2022 AMC 10B 详解 | LIVE by Po-Shen Loh

1.

对所有实数 $x$ 和 $y$ ，定义 $x~\diamondsuit~ y$ 为 $|x-y|$ 。求 $(1~\diamondsuit~(2~\diamondsuit~3))-((1~\diamondsuit~2)~\diamondsuit~3)$

Define $x~\diamondsuit~ y$ to be $|x-y|$ for all real numbers $x$ and $y.$ What is the value of $(1~\diamondsuit~(2~\diamondsuit~3))-((1~\diamondsuit~2)~\diamondsuit~3)?$

$-2$

$-1$

$0$

$1$

$2$

知识点：自定义运算绝对值

难度评级：560

解答：

$(1~\diamondsuit~(2~\diamondsuit~3))-((1~\diamondsuit~2)~\diamondsuit~3) =$ $|1-|2-3|| - ||1-2|-3| =$ $|1-1| - |1-3| = 0-2 = -2.$

先计算最内层括号，再依次向外计算，即得上述结果。所以正确答案是 A。

$(1~\diamondsuit~(2~\diamondsuit~3))-((1~\diamondsuit~2)~\diamondsuit~3) =$ $|1-|2-3|| - ||1-2|-3| =$ $|1-1| - |1-3| = 0-2 = -2.$

Thus, the answer is A .

2.

在菱形 $ABCD$ 中，点 $P$ 在边 $\overline{AD}$ 上，且 $\overline{BP} \perp \overline{AD}$ 、 $AP = 3$ 、 $PD = 2$ 。求 $ABCD$ 的面积。（注：图未按比例绘制。）

In rhombus $ABCD,$ point $P$ lies on segment $\overline{AD}$ so that $\overline{BP} \perp \overline{AD},$ $AP = 3,$ and $PD = 2.$ What is the area of $ABCD?$ (Note: The figure is not drawn to scale.)

$3\sqrt 5$

$10$

$6\sqrt 5$

$20$

$25$

知识点：菱形勾股定理面积

难度评级：870

解答：

因为这是菱形，所以 $AB = AD = AP + PD = 5.$ 由勾股定理， $AP^2 + BP^2 = AB^2$ 也就是 $9+ BP^2 = 25.$ 因此 $BP = 4.$ 菱形面积为 $bh$ ，所以面积为 $5\cdot 4 = 20$ 。

所以正确答案是 D。

Since we have a rhombus, we know $AB = AD = AP + PD = 5.$ And by the Pythagorean Theorem, $AP^2 + BP^2 = AB^2$ We know that $9+ BP^2 = 25.$ $BP = 4.$ The area of a rhombus is $bh,$ so the area is $5\cdot 4 = 20.$

Thus, the answer is D .

3.

有多少个三位正整数含有奇数个偶数数字？

How many three-digit positive integers have an odd number of even digits?

$150$

$250$

$350$

$450$

$550$

知识点：数字奇偶性乘法原理

难度评级：1100

解答：

前三位中的前两位共有 $9\cdot 10 = 90$ 种选择。

如果前两位中偶数数字的个数已经是奇数，则个位必须是奇数，有 $5$ 种选择。否则个位必须是偶数，也有 $5$ 种选择。无论前两位如何选择，个位都有 $5$ 种选择。

因此，前两位有 $90$ 种选法，个位有 $5$ 种选法，总数为 $90\cdot 5 = 450$ 。

所以正确答案是 D。

First, we can choose any combination for the first two digits. This would have $9\cdot 10 = 90$ choices.

Then, if there are an odd number of even digits among them, I make the units digit odd, which can be done in $5$ ways. Otherwise, I make the units digit even, which can be done in $5$ ways. Regardless of my choice of the first two digits, I have $5$ ways to choose the units digit.

Therefore, there are $90$ ways to choose the first two digits, and $5$ ways to choose the last digit, so the total number of ways is $90\cdot 5 = 450.$

Thus, the answer is D .

4.

一头驴突然打嗝，第一次打嗝发生在某天下午 $4:00$ 。假设它每隔 $5$ 秒规律地打一次嗝。它第 $700$ 次打嗝发生在什么时间？

A donkey suffers an attack of hiccups and the first hiccup happens at $4:00$ one afternoon. Suppose that the donkey hiccups regularly every $5$ seconds. At what time does the donkey’s $700$ th hiccup occur?

$4:58$ 后十五秒

15 seconds after $4:58$

$4:58$ 后二十秒

20 seconds after $4:58$

$4:58$ 后二十五秒

25 seconds after $4:58$

$4:58$ 后三十秒

30 seconds after $4:58$

$4:58$ 后三十五秒

35 seconds after $4:58$

知识点：等差数列单位换算

难度评级：870

解答：

要求第 $700$ 次打嗝，就要看它比第一次晚了 $699$ 个间隔。

所以时间是 $4:58$ 后 $15$ 秒，因为 $699\cdot 5 = 3495$ ，且 $3495 = 60\cdot 58+15$ ，即经过 $58$ 分 $15$ 秒。

所以答案是 A。

Since we want to look at the $700$ th hiccup, we need to look at time that is $699$ hiccups after the first one.

This would be $699\cdot 5 = 3495$ seconds. Note that $3495 = 60\cdot 58+15,$ so the time would be $58$ minutes and $15$ seconds after the first hiccup. This would therefore be $4:58$ and $15$ seconds.

Thus, the answer is A .

5.

求 $\frac{\left(1+\frac{1}{3}\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{5^2}\right)\left(1-\frac{1}{7^2}\right)}}$ 的值。

What is the value of $\frac{\left(1+\frac{1}{3}\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{5^2}\right)\left(1-\frac{1}{7^2}\right)}}?$

$\sqrt3$

$2$

$\sqrt{15}$

$4$

$\sqrt{105}$

知识点：平方差根式

难度评级：1280

解答：

对分母中的各因子使用平方差公式，可得 $\sqrt{(1-\frac{1}{3^2})(1-\frac{1}{5^2})(1-\frac{1}{7^2})}$ $=\sqrt{(1-\frac{1}{3})(1-\frac{1}{5})(1-\frac{1}{7})}$ $\cdot\sqrt{(1+\frac{1}{3})(1+\frac{1}{5})(1+\frac{1}{7})}$ $=\sqrt{ (\frac 23)(\frac 45)(\frac 67) } \sqrt{ (\frac 43)(\frac 65)(\frac 87) }.$

分子则可写成 $(1+\frac{1}{3})(1+\frac{1}{5})(1+\frac{1}{7})$ $= (\frac 43)(\frac 65)(\frac 87) .$

$\begin{align*}&\frac{(\frac 43)(\frac 65)(\frac 87)}{\sqrt{ (\frac 23)(\frac 45)(\frac 67) } \sqrt{ (\frac 43)(\frac 65)(\frac 87) }} \\&=\frac{\sqrt{ (\frac 43)(\frac 65)(\frac 87) }} {\sqrt{ (\frac 23)(\frac 45)(\frac 67) }} \\&= \frac{\sqrt 8 } { \sqrt 2} \\&= 2.\end{align*}$

所以答案是 B。

Lets work with the denominator first. By using difference of two squares on each term, we get the denominator as $\sqrt{(1-\frac{1}{3^2})(1-\frac{1}{5^2})(1-\frac{1}{7^2})}$ $=\sqrt{(1-\frac{1}{3})(1-\frac{1}{5})(1-\frac{1}{7})}$ $\cdot\sqrt{(1+\frac{1}{3})(1+\frac{1}{5})(1+\frac{1}{7})}$ $=\sqrt{ (\frac 23)(\frac 45)(\frac 67) } \sqrt{ (\frac 43)(\frac 65)(\frac 87) }.$

Furthermore, we simplify the numerator as follows: $(1+\frac{1}{3})(1+\frac{1}{5})(1+\frac{1}{7})$ $= (\frac 43)(\frac 65)(\frac 87) .$

Dividing the numerator and denominator yields $\begin{align*}&\frac{(\frac 43)(\frac 65)(\frac 87)}{\sqrt{ (\frac 23)(\frac 45)(\frac 67) } \sqrt{ (\frac 43)(\frac 65)(\frac 87) }} \\&=\frac{\sqrt{ (\frac 43)(\frac 65)(\frac 87) }} {\sqrt{ (\frac 23)(\frac 45)(\frac 67) }} \\&= \frac{\sqrt 8 } { \sqrt 2} \\&= 2.\end{align*}$

Thus, the answer is B .

6.

数列 $121, 11211, 1112111, \ldots$ 的前十项中有多少个质数？

How many of the first ten numbers of the sequence $121, 11211, 1112111, \ldots$ are prime numbers?

$0$

$1$

$2$

$3$

$4$

知识点：质数因式分解

难度评级：1140

解答：

我们断言这些数都不可能是质数。

第 $n$ 项可写成 $\sum_{k=0}^{2n} 10^k + 10^n = \sum_{k=0}^{n} 10^k +$ $\sum_{k=n}^{2n} 10^k =\sum_{k=0}^{n} 10^k + \sum_{k=0}^{n} 10^k \cdot 10^n$ $= (10^n+1)(\sum_{k=0}^{n} 10^k).$ 这说明该数可写成两个大于 $1$ 的整数的乘积，所以没有质数。

所以正确答案是 A。

We claim that none of these numbers can ever be prime.

We prove this claim by noticing that the $n$ th number is $\sum_{k=0}^{2n} 10^k + 10^n = \sum_{k=0}^{n} 10^k +$ $\sum_{k=n}^{2n} 10^k =\sum_{k=0}^{n} 10^k + \sum_{k=0}^{n} 10^k \cdot 10^n$ $= (10^n+1)(\sum_{k=0}^{n} 10^k).$ This shows that the number can be written as the product of two numbers greater than $1,$ so there are no primes.

Thus, the answer is A .

7.

对多少个常数 $k$ ，多项式 $x^{2}+kx+36$ 有两个不同的整数根？

For how many values of the constant $k$ will the polynomial $x^{2}+kx+36$ have two distinct integer roots?

$\ 6$

$\ 8$

$\ 9$

$\ 14$

$\ 16$

知识点：韦达定理因数系统列举

难度评级：1070

解答：

设两个根为 $r,s$ 。由韦达定理， $rs = 36$ ，且 $r + s = -k$ 。相应的因式分解为 $\begin{align*}&x^2+kx+36 \\ &= (x-r)(x-s) \\&= x^2-(r+s)x+rs \end{align*}$

因此需要 $r$ 和 $s$ 是互不相同的整数，并满足 $rs = 36$ 。所有可能的因数对为 $\pm\{1,36\},\pm\{2,18\},\pm\{3,12\}$ 以及 $\pm\{4,9\}.$

不同整数根对应不同的无序因数对。共有 $8$ 个因数对，每一对都给出一个不同的 $k$ ，所以符合条件的系数共有八个。这里的未知系数就是 $k$ 。

所以正确答案是 B。

Let the roots be $r,s.$ Then: $\begin{align*}&x^2+kx+36 \\ &= (x-r)(x-s) \\&= x^2-(r+s)x+rs \end{align*}$ And so, $rs = 36$ and $r + s = -k.$

Therefore, we need $r$ and $s$ distinct such that $rs = 36.$ All the possible factor pairs are $\pm\{1,36\},\pm\{2,18\},\pm\{3,12\}$ and $\pm\{4,9\}.$

Each of these unordered pairs produces a unique value for $k,$ so there are $8$ possible values for $k.$

Thus, B is the correct answer.

8.

考虑下面 $100$ 个集合，每个集合含 $10$ 个元素： $\begin{align*} &\{1,2,3,\ldots,10\}, \\ &\{11,12,13,\ldots,20\},\\ &\{21,22,23,\ldots,30\},\\ &\vdots\\ &\{991,992,993,\ldots,1000\}. \end{align*}$ 其中有多少个集合恰好含有两个 $7$ 的倍数？

Consider the following $100$ sets of $10$ elements each: $\begin{align*} &\{1,2,3,\ldots,10\}, \\ &\{11,12,13,\ldots,20\},\\ &\{21,22,23,\ldots,30\},\\ &\vdots\\ &\{991,992,993,\ldots,1000\}. \end{align*}$ How many of these sets contain exactly two multiples of $7?$

$\ 40$

$\ 42$

$\ 43$

$\ 49$

$\ 50$

知识点：倍数模运算分类讨论

难度评级：1370

解答：

若某个集合中第一个 $7$ 的倍数个位为一、二或三，再加七仍在同一个十数段内，于是该集合恰好有两个七的倍数。

若个位为 $4,5,6,7$ ，再加 $7$ 会进入下一个十数段；所以个位为 $4,5,6,7$ 时，该集合只有一个 $7$ 的倍数。

如果最后一位是 $1,2,3$ ，加上 $7$ 后会在同一集合中得到个位为 $8,9,0$ 的数。

在前 $98$ 个集合中，第一个 $7$ 的倍数的个位数会在 $1,2,3,4,5,6,7$ 中循环，每 $7$ 个集合重复一次。这会给出 $42$ 个集合；在其中，第一个 $7$ 的倍数之后还有另一个 $7$ 的倍数。

第 $99$ 个和第 $100$ 个集合不满足，因为它们给出的第一个倍数分别是 $987$ 和 $994$ ，不可能含有 $2$ 个 $7$ 的倍数。因此共有 $42$ 个集合满足条件。

所以答案是 B。

We can analyze the units digit of the first multiple of $7$ in each set.

If the last digit is $4,5,6,7,$ then adding $7$ yields numbers outside the set, so the sets with a multiple of $7$ such that its units digit is $4,5,6,7$ would have only one multiple.

If the last digit is $1,2,3$ then adding $7$ would yield a units digit of $8,9,0$ in the same set.

Out of the first $98$ sets, there are an equal number of occurrences of sets such that the first multiple of $7$ has each of the units digit of $1,2,3,4,5,6,7$ since they cycle every $7$ sets. These yield $42$ sets whose first multiple of $7$ contains two multiples of $7.$

The $99$ th and $100$ th set wouldn't work since they yield a set whose first multiples are $987$ and $994$ respectively, which can't have $2$ multiples of $7.$ This means we have $42$ sets that work.

Thus, the answer is B .

9.

和式 $\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+\dots+\dfrac{2021}{2022!}$ 可写成 $a-\dfrac{1}{b!}$ ，其中 $a$ 和 $b$ 为正整数。求 $a+b$ 。

The sum $\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+\dots+\dfrac{2021}{2022!}$ can be expressed as $a-\dfrac{1}{b!},$ where $a$ and $b$ are positive integers. What is $a+b?$

$\ 2020$

$\ 2021$

$\ 2022$

$\ 2023$

$\ 2024$

知识点：裂项相消阶乘数学归纳法

难度评级：1220

解答：

我们先证明 $\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+\dots+\dfrac{n-1}{n!}$ $= 1- \dfrac 1{n!}.$

可以用归纳法证明这一恒等式。

$n=2$ ， $\dfrac 1{2!} = 1-\dfrac 1{2!}$ 。

假设结论对 $n-1$ 成立，则下一步为 $\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+\dots+\dfrac{n-1}{n!}$ $= 1- \dfrac 1{(n-1)!}+\dfrac{n-1}{n!}$ $= 1- \dfrac n{n!}+\dfrac{n-1}{n!} = 1 - \dfrac 1{n!}.$

代入 $n=2022$ ，得到 $1- \dfrac 1{2022!}$ ，所以 $a=1,b=2022$ ，两数之和为 $2023$ 。

所以答案是 D。

We claim $\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+\dots+\dfrac{n-1}{n!}$ $= 1- \dfrac 1{n!}.$

To prove this, we can use induction.

If $n=2,$ then the sum is $\dfrac 1{2!} = 1-\dfrac 1{2!} .$

If it works for $n-1,$ then $\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+\dots+\dfrac{n-1}{n!}$ $= 1- \dfrac 1{(n-1)!}+\dfrac{n-1}{n!}$ $= 1- \dfrac n{n!}+\dfrac{n-1}{n!} = 1 - \dfrac 1{n!}.$

This means our formula is proven, so we can get our answer by plugging in $n=2022.$ Our answer is $1- \dfrac 1{2022!},$ yielding $a=1,b=2022,$ making our answer $2023.$

Thus, our answer is D .

10.

Camila 写下五个正整数。这些整数的唯一众数比中位数大 $2$ ，且中位数比算术平均数大 $2$ 。众数的最小可能值是多少？

Camila writes down five positive integers. The unique mode of these integers is $2$ greater than their median, and the median is $2$ greater than their arithmetic mean. What is the least possible value for the mode?

$\ 5$

$\ 7$

$\ 9$

$\ 11$

$\ 13$

知识点：平均数中位数（数据）众数最优化

难度评级：1660

解答：

设五个数按从小到大为 $a,b,c,d,e$ 。中位数为 $c$ ，众数为 $c+2$ 。

因为众数大于中位数且唯一，最后两个数必须都是 $c+2$ ，所以数列为 $a,b,c,c+2,c+2$ 。

平均数为 $c-2$ ，于是 $\begin{aligned} &\frac{a+b+c+(c+2)+(c+2)}{5} \\ &\quad = c-2. \end{aligned}$ 化简得 $a+b+3c+4=5c-10$ ，所以 $a+b=2c-14$ 。

为了让众数唯一， $a$ 和 $b$ 必须是不同的正整数，且都小于 $c$ 。由于 $a+b=2c-14$ 为偶数，最小可行和为 $1+3=4$ ，所以 $2c-14\ge4$ ，即 $c\ge9$ 。

因此最小可能众数为 $c+2=11$ ，并且 $1,3,9,11,11$ 确实可行。

所以正确答案是 D。

Let the integers in increasing order be $a,b,c,d,e.$ The median is $c,$ and the unique mode is $c+2.$

Because the mode is larger than the median and is unique, the last two entries must both be $c+2,$ so the list is $a,b,c,c+2,c+2.$

The mean is $c-2,$ so $\begin{aligned} &\frac{a+b+c+(c+2)+(c+2)}{5} \\ &\quad = c-2. \end{aligned}$ Hence $a+b+3c+4=5c-10,$ so $a+b=2c-14.$

To keep the mode unique, $a$ and $b$ must be distinct positive integers, both less than $c.$ Since $a+b=2c-14$ is even, the smallest such sum is $1+3=4,$ so $2c-14\ge4,$ giving $c\ge9.$

The smallest possible mode is therefore $c+2=11,$ and it is attainable with $1,3,9,11,11.$

Thus, the answer is D .

11.

一个大型学区的所有高中都参加了卖 T 恤的募款活动。下面哪个选项在逻辑上等价于这句话：“没有比 Euclid HS 更大的学校卖出的 T 恤比 Euclid HS 更多”？

All the high schools in a large school district are involved in a fundraiser selling T-shirts. Which of the choices below is logically equivalent to the statement "No school bigger than Euclid HS sold more T-shirts than Euclid HS"?

所有比 Euclid HS 小的学校卖出的 T 恤都比 Euclid HS 少。

All schools smaller than Euclid HS sold fewer T-shirts than Euclid HS.

没有任何卖出 T 恤比 Euclid HS 更多的学校比 Euclid HS 更大。

No school that sold more T-shirts than Euclid HS is bigger than Euclid HS.

所有比 Euclid HS 大的学校卖出的 T 恤都比 Euclid HS 少。

All schools bigger than Euclid HS sold fewer T-shirts than Euclid HS.

所有卖出 T 恤比 Euclid HS 少的学校都比 Euclid HS 小。

All schools that sold fewer T-shirts than Euclid HS are smaller than Euclid HS.

所有比 Euclid HS 小的学校卖出的 T 恤都比 Euclid HS 多。

All schools smaller than Euclid HS sold more T-shirts than Euclid HS.

知识点：逻辑推理

难度评级：900

解答：

原命题可写为：如果一所学校比 Euclid HS 更大，那么它卖出的 T 恤不比 Euclid HS 更多。

它的逆否命题是：如果一所学校卖出的 T 恤比 Euclid HS 更多，那么它不比 Euclid HS 更大。

这正是选项 B 的表述，所以正确答案是 B。

First, we have no information about schools that are smaller than Euclid HS, so we can eliminate all the choices that mention smaller schools. This leaves just B and C to look at.

Now, given our statement, we know that if a school is bigger than Euclid, then it couldn't have sold more than Euclid. This means if a school sold more, it couldn't have been bigger, which corresponds to choice B .

Thus, the answer is B .

12.

一对公平的 $6$ 面骰子掷 $n$ 次。使得至少有一次掷出的点数和为 $7$ 的概率大于 $\dfrac{1}{2}$ 的最小 $n$ 是多少？

A pair of fair $6$ -sided dice is rolled $n$ times. What is the least value of $n$ such that the probability that the sum of the numbers face up on a roll equals $7$ at least once is greater than $\dfrac{1}{2}?$

$2$

$3$

$4$

$5$

$6$

知识点：骰子（概率）对立事件概率

难度评级：960

解答：

也可以求最小的 $n$ ，使没有掷出和为 $7$ 的概率小于 $\dfrac 12.$ 每次以 $\dfrac 16$ 的概率掷出和为 $7,$ 因此以 $\dfrac 56$ 的概率不掷出 $7.$

因此 $7$ 一次也没有出现的概率是 $\left(\dfrac 56\right)^n.$ 要找最小的 $n$ ，使 $\left(\dfrac 56\right)^n < \dfrac 12$ 。

当 $n=3,$ 时，概率为 $\dfrac{125}{216}$ ，大于 $\dfrac 12.$

当 $n=4,$ 时，概率为 $\dfrac{625}{1296}$ ，小于 $\dfrac 12.$ 因此答案是 $4.$

所以正确答案是 C。

To compute this, we can also find the least $n$ such that the probability of not rolling a $7$ is less than $\dfrac 12.$ Each roll has an independent probability of $\dfrac 16$ of getting $7,$ so it has a $\dfrac 56$ probability of not landing on $7.$

Thus, the probability of none of the rolls being $7$ is $\left(\dfrac 56\right)^n.$ We must find the least $n$ such that $\left(\dfrac 56\right)^n < \dfrac 12.$

If $n=3,$ then the probability is $\dfrac{125}{216},$ which is greater than $\dfrac 12.$

If $n=4,$ then the probability is $\dfrac{625}{1296},$ which is less than $\dfrac 12.$ This makes the answer $4.$

Thus, the answer is C .

13.

一对质数的正差为 $2$ ，它们的立方的正差为 $31106$ 。大于这两个质数的最小质数的各位数字之和是多少？

The positive difference between a pair of primes is equal to $2,$ and the positive difference between the cubes of the two primes is $31106.$ What is the sum of the digits of the least prime that is greater than those two primes?

$\ 8$

$\ 10$

$\ 11$

$\ 13$

$\ 16$

知识点：质数立方和与立方差数字

难度评级：1140

解答：

设两个质数相差 $2$ ，分别为 $m-1,m+1$ ，中间数为 $m$ 。

于是 $(m+1)^3-(m-1)^3 = 31106 ,$ 也就是 $m^3+3m^2+3m+1$ $-(m^3-3m^2+3m-1)$ $= 6m^2+2 = 31106.$

因此 $m^2= 5184$ ，所以 $m=72$ 。

因此两个质数为 $71,73$ 。大于它们的最小质数是 $79$ ，其数字和为 $16$ 。

所以答案是 E。

Since the primes are $2$ away from each other, we can make them equal to $m-1,m+1,$ where $m$ is their average.

Then, $(m+1)^3-(m-1)^3 = 31106 ,$ making $m^3+3m^2+3m+1$ $-(m^3-3m^2+3m-1)$ $= 6m^2+2 = 31106.$

Therefore, $m^2= 5184,$ so $m=72.$

The primes are therefore $71,73.$ The least prime greater than both of those is $79,$ and its digit sum is $16.$

Thus, the answer is E .

14.

设 $S$ 是 $\left\{ 1, 2, 3, \cdots , 25 \right\}$ 的子集，并且 $S$ 中任意两个元素（可以相同）的和都不是 $S$ 的元素。 $S$ 最多可以含有多少个元素？

Suppose that $S$ is a subset of $\left\{ 1, 2, 3, \cdots , 25 \right\}$ such that the sum of any two (not necessarily distinct) elements of $S$ is never an element of $S.$ What is the maximum number of elements $S$ may contain?

$\ 12$

$\ 13$

$\ 14$

$\ 15$

$\ 16$

知识点：子集极端原理配对与分组

难度评级：1600

解答：

集合 $S = \{13,14 \cdots ,25\}$ 有 $13$ 个元素，且任意两个元素之和都大于二十五，所以这个大小可以达到。

反过来，设 $m$ 是 $S$ 的最大元素。对 $S$ 中每个满足 $i<m$ 的 $i$ ，数 $m-i$ 不能也属于 $S$ 。

因此，在小于 $m$ 的数中，和为 $m$ 的每一对至多选一个；若 $m$ 为偶数，中间的数也不能选。于是小于 $m$ 的元素至多有 $\lfloor \dfrac {m-1}2 \rfloor$ 个，计入 $m$ 本身后总数至多为 $\lfloor \dfrac{m-1}2 \rfloor +1$ 。

这个上界在 $m=25$ 时最大，等于 $13$ 。

所以正确答案是 B。

The set $S = \{13,14 \cdots ,25\}$ has $13$ elements, and every pair has sum greater than 25, so this size is attainable.

Conversely, let $m$ be the maximum element of $S$ . For every element of $S$ satisfying $i<m$ , the number $i$ and the number $m-i$ cannot both belong to $S$ .

Thus, among the numbers below $m$ , at most one number can be chosen from each pair with sum $m$ ; if $m$ is even, the middle number cannot be chosen either. Hence at most $\lfloor \dfrac {m-1}2 \rfloor$ elements lie below $m$ , and including $m$ gives at most $\lfloor \dfrac{m-1}2 \rfloor +1$ elements.

The maximum value of this has $m=25,$ yielding $13.$

Thus, the answer is B .

15.

设 $S_n$ 是一个公差为 $2$ 的等差数列的前 $n$ 项和。商 $\dfrac{S_{3n}}{S_n}$ 与 $n$ 无关。求 $S_{20}$ 。

Let $S_n$ be the sum of the first $n$ term of an arithmetic sequence that has a common difference of $2.$ The quotient $\dfrac{S_{3n}}{S_n}$ does not depend on $n.$ What is $S_{20}?$

$340$

$360$

$380$

$400$

$420$

知识点：等差数列求和

难度评级：1820

解答：

设数列为 $a_1, a_2, \cdots$ 。令 $a$ 为 $a_1$ 前一项的值，则 $a_n = a+2n$ 。

因此 $S_n = \sum_{i=1}^n (a+2i) =$ $a\cdot n + 2\sum_{i=1}^ni = an + n^2+n.$

于是 $\dfrac{S_{3n}}{S_n} = \dfrac{3an + 9n^2+3n}{an + n^2+n} =$ $\dfrac{3a + 9n+3}{a + n+1}.$ 需要选择 $a$ ，使这个值为常数。

若这个值为常数，则它减去 $9$ 后也为常数： $\dfrac{3a + 9n+3}{a + n+1}-9 = \dfrac{-6a-6}{a+n+1}$ 分母随 $n$ 变化，因此分子必须为 $0$ 。

于是 $-6a-6 = 0$ ， $a=-1$ 。

$S_{20} = -1(20)+20^2+20 = 400.$

所以答案是 D。

Let the sequence be $a_1, a_2, \cdots.$ Create $a$ by making it the term before $a_1$ in the sequence. This would make $a_n = a+2n.$

This would make $S_n = \sum_{i=1}^n (a+2i) =$ $a\cdot n + 2\sum_{i=1}^ni = an + n^2+n.$

This makes $\dfrac{S_{3n}}{S_n} = \dfrac{3an + 9n^2+3n}{an + n^2+n} =$ $\dfrac{3a + 9n+3}{a + n+1}.$ Thus, we must find $a$ such that this value is constant.

If our given value is constant, than the given value minus $9$ is constant, so $\dfrac{3a + 9n+3}{a + n+1}-9 = \dfrac{-6a-6}{a+n+1}$ is constant. As $n$ increases, the numerator is constant and the denominator is increasing. Therefore, if the number is constant, the numerator must be $0.$

Since $-6a-6 = 0,$ we have $a=-1.$

With our formula from before, we have $S_{20} = -1(20)+20^2+20 = 400.$

Thus, the answer is D .

16.

下图显示一个边长为 $4$ 和 $8$ 的矩形，以及一个边长为 $5$ 的正方形。正方形的三个顶点分别在矩形的三条不同边上，如图所示。求同时位于正方形和矩形内部的区域面积。

The diagram below shows a rectangle with side lengths $4$ and $8$ and a square with side length $5.$ Three vertices of the square lie on three different sides of the rectangle, as shown. What is the area of the region inside both the square and the rectangle?

$15\dfrac{1}{8}$

$15\dfrac{3}{8}$

$15\dfrac{1}{2}$

$15\dfrac{5}{8}$

$15\dfrac{7}{8}$

知识点：全等（几何）相似梯形

难度评级：2150

解答：

先按下图标记各点。

因为矩形高为四，正方形边长为五，所以 $AB = 4$ ，由勾股定理可得 $AC = 3$ 。又因为 $\angle ACB$ 与 $\angle DCE$ 互余、 $\angle BAC = \angle CDE = 90^\circ$ ，并且 $BC = CE$ ，可知 $ABC \cong CDE$ 。因此 $ED = 3$ ， $EF = 1$ 。

因为 $\angle CED$ 与 $\angle GEF$ 互余，且 $\angle EFG = \angle CDE = 90^\circ$ ，可知 $EFG$ 与 $CDE$ 相似。于是 $\dfrac{EC}{CD} = \dfrac{EG}{EF}$ ，所以 $EG = 1.25$ 。阴影部分是梯形，面积为 $\dfrac{(GE+BC)EC}2 = \dfrac{ 5(5+1.25)}2$ $= \dfrac{5\cdot 6.25}{2} = 15.625.$

这等于 $15 \dfrac 58$ 。

所以正确答案是 D。

Firstly, let's label the points as follows:

Since we have a rectangle, $AB = 4.$ By the Pythagorean Theorem, we have $AC = 3.$ Then, since $\angle ACB$ and $\angle DCE$ are complementary, $\angle BAC = \angle CDE = 90^\circ,$ and $BC = CE$ we know $ABC \cong CDE.$ Therefore, $ED = 3$ and $EF = 1.$

Since $\angle CED$ and $\angle GEF$ are complementary and $\angle EFG = \angle CDE = 90^\circ,$ we know $EFG$ and $CDE$ are similar. This means $\dfrac{EC}{CD} = \dfrac{EG}{EF},$ so $EG = 1.25.$ Since the shaded region is a trapezoid, we can get the area as $\dfrac{(GE+BC)EC}2 = \dfrac{ 5(5+1.25)}2$ $= \dfrac{5\cdot 6.25}{2} = 15.625.$

This is equal to $15 \dfrac 58.$

Thus, the answer is D .

17.

下列数中，有一个不能被任何小于 $10$ 的质数整除。是哪一个？

One of the following numbers is not divisible by any prime number less than $10.$ Which is it?

$2^{606}-1$

$2^{606}+1$

$2^{607}-1$

$2^{607}+1$

$2^{607}+3^{607}$

知识点：因式分解模运算整除性

难度评级：1820

解答：

使用事实： $a^n-b^n$ 能被 $a-b$ 整除。

选项 A 是 $2^{606}-1=4^{303}-1$ ，能被 $4-1=3$ 整除。

选项 B 是 $2^{606}+1=4^{303}-(-1)^{303}$ ，能被 $4-(-1)=5$ 整除。

选项 D 是 $2^{607}+1$ 。因为 $2^{606}-1$ 能被 $3$ 整除，乘以 $2$ 得到 $2^{607}-2$ 能被 $3$ 整除，从而 $2^{607}+1$ 也能被 $3$ 整除。

选项 E 是 $3^{607}+2^{607}=3^{607}-(-2)^{607}$ ，能被 $3-(-2)=5$ 整除。

对于选项 C， $2^{607}-1$ 是奇数。并且 $2^{607}\equiv2\pmod3$ 、 $2^{607}\equiv3\pmod5$ 、 $2^{607}\equiv2\pmod7$ ，所以 $2^{607}-1$ 不能被 $3,5,$ 或 $7$ 整除。

所以答案是 C。

Use the fact that $a^n-b^n$ is divisible by $a-b.$

Choice A is $2^{606}-1=4^{303}-1,$ which is divisible by $4-1=3.$

Choice B is $2^{606}+1=4^{303}-(-1)^{303},$ which is divisible by $4-(-1)=5.$

Choice D is $2^{607}+1.$ Since $2^{606}-1$ is divisible by $3,$ multiplying by $2$ gives $2^{607}-2$ divisible by $3,$ so $2^{607}+1$ is divisible by $3.$

Choice E is $3^{607}+2^{607}=3^{607}-(-2)^{607},$ which is divisible by $3-(-2)=5.$

For choice C, $2^{607}-1$ is odd. Also $2^{607}\equiv2\pmod3,$ $2^{607}\equiv3\pmod5,$ and $2^{607}\equiv2\pmod7,$ so $2^{607}-1$ is not divisible by $3,5,$ or $7.$

Thus, our answer is C .

18.

考虑未知数为 $x$ 、 $y$ 、 $z$ 的三元一次方程组： $\begin{cases} a_1 x + b_1 y + c_1 z & = 0 \\ a_2 x + b_2 y + c_2 z & = 0 \\ a_3 x + b_3 y + c_3 z & = 0 \end{cases}$ 其中每个系数都是 $0$ 或 $1$ ，且方程组有不同于 $x=y=z=0$ 的解。例如，下面就是一个这样的方程组： $\begin{cases} 1 x + 1 y + 0 z & = 0 \\ 0 x + 1 y + 1 z & = 0 \\ 0 x + 0 y + 0 z & = 0 \end{cases}$ 它有非零解 $(x,y,z) = (1, -1, 1)$ 。这样的方程组共有多少个？（一个方程组中的方程不必互不相同；同样的方程以不同顺序出现时，视为不同方程组。）

Consider systems of three linear equations with unknowns $x,$ $y,$ and $z,$ $\begin{cases} a_1 x + b_1 y + c_1 z & = 0 \\ a_2 x + b_2 y + c_2 z & = 0 \\ a_3 x + b_3 y + c_3 z & = 0 \end{cases}$ where each of the coefficients is either $0$ or $1$ and the system has a solution other than $x=y=z=0.$ For example, one such system is $\begin{cases} 1 x + 1 y + 0 z & = 0 \\ 0 x + 1 y + 1 z & = 0 \\ 0 x + 0 y + 0 z & = 0 \end{cases}$ with a nonzero solution of $(x,y,z) = (1, -1, 1).$ How many such systems of equations are there? (The equations in a system need not be distinct, and two systems containing the same equations in a different order are considered different.)

$\ 302$

$\ 338$

$\ 340$

$\ 343$

$\ 344$

知识点：方程组补集计数分类讨论

难度评级：1970

解答：

共有 $2^9 =512$ 种配置。现在用补集计数来判断其中有多少种有多于一个解。

如果某个配置有 $3$ 个不包含冗余信息的方程，那么它只有一个解。

零向量不能出现，也就是不能有 $a,b,c =0$ 的行，且三行必须互不相同，所以有 $7$ 、 $6$ 、 $5$ 种有序选择。

这给出 $210$ 种配置。不过有些配置仍可能有冗余信息。如果两个方程相加得到另一个方程，就有冗余。

其中还要去掉三行线性相关的情况。

情况 $1:$ $1$ 个方程满足 $a,b,c=1$ ，另一个方程恰有 $1$ 个变量系数为 $1$ ，第三个方程恰有 $2$ 个变量系数为 $1$ 。有 $3$ 种方法选择哪个方程所有变量系数都是 $1$ 。然后有 $2$ 种方法选择哪个方程只有一个变量系数为 $1$ ，而这个方程有 $3$ 种方法选择哪个变量系数为 $1$ 。这种情况要排除 $3\cdot 2\cdot 3=18$ 种配置。

情况 $2:$ $1$ 个方程有 $2$ 个变量系数为 $1$ ，另外两个方程各自只有一个变量系数为 $1$ ，这两个变量彼此不同，并且其中一个变量也出现在第一个方程中。有 $3$ 种方法选择哪个方程有 $2$ 个变量系数为 $1$ ，有 $3$ 种方法选择哪些变量系数为 $1$ ，还有 $2$ 种方法安排另外两个方程的顺序。这种情况也要排除 $3\cdot 2\cdot 3=18$ 种配置。

因而只有零解的方程组有 $210-18-18=174$ 个，所以有非零解的方程组数为 $512-174=338$ 。

所以正确答案是 B。

There are $2^9 =512$ total configurations. Now, we can use complementary counting to determine how many have more than one solution.

If a configuration has $3$ equations which don't contain redundant information, then it has only one solution.

This means every equation has to be different. Also, if any equation has $a,b,c =0,$ then it doesn't provide any information, making it redundant. This means we have $7$ choices for the first equation, $6$ choices for the second, and $5$ choices for the third.

This yields $210$ configurations. However, some configurations may still yield redundant information. If two equations add to the other equation, then there is a redundancy.

There are two cases for this to happen.

Case $1:$ $1$ of the equations has $a,b,c=1,$ another equation has $1$ of the variables being $1$ and the other equation has $2$ variables being $1.$ There are $3$ ways to choose which equation has every variable as $1.$ Then, there are $2$ ways to choose which variables have one variable being $1,$ and this equation has $3$ ways to choose which variable is $1.$ This case has $3\cdot 2\cdot 3=18$ configurations to exclude.

Case $2:$ $1$ of the equations has $2$ variables being $1,$ and the other two equations have only one variable being $1,$ with those variables being different from each other, but one of the variables chosen in the first equation. There are $3$ ways to choose the equation with $2$ variables being $1,$ there are $3$ ways to choose which variables are $1,$ and $2$ ways to choose the order of the other equations. This case has $3\cdot 2\cdot 3=18$ configurations to exclude.

There are a total of $210-18-18=174$ cases which have only one solution. This means $512-174=338$ configurations have multiple solutions, making at least one nonzero.

Thus, the answer is B .

19.

$5 \times 5$ 方格中的每个小方格要么被填充，要么为空；每个小方格最多有八个相邻小方格，相邻表示共边或共顶点。按以下规则变换方格：

• 任意一个已填充的小方格，如果有两个或三个已填充的相邻小方格，则保持填充。

• 任意一个空小方格，如果恰有三个已填充的相邻小方格，则变为填充。

• 所有其他小方格保持为空或变为空。下图显示一个变换示例。

假设这个 $5 \times 5$ 方格有一圈空边框，围住一个 $3 \times 3$ 子方格。经过一次变换后，最终方格只在中心有一个已填充小方格。有多少种初始配置会产生这种结果？（旋转或翻折后相同的配置仍视为不同。）

Each square in a $5 \times 5$ grid is either filled or empty, and has up to eight adjacent neighboring squares, where neighboring squares share either a side or a corner. The grid is transformed by the following rules:

• Any filled square with two or three filled neighbors remains filled.

• Any empty square with exactly three filled neighbors becomes a filled square.

• All other squares remain empty or become empty. A sample transformation is shown in the figure below.

Suppose the $5 \times 5$ grid has a border of empty squares surrounding a $3 \times 3$ subgrid. How many initial configurations will lead to a transformed grid consisting of a single filled square in the center after a single transformation? (Rotations and reflections of the same configuration are considered different.)

$\ 14$

$\ 18$

$\ 22$

$\ 26$

$\ 30$

知识点：过程模拟分类讨论对称性

难度评级：2390

解答：

若中心格初始已填充，则它最终保持填充需要另外有 $2$ 或 $3$ 个已填充邻格，也就是相邻填充数为 $2$ 或 $3$ 。

这些邻格自己不能有过多已填充邻格。检查 $4$ 个角和 $3$ 类相邻关系时，要避免每个邻格拥有 $2$ 或 $3$ 个已填充邻格；因此可行时每个外侧填充格至多只有 $1$ 个相关邻格。

因此除中心外的填充格彼此不能相邻。

可行情况只有选取一对相对角格，共 $2$ 种。

假设中心格初始为空。那么中心格有 $3$ 个已填充邻格，而这些邻格每个最多只有 $1$ 个邻格。这说明没有小方格有两个已填充邻格。因此只可能有以下几种方式：

与此同时，每个被选中的邻格最终必须变空，所以它们彼此之间不能使任何一个拥有两个或三个已填充邻格。图中前三类各有 $4$ 个旋转，最后一类有 $8$ 个旋转或翻折，共 $20$ 种；加上前面的两种，总数为 $20+2=22$ 。

所以正确答案是 C。

Suppose the center is initially filled. Then, there are either either $2$ or $3$ other filled squares, each of which can't have $2$ or $3$ filled neighbors.

This means that there are at most $4$ filled squares, so each square has at most $3$ neighbors. Since they don't have $2$ or $3$ neighbors, they must have at most $1$ neighbor. The center square is a neighbor, so they can't have any other neighbor.

Suppose I have a filled square on an edge. Since there is some filled square that isn't a neighbor of the square, we can examine the two edges which are neighbors of the filled edge. If I have a filled edge on the corner, the edge on the same side as the corner would have three neighbors. If I choose the opposite edge, the adjacent edges would have three neighbors.

Suppose I choose a corner. Then, I need to choose another corner. If I choose the adjacent corner, then the edge between would have three neighbors, making it filled. Therefore, it must be the opposite corner. This has $2$ configurations.

Suppose the center is initially empty. Then, there are $3$ filled neighbors of the center, each with at most $1$ neighbors. This means no square has two filled neighbors. This makes it only possible to do in the following ways:

The first three can rotated making $4$ configurations, and the last one can be rotated and reflected making $8$ configurations. There are $20$ configurations with the center being empty. This means there are $20+2=22$ different configurations.

Thus, the answer is C .

20.

设 $ABCD$ 是一个菱形，且 $\angle ADC = 46^\circ$ 。设 $E$ 是 $\overline{CD}$ 的中点， $F$ 是 $\overline{BE}$ 上的一点，并且 $\overline{AF}$ 垂直于 $\overline{BE}$ 。求 $\angle BFC$ 的度数。

Let $ABCD$ be a rhombus with $\angle ADC = 46^\circ.$ Let $E$ be the midpoint of $\overline{CD},$ and let $F$ be the point on $\overline{BE}$ such that $\overline{AF}$ is perpendicular to $\overline{BE}.$ What is the degree measure of $\angle BFC?$

$\ 110$

$\ 111$

$\ 112$

$\ 113$

$\ 114$

知识点：导角圆内接四边形圆周角

难度评级：2150

解答：

延长 $BE$ 与 $AD$ 交于 $G$ 。因为 $\angle GDE = \angle ECB,$ $\angle GED = \angle BEC$ 且 $DE = EC,$ 可知 $GDE \cong BCE$ 。因此 $DG =BC = AD$ 。这说明以 $D$ 为圆心、过 $A$ 的圆也经过 $C,G$ 。

又因为 $AFG$ 是直角三角形，这个圆也是它的外接圆，所以 $F$ 也在圆上。

因为 $\angle GDC = 134^\circ$ ，所以 $\angle CFE = \angle CFG = \dfrac {\overset{\Large\frown}{CG}} 2$ $= \dfrac{134^\circ}{2} = 67^\circ.$ 因此 $\angle BFC = 180^\circ - 67^\circ = 113^\circ.$

所以正确答案是 D。

First, we extend $BE$ and $AD$ such that they meet at $G.$ Since $\angle GDE = \angle ECB,$ $\angle GED = \angle BEC$ and $DE = EC,$ we know $GDE \cong BCE.$ Therefore, $DG =BC = AD.$ This means that if we construct a circle with center $D$ that includes $A,$ $C,G$ are also on it.

Also, since $AFG$ is a right triangle, the drawn circle would be its circumcircle, placing $F$ on the circle.

Since $\angle GDC = 134^\circ,$ we can get $\angle CFE = \angle CFG = \dfrac {\overset{\Large\frown}{CG}} 2$ $= \dfrac{134^\circ}{2} = 67^\circ.$ Therefore, $\angle BFC = 180^\circ - 67^\circ = 113^\circ.$

Thus, the answer is D .

21.

设 $P(x)$ 是一个有理系数多项式。 $P(x)$ 除以 $x^2 + x + 1$ 的余式为 $x+2$ ， $P(x)$ 除以 $x^2+1$ 的余式为 $2x+1$ 。满足这两个条件且次数最小的多项式唯一。求这个多项式各系数平方和。

Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1,$ the remainder is $x+2,$ and when $P(x)$ is divided by the polynomial $x^2+1,$ the remainder is $2x+1.$ There is a unique polynomial of least degree with these two properties. What is the sum of the squares of the coefficients of that polynomial?

$\ 10$

$\ 13$

$\ 19$

$\ 20$

$\ 23$

知识点：多项式方程组

难度评级：2150

解答：

因为 $P(x)$ 除以 $x^2+x+1$ 的余式为 $x+2$ ，可写成 $P(x) =$ $(x^2+x+1)Q(x)+x+2$ 其中 $Q(x)$ 是某个多项式。注意 $P(x)$ 除以 $x^2+1$ 的余式等于 $xQ(x) + x+2$ 的余式。

若 $Q(x)=c$ 是常数，则这个余式为 $(c+1)x+2$ ，不可能等于 $2x+1$ ，因为常数项 $2$ 与 $1$ 不同。所以 $Q$ 的次数至少为 $1$ 。

试令 $Q(x) = ax+b$ 。则 $P(x)$ 模 $x^2+1$ 的余式等于 $\begin{aligned} &(ax+b)x+x+2 \\ &\quad = ax^2 + (b+1)x+2 \end{aligned}$ 模 $x^2+1$ 的余式。减去 $a(x^2+1)$ 后，余式为 $(b+1)x +(2-a)$ 。它应等于 $2x+1$ 。因此 $b+1=2$ 且 $2-a=1$ ，所以 $a=b=1$ 。

这说明 $P(x) =$ $(x+1)(x^2+x+1)+x+2 =$ $x^3+2x^2+3x+3.$ 系数平方和为 $23$ 。

所以正确答案是 E。

Since $P(x)$ has a remainder of $x+2$ when divided by $x^2+x+1,$ it must be able to be written as $P(x) =$ $(x^2+x+1)Q(x)+x+2$ for some polynomial $Q(x).$ Note that the remainder of $P(x)$ when divided by $x^2+1$ is equal to the remainder when $xQ(x) + x+2.$

If $Q(x)=c$ is constant, this remainder is $(c+1)x+2,$ which can never equal $2x+1$ because the constant terms $2$ and $1$ differ. So $Q$ must have degree at least $1.$

Try $Q(x) = ax+b.$ Then the remainder of $P(x)$ modulo $x^2+1$ equals the remainder of $\begin{aligned} &(ax+b)x+x+2 \\ &\quad = ax^2 + (b+1)x+2 \end{aligned}$ modulo $x^2+1.$ Subtracting $a(x^2+1)$ gives $(b+1)x +(2-a),$ which must equal $2x+1.$ Thus $b+1=2$ and $2-a=1,$ so $a=b=1.$

This means $P(x) =$ $(x+1)(x^2+x+1)+x+2 =$ $x^3+2x^2+3x+3.$ The sum of the squares of the coefficients is $23.$

Thus, the answer is E .

22.

设 $S$ 为坐标平面中所有同时与下面三个圆相切的圆的集合： $\begin{align*}x^{2}+y^{2}&=4,\\ x^{2}+y^{2}&=64, \\ (x-5)^{2}+y^{2}&=3.\end{align*}$ 求 $S$ 中所有圆的面积之和。

Let $S$ be the set of circles in the coordinate plane that are tangent to each of the three circles with equations $\begin{align*}x^{2}+y^{2}&=4,\\ x^{2}+y^{2}&=64, \\ (x-5)^{2}+y^{2}&=3.\end{align*}$ What is the sum of the areas of all circles in $S?$

$\ 48 \pi$

$\ 68 \pi$

$\ 96 \pi$

$\ 102 \pi$

$\ 136 \pi$

知识点：相切圆圆面积分类讨论

难度评级：2390

解答：

记 $x^2 + y^2 = 64$ 为圆 $O$ ， $x^2 + y^2 = 4$ 为圆 $P$ ， $(x - 5)^2 + y^2 = 3$ 为圆 $Q$ 。

首先注意， $S$ 中任意圆 $R$ 都必须与 $O$ 内切。接着按照 $R$ 与 $P$ 和 $Q$ 的相切方式分类。

情况 $1:$ 这对应粉色圆。此时 $P$ 和 $Q$ 都与 $R$ 内切。

情况 $2:$ 这对应蓝色圆。此时 $P$ 和 $Q$ 都与 $R$ 外切。

情况 $3:$ 这对应绿色圆。此时 $P$ 外切、 $Q$ 内切于 $R$ 。

情况 $4:$ 这对应红色圆。此时 $P$ 内切、 $Q$ 外切于 $R$ 。

可以把情况 $1$ 和 $4$ 放在一起考虑。注意 $O$ 和 $P$ 同心。连接 $R$ 与 $O$ 圆心的直线会经过 $R$ 与 $O$ 的切点以及 $R$ 与 $P$ 的切点。

这条线是 $R$ 的直径，其长度为 $r_P + r_O = 2 + 8 = 10.$ 因此 $R$ 的半径是 $5$ 。

再把情况 $2$ 和 $3$ 放在一起考虑。与上面类似，连接 $R$ 与 $O$ 圆心的直线会经过切点。

不过这一次， $R$ 的直径为 $r_O - r_P = 8 - 2 = 6.$ 因此 $R$ 的半径是 $3$ 。

$S$ 中共有 $8$ 个圆：其中 $4$ 个半径为 $5$ ，另外 $4$ 个半径为 $3$ ；这是因为把图中的所有圆关于横轴翻折，还能得到另外 $4$ 个圆。

$S$ 中所有圆的总面积为 $4 (5^2 \pi + 3^2 \pi) = 136 \pi.$ 所以正确答案是 E。

Let $x^2 + y^2 = 64$ be circle $O,$ $x^2 + y^2 = 4$ be circle $P,$ and $(x - 5)^2 + y^2 = 3$ be circle $Q.$

First note that every circle, $R,$ in $S$ is internally tangent to $O.$ Then we case on the tangency of $R$ with $P$ and $Q.$

Case $1:$ This corresponds to the pink circle. This is where $P$ and $Q$ are internally tangent to $R.$

Case $2:$ This corresponds to the bluish circle. This is where $P$ and $Q$ are externally tangent to $R.$

Case $3:$ This corresponds to the green circle. This is where $P$ is externally and $Q$ is internally tangent to $R.$

Case $4:$ This corresponds to the red circle. This is where $P$ is internally and $Q$ is externally tangent to $R.$

We can consider cases $1$ and $4$ together. Note that $O$ and $P$ have the same center. This means that the line connecting the center of $R$ and $O$ passes through the tangency points of both $R$ and $O$ and $R$ and $P.$

This line is the diameter of $R,$ and it has length $r_P + r_O = 2 + 8 = 10.$ Therefore, the radius of $R$ is $5.$

Consider cases $2$ and $3$ together. Similarly to above, the line connecting the center of $R$ and $O$ will pass through the tangency points.

This time, however, the diameter of $R$ is $r_O - r_P = 8 - 2 = 6.$ This makes the radius of $R$ $3.$

$S$ contains $8$ circles: $4$ of which have radius $5$ and $4$ of which have radius $3$ (this is because we can flip all the circles in the diagram over the x-axis to get $4$ more circles).

The total area of the circles in $S$ is therefore $4 (5^2 \pi + 3^2 \pi) = 136 \pi.$ Thus, E is the correct answer.

23.

蚂蚁 Amelia 从数轴上的 $0$ 出发，并按如下方式爬行。对于 $n=1,2,3,$ Amelia 独立且均匀地从区间 $(0,1).$ 中随机选择一个持续时间 $t_n$ 和一个增量 $x_n$ 。在第 $n$ 步中，Amelia 沿正方向移动 $x_n$ 个单位，用时 $t_n$ 分钟。如果总用时在第 $n$ 步期间已经超过 $1$ 分钟，她会在该步结束时停止；否则继续下一步，最多走 $3$ 步。Amelia 停止时位置大于 $1$ 的概率是多少？

Ant Amelia starts on the number line at $0$ and crawls in the following manner. For $n=1,2,3,$ Amelia chooses a time duration $t_n$ and an increment $x_n$ independently and uniformly at random from the interval $(0,1).$ During the $n$ th step of the process, Amelia moves $x_n$ units in the positive direction, using up $t_n$ minutes. If the total elapsed time has exceeded $1$ minute during the $n$ th step, she stops at the end of that step; otherwise, she continues with the next step, taking at most $3$ steps in all. What is the probability that Amelia’s position when she stops will be greater than $1?$

$\dfrac 13$

$\dfrac 12$

$\dfrac 23$

$\dfrac 34$

$\dfrac 56$

知识点：几何概率独立事件分类讨论

难度评级：2150

解答：

停止时间只依赖时间变量，而最终位置只依赖距离变量，所以相应概率可以相乘。

两个独立 $(0,1)$ 数之和小于 $1$ 的概率是单位正方形中一个直角三角形的面积，即 $\frac12$ 。

三个独立 $(0,1)$ 数之和小于 $1$ 的概率是截距为 $1$ 的四面体体积，即 $\frac16$ 。

若 $t_1+t_2>1$ ，Amelia 走两步后停止。其概率为 $\frac12$ ，且独立地有 $x_1+x_2>1$ 的概率为 $\frac12$ ，贡献 $\frac14$ 。

若 $t_1+t_2<1$ ，Amelia 会走第三步。其概率为 $\frac12$ ，且独立地有 $x_1+x_2+x_3>1$ 的概率为 $1-\frac16=\frac56$ ，贡献 $\frac5{12}$ 。

总概率为 $\frac14+\frac5{12}=\frac23$ 。

所以正确答案是 C。

The stopping time depends only on the time variables, while the final position depends only on the distance variables, so the corresponding probabilities multiply.

For two independent numbers in $(0,1),$ the probability that their sum is less than $1$ is the area of a right triangle, namely $\frac12.$

For three independent numbers in $(0,1),$ the probability that their sum is less than $1$ is the volume of a tetrahedron with side intercepts $1,$ namely $\frac16.$

If $t_1+t_2>1,$ Amelia stops after two steps. This has probability $\frac12,$ and independently $x_1+x_2>1$ has probability $\frac12,$ contributing $\frac14.$

If $t_1+t_2<1,$ Amelia takes the third step. This has probability $\frac12,$ and independently $x_1+x_2+x_3>1$ has probability $1-\frac16=\frac56,$ contributing $\frac5{12}.$

The total probability is $\frac14+\frac5{12}=\frac23.$

Thus, the answer is C .

24.

考虑满足下列条件的函数 $f$ ：对所有实数 $x$ 和 $y$ ，都有 $|f(x)-f(y)|\leq \dfrac{1}{2}|x-y|$ 。在这些函数中，再要求 $f(300) = f(900)$ 。求 $f(f(800))-f(f(400))$ 的最大可能值。

Consider functions $f$ that satisfy $|f(x)-f(y)|\leq \dfrac{1}{2}|x-y|$ for all real numbers $x$ and $y.$ Of all such functions that also satisfy the equation $f(300) = f(900),$ what is the greatest possible value of $f(f(800))-f(f(400))?$

$25$

$50$

$100$

$150$

$200$

知识点：函数不等式极端原理

难度评级：2390

解答：

由收缩条件可得 $|f(f(400))-f(f(300))|$ $\leq \dfrac 12 | f(400) - f(300)|$ $\leq \dfrac 14 |400-300| = 25,$ 同理， $|f(f(900))-f(f(800))|$ $\leq \dfrac 12 | f(900) - f(800)|$ $\leq \dfrac 14 |900-800| = 25.$

因为 $f(900) = f(300)$ ，由三角不等式， $|f(f(800)) - f(f(400))| =$ $|(f(f(800)) - f(f(900))) -$ $(f(f(400))-f(f(300)))| \leq$ $|(f(f(800)) - f(f(900)))| +$ $|(f(f(400))-f(f(300)))|$ $\leq 50.$

为使上界能够达到，定义 $f$ 为依次经过下列各点的分段线性函数： $\begin{gathered}(300,600),(400,550),(550,575),\\ (650,625),(800,650),(900,600).\end{gathered}$ 并规定当 $x\leq300$ 或 $x\geq900$ 时 $f(x)=600$ 。每一段斜率的绝对值都不超过 $\dfrac12$ ，因此满足收缩条件。特别地， $f(300)=f(900)=600$ 、 $f(400)=550$ 、 $f(800)=650$ 。所以 $f(f(400))=f(550)=575$ ，而 $f(f(800))=f(650)=625$ ，两者之差为 $50$ 。

上述分段线性函数满足所需条件并达到该上界，因此这个上界就是最大值。所以正确答案是 B。

Note that $|f(f(400))-f(f(300))|$ $\leq \dfrac 12 | f(400) - f(300)|$ $\leq \dfrac 14 |400-300| = 25,$ and $|f(f(900))-f(f(800))|$ $\leq \dfrac 12 | f(900) - f(800)|$ $\leq \dfrac 14 |900-800| = 25.$

Since $f(900) = f(300),$ by the triangle inequality, we know $|f(f(800)) - f(f(400))| =$ $|(f(f(800)) - f(f(900))) -$ $(f(f(400))-f(f(300)))| \leq$ $|(f(f(800)) - f(f(900)))| +$ $|(f(f(400))-f(f(300)))|$ $\leq 50.$

To attain the bound, define $f$ by linear interpolation through the points $\begin{gathered}(300,600),(400,550),(550,575),\\ (650,625),(800,650),(900,600).\end{gathered}$ and set $f(x)=600$ for $x\leq300$ or $x\geq900$ . Every segment has slope with absolute value at most $\dfrac12$ , so the contraction condition holds. In particular, $f(300)=f(900)=600$ , $f(400)=550$ , and $f(800)=650$ . Hence $f(f(400))=f(550)=575$ , while $f(f(800))=f(650)=625$ , giving the difference $50$ .

Thus, the answer is B .

25.

设 $x_0,x_1,x_2,\dotsc$ 是一个数列，其中每个 $x_k$ 都是 $0$ 或 $1$ 。对每个正整数 $n$ ，定义 $S_n = \sum_{k=0}^{n-1} x_k 2^k$ 。假设对所有 $n \geq 1$ ，都有 $7S_n \equiv 1 \pmod{2^n}$ 。求 $x_{2019} + 2x_{2020} +$ $4x_{2021} + 8x_{2022}$ 的值。

Let $x_0,x_1,x_2,\dotsc$ be a sequence of numbers, where each $x_k$ is either $0$ or $1.$ For each positive integer $n,$ define $S_n = \sum_{k=0}^{n-1} x_k 2^k$ Suppose $7S_n \equiv 1 \pmod{2^n}$ for all $n \geq 1.$ What is the value of the sum $x_{2019} + 2x_{2020} +$ $4x_{2021} + 8x_{2022}?$

$6$

$7$

$12$

$14$

$15$

知识点：模运算进制 2的幂

难度评级：2390

解答：

首先注意， $\dfrac{ S_{2023} - S_{2019}}{2^{2019}} = x_{2019} +$ $2x_{2020} + 4x_{2021} + 8x_{2022}.$ 因此只需求出 $S_{2019}, S_{2023}$ 。另外， $0 \leq S_n \leq \sum_{k=0}^{n-1} 2^k < 2^n.$

由 $7S_n \equiv 1 \pmod{2^n}$ ，可取某个整数 $m$ ，使 $7S_n = m2^n +1 \implies$ $S_n = \dfrac {m2^n+1}{7}$ 。又由 $0 \leq S_n < 2^n$ ，有 $0 \leq \dfrac {m2^n+1}7 < 2^n.$ 因而 $0 \leq m < 7$ 。接下来只需寻找使 $m2^n + 1$ 能被 $7$ 整除的 $m$ ，即满足 $m2^n + 1 \equiv 0 \mod 7$ 的整数。

当 $n=2019,$ 时， $0 \equiv m2^{2019} + 1 \equiv$ $m(2^3)^{673}+1 \equiv$ $m8^{673} + 1\equiv m+1,$ 所以 $m=6.$ 因此 $S_{2019} = \dfrac {6(2^{2019})+1}{7}.$

当 $n=2023,$ 时， $0 \equiv m2^{2023} + 1$ $\equiv m(2^3)^{674}\cdot 2+1$ $\equiv m8^{674}\cdot 2 + 1$ $\equiv 2m+1,$ 所以 $m=3.$ 因此 $S_{2023} = \dfrac {3(2^{2023})+1}{7}.$

$\begin{gathered} \frac{S_{2023}-S_{2019}}{2^{2019}} = \frac{1}{2^{2019}} \\ {}\cdot \small \left(\frac{3\cdot2^{2023}+1}{7}-\frac{6\cdot2^{2019}+1}{7}\right) \\ = \frac{42\cdot2^{2019}}{7\cdot2^{2019}} = 6. \end{gathered}$

所以正确答案是 A。

Note first that $\dfrac{ S_{2023} - S_{2019}}{2^{2019}} = x_{2019} +$ $2x_{2020} + 4x_{2021} + 8x_{2022}.$ Therefore, we should attempt to find $S_{2019}, S_{2023}.$ Also, note that $0 \leq S_n \leq \sum_{k=0}^{n-1} 2^k < 2^n.$

Now, since $7S_n \equiv 1 \pmod{2^n},$ we know $7S_n = m2^n +1 \implies$ $S_n = \dfrac {m2^n+1}{7}$ for some integer $m.$ Also, since $0 \leq S_n < 2^n,$ we know $0 \leq \dfrac {m2^n+1}7 < 2^n.$ This means $0 \leq m < 7.$ Now, we find $m$ such that $m2^n + 1$ is divisible by $7.$ This makes $m2^n + 1 \equiv 0 \mod 7.$

If $n=2019,$ then $0 \equiv m2^{2019} + 1 \equiv$ $m(2^3)^{673}+1 \equiv$ $m8^{673} + 1\equiv m+1,$ so $m=6.$ This makes $S_{2019} = \dfrac {6(2^{2019})+1}{7}.$

If $n=2023,$ then $0 \equiv m2^{2023} + 1$ $\equiv m(2^3)^{674}\cdot 2+1$ $\equiv m8^{674}\cdot 2 + 1$ $\equiv 2m+1,$ so $m=3.$ This makes $S_{2023} = \dfrac {3(2^{2023})+1}{7}.$

Our answer is $\begin{gathered} \frac{S_{2023}-S_{2019}}{2^{2019}} = \frac{1}{2^{2019}} \\ {}\cdot \small \left(\frac{3\cdot2^{2023}+1}{7}-\frac{6\cdot2^{2019}+1}{7}\right) \\ = \frac{42\cdot2^{2019}}{7\cdot2^{2019}} = 6. \end{gathered}$

Thus, the correct answer is A .