辑推理

难度评级：1540

小提示：

Chris 和 LeRoy 不可能是同一物种。

Chris and LeRoy cannot have the same species

大提示：

如果 Brian 是蟾蜍，Mike 的陈述会导致矛盾。

If Brian were a toad, Mike’s statement would create a contradiction

解答：

Chris 和 LeRoy 不可能都是青蛙，否则他们的陈述都会是真的；也不可能都是蟾蜍，否则他们的陈述都会是假的。因此两者中恰有一只是蟾蜍。

如果 Brian 是蟾蜍，他的陈述会说明 Mike 是青蛙。这样 Brian 与 Chris、LeRoy 中的那只蟾蜍就使 Mike 的陈述为真，这与青蛙总说假话矛盾。因此 Brian 是青蛙。他的陈述为假，所以 Mike 也是青蛙。再加上 Chris、LeRoy 中的那只青蛙，共有 $3$ 只青蛙。

所以正确答案是 D。

Chris and LeRoy cannot both be frogs, because then both of their statements would be true. They cannot both be toads either, because then both statements would be false. Thus exactly one of them is a toad.

If Brian were a toad, his statement would make Mike a frog. Brian and the one toad among Chris and LeRoy would then make Mike’s statement true, which is impossible for a frog. Therefore Brian is a frog. His statement is false, so Mike is also a frog. Along with the one frog among Chris and LeRoy, there are $3$ frogs.

Thus, D is the correct answer.

16.

非退化 $\triangle ABC$ 的边长都是整数， $\overline{BD}$ 是角平分线， $AD = 3$ ， $DC = 8$ 。周长的最小可能值是多少？

Nondegenerate $\triangle ABC$ has integer side lengths, $\overline{BD}$ is an angle bisector, $AD = 3,$ and $DC = 8.$ What is the smallest possible value of the perimeter?

$30$

$33$

$35$

$36$

$37$

知识点：角平分线定理三角不等式最优化

难度评级：1600

小提示：

使用角平分线定理。

Use the Angle Bisector Theorem

大提示：

$AB:BC=3:8$ ，并且 $AC=11$ 。

$AB:BC=3:8$ , and $AC=11$

解答：

由角平分线定理， $\dfrac{AB}{3} = \dfrac{BC}{8}$ $AB = \dfrac{3}{8} BC\text{。}$

为使 $AB$ 和 $BC$ 都是整数， $BC$ 必须是 $8$ 的倍数。

若为了最小化周长取 $BC = 8$ 、 $AB = 3$ ，三角形会退化。

因此 $BC$ 必须取 $16$ ，此时 $AB = 6$ 。又因为 $AC = AD + DC = 11$ ，所以周长为 $16 + 6 + 11 = 33\text{。}$

所以正确答案是 B。

Using the Angle Bisector Theorem, we have that $\dfrac{AB}{3} = \dfrac{BC}{8}$ $AB = \dfrac{3}{8} BC.$

For $AB$ and $BC$ to be integers, we must have that $BC$ is a multiple of $8.$

To minimize the perimeter, we can set $BC = 8$ and $AB = 3.$ This, however, makes the triangle degenerate.

$BC$ must then be $16$ and $AB = 6.$ Since $AC = AD + DC = 11,$ the perimeter is $16 + 6 + 11 = 33.$

Thus, B is the correct answer.

17.

一个实心立方体边长为 $3$ 英寸。在每个面的中心切出一个 $2$ 英寸乘 $2$ 英寸的正方形孔。每个切口的边都与立方体的边平行，并且每个孔都贯穿整个立方体。剩余立体的体积是多少立方英寸？

A solid cube has side length $3$ inches. A $2$ -inch by $2$ -inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?

$7$

$8$

$10$

$12$

$15$

知识点：体积容斥原理立体几何

难度评级：1790

小提示：

对三个长方体孔使用容斥。

Use inclusion-exclusion for the three rectangular holes

大提示：

三个 $2\times2\times3$ 的孔在中心的 $2\times2\times2$ 立方体中重叠。

The three $2\times2\times3$ holes overlap in the central $2\times2\times2$ cube

解答：

三个被切出的长方体都在立方体中心相交。

交集是边长为 $2$ 的立方体。因此被切除区域的体积为 $\begin{aligned}3 \cdot 2 \cdot 2 \cdot 3 - 2 \cdot 2^3 &= 36 - 16 \\&= 20\end{aligned}\text{。}$

中心区域被所有 $3$ 个孔都包含，所以要减去重复计入的两次。

剩余体积为 $3^3 - 20 = 27 - 20 = 7\text{。}$

所以正确答案是 A。

Note that all the cut out solids intersect in the middle of the cube.

This region of intersection is a cube with side length $2.$ Then the volume of the cutout region is $\begin{aligned}3 \cdot 2 \cdot 2 \cdot 3 - 2 \cdot 2^3 &= 36 - 16 \\&= 20.\end{aligned}$

We have to subtract out the center region twice since it is included in all $3$ regions.

The remaining volume is then $3^3 - 20 = 27 - 20 = 7.$

Thus, A is the correct answer.

18.

Bernardo 从集合 $\{1,2,3,4,5,6,7,8,9\}$ 中随机选出 $3$ 个不同的数，并按降序排列形成一个 $3$ 位数。Silvia 从集合 $\{1,2,3,4,5,6,7,8\}$ 中随机选出 $3$ 个不同的数，也按降序排列形成一个 $3$ 位数。Bernardo 的数大于 Silvia 的数的概率是多少？

Bernardo randomly picks $3$ distinct numbers from the set $\{1,2,3,4,5,6,7,8,9\}$ and arranges them in descending order to form a $3$ -digit number. Silvia randomly picks $3$ distinct numbers from the set $\{1,2,3,4,5,6,7,8\}$ and also arranges them in descending order to form a $3$ -digit number. What is the probability that Bernardo’s number is larger than Silvia’s number?

$\dfrac{47}{72}$

$\dfrac{37}{56}$

$\dfrac{2}{3}$

$\dfrac{49}{72}$

$\dfrac{39}{56}$

知识点：基本概率分类讨论对称性

难度评级：1900

小提示：

按 Bernardo 是否选到 $9$ 分情况。

Separate cases according to whether Bernardo picks $9$

大提示：

若没有 $9$ ，两人的数对称，除了选到相同集合的情况。

Without a $9$ , the two numbers are symmetric except when the chosen sets match

解答：

分两种情况：Bernardo 选到 $9$ ，或没有选到这个数。

情况 $1$ ：Bernardo 选到 $9$ 。

由于一个数字已经固定，另外两个数有 $\binom{8}{2} = 28$ 种选法。

总选法为 $\binom{9}{3} = 84$ ，概率为 $\dfrac{28}{84} = \dfrac{1}{3}\text{。}$

注意，如果 Bernardo 选到 $9$ ，他一定比 Silvia 的数大。

因此 Bernardo 在这种情况下总是获胜。

情况 $2$ ：Bernardo 没有选到 $9$ 。

这种情况发生的概率为 $1 - \frac{1}{3} = \frac{2}{3}$ 。由于两人此时从相同的数字中选择，除了平局以外，双方获胜的概率相同。

Silvia 与 Bernardo 选到同一组数字的概率为 $\dfrac{1}{\binom{8}{3}} = \dfrac{1}{56}\text{。}$ 所以在这种情况下，Bernardo 的数较大的概率为 $\dfrac{1 - \frac{1}{56}}{2} = \dfrac{55}{112}\text{。}$

因此总概率为 $\dfrac{1}{3} + \dfrac{2}{3} \cdot \dfrac{55}{112} = \dfrac{37}{56}\text{。}$

所以正确答案是 B。

There are two cases: Bernardo picks a $9$ or he doesn’t.

Case $1:$ Bernardo picks a $9$

Since a number is fixed, there are $\binom{8}{2} = 28$ ways to choose the other two numbers.

There are a total of $\binom{9}{3} = 84$ ways to pick all three numbers. The probability is then $\dfrac{28}{84} = \dfrac{1}{3}.$

Note that if Bernardo picks a $9,$ he automatically has a greater number than Silvia.

This means that Bernardo always wins in this case.

Case $2:$ Bernardo doesn’t pick a $9$

There is a $1 - \frac{1}{3} = \frac{2}{3}$ chance of this happening. Since both people are choosing from the same numbers, they have an equal chance of winning.

We still need to find the probability that the numbers are the same. There is a $\dfrac{1}{\binom{8}{3}} = \dfrac{1}{56}$ chance that Silvia chooses the same numbers as Bernardo. The probability that Bernardo gets a higher number is then $\dfrac{1 - \frac{1}{56}}{2} = \dfrac{55}{112}.$

The total probability of Bernardo getting a higher number is then $\dfrac{1}{3} + \dfrac{2}{3} \cdot \dfrac{55}{112} = \dfrac{37}{56}.$

Thus, B is the correct answer.

19.

等角六边形 $ABCDEF$ 的边长满足 $AB=CD=EF=1$ 和 $BC=DE=FA=r\text{。}$ $\triangle ACE$ 的面积是六边形面积的 $70\%$ 。 $r$ 的所有可能值之和是多少？

Equiangular hexagon $ABCDEF$ has side lengths $AB=CD=EF=1$ and $BC=DE=FA=r.$ The area of $\triangle ACE$ is $70\%$ of the area of the hexagon. What is the sum of all possible values of $r?$

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

$\dfrac{10}{3}$

$4$

$\dfrac{17}{4}$

$6$

知识点：等角多边形面积比余弦定理韦达定理

难度评级：1960

小提示：

将六边形分成 $\triangle ACE$ 和三个角上的三角形。

Split the hexagon into $\triangle ACE$ and three corner triangles

大提示：

用 $r^2+r+1$ 和 $r$ 表示两个面积。

Express both areas using $r^2+r+1$ and $r$

解答：

注意 $\triangle ACE$ 是等边三角形。在 $\triangle ABC$ 中用余弦定理，得到 $AC^2=r^2+1^2$ $-2r\cos120^\circ$ $=r^2+r+1$ 。

因此 $\triangle ACE$ 的面积为 $\dfrac{\sqrt3}{4} (r^2 + r + 1)\text{。}$

三个角上的三角形 $\triangle ABC$ 、 $\triangle CDE$ 和 $\triangle EFA$ 各自面积为 $\frac12\cdot1\cdot r\cdot\sin120^\circ=\frac{r\sqrt3}{4}$ 。

因此六边形的面积为 $\dfrac{\sqrt3}{4}(r^2+r+1)$ $+3\cdot\dfrac{r\sqrt3}{4}$ $=\dfrac{\sqrt3}{4}(r^2+4r+1)$ 。

条件 $[ACE]=70\%\cdot[ABCDEF]$ 给出 $r^2+r+1=\dfrac{7}{10}(r^2+4r+1)\text{，}$ 所以 $r^2-6r+1=0$ 。

由韦达定理，所有可能 $r$ 的和为 $6$ 。

所以正确答案是 E。

Note that $\triangle ACE$ is equilateral. Using the Law of Cosines in $\triangle ABC,$ we get $AC^2=r^2+1^2$ $-2r\cos120^\circ$ $=r^2+r+1.$

The area of $\triangle ACE$ is then $\dfrac{\sqrt3}{4} (r^2 + r + 1).$

The three corner triangles $\triangle ABC,$ $\triangle CDE,$ and $\triangle EFA$ each have area $\frac12\cdot1\cdot r\cdot\sin120^\circ=\frac{r\sqrt3}{4}.$

Thus the hexagon has area $\dfrac{\sqrt3}{4}(r^2+r+1)$ $+3\cdot\dfrac{r\sqrt3}{4}$ $=\dfrac{\sqrt3}{4}(r^2+4r+1).$

The condition $[ACE]=70\%\cdot[ABCDEF]$ gives $r^2+r+1=\dfrac{7}{10}(r^2+4r+1),$ so $r^2-6r+1=0.$

By Vieta’s formulas, the sum of the possible values of $r$ is $6.$

Thus, E is the correct answer.

20.

一只被困在边长为 $1$ 米的立方体盒子内的苍蝇，决定通过访问盒子的每个角来打发无聊。它从某个角出发并回到同一个角，且其他每个角都恰好访问一次。它从一个角到另一个角时，要么飞行，要么爬行，路径都是直线。它的路径最大可能长度是多少米？

A fly trapped inside a cubical box with side length $1$ meter decides to relieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once. To get from a corner to any other corner, it will either fly or crawl in a straight line. What is the maximum possible length, in meters, of its path?

$4+4\sqrt{2}$

$2+4\sqrt{2}+2\sqrt{3}$

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

$4\sqrt{2}+4\sqrt{3}$

$3\sqrt{2}+5\sqrt{3}$

知识点：正方体图论最优化

难度评级：2070

小提示：

每一步长度为 $1,\sqrt2$ ，或 $\sqrt3$ 。

Each move has length $1,\sqrt2,$ or $\sqrt3$

大提示：

立方体中只有 $4$ 条空间对角线，所以剩余步长至多为面对角线。

There are only $4$ space diagonals, so the remaining moves are at most face diagonals

解答：

苍蝇每一步可能的长度只有 $1, \sqrt2$ 或 $\sqrt3$ 。

立方体只有 $4$ 条空间对角线，所以最多 $4$ 步长度为 $\sqrt3$ 。另外 $4$ 步长度最多为 $\sqrt2$ 。

这个上界可以达到，例如在顶点间交替使用空间对角线和面对角线。

因此路径长度为 $4\sqrt2 + 4\sqrt3\text{。}$

所以正确答案是 D。

Note that all the paths the fly can take have lengths of $1, \sqrt2,$ or $\sqrt3.$

There are only $4$ space diagonals in the cube, so at most $4$ moves can have length $\sqrt3.$ The other $4$ moves have length at most $\sqrt2.$

This upper bound is attainable, for example by alternating space diagonals and face diagonals around the vertices.

The path has length $4\sqrt2 + 4\sqrt3.$

Thus, D is the correct answer.

21.

多项式 $x^3-ax^2+bx-2010$ 有三个正整数零点。 $a$ 的最小可能值是多少？

The polynomial $x^3-ax^2+bx-2010$ has three positive integer zeros. What is the smallest possible value of $a?$

$78$

$88$

$98$

$108$

$118$

知识点：韦达定理质因数分解最优化

难度评级：2070

小提示：

根的乘积为 $2010$ ，和为 $a$ 。

The roots multiply to $2010$ and sum to $a$

大提示：

把 $67$ 单独放入一个根，再把 $30$ 分成两个因数使和最小。

Put $67$ alone, then split $30$ into two factors with smallest sum

解答：

设三个正整数根为 $r\le s\le t$ 。由韦达定理， $rst=2010$ 且 $a=r+s+t$ 。

因为 $2010=2\cdot3\cdot5\cdot67$ ，某个根必须能被 $67$ 整除。如果这个根大于 $67$ ，则至少为 $134$ ，已经比下面构造的和更大。

因此取 $t=67$ ，并在 $rs=30$ 下最小化 $r+s$ 。和最小的因数对是 $5$ 和 $6$ ，所以 $a=5+6+67=78$ 。

所以正确答案是 A。

Let the roots be positive integers $r\le s\le t.$ By Vieta’s formulas, $rst=2010$ and $a=r+s+t.$

Since $2010=2\cdot3\cdot5\cdot67,$ one root must be divisible by $67.$ If that root is larger than $67,$ then it is at least $134,$ which is already worse than the construction below.

Thus take $t=67$ and minimize $r+s$ with $rs=30.$ The factor pair with smallest sum is $5$ and $6,$ so $a=5+6+67=78.$

Thus, A is the correct answer.

22.

在一个圆上选取八个点，并连接每一对点得到弦。没有三条弦在圆内部同一点相交。会形成多少个三个顶点都在圆内部的三角形？

Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point inside the circle. How many triangles with all three vertices in the interior of the circle are created?

$28$

$56$

$70$

$84$

$140$

知识点：交点计数组合双射

难度评级：2160

小提示：

先选择作为三条弦端点的六个圆上点。

Choose the six circle points used as endpoints of the three chords

大提示：

对按圆周顺序排列的六个点，只有连接三对相对端点的弦能形成内部三角形。

For six points in order, only the three opposite-pair chords make the interior triangle

解答：

一个内部三角形由三条在圆内部两两相交的弦形成。这样的三角形使用圆上的六个不同端点。

反过来，对任意选出的六个圆上点，按圆周顺序看，只有一种方法把相对端点配对成三条弦，使这三条弦在圆内两两相交。

因此三角形的数量为 $\binom{8}{6}=\binom{8}{2}=28$ 。

所以正确答案是 A。

An interior triangle is formed by three chords that pairwise intersect inside the circle. Such a triangle uses six distinct endpoints on the circle.

Conversely, for any six chosen points in circular order, exactly one set of three chords pairs opposite endpoints so that the three chords intersect pairwise inside the circle.

Therefore the number of triangles is $\binom{8}{6}=\binom{8}{2}=28.$

Thus, A is the correct answer.

23.

一排有 $2010$ 个盒子，每个盒子里有一颗红弹珠；对 $1 \le k \le 2010$ ，第 $k$ 个盒子还含有 $k$ 颗白弹珠。Isabella 从第一个盒子开始，依次从每个盒子中随机抽取一颗弹珠。她第一次抽到红弹珠时停止。令 $P(n)$ 为 Isabella 正好抽取 $n$ 颗弹珠后停止的概率。使 $P(n) \lt \dfrac{1}{2010}$ 的最小 $n$ 是多少？

Each of $2010$ boxes in a line contains a single red marble, and for $1 \le k \le 2010,$ the box in the $k$ th position also contains $k$ white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let $P(n)$ be the probability that Isabella stops after drawing exactly $n$ marbles. What is the smallest value of $n$ for which $P(n) \lt \dfrac{1}{2010}?$

$45$

$63$

$64$

$201$

$1005$

知识点：基本概率裂项相消不等式

难度评级：2240

小提示：

前 $n-1$ 次必须抽到白弹珠，第 $n$ 次抽到红弹珠。

The first $n-1$ draws must be white, then the $n$ th draw red

大提示：

抽到白弹珠的概率连乘后会逐项相消： $\dfrac12\cdot\dfrac23\cdots\dfrac{n-1}{n}$ 。

The white probabilities telescope: $\dfrac12\cdot\dfrac23\cdots\dfrac{n-1}{n}$

解答：

第 $k$ 个盒子共有 $k + 1$ 颗弹珠，所以从中抽到白弹珠的概率为 $\dfrac{k}{k + 1}$ 。

抽到红弹珠的概率为 $\dfrac{1}{k + 1}$ 。要在第 $n$ 次抽取后停止，前 $n - 1$ 次必须抽到白弹珠，最后一次抽到红弹珠。

概率为 $\dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \ldots \cdot \dfrac{n - 1}{n} \cdot \dfrac{1}{n + 1}\text{。}$

分子与相邻分母相消，得到概率为 $\dfrac{1}{n(n + 1)}$ 。

我们需要找出使下式成立的最小 $n$ ： $\dfrac{1}{n(n + 1)} \lt \dfrac{1}{2010}$ $n(n + 1) \gt 2010\text{。}$

逐一试算可知，满足条件的最小 $n$ 为 $45$ 。

所以正确答案是 A。

Since there are $k + 1$ marbles in the $k$ th box, there is a $\dfrac{k}{k + 1}$ chance Isabella draws a white marble from it.

The probability of drawing a red marble is then $\dfrac{1}{k + 1}.$ To stop after drawing the $n$ th marble, the first $n - 1$ marbles must have been white.

This happens with a probability of $\dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \ldots \cdot \dfrac{n - 1}{n} \cdot \dfrac{1}{n + 1}.$

Note that all the numerators cancel with the adjacent denominator, which means that this expression reduces to $\dfrac{1}{n(n + 1)}.$

We have to find the smallest $n$ such that $\dfrac{1}{n(n + 1)} \lt \dfrac{1}{2010}$ $n(n + 1) \gt 2010.$

Guessing and checking gives us that the smallest $n$ that works is $45.$

Thus, A is the correct answer.

24.

$90!$ 的最后两个非零数字组成的数等于 $n$ 。 $n$ 是多少？

The number obtained from the last two nonzero digits of $90!$ is equal to $n.$ What is $n?$

$12$

$32$

$48$

$52$

$68$

知识点：阶乘模运算中国剩余定理末尾零

难度评级：2390

小提示：

去掉产生末尾零的因数，然后模 $100$ 计算。

Remove the factors making trailing zeroes, then work modulo $100$

大提示：

用模 $4$ 和模 $25$ 的同余确定最后两位。

Use congruences modulo $4$ and modulo $25$ to pin down the last two digits

解答：

$90!$ 末尾零的个数为 $\left\lfloor\dfrac{90}{5}\right\rfloor+\left\lfloor\dfrac{90}{25}\right\rfloor=21$ 。令 $N=\dfrac{90!}{10^{21}}$ 。

去掉 $10^{21}$ 后仍剩下超过两个因数 $2$ ，所以 $N\equiv0 \pmod4$ 。

设 $A$ 为 $90!$ 中不被 $5$ 整除的因数之积， $B$ 为被 $5$ 整除的因数之积。按模 $25$ 的余数分组，得到 $A\equiv1\pmod{25}$ 且 $\dfrac{B}{5^{21}}\equiv-1\pmod{25}$ 。

因此 $\dfrac{90!}{5^{21}}\equiv-1\pmod{25}$ 。又因为 $2^{21}\equiv2\pmod{25}$ ，所以 $N=\dfrac{90!}{5^{21}\cdot2^{21}}$ $\equiv-13$ $\equiv12\pmod{25}$ 。

同时满足 $0\pmod4$ 和 $12\pmod{25}$ 的数为 $12\pmod{100}$ ，所以最后两个非零数字组成 $12$ 。

所以正确答案是 A。

The number of trailing zeroes in $90!$ is $\left\lfloor\dfrac{90}{5}\right\rfloor+\left\lfloor\dfrac{90}{25}\right\rfloor=21.$ Let $N=\dfrac{90!}{10^{21}}.$

There are still more than two factors of $2$ left after removing $10^{21},$ so $N\equiv0 \pmod4.$

Let $A$ be the product of factors of $90!$ not divisible by $5,$ and let $B$ be the product of the factors divisible by $5.$ Grouping residues modulo $25$ gives $A\equiv1\pmod{25}$ and $\dfrac{B}{5^{21}}\equiv-1\pmod{25}.$

Therefore $\dfrac{90!}{5^{21}}\equiv-1\pmod{25}.$ Since $2^{21}\equiv2\pmod{25},$ $N=\dfrac{90!}{5^{21}\cdot2^{21}}$ $\equiv-13$ $\equiv12\pmod{25}.$

The number congruent to $0\pmod4$ and $12\pmod{25}$ is $12\pmod{100},$ so the last two nonzero digits form $12.$

Thus, A is the correct answer.

25.

Jim 从一个正整数 $n$ 开始构造一个数列。每个后续数都由当前数减去不超过当前数的最大整数平方得到，直到达到零为止。例如，如果 Jim 从 $n = 55$ 开始，则他的数列包含 $5$ 个数： $\begin{array}{ccccc} {}&{}&{}&{}&55\\ 55&-&7^2&=&6\\ 6&-&2^2&=&2\\ 2&-&1^2&=&1\\ 1&-&1^2&=&0\\ \end{array}$ 设 $N$ 为使 Jim 的数列有 $8$ 个数的最小数。 $N$ 的个位数字是多少？

Jim starts with a positive integer $n$ and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with $n = 55,$ then his sequence contains $5$ numbers: $\begin{array}{ccccc} {}&{}&{}&{}&55\\ 55&-&7^2&=&6\\ 6&-&2^2&=&2\\ 2&-&1^2&=&1\\ 1&-&1^2&=&0\\ \end{array}$ Let $N$ be the smallest number for which Jim’s sequence has $8$ numbers. What is the units digit of $N?$

$1$

$3$

$5$

$7$

$9$

知识点：完全平方数递推逆推法

难度评级：2440

小提示：

从 $0$ 开始反向构造最小可能数列。

Build the smallest possible sequence backwards from $0$

大提示：

每一步反向添加一个最小的平方，同时保证贪心规则会减去它。

At each reverse step, add the smallest square that the greedy rule would subtract

解答：

从 $0$ 开始反向工作。加上 $1^2$ 得到 $1$ 。

再次加上 $1^2$ 得到 $2$ ，再加 $1^2$ 得到 $3$ 。

若此时再加 $1^2$ ，会得到 $4$ ，但对于 $4$ ， $1^2$ 不是不超过它的最大平方。

因此下一步加 $2^2$ 得到 $7$ 。继续反向构造，直到正向数列共有八项，得到 $7223$ 。

$\begin{array}{ccccc} {}&{}&{}&{}&7223\\ 7223&-&84^2&=&167\\ 167&-&12^2&=&23\\ 23&-&4^2&=&7\\ 7&-&2^2&=&3\\ 3&-&1^2&=&2\\2&-&1^2&=&1\\1&-&1^2&=&0\end{array}$

所以正确答案是 B。

We can work backwards starting with $0.$ From this, we can add on $1^2$ to get $1.$

We can again add on $1^2$ to get $2.$ Again, adding on $1^2$ gives us $3.$

If we add on $1^2$ now, we get $4,$ but then $1^2$ is not the greatest square less than or equal to $4.$

Then adding on $2^2$ gives us $7.$ Continuing until there are eight terms in the forward sequence gives $7223.$

$\begin{array}{ccccc} {}&{}&{}&{}&7223\\ 7223&-&84^2&=&167\\ 167&-&12^2&=&23\\ 23&-&4^2&=&7\\ 7&-&2^2&=&3\\ 3&-&1^2&=&2\\2&-&1^2&=&1\\1&-&1^2&=&0\end{array}$

Thus, B is the correct answer.