2017 AMC 10B 详解 | LIVE by Po-Shen Loh

1.

Mary 想了一个正的两位数。她把它乘以 $3$ ，再加上 $11$ 。然后她交换所得结果的两个数字，得到一个介于 $71$ 和 $75$ 之间（含端点）的数。Mary 原来的数是多少？

Mary thought of a positive two-digit number. She multiplied it by $3$ and added $11.$ Then she switched the digits of the result, obtaining a number between $71$ and $75,$ inclusive. What was Mary’s number?

$11$

$12$

$13$

$14$

$15$

知识点：数字整除性逆推法

难度评级：960

小提示：

从可能的交换后结果倒推。

Work backward from the possible reversed results

大提示：

交换回去并减去 $11$ 后，检查是否能被 $3$ 整除。

After reversing and subtracting $11$ , check divisibility by $3$

解答：

从可能的结果倒推。把 $71,72,73,74,75$ 的两个数字换回去，分别得到 $17,27,37,47,57$ 。再减去 $11$ ，依次得到 $6,16,26,36,46$ 。其中只有 $6$ 和 $36$ 能被 $3$ 整除，而其中只有 $36\div3=12$ 是两位数。

所以正确答案是 B。

Work backward from the possible results. Reversing $71,72,73,74,75$ gives $17,27,37,47,57,$ respectively. Subtracting $11$ gives $6,16,26,36,46.$ Only $6$ and $36$ are divisible by $3,$ and only $36\div3=12$ is a two-digit number.

Thus, the correct answer is B .

2.

Sofia 在学校的 $400$ 米跑道上跑了 $5$ 圈。每一圈中，她前 $100$ 米的平均速度为每秒 $4$ 米，剩下 $300$ 米的平均速度为每秒 $5$ 米。Sofia 跑完 $5$ 圈共花了多少时间？

Sofia ran $5$ laps around the $400$ -meter track at her school. For each lap, she ran the first $100$ meters at an average speed of $4$ meters per second and the remaining $300$ meters at an average speed of $5$ meters per second. How much time did Sofia take running the $5$ laps?

$5$ 分 $35$ 秒

$5$ minutes and $35$ seconds

$6$ 分 $40$ 秒

$6$ minutes and $40$ seconds

$7$ 分 $5$ 秒

$7$ minutes and $5$ seconds

$7$ 分 $25$ 秒

$7$ minutes and $25$ seconds

$8$ 分 $10$ 秒

$8$ minutes and $10$ seconds

知识点：路程、速度与时间单位换算

难度评级：870

小提示：

先求一圈中前 $100$ 米和后 $300$ 米各用多少时间。

Find the time for the first $100$ meters and the last $300$ meters of one lap

大提示：

将一圈用时乘以 $5$ ，再把秒换算成分钟。

Multiply one-lap time by $5$ and convert seconds to minutes

解答：

Sofia 总共以每秒 $4$ 米跑了 $5\cdot 100=500$ 米，以每秒 $5$ 米跑了 $5\cdot 300=1500$ 米。

因此总时间为 $\frac{500}4 + \frac{1500}5 = 425$ 秒。

这个时间等于 $7$ 分 $5$ 秒。

所以正确答案是 C。

She ran a total of $5\cdot 100=500$ meters at $4$ meters per second and $5\cdot 300=1500$ meters at $5$ meters per second.

Therefore, her time is $\frac{500}4 + \frac{1500}5 = 425$ seconds.

This is equal to a total of $7$ minutes and $5$ seconds.

Thus, the correct answer is C .

3.

实数 $x$ 、 $y$ 、 $z$ 满足 $0 < x < 1$ 、 $-1 < y < 0$ 、 $1 < z < 2$ 。

下面哪个数一定为正？

Real numbers $x,$ $y,$ and $z$ satisfy the inequalities $0 < x < 1,$ $-1 < y < 0,$ and $1 < z < 2.$

Which of the following numbers is necessarily positive?

$y+x^2$

$y+xz$

$y+y^2$

$y+2y^2$

$y+z$

知识点：不等式反例

难度评级：960

小提示：

选项 $y+z$ 可以直接由给定范围判断。

The choice $y+z$ can be tested directly from the given bounds

大提示：

其余表达式各可用一个小反例排除。

Use one small counterexample to reject each other expression

解答：

因为 $-1 < y$ 且 $1 < z$ ，所以 $0 < y+z$ ，因此选项 E 一定正确。

此外，用下面这组取值可以排除其余每个选项： $x=0.1\text{，}$ $y=-0.25\text{，}$ $z=1.25\text{。}$

所以正确答案是 E。

Since $-1 < y$ and $1 < z,$ we can add the inequalities to see that $0 < y+z.$ This naturally proves choice E correct.

Furthermore, we can eliminate every other choice with the following values: $x=0.1,$ $y=-0.25,$ $z=1.25.$

Thus, the correct answer is E .

4.

设 $x$ 和 $y$ 是非零实数，且满足 $\frac{3x+y}{x-3y}=-2\text{。}$ 求下式的值： $\frac{x+3y}{3x-y}\text{。}$

Suppose that $x$ and $y$ are nonzero real numbers such that $\frac{3x+y}{x-3y}=-2.$ What is the value of $\frac{x+3y}{3x-y}?$

$-3$

$-1$

$1$

$2$

$3$

知识点：分式方程代数变形

难度评级：900

小提示：

将给定方程清除分母。

Clear the denominator in the given equation

大提示：

这个方程很快会推出 $x$ 与 $y$ 的关系。

The equation quickly forces a relation between $x$ and $y$

解答：

由 $\frac{3x+y}{x-3y}=-2\text{，}$ 两边同乘分母，得到 $3x+y = 6y-2x\text{。}$ 整理可得 $x=y$ 。

因此 $\frac{x+3y}{3x-y} = \frac{x+3x}{3x-x}=2\text{。}$

所以正确答案是 D。

Given that $\frac{3x+y}{x-3y}=-2,$ we can multiply by the denominator to get $3x+y = 6y-2x.$ Solving, we can see that $x=y.$

Therefore, $\frac{x+3y}{3x-y} = \frac{x+3x}{3x-x}=2.$

Thus, the correct answer is D .

5.

Camilla 原来拥有的蓝莓味软糖数量是樱桃味软糖数量的两倍。她每种各吃了 $10$ 颗后，蓝莓味软糖数量变成樱桃味软糖数量的三倍。她原来有多少颗蓝莓味软糖？

Camilla had twice as many blueberry jelly beans as cherry jelly beans. After eating $10$ pieces of each kind, she now has three times as many blueberry jelly beans as cherry jelly beans. How many blueberry jelly beans did she originally have?

$10$

$20$

$30$

$40$

$50$

知识点：方程组一次方程

难度评级：900

小提示：

设原来的樱桃味和蓝莓味软糖数量为变量。

Let the original cherry and blueberry counts be variables

大提示：

用吃之前和吃之后的两个关系式。

Use one equation before and one equation after eating the jelly beans

解答：

设原来樱桃味有 $c$ 颗，蓝莓味有 $b$ 颗。

由第一句和第二句分别可得 $b = 2c$ $b-10=3(c-10)$ 这两个方程。

因此 $2c-10 = 3c-30$ $c=20\text{。}$ 这说明 $b = 2\cdot 20=40\text{。}$

所以正确答案是 D。

Let the number of cherry jelly beans be $c$ and let the number of blueberry jelly beans be $b.$

Then, we know $b = 2c$ $b-10=3(c-10)$ from the first and second statements respectively.

Therefore, $2c-10 = 3c-30$ $c=20.$ This means that $b = 2\cdot 20=40.$

Thus, the correct answer is D .

6.

一个 $3$ 英寸 $\times$ $2$ 英寸 $\times$ $3$ 英寸的盒子中，最多能放入多少个实心的 $2$ 英寸 $\times$ $2$ 英寸 $\times$ $1$ 英寸长方体？

What is the largest number of solid $2$ -in $\times$ $2$ -in $\times$ $1$ -in blocks that can fit in a $3$ -in $\times$ $2$ -in $\times$ $3$ -in box?

$3$

$4$

$5$

$6$

$7$

知识点：体积长方体极限情形界定

难度评级：1070

小提示：

先用体积给出上界。

Start with the volume upper bound

大提示：

再检查四块积木确实能排进盒子。

Then check that four blocks really can be arranged

解答：

大盒子的体积是 $3\cdot 3\cdot 2 = 18$ ，每块小积木体积是 $2\cdot 2\cdot 1=4$ 。所以最多能放 $4$ 块。

这个上界可以达到：先把 $3$ 块小积木放入一个 $3 \times 2 \times 2$ 的长方体，再在剩下的 $3 \times 2 \times 1$ 空间中放入一块。

所以正确答案是 B。

The volume of the large solid object is $3\cdot 3\cdot 2 = 18$ and volume of the smaller object is $2\cdot 2\cdot 1=4.$ This means we can fit at most $4$ of the small objects.

We can make this happen by putting $3$ of the small objects in a $3 \times 2 \times 2$ rectangular prism, and then we have a $3 \times 2 \times 1$ space left where we can place one small object.

Thus, the correct answer is B .

7.

Samia 骑自行车去拜访朋友，平均速度为每小时 $17$ 千米。当她走完到朋友家的路程的一半时，轮胎漏气了，于是她以每小时 $5$ 千米的速度走完剩下的一半路程。

她总共花了 $44$ 分钟到达朋友家。四舍五入到最接近的十分之一千米，Samia 走了多远？

Samia set off on her bicycle to visit her friend, traveling at an average speed of $17$ kilometers per hour. When she had gone half the distance to her friend’s house, a tire went flat, and she walked the rest of the way at $5$ kilometers per hour.

In all, it took her $44$ minutes to reach her friend’s house. In kilometers rounded to the nearest tenth, how far did Samia walk?

$2.0$

$2.2$

$2.8$

$3.4$

$4.4$

知识点：路程、速度与时间一次方程

难度评级：1220

小提示：

设她步行的距离为 $d$ ，则骑车距离也是 $d$ 。

Let the walking distance be $d$ , so the biking distance is also $d$

大提示：

骑车时间加步行时间等于 $44$ 分钟。

Add the biking time and walking time to get $44$ minutes

解答：

设 $d$ 为 Samia 步行的距离。她骑车走过的距离与此相同，所以由总用时可得 $\dfrac{d}{17}+\dfrac{d}{5}=\dfrac{44}{60}\text{。}$ 解得 $d=\dfrac{17}{6}=2.833\ldots$ ，四舍五入到十分位约为 $2.8$ 千米。

所以正确答案是 C。

Let $d$ be the distance Samia walked. She bicycled the same distance, so her total travel time gives $\dfrac{d}{17}+\dfrac{d}{5}=\dfrac{44}{60}.$ Solving yields $d=\dfrac{17}{6}=2.833\ldots,$ which rounds to $2.8$ kilometers.

Thus, the correct answer is C .

8.

点 $A(11, 9)$ 和 $B(2, -3)$ 是 $\triangle ABC$ 的顶点，且 $AB=AC$ 。从 $A$ 作到底边的高，与对边交于 $D(-1, 3)$ 。点 $C$ 的坐标是什么？

Points $A(11, 9)$ and $B(2, -3)$ are vertices of $\triangle ABC$ with $AB=AC.$ The altitude from $A$ meets the opposite side at $D(-1, 3).$ What are the coordinates of point $C?$

$(-8, 9)$

$(-4, 8)$

$(-4, 9)$

$(-2, 3)$

$(-1, 0)$

知识点：坐标几何中点等腰三角形

难度评级：1070

小提示：

在等腰三角形中，顶角作出的高也平分底边。

In an isosceles triangle, the altitude from the vertex also bisects the base

大提示：

把 $D$ 当作 $BC$ 的中点。

Use $D$ as the midpoint of $BC$

解答：

因为 $AB=AC$ ，所以从 $A$ 作出的高也平分底边 $BC$ 。因此 $D$ 是 $BC$ 的中点。若 $C=(x,y)$ ，则 $\frac{x+2}2 = -1\text{，}$ $\frac{y-3}2=3\text{。}$ 于是 $C=(x,y)=(-4,9)\text{。}$

所以正确答案是 C。

Since $AB=AC,$ the altitude from $A$ also bisects the base $BC.$ Therefore, $D$ is the midpoint of $BC.$ If $C=(x,y),$ then we have $\frac{x+2}2 = -1,$ $\frac{y-3}2=3.$ As such, $C=(x,y)=(-4,9).$

Thus, the correct answer is C .

9.

一个广播节目有一个由 $3$ 道选择题组成的测验，每题有 $3$ 个选项。参赛者答对 $2$ 题或更多即获胜。参赛者随机回答每一道题。获胜的概率是多少？

A radio program has a quiz consisting of $3$ multiple-choice questions, each with $3$ choices. A contestant wins if he or she gets $2$ or more of the questions right. The contestant answers randomly to each question. What is the probability of winning?

$\dfrac{1}{27}$

$\dfrac{1}{9}$

$\dfrac{2}{9}$

$\dfrac{7}{27}$

$\dfrac{1}{2}$

知识点：二项概率分类讨论

难度评级：1140

小提示：

分别计算恰好答对三题和恰好答对两题。

Count exactly three correct plus exactly two correct

大提示：

恰好答对两题时，先选择哪一题答错。

For exactly two correct, choose which question is missed

解答：

全部 $3$ 题答对的概率为 $\left(\dfrac13\right)^3=\dfrac1{27}\text{。}$ 恰好答对 $2$ 题的概率为 $\binom32\left(\dfrac13\right)^2\left(\dfrac23\right)=\dfrac6{27}\text{。}$ 合计概率为 $\dfrac6{27}+\dfrac1{27}=\dfrac7{27}\text{。}$

所以正确答案是 D。

The probability that a contestant gets all $3$ correct is $\left(\dfrac13\right)^3=\dfrac1{27}.$ The probability of getting exactly $2$ correct is $\binom32\left(\dfrac13\right)^2\left(\dfrac23\right)=\dfrac6{27}.$ The combined probability is $\dfrac6{27}+\dfrac1{27}=\dfrac7{27}.$

Thus, the correct answer is D .

10.

方程为 $ax-2y=c$ 和 $2x+by=-c$ 的两条直线互相垂直，并且相交于 $(1, -5)$ 。 $c$ 的值是多少？

The lines with equations $ax-2y=c$ and $2x+by=-c$ are perpendicular and intersect at $(1, -5).$ What is $c?$

$-13$

$-8$

$2$

$8$

$13$

知识点：坐标几何斜率方程组

难度评级：1370

小提示：

把两条直线都写成斜截式。

Rewrite both lines in slope-intercept form

大提示：

垂直斜率和交点 $(1,-5)$ 会决定参数。

Perpendicular slopes and the point $(1,-5)$ determine the parameters

解答：

第一个方程可改写为 $y=\dfrac a2x-\dfrac c2$ ，第二个可改写为 $y=-\dfrac2b x-\dfrac cb$ 。因为两条直线垂直，它们的斜率之积为 $-1$ ，所以 $a=b$ 。

把 $(x,y)=(1,-5)$ 代入原来的两个方程，得到 $a+10=c$ 和 $2-5a=-c$ 。两式相加得 $12-4a=0$ ，所以 $a=3$ ， $c=13$ 。

所以正确答案是 E。

The first equation can be rewritten as $y=\dfrac a2x-\dfrac c2,$ and the second as $y=-\dfrac2b x-\dfrac cb.$ Because the lines are perpendicular, their slopes multiply to $-1,$ so $a=b.$

Substituting $(x,y)=(1,-5)$ into the two original equations gives $a+10=c$ and $2-5a=-c.$ Adding these equations yields $12-4a=0,$ so $a=3$ and $c=13.$

Thus, the correct answer is E .

11.

在 Typico 高中， $60\%$ 的学生喜欢跳舞，其余学生不喜欢。在真正喜欢跳舞的人中， $80\%$ 说他们喜欢，剩下的人说他们不喜欢。在真正不喜欢跳舞的人中， $90\%$ 说他们不喜欢，剩下的人说他们喜欢。那些说自己不喜欢跳舞的学生中，实际喜欢跳舞的占多少比例？

At Typico High School, $60\%$ of the students like dancing, and the rest dislike it. Of those who like dancing, $80\%$ say that they like it, and the rest say that they dislike it. Of those who dislike dancing, $90\%$ say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?

$10\%$

$12\%$

$20\%$

$25\%$

$33\frac{1}{3}\%$

知识点：条件概率百分数

难度评级：1370

小提示：

区分真实喜好和口头回答。

Separate actual preference from reported preference

大提示：

分母是所有说自己不喜欢跳舞的人。

The denominator is everyone who says they dislike dancing

解答：

真正喜欢跳舞的人占 $60\%$ ，其中只有 $80\%$ 说自己喜欢跳舞。因此全体学生中有 $48\%$ 说自己喜欢跳舞，而真正喜欢但说不喜欢的占 $60\%-48\% = 12\%$ 。

真正不喜欢跳舞的人占 $40\%$ ，其中 $90\%$ 说自己不喜欢，因而这部分占全体的 $36\%$ 。

所以说自己不喜欢跳舞的人共占 $12\%+36\% = 48\%$ 。

因此，说自己不喜欢、但实际上喜欢跳舞的人所占比例为 $\frac{12}{48} = \frac 14 = 25\%\text{。}$

所以正确答案是 D。

Observe that of the $60\%$ of people that actually like dancing, only $80\%$ say they like dancing. This suggests that $48\%$ of the students say that they like dancing, and as such, $60\%-48\% = 12\%$ of the students who like dancing say they don’t like it.

Then, we know that $90\%$ of the $40\%$ of people who don’t like dancing say they don’t like it, which is $36\%$ of the total student population.

This means the total amount of people who say they don’t like dancing is $12\%+36\% = 48\%.$

We know then that the fraction of people who say they dislike dancing but actually like it is equal to: $\frac{12}{48} = \frac 14 = 25\% .$

Thus, the correct answer is D .

12.

Elmer 的新车燃油效率（以每升行驶千米数衡量）比旧车高 $50\%$ 。不过新车使用柴油，柴油每升价格比旧车使用的汽油贵 $20\%$ 。如果 Elmer 长途旅行时使用新车而不是旧车，他会省下百分之多少的钱？

Elmer’s new car gets $50\%$ better fuel efficiency, measured in kilometers per liter, than his old car. However, his new car uses diesel fuel, which is $20\%$ more expensive per liter than the gasoline his old car uses. By what percent will Elmer save money if he uses his new car instead of his old car for a long trip?

$20\%$

$26\tfrac23\%$

$27\tfrac79\%$

$33\tfrac13\%$

$41\tfrac23\%$

知识点：速率百分数

难度评级：1280

小提示：

比较每千米成本，而不是每升成本。

Compare cost per kilometer, not cost per liter

大提示：

新燃料用的升数更少，但每升更贵。

New fuel uses fewer liters but each liter costs more

解答：

设旧车的燃油效率为每升 $M$ 千米，汽油价格为每升 $C$ 美元。于是旧车每千米的燃油费用为 $\dfrac CM$ 美元。

新车每升可行驶 $1.5M$ 千米，而它所用燃料的价格为每升 $1.2C$ 美元，所以新车每千米的燃油费用为 $\dfrac{1.2C}{1.5M}=0.8\dfrac CM\text{。}$

新车的燃油费用是旧车的 $80\%$ ，所以 Elmer 节省了 $20\%$ 。

所以正确答案是 A。

Let the old car’s fuel efficiency be $M$ kilometers per liter and let gasoline cost $C$ dollars per liter. The old car therefore costs $\dfrac CM$ dollars per kilometer to fuel.

The new car gets $1.5M$ kilometers per liter and its fuel costs $1.2C$ dollars per liter, so its fuel cost per kilometer is $\dfrac{1.2C}{1.5M}=0.8\dfrac CM.$

The new fuel cost is $80\%$ of the old one, so Elmer saves $20\%.$

Thus, the correct answer is A .

13.

有 $20$ 名学生参加课后项目，该项目提供瑜伽、桥牌和绘画课程。每名学生至少选一门课，也可以选两门或三门。

有 $10$ 人选瑜伽， $13$ 人选桥牌， $9$ 人选绘画。有 $9$ 人至少选了两门课。选了三门课的学生有多少人？

There are $20$ students participating in an after-school program offering classes in yoga, bridge, and painting. Each student must take at least one of these three classes, but may take two or all three.

There are $10$ students taking yoga, $13$ taking bridge, and $9$ taking painting. There are $9$ students taking at least two classes. How many students are taking all three classes?

$1$

$2$

$3$

$4$

$5$

知识点：容斥原理方程组

难度评级：1280

小提示：

设选恰好一门、两门、三门课的人数分别为 $x,y,z$ 。

Let $x,y,z$ count students taking exactly one, two, and three classes

大提示：

总选课人次会以权重 $1,2,3$ 计入这些组。

The total class enrollment counts these groups with weights $1,2,3$

解答：

总选课人次为 $10+13+9=32$ 。

设 $x$ 表示选 $1$ 门课的人数， $y$ 表示选 $2$ 门课的人数， $z$ 表示选 $3$ 门课的人数。

于是 $x+2y+3z = 32$ 。

因此总人数为 $20$ ，所以 $x+y+z = 20$ ，从而 $y+2z=12$ 。

至少选两门课的人数为 $9$ ，所以 $y+z = 9$ 。

因此 $z=3$ ，这就是答案。

所以正确答案是 C。

The number of classes taken total is $10+13+9=32.$

Let $x$ represent the number of people who take $1,$ let $y$ represent the number of people who take $2$ classes, and let $z$ represent the number of people who take $3$ classes.

Then, we know $x+2y+3z = 32.$

As such, the total number of people is $20,$ so $x+y+z = 20.$ This makes $y+2z=12.$

The number of people who take at least two classes is $9,$ so $y+z = 9.$

Therefore, $z=3,$ making that the answer.

Thus, the correct answer is C .

14.

从 $1\leq N \leq 2020$ 的整数中随机选一个整数 $N$ 。 $N^{16}$ 除以 $5$ 的余数为 $1$ 的概率是多少？

An integer $N$ is selected at random in the range $1\leq N \leq 2020$ . What is the probability that the remainder when $N^{16}$ is divided by $5$ is $1?$

$\dfrac{1}{5}$

$\dfrac{2}{5}$

$\dfrac{3}{5}$

$\dfrac{4}{5}$

$1$

知识点：费马小定理模幂运算基本概率

难度评级：1370

小提示：

模 $5$ 时，只需要看 $N$ 是否能被 $5$ 整除。

Modulo $5$ , only whether $N$ is divisible by $5$ matters

大提示：

使用费马小定理，或直接检查模 $5$ 的剩余类。

Use Fermat’s little theorem or check residue classes modulo $5$

解答：

根据费马小定理，只要 $N$ 不能被 $5$ 整除，就有 $N^4\equiv1\pmod5$ 。因此 $N^{16}\equiv(N^4)^4\equiv1\pmod5\text{。}$

$5$ 的倍数共有 $\frac{2020}{5}=404$ 个，所以 $N$ 的可取值共有 $2020-404=1616$ 个。

$5$ 的倍数满足 $N^{16}\equiv0\pmod5$ ，所以其余的值都不符合要求。因此所求概率为 $\dfrac{1616}{2020}=\dfrac45\text{。}$

所以正确答案是 D。

By Fermat’s Little Theorem, $N^4\equiv1\pmod5$ whenever $N$ is not divisible by $5.$ Therefore, $N^{16}\equiv(N^4)^4\equiv1\pmod5.$

There are $\frac{2020}{5}=404$ multiples of $5,$ so there are $2020-404=1616$ allowable values of $N.$

A multiple of $5$ has $N^{16}\equiv0\pmod5,$ so no other values work. Thus the probability is $\dfrac{1616}{2020}=\dfrac45.$

Thus, the correct answer is D .

15.

长方形 $ABCD$ 中， $AB=3$ ， $BC=4$ 。点 $E$ 是从 $B$ 向对角线 $\overline{AC}$ 所作垂线的垂足。求 $\triangle ADE$ 的面积。

Rectangle $ABCD$ has $AB=3$ and $BC=4.$ Point $E$ is the foot of the perpendicular from $B$ to diagonal $\overline{AC}.$ What is the area of $\triangle ADE?$

$1$

$\dfrac{42}{25}$

$\dfrac{28}{15}$

$2$

$\dfrac{54}{25}$

知识点：相似面积比直角三角形

难度评级：1600

小提示：

使用 $\triangle ABE$ 与 $\triangle ACB$ 的相似。

Use similarity between $\triangle ABE$ and $\triangle ACB$

大提示：

再利用 $\triangle EAD$ 与 $\triangle CDA$ 在 $AC$ 上的底边来比较它们。

Then compare $\triangle EAD$ to $\triangle CDA$ using their bases on $AC$

解答：

$\triangle CDA$ 的面积是 $\dfrac{3\cdot4}{2}=6$ 。因为 $E$ 在 $AC$ 上，三角形 $EAD$ 和 $CDA$ 共用从 $D$ 作出的同一条高，所以 $[EAD]=[CDA]\cdot \dfrac{AE}{AC}$ 。

由勾股定理， $AC=5$ 。又由 $\triangle ABE\sim \triangle ACB$ ， $\dfrac{AE}{AB}=\dfrac{AB}{AC}=\dfrac35$ ，所以 $AE=\dfrac95$ ， $\dfrac{AE}{AC}=\dfrac{9}{25}$ 。

因此 $[EAD]=6\cdot \dfrac{9}{25}=\dfrac{54}{25}$ 。所以正确答案是 E。

The area of $\triangle CDA$ is $\dfrac{3\cdot4}{2}=6$ . Since $E$ lies on $AC$ , triangles $EAD$ and $CDA$ share the same altitude from $D$ , so $[EAD]=[CDA]\cdot \dfrac{AE}{AC}$ .

By the Pythagorean Theorem, $AC=5$ . Also $\triangle ABE\sim \triangle ACB$ , so $\dfrac{AE}{AB}=\dfrac{AB}{AC}=\dfrac35$ , giving $AE=\dfrac95$ . Thus $\dfrac{AE}{AC}=\dfrac{9}{25}$ .

Therefore $[EAD]=6\cdot \dfrac{9}{25}=\dfrac{54}{25}$ . Thus, E is the correct answer.

16.

小于或等于 $2017$ 的正整数中，十进制表示含有数字 $0$ 的有多少个？

How many of the base-ten numerals for the positive integers less than or equal to $2017$ contain the digit $0?$

$469$

$471$

$475$

$478$

$481$

知识点：数字补集计数

难度评级：1480

小提示：

数补集：不含数字 $0$ 的数。

Count the complement: numbers with no digit $0$

大提示：

按一位数、两位数、三位数和到 $2017$ 为止的四位数分开处理。

Split by one-, two-, three-, and four-digit numbers up to $2017$

解答：

小于 $100$ 的数中，只有 $10$ 的倍数才含有数字 $0$ ，这样的数有 $9$ 个。

对 $100$ 到 $999$ （含端点）的数用补集计数。这个范围共有 $900$ 个数。没有 $0$ 的数有 $9\cdot 9\cdot 9=729$ 个，因为每一位都有 $9$ 种选择使它不是 $0$ 。因此这个范围内共有 $171$ 个。

对 $1000$ 到 $1999$ （含端点）的数再次用补集计数。这个范围共有 $1000$ 个数。没有 $0$ 的数有 $1\cdot 9\cdot 9\cdot 9=729$ 个，因为后 $3$ 位各有 $9$ 种选择使它不是 $0$ ，首位必须是 $1$ 。因此这个范围内共有 $271$ 个。

从 $2000$ 到 $2017$ （含端点）共有 $18$ 个数，每个数从左数第二位都是 $0$ 。

总数为 $9+171+271+18=469\text{。}$

所以正确答案是 A。

For numbers less than $100,$ we only have a $0$ if it is a multiple of $10,$ of which there are $9.$

For numbers between $100$ and $999$ inclusive, we will use complementary counting. There are $900$ total numbers in this range. Also, there are $9\cdot 9\cdot 9=729$ numbers in this range with no $0$ since there are $9$ ways to choose each digit to not be $0.$ Thus, the total in this range is $171.$

For numbers between $1000$ and $1999$ inclusive, we will use complementary counting again. There are $1000$ total numbers in this range. Also, there are $1\cdot 9\cdot 9\cdot 9=729$ numbers in this range with no $0$ since there are $9$ ways to choose each of the last $3$ digits to not be $0$ and the first digit must be $1.$ Thus, the total in this range is $271.$

There are $18$ numbers between $2000$ and $2017$ inclusive, each with a $0$ in the second digit from the left.

This makes the total $9+171+271+18=469.$

Thus, the correct answer is A .

17.

若一个正整数是一位数，或者它的数字从左到右读时构成严格递增或严格递减的序列，则称它为 单调数。例如 $3$ 、 $23578$ 和 $987620$ 是单调数，但 $88$ 、 $7434$ 和 $23557$ 不是。共有多少个单调正整数？

Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3,$ $23578,$ and $987620$ are monotonous, but $88,$ $7434,$ and $23557$ are not. How many monotonous positive integers are there?

$1024$

$1524$

$1533$

$1536$

$2048$

知识点：子集数字双射

难度评级：1660

小提示：

严格递增的数由它的数字集合唯一决定。

Increasing numbers are determined by their digit set

大提示：

严格递减的数可以含 $0$ ，但数 $0$ 本身不是正整数。

Decreasing numbers may include $0$ , but the number $0$ itself is not positive

解答：

严格递增的正整数对应于 $\{1,2,3,4,5,6,7,8,9\}$ 的非空子集，把选出的数字按递增顺序写出。因此有 $2^9-1=511$ 个。

严格递减的正整数对应于 $\{0,1,2,3,4,5,6,7,8,9\}$ 的子集，把选出的数字按递减顺序写出，但要排除空集和 $\{0\}$ 。因此有 $2^{10}-2=1022$ 个。

一位数 $1$ 到 $9$ 被两边都算了一次，所以总数为 $511+1022-9=1524$ 。所以正确答案是 B。

The strictly increasing positive integers correspond to the nonempty subsets of $\{1,2,3,4,5,6,7,8,9\}$ , written in increasing order. There are $2^9-1=511$ of these.

The strictly decreasing positive integers correspond to subsets of $\{0,1,2,3,4,5,6,7,8,9\}$ , written in decreasing order, except for the empty set and $\{0\}$ . There are $2^{10}-2=1022$ of these.

The one-digit numbers $1$ through $9$ were counted in both groups, so the total is $511+1022-9=1524$ . Thus, B is the correct answer.

18.

在下图中， $6$ 个圆盘中有 $3$ 个要涂成蓝色， $2$ 个要涂成红色， $1$ 个要涂成绿色。如果两个涂色方案可以通过整个图形的旋转或反射相互得到，则视为相同。共有多少种不同涂法？

In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?

$6$

$8$

$9$

$12$

$15$

知识点：组合对称性分类讨论

难度评级：2010

小提示：

利用对称性，把绿色圆盘的位置化成两种情形。

Use symmetry to reduce the green disk to two cases

大提示：

对每个绿色位置，在对称意义下数红色圆盘的放法。

For each green position, count possible red-disk placements up to symmetry

解答：

由对称性，绿色圆盘的位置只有两类：角上的位置或边中点的位置。每类各固定一个代表位置。此时选出两个红色圆盘的方法有 $\binom52=10$ 种。

保持绿色位置不动的那个反射会固定其余圆盘中的一个，并把另外四个圆盘两两交换。恰好有 $2$ 种红色圆盘的选法在这个反射下不变，即选中被交换的某一对。其余 $8$ 种选法两两互为镜像，构成 $4$ 对。因此绿色圆盘的每一类位置都给出 $2+4=6$ 种涂色方案。

于是两类位置合起来给出 $6+6=12$ 种涂色方案。

所以正确答案是 D。

By symmetry, the green disk has two possible types of position: a corner or a side midpoint. Fix one representative of either type. There are $\binom52=10$ ways to choose the two red disks.

The reflection that fixes the green position fixes one of the other disks and exchanges the other four disks in two pairs. Exactly $2$ red-disk choices are unchanged by this reflection: choosing either exchanged pair. The other $8$ choices form $4$ mirror-image pairs. Hence there are $2+4=6$ paintings for each type of green position.

The two types therefore give $6+6=12$ paintings.

Thus, the correct answer is D .

19.

设 $ABC$ 是等边三角形。将边 $\overline{AB}$ 过 $B$ 延长到点 $B'$ ，使 $BB'=3 \cdot AB$ 。类似地，将 $\overline{BC}$ 过 $C$ 延长到 $C'$ ，使 $CC'=3 \cdot BC$ ，将 $\overline{CA}$ 过 $A$ 延长到 $A'$ ，使 $AA'=3 \cdot CA$ 。

$\triangle A'B'C'$ 的面积与 $\triangle ABC$ 的面积之比是多少？

Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB.$ Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC,$ and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA.$

What is the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC?$

$9:1$

$16:1$

$25:1$

$36:1$

$37:1$

知识点：面积比等边三角形三角形面积

难度评级：1860

小提示：

把大三角形分成原三角形和周围的六个三角形。

Break the large triangle into the original triangle and six surrounding triangles

大提示：

用底和高比较每个周围三角形与原三角形的面积。

Compare each surrounding triangle area to the original using base and height

解答：

设 $\triangle ABC$ 的面积为 $X$ 。 $\triangle BB'C$ 、 $\triangle CC'A$ 和 $\triangle AA'B$ 的底边都是 $\triangle ABC$ 对应边的三倍，而对应的高相同。因此它们的面积都是 $3X$ 。

接着， $\triangle AA'C'$ 的底是 $\triangle ACC'$ 的三倍，高相同；后者面积为 $3X$ 。所以 $\triangle AA'C'$ 的面积为 $9X$ 。同理， $\triangle CC'B'$ 和 $\triangle BB'A'$ 的面积也都是 $9X$ 。

这七个区域恰好分割大三角形，所以 $\begin{aligned}[A'B'C']&=X+3(3X)+3(9X)\\&=37X\end{aligned}\text{。}$ 所求比为 $37:1$ 。

所以正确答案是 E。

Let $X$ be the area of $\triangle ABC.$ Each of $\triangle BB'C,$ $\triangle CC'A,$ and $\triangle AA'B$ has a base three times as long as a side of $\triangle ABC$ and the same corresponding altitude. Each therefore has area $3X.$

Next, $\triangle AA'C'$ has three times the base and the same altitude as $\triangle ACC',$ whose area is $3X.$ Thus $\triangle AA'C'$ has area $9X.$ Similarly, $\triangle CC'B'$ and $\triangle BB'A'$ each have area $9X.$

These seven regions partition the large triangle, so $\begin{aligned}[A'B'C']&=X+3(3X)+3(9X)\\&=37X.\end{aligned}$ The requested ratio is $37:1.$

Thus, the correct answer is E .

20.

数 $21!$ $=51{,}090{,}942{,}171{,}709{,}440{,}000$ 有超过 $60{,}000$ 个正整数因子。随机选择其中一个因子。它是奇数的概率是多少？

The number $21!$ $=51{,}090{,}942{,}171{,}709{,}440{,}000$ has over $60{,}000$ positive integer divisors. One of them is chosen at random. What is the probability that it is odd?

$\dfrac{1}{21}$

$\dfrac{1}{19}$

$\dfrac{1}{18}$

$\dfrac{1}{2}$

$\dfrac{11}{21}$

知识点：因数个数质因数分解基本概率

难度评级：1540

小提示：

计算 $21!$ 中因子 $2$ 的指数。

Count the exponent of $2$ in $21!$

大提示：

对每个奇数部分，只有一种 $2$ 的指数会得到奇因子。

For each odd part of a divisor, only one exponent of $2$ gives an odd divisor

解答：

$21!$ 中 $2$ 的指数为 $\begin{aligned} &\left\lfloor\dfrac{21}{2}\right\rfloor+ \left\lfloor\dfrac{21}{4}\right\rfloor\\ &\quad{}+\left\lfloor\dfrac{21}{8}\right\rfloor+ \left\lfloor\dfrac{21}{16}\right\rfloor\\ &=10+5+2+1=18 \end{aligned}\text{。}$ 因此 $21!=2^{18}d$ ，其中 $d$ 是某个奇整数。

对 $d$ 的每个因子 $x$ ， $21!$ 中奇数部分为 $x$ 的因子恰好是 $x,2x,\ldots,2^{18}x$ 。这 $19$ 个因子中只有一个是奇数，所以所求概率为 $\dfrac1{19}$ 。

所以正确答案是 B。

The exponent of $2$ in $21!$ is $\begin{aligned} &\left\lfloor\dfrac{21}{2}\right\rfloor+ \left\lfloor\dfrac{21}{4}\right\rfloor\\ &\quad{}+\left\lfloor\dfrac{21}{8}\right\rfloor+ \left\lfloor\dfrac{21}{16}\right\rfloor\\ &=10+5+2+1=18. \end{aligned}$ Thus $21!=2^{18}d$ for some odd integer $d.$

For every divisor $x$ of $d,$ the divisors of $21!$ with odd part $x$ are $x,2x,\ldots,2^{18}x.$ Exactly one of these $19$ divisors is odd, so the probability is $\dfrac1{19}.$

Thus, the correct answer is B .

21.

在 $\triangle ABC$ 中， $AB=6$ ， $AC=8$ ， $BC=10$ ，且 $D$ 是 $\overline{BC}$ 的中点。求 $\triangle ADB$ 与 $\triangle ADC$ 内切圆半径之和。

In $\triangle ABC,$ $AB=6,$ $AC=8,$ $BC=10,$ and $D$ is the midpoint of $\overline{BC}.$ What is the sum of the radii of the circles inscribed in $\triangle ADB$ and $\triangle ADC?$

$\sqrt{5}$

$\dfrac{11}{4}$

$2\sqrt{2}$

$\dfrac{17}{6}$

$3$

知识点：内切圆、内心与内切圆半径直角三角形三角形面积

难度评级：1720

小提示：

先识别 $6,8,10$ 三角形是直角三角形。

First recognize the $6,8,10$ triangle as right

大提示：

对两个小三角形使用面积等于内切圆半径乘以半周长。

Use area equals inradius times semiperimeter for the two smaller triangles

解答：

三角形 $ABC$ 是在 $A$ 处为直角的直角三角形。由于 $D$ 是斜边中点，所以它是这个三角形的外心。

因此 $AD = BD = DC = 5\text{。}$ 同时， $ABC$ 的面积为 $\dfrac{6\cdot 8}2 = 24\text{。}$

底边 $BD$ 和 $DC$ 相等，两个三角形从 $A$ 作出的高也相同。因此， $\triangle ABD$ 和 $\triangle ACD$ 的面积都为 $12$ 。

对每个三角形，都有 $A = rs$ ，其中 $A$ 是面积， $r$ 是内切圆半径， $s$ 是半周长。等价地， $12=\dfrac12rP$ ，其中 $P$ 是周长，所以 $r=\dfrac{24}{P}$ 。对 $\triangle ABD$ ，内切圆半径为 $\dfrac{24}{5+5+6}=\dfrac32$ 。对 $\triangle ACD$ ，内切圆半径为 $\dfrac{24}{5+5+8}=\dfrac43$ 。

两者之和为 $\dfrac 32 + \dfrac 43 = \dfrac{17}6$ 。

所以正确答案是 D。

The triangle $ABC$ is a right triangle with a right angle at $A.$ This makes $D$ the circumcenter of the triangle since it is the midpoint of the hypotenuse.

Therefore, $AD = BD = DC = 5.$ Also, the area of $ABC$ is $\dfrac{6\cdot 8}2 = 24.$

The bases $BD$ and $DC$ are equal, and the two triangles share the same altitude from $A.$ Therefore, $\triangle ABD$ and $\triangle ACD$ each have area $12.$

Then, for each triangle, we have $A = rs$ where $A$ is the area, $r$ is the inradius, and $s$ is the semiperimeter. Equivalently, $12=\dfrac12rP,$ where $P$ is the perimeter, so $r=\dfrac{24}{P}.$ For $\triangle ABD,$ the inradius is $\dfrac{24}{5+5+6}=\dfrac32.$ For $\triangle ACD,$ it is $\dfrac{24}{5+5+8}=\dfrac43.$

Their sum is $\dfrac 32 + \dfrac 43 = \dfrac{17}6 .$

Thus, the correct answer is D .

22.

半径为 $2$ 的圆的直径 $\overline{AB}$ 延长到圆外一点 $D$ ，使 $BD=3$ 。点 $E$ 满足 $ED=5$ ，且直线 $ED$ 垂直于直线 $AD$ 。线段 $\overline{AE}$ 与圆交于位于 $A$ 和 $E$ 之间的点 $C$ 。求 $\triangle ABC$ 的面积。

The diameter $\overline{AB}$ of a circle of radius $2$ is extended to a point $D$ outside the circle so that $BD=3.$ Point $E$ is chosen so that $ED=5$ and line $ED$ is perpendicular to line $AD.$ Segment $\overline{AE}$ intersects the circle at a point $C$ between $A$ and $E.$ What is the area of $\triangle ABC?$

$\dfrac{120}{37}$

$\dfrac{140}{39}$

$\dfrac{145}{39}$

$\dfrac{140}{37}$

$\dfrac{120}{31}$

知识点：圆周角相似面积比

难度评级：1900

小提示：

利用半圆所对的圆周角得到一个直角三角形。

Use the semicircle angle to get a right triangle

大提示：

通过相似比较 $\triangle ABC$ 和 $\triangle AED$ 。

Compare $\triangle ABC$ and $\triangle AED$ by similarity

解答：

因为半径为 $2$ ，且 $BD =3$ ，所以 $AD = 7$ 。由于 $ED = 5$ ，且 $D$ 处角为直角， $ADE$ 的面积为 $\dfrac{5\cdot 7}{2} = \dfrac{35}{2}$ 。

由勾股定理， $AE=\sqrt{5^2+7^2}=\sqrt{74}$ 。又因为 $AB$ 是直径，所以 $\angle ACB$ 是直角。两个三角形共用 $A$ 处的角，因此由角角相似可知 $\triangle ABC\sim\triangle AED$ 。

它们对应的斜边分别是 $AB=4$ 和 $AE=\sqrt{74}$ ，所以面积之比为 $\dfrac{[ABC]}{[AED]}=\left(\dfrac4{\sqrt{74}}\right)^2=\dfrac8{37}\text{。}$ 因此 $[ABC]=\dfrac8{37}\cdot\dfrac{35}{2}=\dfrac{140}{37}\text{。}$

所以正确答案是 D。

Since the radius is $2$ and $BD =3,$ we have $AD = 7.$ Since $ED = 5$ and the angle at $D$ is a right angle, the area of $ADE$ is $\dfrac{5\cdot 7}{2} = \dfrac{35}{2} .$

By the Pythagorean Theorem, $AE=\sqrt{5^2+7^2}=\sqrt{74}.$ Also, $\angle ACB$ is a right angle because $AB$ is a diameter. The triangles share the angle at $A,$ so $\triangle ABC\sim\triangle AED$ by angle-angle similarity.

Their corresponding hypotenuses are $AB=4$ and $AE=\sqrt{74},$ so their area ratio is $\dfrac{[ABC]}{[AED]}=\left(\dfrac4{\sqrt{74}}\right)^2=\dfrac8{37}.$ Therefore, $[ABC]=\dfrac8{37}\cdot\dfrac{35}{2}=\dfrac{140}{37}.$

Thus, the correct answer is D .

23.

令 $N=123456789101112\dots4344$ 为把整数 $1$ 到 $44$ 依次写在一起形成的 $79$ 位数。 $N$ 除以 $45$ 的余数是多少？

Let $N=123456789101112\dots4344$ be the $79$ -digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45?$

$1$

$4$

$9$

$18$

$44$

知识点：模运算中国剩余定理数字

难度评级：1660

小提示：

分别求模 $5$ 和模 $9$ 的余数。

Find the remainder modulo $5$ and modulo $9$

大提示：

再把这两个条件合并成模 $45$ 的余数。

Combine them to get the remainder modulo $45$

解答：

要找除以 $45$ 的余数，必须分别找除以 $5$ 和 $9$ 的余数。除以 $5$ 的余数就是个位数字除以 $5$ 的余数，所以是 $4$ 。

要找除以 $9$ 的余数，通常找数字和。不过每个两位数除以 $9$ 的余数与它的数字和相同，所以可以直接把 $1$ 到 $44$ 的整数相加，因为每个整数与它自己的数字和同余。这个和为 $\dfrac{44\cdot45}{2}=990$ ，它是 $9$ 的倍数。因此 $N$ 是 $9$ 的倍数。

这个数是 $9$ 的倍数，并且除以 $5$ 的余数为 $4$ ，所以除以 $45$ 的余数为 $9$ 。

所以正确答案是 C。

To find the remainder when divided by $45,$ we must find the remainder when divided by $5$ and $9.$ The remainder when divided by $5$ is the remainder when the units digit is divided by $5,$ making it $4.$

To find the remainder when divided by $9,$ we usually find the sum of the digits. However, each double digit number has the same remainder when divided by $9$ as its digit sum, so we can just sum the integers from $1$ to $44,$ because each integer is congruent to its own digit sum. This sum is $\dfrac{44\cdot45}{2}=990,$ which is a multiple of $9.$ Thus, $N$ is a multiple of $9.$

Since it is a multiple of $9$ and has a remainder of $4$ when divided by $5,$ the remainder when divided by $45$ is $9.$

Thus, the correct answer is C .

24.

一个等边三角形的三个顶点都在双曲线 $xy=1$ 上，并且该双曲线的一个顶点是这个三角形的重心。这个三角形面积的平方是多少？

The vertices of an equilateral triangle lie on the hyperbola $xy=1,$ and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?

$48$

$60$

$108$

$120$

$169$

知识点：双曲线等边三角形重心对称性

难度评级：2380

小提示：

利用双曲线关于 $y=x$ 的对称性。

Use symmetry of the hyperbola about $y=x$

大提示：

把重心到顶点的距离与等边三角形的外接圆半径联系起来。

Relate the centroid-to-vertex distance to the circumradius of the equilateral triangle

解答：

由对称性，不妨设重心是双曲线顶点 $G=(1,1)$ 。三角形至少有两个顶点位于双曲线的同一支上。它们不能都位于负支：如果其中两个顶点的 $x$ 坐标为负，第三个顶点的横坐标就会大于 $3$ ，而三个 $y$ 坐标之和会小于 $\frac{1}{3}$ ，这与重心是 $(1,1)$ 矛盾。因此两个顶点位于正支。

把这两个顶点写成 $P=(a,\frac{1}{a})$ 和 $Q=(b,\frac{1}{b})$ ，其中 $a,b>0$ 。等边三角形的重心也是外心，所以 $P$ 和 $Q$ 到 $G$ 的距离相等。对 $t>0$ ，点 $(t,\frac{1}{t})$ 到 $G$ 的距离平方为 $\left(t+\dfrac1t\right)\left(t+\dfrac1t-2\right)\text{。}$ 当 $t+\frac{1}{t}$ 从 $2$ 开始增大时，此式严格递增，所以两个不同的点必须满足 $b=\frac{1}{a}$ 。因此 $P$ 和 $Q$ 关于 $y=x$ 对称。

第三个顶点位于垂直平分线 $y=x$ 上。它的坐标还满足 $xy=1$ ，所以它是 $(1,1)$ 或 $(-1,-1)$ 。它不能与重心重合，所以是 $(-1,-1)$ 。因此外接圆半径就是从 $(1,1)$ 到 $(-1,-1)$ 的距离，即 $2\sqrt2$ 。

从中心把等边三角形分成三个三角形，其面积为 $3\cdot\dfrac12(2\sqrt2)^2\sin120^\circ=6\sqrt3\text{。}$ 面积的平方为 $(6\sqrt3)^2=108$ 。

所以正确答案是 C。

By symmetry, assume that the centroid is the hyperbola vertex $G=(1,1).$ At least two triangle vertices lie on the same branch of the hyperbola. They cannot both lie on the negative branch: if two of their $x$ -coordinates were negative, the third would exceed $3,$ while the sum of the three $y$ -coordinates would be less than $\frac{1}{3},$ contradicting that their centroid is $(1,1).$ Thus two vertices lie on the positive branch.

Write these vertices as $P=(a,\frac{1}{a})$ and $Q=(b,\frac{1}{b}),$ where $a,b>0.$ The centroid of an equilateral triangle is also its circumcenter, so $P$ and $Q$ are equidistant from $G.$ For $t>0,$ the squared distance from $(t,\frac{1}{t})$ to $G$ is $\left(t+\dfrac1t\right)\left(t+\dfrac1t-2\right).$ This is strictly increasing as $t+\frac{1}{t}$ increases from $2,$ so the distinct points must satisfy $b=\frac{1}{a}.$ Hence $P$ and $Q$ are reflections across $y=x.$

The third vertex lies on the perpendicular bisector $y=x.$ Its coordinates also satisfy $xy=1,$ so it is either $(1,1)$ or $(-1,-1).$ It cannot equal the centroid, so it is $(-1,-1).$ Therefore, the circumradius is the distance from $(1,1)$ to $(-1,-1),$ namely $2\sqrt2.$

Dividing the equilateral triangle into three triangles at its center gives its area as $3\cdot\dfrac12(2\sqrt2)^2\sin120^\circ=6\sqrt3.$ The square of the area is $(6\sqrt3)^2=108.$

Thus, the correct answer is C .

25.

去年 Isabella 参加了 $7$ 次数学考试，得到了 $7$ 个不同的分数，每个分数都是 $91$ 到 $100$ 之间（含端点）的整数。每次考试后，她都注意到到目前为止考试分数的平均数是整数。她第七次考试得了 $95$ 分。她第六次考试得了多少分？

Last year Isabella took $7$ math tests and received $7$ different scores, each an integer between $91$ and $100,$ inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was $95.$ What was her score on the sixth test?

$92$

$94$

$96$

$98$

$100$

知识点：整除性平均数极限情形界定

难度评级：2300

小提示：

最终的总分必须是 $7$ 的倍数。

The final total score must be a multiple of $7$

大提示：

用第七次的分数确定前六次分数之和，再利用能被 $5$ 整除的条件。

Use the seventh score to force the first six-score sum, then use divisibility by $5$

解答：

设七次总分为 $S$ 。因为七次后的平均数是整数， $S$ 可被 $7$ 整除。七个不同分数都在 $91$ 到 $100$ 之间，所以 $91+92+\cdots+97$ $\le S\le$ $94+95+\cdots+100$ 。

因此 $658\le S\le 679$ ，可能的 $7$ 的倍数为 $658,665,672,679$ 。由于第七次分数是 $95$ ，前六次总分为 $S-95$ ，它必须能被 $6$ 整除。这迫使 $S=665$ 。

因此前六次总分为 $570$ 。前五次平均数也是整数，所以前五次总分能被 $5$ 整除。因此第六次分数也必须能被 $5$ 整除。由于第七次已经是 $95$ ，且所有分数不同，第六次分数只能是 $100$ 。所以正确答案是 E。

Let $S$ be the sum of all seven scores. Since all seven averages were integers, $S$ is divisible by $7$ . Also the seven distinct scores are between $91$ and $100$ , so $91+92+\cdots+97$ $\le S\le$ $94+95+\cdots+100$ .

Thus $658\le S\le 679$ , and the possible multiples of $7$ are $658,665,672,679$ . Since the seventh score is $95$ , the first six scores sum to $S-95$ , which must be divisible by $6$ . This forces $S=665$ .

The first six scores sum to $570$ . The first five-score average was also an integer, so the sum of the first five scores is divisible by $5$ . Therefore the sixth score is divisible by $5$ . Since the seventh score is already $95$ and all scores are distinct, the sixth score is $100$ . Thus, E is the correct answer.