下列表达式等价于哪一个？ \beginaligned &(2+3)(2^2+3^2)\ &\quad×(2^4+3^4)(2^8+3^8)\ &\quad×(2^16+3^16)(2^32+3^32)\ &\quad×(2^64+3^64) \endaligned？

(B) 3^127 + 2^127 + 2 × 3^63 + 3 × 2^63

2021 AMC 10A Spring 真题与在线限时测验

计时

1:15:00

1.

求下式的值： $(2^2-2)-(3^2-3)+(4^2-4)\text{？}$

What is the value of $(2^2-2)-(3^2-3)+(4^2-4)?$

$1$

$2$

$5$

$8$

$12$

答案：D

知识点：运算顺序

难度评级：450

小提示：

先分别计算括号里的每一项，再处理前面的正负号。

Evaluate each parenthesized expression before combining the signs

大提示：

注意每一项都有 $n^2-n$ 的形式。

Notice that each term has the form $n^2-n$

解答：

$\begin{aligned} (2^2 - 2) - &(3^2 - 3) + (4^2 - 4) \\ &= 2 - 6 + 12 \\ &= 8 \end{aligned}\text{。}$

所以正确答案是 D。

$\begin{aligned} (2^2 - 2) - &(3^2 - 3) + (4^2 - 4) \\ &= 2 - 6 + 12 \\ &= 8. \end{aligned}$

Thus, D is the correct answer.

2.

Portia 的高中学生数是 Lara 的高中的 $3$ 倍。两所高中一共有 $2600$ 名学生。Portia 的高中有多少名学生？

Portia’s high school has $3$ times as many students as Lara’s high school. The two high schools have a total of $2600$ students. How many students does Portia’s high school have?

$600$

$650$

$1950$

$2000$

$2050$

答案：C

知识点：比与比例一次方程

难度评级：560

小提示：

把 Lara 学校的人数看作一份。

Let Lara’s enrollment be one part

大提示：

Portia 学校占总人数四份中的三份。

Portia has three of the four equal parts of the total

解答：

设 Lara 的学校有 $x$ 名学生，则 Portia 的学校有 $3x$ 名学生。

因此 $\begin{aligned} 3x + x &= 2600 \\ x &= 650 \end{aligned}\text{。}$ 所以 $3x = 1950$ 。

所以正确答案是 C。

Let $x$ be the number of students in Lara’s high school. Then Portia’s high school has $3x$ students.

Therefore, $\begin{aligned} 3x + x &= 2600 \\ x &= 650. \end{aligned}$ Then $3x = 1950.$

Thus, C is the correct answer.

3.

两个自然数的和为 $17{,}402$ 。其中一个数能被 $10$ 整除。如果擦去这个数的个位数字，就得到另一个数。这两个数的差是多少？

The sum of two natural numbers is $17{,}402.$ One of the two numbers is divisible by $10.$ If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?

$10{,}272$

$11{,}700$

$13{,}362$

$14{,}238$

$15{,}426$

答案：D

知识点：位值一次方程

难度评级：900

小提示：

如果删去末尾的零得到另一个数，那么较大的数是较小数的十倍。

If deleting the final zero gives the other number, the larger number is ten times the smaller one

大提示：

先用总和求出较小的那个数。

Use the sum to find the smaller number first

视频讲解：

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文字解答：

设两个数为 $x$ 和 $y$ 。不妨设 $x$ 能被 $10$ 整除，那么 $x$ 的个位数字是 $0$ 。

擦去 $x$ 的个位数字，相当于把它除以 $10$ 。题意还给出 $\dfrac{x}{10} = y$ 。

由两数之和可得 $\begin{aligned} \dfrac{x}{10} + x &= 17,402 \\ \dfrac{11x}{10} &= 17,402 \\ x &= 15,820 \end{aligned}\text{。}$ 因此两数之差为 $x - \dfrac{x}{10} = 14,238\text{。}$

所以正确答案是 D。

Let $x$ and $y$ be the two numbers. WLOG, let $x$ be divisible by $10.$ Then the units digit of $x$ is $0.$

If we erase the units digit, then we are essentially dividing $x$ by $10.$ The problem statement also gives us that $\dfrac{x}{10} = y.$

Therefore, $\begin{aligned} \dfrac{x}{10} + x &= 17,402 \\ \dfrac{11x}{10} &= 17,402 \\ x &= 15,820. \end{aligned}$ Then $x - \dfrac{x}{10} = 14,238.$

Thus, D is the correct answer.

4.

一辆小车沿山坡向下滚动，第一秒行驶 $5$ 英寸，并且不断加速，使得之后每个连续的 $1$ 秒时间段内，它都比前一个 $1$ 秒时间段多行驶 $7$ 英寸。小车用 $30$ 秒到达山脚。它一共行驶了多少英寸？

A cart rolls down a hill, traveling $5$ inches the first second and accelerating so that during each successive $1$ -second time interval, it travels $7$ inches more than during the previous $1$ -second interval. The cart takes $30$ seconds to reach the bottom of the hill. How far, in inches, does it travel?

$215$

$360$

$2992$

$3195$

$3242$

答案：D

知识点：等差数列求和

难度评级：870

小提示：

每秒行驶的距离组成一个等差数列。

The distances traveled each second form an arithmetic sequence

大提示：

先求第 $30$ 项，再用首项和末项的平均数求和。

Find the $30$ th term, then use the average of the first and last terms

视频讲解：

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文字解答：

每秒行驶的距离构成等差数列： $5, 5 + 7, 5 + 2 \cdot 7, \ldots$

等差数列求和的标准公式为 $n\cdot\dfrac{a_1+a_n}{2}\text{。}$ 已知项数为 $30$ ，首项为 $5$ 。末项为 $5 + 29 \cdot 7 = 208\text{。}$

把这些值代入公式，得到 $30 \cdot \dfrac{5 + 208}{2} = 15 \cdot 213 = 3195\text{。}$

所以正确答案是 D。

The distance travelled every second forms an arithmetic sequence: $5, 5 + 7, 5 + 2 \cdot 7, \ldots$

The standard arithmetic-sequence sum formula is $n\cdot\dfrac{a_1+a_n}{2}.$ We know the number of terms is $30$ and the first term is $5.$ The last term is $5 + 29 \cdot 7 = 208.$

Plugging these values into the expression yields $30 \cdot \dfrac{5 + 208}{2} = 15 \cdot 213 = 3195.$

Thus, D is the correct answer.

5.

一个有 $k > 12$ 名学生的班级，测验分数的平均数为 $8$ 。其中 $12$ 个测验分数的平均数为 $14$ 。其余测验分数的平均数用 $k$ 表示是多少？

The quiz scores of a class with $k > 12$ students have a mean of $8.$ The mean of a collection of $12$ of these quiz scores is $14.$ What is the mean of the remaining quiz scores in terms of $k?$

$\dfrac{14-8}{k-12}$

$\dfrac{8k-168}{k-12}$

$\dfrac{14}{12} - \dfrac{8}{k}$

$\dfrac{14(k-12)}{k^2}$

$\dfrac{14(k-12)}{8k}$

答案：B

知识点：平均数代数变形

难度评级：900

小提示：

把每个平均数都转化为总分。

Convert each mean into a total score

大提示：

从全班总分中减去那 $12$ 个分数的总分。

Subtract the known total for the chosen $12$ scores from the class total

视频讲解：

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文字解答：

全班所有人的总分为 $8k$ 。这 $12$ 个分数的总分为 $12 \cdot 14 = 168$ 。

其余学生的总分为 $8k - 168$ ，人数为 $k - 12$ 。因此平均数为 $\dfrac{8k - 168}{k - 12}\text{。}$

所以正确答案是 B。

The sum of the scores of everyone in the class is $8k.$ The sum of the scores in the collection of $12$ is $12 \cdot 14 = 168.$

This means that the sum of the scores of everyone not in the collection is $8k - 168.$ There are also $k - 12$ people not in the collection. Therefore, the average is $\dfrac{8k - 168}{k - 12}.$

Thus, B is the correct answer.

6.

Chantal 和 Jean 从登山口出发，沿小路走向一座消防塔。Jean 背着很重的背包，走得较慢。Chantal 开始时以每小时 $4$ 英里的速度行走。到消防塔的一半路程处，小路变得很陡，她放慢到每小时 $2$ 英里。到达消防塔后，她立即掉头，并以每小时 $3$ 英里的速度下行陡峭的那一段。她在半程点遇到 Jean。到他们相遇为止，Jean 的平均速度是多少英里每小时？

Chantal and Jean start hiking from a trailhead toward a fire tower. Jean is wearing a heavy backpack and walks slower. Chantal starts walking at $4$ miles per hour. Halfway to the tower, the trail becomes really steep, and Chantal slows down to $2$ miles per hour. After reaching the tower, she immediately turns around and descends the steep part of the trail at $3$ miles per hour. She meets Jean at the halfway point. What was Jean’s average speed, in miles per hour, until they meet?

$\dfrac{12}{13}$

$1$

$\dfrac{13}{12}$

$\dfrac{24}{13}$

$2$

答案：A

知识点：路程、速度与时间

难度评级：1140

小提示：

设半条路的长度为 $d$ 。

Let half the trail length be $d$

大提示：

Jean 到达半程点所用的时间，正好等于 Chantal 三段行程的总时间。

Jean reaches the halfway point in exactly the same time Chantal takes for her three hiking segments

视频讲解：

Click to load, then click again to play

文字解答：

设到消防塔的全程距离为 $2d$ ，其中 $d > 0$ 。

于是 Chantal 徒步用了 $\dfrac{d}{4} + \dfrac{d}{2} + \dfrac{d}{3} = \dfrac{13d}{12}$ 小时。

Jean 在 $\dfrac{13d}{12}$ 小时内走了 $d$ 英里，所以他的速度为 $d \div \dfrac{13d}{12} = \dfrac{12}{13}$ 英里每小时。

所以正确答案是 A。

Let $2d$ be the distance from the trailhead to the fire tower, where $d > 0.$

Then Chantal hiked for $\dfrac{d}{4} + \dfrac{d}{2} + \dfrac{d}{3} = \dfrac{13d}{12}$ hours.

If Jean travelled $d$ miles in $\dfrac{13d}{12}$ hours, then his speed was $d \div \dfrac{13d}{12} = \dfrac{12}{13}$ miles per hour.

Thus, A is the correct answer.

7.

Tom 收藏了 $13$ 条蛇，其中 $4$ 条是紫色的， $5$ 条是开心的。他观察到

• 所有开心的蛇都会加法，

• 没有紫色的蛇会减法，并且

• 所有不会减法的蛇也不会加法。

关于 Tom 的蛇，可以推出下列哪个结论？

Tom has a collection of $13$ snakes, $4$ of which are purple and $5$ of which are happy. He observes that

• all of his happy snakes can add,

• none of his purple snakes can subtract, and

• all of his snakes that can’t subtract also can’t add.

Which of these conclusions can be drawn about Tom’s snakes?

紫色的蛇会加法。

Purple snakes can add.

紫色的蛇是开心的。

Purple snakes are happy.

会加法的蛇是紫色的。

Snakes that can add are purple.

开心的蛇不是紫色的。

Happy snakes are not purple.

开心的蛇不会减法。

Happy snakes can’t subtract.

答案：D

知识点：逻辑推理

难度评级：960

小提示：

把每句话翻译成一个蕴含关系。

Translate each statement into an implication

大提示：

把“紫色推出不会减法”和“不会减法推出不会加法”连起来。

Combine purple implies cannot subtract with cannot subtract implies cannot add

视频讲解：

Click to load, then click again to play

文字解答：

紫色的蛇不会减法，而不会减法的蛇也不会加法，所以紫色的蛇不会加法。

另一方面，开心的蛇都会加法，所以开心的蛇不可能是紫色的。

所以正确答案是 D。

Note that the third condition ensures that purple snakes can’t add.

We also know that all happy snakes can add, which means that happy snakes can’t be purple as well.

Thus, D is the correct answer.

8.

一名学生把数 $66$ 乘以循环小数 $\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}}\text{，}$ 其中 $a$ 和 $b$ 是数字。他没有注意到循环记号，只是计算了 $66$ 乘以 $\underline{1}.\underline{a} \ \underline{b}$ 。后来他发现自己的答案比正确答案小 $0.5$ 。 $2$ 位数 $\underline{a} \ \underline{b}$ 是多少？

When a student multiplied the number $66$ by the repeating decimal, $\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}},$ where $a$ and $b$ are digits, he did not notice the notation and just multiplied $66$ times $\underline{1}.\underline{a} \ \underline{b}.$ Later he found that his answer is $0.5$ less than the correct answer. What is the $2$ -digit integer $\underline{a} \ \underline{b}?$

$15$

$30$

$45$

$60$

$75$

答案：E

知识点：循环小数等比数列

难度评级：1370

小提示：

比较 $1.\overline{ab}$ 和 $1.ab$ 。

Compare $1.\overline{ab}$ with $1.ab$

大提示：

小数点后百分位之后的循环部分造成了误差。

The repeating tail after the hundredths place accounts for the error

视频讲解：

Click to load, then click again to play

文字解答：

设 $N=10a+b$ ，即由数字 $a$ 和 $b$ 组成的两位数。于是 $1.\overline{ab}=1+\frac{N}{99}$ 而学生所用的有限小数是 $1.ab=1+\frac{N}{100}\text{。}$

正确的乘积比学生算出的乘积大 $0.5$ ，所以 $66\left(\frac{N}{99}-\frac{N}{100}\right) =\frac{N}{150}=0.5\text{。}$ 因此 $N=75$ 。

所以正确答案是 E。

Let $N=10a+b,$ the two-digit integer formed by the digits $a$ and $b.$ Then $1.\overline{ab}=1+\frac{N}{99}$ while the terminating decimal the student used is $1.ab=1+\frac{N}{100}.$

The correct product exceeds the student’s product by $0.5,$ so $66\left(\frac{N}{99}-\frac{N}{100}\right) =\frac{N}{150}=0.5.$ Hence $N=75.$

Thus, E is the correct answer.

9.

对实数 $x$ 和 $y$ ， $(xy-1)^2+(x+y)^2$ 的最小可能值是多少？

What is the least possible value of $(xy-1)^2+(x+y)^2$ for real numbers $x$ and $y?$

$0$

$\dfrac{1}{4}$

$\dfrac{1}{2}$

$1$

$2$

答案：D

知识点：最优化代数变形

难度评级：770

小提示：

展开表达式，寻找抵消。

Expand the expression and look for cancellation

大提示：

展开后，除常数项外每一项都是非负的。

After expansion, every term except the constant is nonnegative

视频讲解：

Click to load, then click again to play

文字解答：

展开得 $\begin{gathered} x^2y^2 - 2xy + 1 + x^2 + 2xy + y^2 \\ = x^2y^2 + x^2 + y^2 + 1 \end{gathered}\text{。}$ 每个平方项都非负，因此除 $1$ 外的各项都取 $0$ 时，总和取得最小值 $1$ 。

当 $x = y = 0$ 时可以取到这个值。

所以正确答案是 D。

Expanding, we get $\begin{gathered} x^2y^2 - 2xy + 1 + x^2 + 2xy + y^2 \\ = x^2y^2 + x^2 + y^2 + 1. \end{gathered}$ Note that every square must be non-negative. Therefore, the minimum value is when all the terms except $1$ are $0,$ making the sum $1.$

This is attainable when $x = y = 0.$

Thus, D is the correct answer.

10.

下列表达式等价于哪一个？ $\begin{aligned} &(2+3)(2^2+3^2)\\ &\quad\cdot(2^4+3^4)(2^8+3^8)\\ &\quad\cdot(2^{16}+3^{16})(2^{32}+3^{32})\\ &\quad\cdot(2^{64}+3^{64}) \end{aligned}\text{？}$

Which of the following is equivalent to $\begin{aligned} &(2+3)(2^2+3^2)\\ &\quad\cdot(2^4+3^4)(2^8+3^8)\\ &\quad\cdot(2^{16}+3^{16})(2^{32}+3^{32})\\ &\quad\cdot(2^{64}+3^{64})? \end{aligned}$

$3^{127} + 2^{127}$

$3^{127} + 2^{127} + 2 \cdot 3^{63}$ $+ 3 \cdot 2^{63}$

$3^{128} - 2^{128}$

$3^{128} + 2^{128}$

$5^{127}$

答案：C

知识点：平方差裂项相消

难度评级：1070

小提示：

乘以 $3-2$ ，它的值等于 $1$ 。

Multiply by $3-2,$ which is equal to $1$

大提示：

每一步都会用一个因子产生下一次平方差。

Each factor then creates the next difference of squares

视频讲解：

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文字解答：

把这个乘积乘以 $3-2=1$ 。反复应用平方差公式，得到 $\begin{aligned} (3-2)(3+2)&=3^2-2^2,\\ (3^2-2^2)(3^2+2^2)&=3^4-2^4 \end{aligned}\text{，}$ 同样的相消一直进行到最后一个因子。因此这个乘积等于 $3^{128}-2^{128}$ 。

所以正确答案是 C。

Multiply the product by $3-2=1.$ Repeatedly applying the difference-of-squares identity gives $\begin{aligned} (3-2)(3+2)&=3^2-2^2,\\ (3^2-2^2)(3^2+2^2)&=3^4-2^4, \end{aligned}$ and the same cancellation continues through the final factor. Therefore the product is $3^{128}-2^{128}.$

Thus, C is the correct answer.

11.

对下列哪个整数 $b$ ， $b$ 进制数 $2021_b - 221_b$ 不能被 $3$ 整除？

For which of the following integers $b$ is the base- $b$ number $2021_b - 221_b$ not divisible by $3?$

$3$

$4$

$6$

$7$

$8$

答案：E

知识点：进制整除性

难度评级：1020

小提示：

把 $2021_b-221_b$ 换算成 $10$ 进制。

Convert $2021_b-221_b$ to base $10$

大提示：

判断 $2b^2(b-1)$ 何时能被 $3$ 整除。

Check when $2b^2(b-1)$ is divisible by $3$

视频讲解：

Click to load, then click again to play

文字解答：

按进制定义，把这个差写成 $10$ 进制表达式： $2b^3 + 2b + 1 - 2b^2 - 2b - 1$ $= 2b^3 - 2b^2 = 2b^2(b - 1)\text{。}$

要使它能被 $3$ 整除， $b$ 或 $b - 1$ 必须能被 $3$ 整除。

唯一两个条件都不满足的选项是 $8$ 。

所以正确答案是 E。

We can express this expression in base $10$ using the definition of bases: $2b^3 + 2b + 1 - 2b^2 - 2b - 1$ $= 2b^3 - 2b^2 = 2b^2(b - 1).$

For this to be divisible by $3,$ either $b$ or $b - 1$ must be divisible by $3.$

The only answer choice that satisfies neither of these conditions is $8.$

Thus, E is the correct answer.

12.

如下图所示，两个顶点朝下的直圆锥中装有相同体积的液体。两个液面的顶部半径分别为 $3$ cm 和 $6$ cm。向每个圆锥中投入一个半径为 $1$ cm 的球形弹珠，弹珠沉到底部且完全浸没，并且没有液体溢出。窄圆锥中液面上升高度与宽圆锥中液面上升高度的比是多少？

Two right circular cones with vertices facing down as shown in the figure below contain the same amount of liquid. The radii of the tops of the liquid surfaces are $3$ cm and $6$ cm. Into each cone is dropped a spherical marble of radius $1$ cm, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level in the narrow cone to the rise of the liquid level in the wide cone?

$1:1$

$47:43$

$2:1$

$40:13$

$4:1$

答案：E

知识点：圆锥相似体积

难度评级：1660

小提示：

相等的液体体积会关联两个初始液面高度。

Equal liquid volumes relate the two initial liquid heights

大提示：

两个圆锥中弹珠排开的体积相同，所以最终液面以下的圆锥体积仍然相等。

The marble adds the same displaced volume in each cone, so the final cone-below-surface volumes are still equal

视频讲解：

Click to load, then click again to play

文字解答：

设窄圆锥和宽圆锥的初始液面高度分别为 $h_1$ 、 $h_2$ 。因为两者的液体体积相等，所以

$\frac13\pi(3)^2h_1=\frac13\pi(6)^2h_2\text{，}$

于是 $h_1=4h_2$ 。

投入相同的弹珠后，两个圆锥中液面以下的最终体积仍然相等，都等于原有液体体积加上一颗弹珠的体积。若新的液面半径分别为 $3x$ 和 $6y$ ，由相似可得新的高度为 $h_1x$ 和 $h_2y$ 。于是

$\frac13\pi(3x)^2h_1x=\frac13\pi(6y)^2h_2y\text{。}$

利用 $h_1=4h_2$ ，可化简为 $x^3=y^3$ ，所以 $x=y$ 。因此液面上升高度之比为

$\begin{aligned} h_1(x-1):h_2(y-1) &= h_1:h_2 \\ &= 4:1 \end{aligned}\text{。}$

所以正确答案是 E。

Let the initial liquid heights in the narrow and wide cones be $h_1$ and $h_2.$ Since the liquid volumes are equal,

$\frac13\pi(3)^2h_1=\frac13\pi(6)^2h_2,$

so $h_1=4h_2.$

After the identical marbles are dropped in, each cone must contain the same final volume below the liquid surface: the original liquid volume plus the volume of one marble. If the new liquid-surface radii are $3x$ and $6y,$ similarity gives new heights $h_1x$ and $h_2y.$ Thus

$\frac13\pi(3x)^2h_1x=\frac13\pi(6y)^2h_2y.$

Using $h_1=4h_2,$ this simplifies to $x^3=y^3,$ so $x=y.$ The rise ratio is therefore

$\begin{aligned} h_1(x-1):h_2(y-1) &= h_1:h_2 \\ &= 4:1. \end{aligned}$

Thus, E is the correct answer.

13.

四面体 $ABCD$ 的边长为 $AB = 2$ 、 $AC = 3$ 、 $AD = 4$ 、 $BC = \sqrt{13}$ 、 $BD = 2\sqrt{5}$ 、 $CD = 5$ 。它的体积是多少？

What is the volume of tetrahedron $ABCD$ with edge lengths $AB = 2,$ $AC = 3,$ $AD = 4,$ $BC = \sqrt{13},$ $BD = 2\sqrt{5},$ and $CD = 5?$

$3$

$2\sqrt{3}$

$4$

$3\sqrt{3}$

$6$

答案：C

知识点：体积立体几何勾股定理

难度评级：1370

小提示：

尝试把 $A$ 放在原点，并让 $AB,AC,AD$ 沿三条互相垂直的坐标轴。

Try placing $A$ at the origin with $AB,AC,AD$ along perpendicular axes

大提示：

检查给出的三条对边长度是否符合这个直角顶点模型。

Check that the given opposite edge lengths match this rectangular-corner model

视频讲解：

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文字解答：

取 $A=(0,0,0)$ 、 $B=(2,0,0)$ 、 $C=(0,3,0)$ 和 $D=(0,0,4)$ 。则

$\begin{aligned} BC &= \sqrt{2^2+3^2}=\sqrt{13}, \\ BD &= \sqrt{2^2+4^2}=2\sqrt5, \\ CD &= \sqrt{3^2+4^2}=5 \end{aligned}\text{，}$

所以这个坐标模型与题目给出的所有棱长相符。三条从 $A$ 出发的棱互相垂直，长度分别为 $2,3,4$ ，因此体积为

$\frac16(2)(3)(4)=4\text{。}$

所以正确答案是 C。

Place $A=(0,0,0),$ $B=(2,0,0),$ $C=(0,3,0),$ and $D=(0,0,4).$ Then

$\begin{aligned} BC &= \sqrt{2^2+3^2}=\sqrt{13}, \\ BD &= \sqrt{2^2+4^2}=2\sqrt5, \\ CD &= \sqrt{3^2+4^2}=5, \end{aligned}$

so this coordinate model matches all the given edge lengths. The tetrahedron is a rectangular-corner tetrahedron with perpendicular edge lengths $2,3,4$ from $A,$ so its volume is

$\frac16(2)(3)(4)=4.$

Thus, C is the correct answer.

14.

多项式 $\begin{aligned} &z^6-10z^5+Az^4+Bz^3\\ &\quad+Cz^2+Dz+16 \end{aligned}$ 的所有根都是正整数，且可以重复。求 $B$ 的值。

All the roots of the polynomial $\begin{aligned} &z^6-10z^5+Az^4+Bz^3\\ &\quad+Cz^2+Dz+16 \end{aligned}$ are positive integers, possibly repeated. What is the value of $B?$

$-88$

$-80$

$-64$

$-41$

$-40$

答案：A

知识点：韦达定理多项式

难度评级：1540

小提示：

用 Vieta 公式确定六个正整数根的和与积。

Use Vieta to determine the sum and product of the six positive integer roots

大提示：

找出积为 $16$ 、和为 $10$ 的唯一六个正整数。

Find the only six positive integers with product $16$ and sum $10$

视频讲解：

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文字解答：

由 Vieta 公式，六个根的和为 $10$ ，积为 $16$ 。因为积是 $2$ 的幂，所以每个正整数根都是 $2$ 的幂。把四个因子 $2$ 分配给六个根，当其中四个根为 $2$ 、两个根为 $1$ 时和最小，而这个和恰好已经等于 $10$ 。因此这些根是 $1,1,2,2,2,2\text{。}$

系数 $B$ 是所有三个根乘积之和的相反数。按取零个、一个还是两个等于 $1$ 的根来分类，得 $\begin{aligned} B&=-\left(\binom43 2^3+2\binom42 2^2\right.\\ &\qquad\left.+\binom41 2\right)\\ &=-88 \end{aligned}\text{。}$

所以正确答案是 A。

By Vieta’s formulas, the six roots have sum $10$ and product $16.$ Because the product is a power of $2,$ every positive integer root is a power of $2.$ Distributing the four factors of $2$ among six roots gives the least possible sum when four roots are $2$ and two roots are $1$ ; that sum is already $10.$ Hence the roots are $1,1,2,2,2,2.$

The coefficient $B$ is the negative of the sum of all products of three roots. Choosing zero, one, or two of the two roots equal to $1$ gives $\begin{aligned} B&=-\left(\binom43 2^3+2\binom42 2^2\right.\\ &\qquad\left.+\binom41 2\right)\\ &=-88. \end{aligned}$

Thus, A is the correct answer.

15.

$A$ ， $B$ ， $C$ 和 $D$ 的值要从 $\{1, 2, 3, 4, 5, 6\}$ 中不重复地选出。共有多少种选择，使得两条曲线 $y=Ax^2+B$ 和 $y=Cx^2+D$ 相交？

两条曲线列出的顺序不重要；例如选择 $A=3$ 、 $B=2$ 、 $C=4$ 、 $D=1$ 与选择 $A=4$ 、 $B=1$ 、 $C=3$ 、 $D=2$ 视为相同。

Values for $A,$ $B,$ $C,$ and $D$ are to be selected from $\{1, 2, 3, 4, 5, 6\}$ without replacement (i.e., no two letters have the same value). How many ways are there to make such choices so that the two curves $y=Ax^2+B$ and $y=Cx^2+D$ intersect?

(The order in which the curves are listed does not matter; for example, the choices $A=3,$ $B=2,$ $C=4,$ $D=1$ is considered the same as the choices $A=4,$ $B=1,$ $C=3,$ $D=2.$ )

$30$

$60$

$90$

$180$

$360$

答案：C

知识点：抛物线组合

难度评级：1540

小提示：

两条抛物线相交，等价于解出的 $x^2$ 非负。

The parabolas intersect exactly when the solved value of $x^2$ is nonnegative

大提示：

$D-B$ 与 $A-C$ 必须同号。

The two differences $D-B$ and $A-C$ must have the same sign

视频讲解：

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文字解答：

令两式相等，得到 $Ax^2 + B = Cx^2 + D$ $x^2(A - C) = D - B$ $x^2 = \dfrac{D - B}{A - C} \geq 0\text{，}$ 因为平方数非负。

因此 $D - B$ 与 $A - C$ 必须同号。

若先为 $(A, C)$ 和 $(B, D)$ 各选两个不同值，则有 $2$ 种安排使两个差同号。

不过还要除以 $2$ ，因为两条曲线列出的顺序不区分。

所以所求选择数为 $\dfrac{1}{2} \binom{6}{2} \binom{4}{2} \cdot 2 = 90\text{。}$

所以正确答案是 C。

Setting the equations equal to each other, we get $Ax^2 + B = Cx^2 + D$ $x^2(A - C) = D - B$ $x^2 = \dfrac{D - B}{A - C} \geq 0$ since squares are non-negative.

This means $D - B$ and $A - C$ must both have the same sign.

If we choose two distinct values for $(A, C)$ and $(B, D),$ there are $2$ ways to arrange them such that the numerator and denominator both have the same sign.

We have to divide by $2,$ however, since the two curves are not considered distinct.

Therefore, the total number of tuples is $\dfrac{1}{2} \binom{6}{2} \binom{4}{2} \cdot 2 = 90.$

Thus, C is the correct answer.

16.

在下面的数列中，对每个 $1\leq n\leq200$ ，整数 $n$ 都出现 $n$ 次： $\begin{aligned} &1,2,2,3,3,3,4,4,4,4,\ldots,\\ &200,200,\ldots,200 \end{aligned}$ 这个数列的中位数是多少？

In the following list of numbers, the integer $n$ appears $n$ times in the list for $1\leq n\leq200.$ $\begin{aligned} &1,2,2,3,3,3,4,4,4,4,\ldots,\\ &200,200,\ldots,200 \end{aligned}$ What is the median of the numbers in this list?

$100.5$

$134$

$142$

$150.5$

$167$

答案：C

知识点：中位数（数据）三角形数

难度评级：1420

小提示：

数列共有 $1+2+\cdots+200$ 项。

The list has $1+2+\cdots+200$ entries

大提示：

用三角数定位中间的两个位置。

Locate the two middle positions using triangular numbers

解答：

这个数列共有 $\begin{aligned} 1+2+\cdots+200&=\frac{200\cdot201}{2}\\ &=20100 \end{aligned}$ 项，所以处于中间的两个位置是第 $10050$ 项和第 $10051$ 项。

到最后一个 $141$ 为止共有 $\frac{141\cdot142}{2}=10011$ 项，到最后一个 $142$ 为止共有 $\frac{142\cdot143}{2}=10153$ 项。因此中间的两项都等于 $142$ ，所以中位数是 $142$ 。

所以正确答案是 C。

The list contains $\begin{aligned} 1+2+\cdots+200&=\frac{200\cdot201}{2}\\ &=20100 \end{aligned}$ entries, so its two middle positions are $10050$ and $10051.$

There are $\frac{141\cdot142}{2}=10011$ entries through the last $141,$ and $\frac{142\cdot143}{2}=10153$ entries through the last $142.$ Thus both middle entries are $142,$ so the median is $142.$

Thus, C is the correct answer.

17.

梯形 $ABCD$ 满足 $\overline{AB}\parallel\overline{CD}$ ， $BC=CD=43$ ，且 $\overline{AD}\perp\overline{BD}$ 。设 $O$ 为对角线 $\overline{AC}$ 与 $\overline{BD}$ 的交点， $P$ 为 $\overline{BD}$ 的中点。

已知 $OP=11$ ，线段 $AD$ 的长度可写成 $m\sqrt{n}$ ，其中 $m$ 和 $n$ 为正整数，且 $n$ 不被任何质数的平方整除。 $m+n$ 是多少？

Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD},$ $BC=CD=43,$ and $\overline{AD}\perp\overline{BD}.$ Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD},$ and let $P$ be the midpoint of $\overline{BD}.$

Given that $OP=11,$ the length of $AD$ can be written in the form $m\sqrt{n},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m+n?$

$65$

$132$

$157$

$194$

$215$

答案：D

知识点：梯形相似勾股定理

难度评级：1950

小提示：

利用等腰三角形，得到一个与 $BD$ 的中点有关的直角三角形。

Use the isosceles triangle to get a right triangle involving the midpoint of $BD$

大提示：

交点 $O$ 按平行底边的比来分割对角线。

The intersection point $O$ splits the diagonals in the ratio of the parallel bases

视频讲解：

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文字解答：

因为 $BC=CD$ ，从 $C$ 到 $BD$ 的中线垂直于 $BD$ ，所以 $\triangle BPC$ 是直角三角形。设 $\angle DBC=\alpha$ 。又因 $AB\parallel CD$ ，有 $\angle ABD=\alpha$ ，因此 $\triangle BPC\sim\triangle BDA$ 。

由于 $P$ 是 $BD$ 的中点， $\frac{BD}{BP}=2$ 。在相似关系中， $BC$ 对应 $AB$ ，所以

$\begin{aligned} \frac{AB}{BC} &=2, \\ AB &=2\cdot43=86 \end{aligned}\text{。}$

另外， $\triangle ABO\sim\triangle CDO$ ，所以

$\frac{BO}{OD}=\frac{AB}{CD}=2\text{。}$

设 $BP=PD=t$ 。由于 $OP=11$ ，且 $P$ 是 $BD$ 的中点，得到 $BO=t+11$ 、 $OD=t-11$ ，所以

$\frac{t+11}{t-11}=2\text{。}$

解得 $t=33$ ，故 $BD=66$ 。最后， $\triangle ABD$ 是直角三角形，所以

$\begin{aligned} AD &= \sqrt{AB^2-BD^2} \\ &= \sqrt{86^2-66^2} \\ &= 4\sqrt{190} \end{aligned}\text{。}$

因此 $m+n=4+190=194$ 。

所以正确答案是 D。

Because $BC=CD,$ the median from $C$ to $BD$ is perpendicular to $BD.$ Thus $\triangle BPC$ is a right triangle. Let $\angle DBC=\alpha.$ Since $AB\parallel CD,$ we also have $\angle ABD=\alpha,$ so $\triangle BPC\sim\triangle BDA.$

Since $P$ is the midpoint of $BD,$ we have $\frac{BD}{BP}=2.$ In the similarity, $BC$ corresponds to $AB,$ so

$\begin{aligned} \frac{AB}{BC} &=2, \\ AB &=2\cdot43=86. \end{aligned}$

Also, $\triangle ABO\sim\triangle CDO,$ so

$\frac{BO}{OD}=\frac{AB}{CD}=2.$

Since $OP=11$ and $P$ is the midpoint of $BD,$ write $BP=PD=t.$ Then $BO=t+11$ and $OD=t-11,$ so

$\frac{t+11}{t-11}=2.$

This gives $t=33,$ hence $BD=66.$ Finally, $\triangle ABD$ is right, so

$\begin{aligned} AD &= \sqrt{AB^2-BD^2} \\ &= \sqrt{86^2-66^2} \\ &= 4\sqrt{190}. \end{aligned}$

Thus $m+n=4+190=194.$

Thus, D is the correct answer.

18.

设 $f$ 是定义在正有理数集上的函数，并且 $f(a\cdot b)=f(a)+f(b)$ 对所有正有理数 $a$ 和 $b$ 都成立。又假设 $f$ 还满足：对每个质数都有 $f(p)=p$ ，其中 $p$ 为质数。下列哪个数 $x$ 满足 $f(x) < 0$ ？

Let $f$ be a function defined on the set of positive rational numbers with the property that $f(a\cdot b)=f(a)+f(b)$ for all positive rational numbers $a$ and $b.$ Suppose that $f$ also has the property that $f(p)=p$ for every prime number $p.$ For which of the following numbers $x$ is $f(x) < 0?$

$\dfrac{17}{32}$

$\dfrac{11}{16}$

$\dfrac{7}{9}$

$\dfrac{7}{6}$

$\dfrac{25}{11}$

答案：E

知识点：函数方程质因数分解

难度评级：1280

小提示：

先确定 $f(p^e)$ 和 $f(\frac{a}{b})$ 的值。

First determine $f(p^e)$ and $f(\frac{a}{b})$

大提示：

用质因数分解逐个计算选项。

Evaluate the choices by prime factorization

视频讲解：

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文字解答：

反复应用该函数方程可得：对每个质数 $p$ 和正整数 $e$ ，都有 $f(p^e)=ef(p)=ep$ 。另外 $f(a)=f\left(\frac ab\right)+f(b)\text{，}$ 所以 $f(\frac{a}{b})=f(a)-f(b)$ 。

用质因数分解逐项计算各个选项，得 $\begin{aligned} f(\frac{17}{32})&=17-5\cdot2=7,\\ f(\frac{11}{16})&=11-4\cdot2=3,\\ f(\frac{7}{9})&=7-2\cdot3=1,\\ f(\frac{7}{6})&=7-2-3=2,\\ f(\frac{25}{11})&=2\cdot5-11=-1 \end{aligned}\text{。}$ 只有最后一个值是负数。

所以正确答案是 E。

Repeated use of the functional equation gives $f(p^e)=ef(p)=ep$ for every prime $p$ and positive integer $e.$ Also, $f(a)=f\left(\frac ab\right)+f(b),$ so $f(\frac{a}{b})=f(a)-f(b).$

Evaluating the choices by prime factorization, $\begin{aligned} f(\frac{17}{32})&=17-5\cdot2=7,\\ f(\frac{11}{16})&=11-4\cdot2=3,\\ f(\frac{7}{9})&=7-2\cdot3=1,\\ f(\frac{7}{6})&=7-2-3=2,\\ f(\frac{25}{11})&=2\cdot5-11=-1. \end{aligned}$ Only the final value is negative.

Thus, E is the correct answer.

19.

由图像 $x^2+y^2 = 3|x-y| + 3|x+y|$ 围成的区域面积为 $m+n\pi$ ，其中 $m$ 、 $n$ 是整数。求 $m + n$ 。

The area of the region bounded by the graph of $x^2+y^2 = 3|x-y| + 3|x+y|$ is $m+n\pi,$ where $m$ and $n$ are integers. What is $m + n?$

$18$

$27$

$36$

$45$

$54$

答案：E

知识点：绝对值圆面积分割

难度评级：2150

小提示：

按 $x-y$ 和 $x+y$ 的符号分成四种情况。

Split the graph by the signs of $x-y$ and $x+y$

大提示：

四种情况会给出围绕中心正方形的半圆弧。

The four cases give semicircle arcs around a central square

解答：

按 $x-y$ 和 $x+y$ 的符号分类。例如在一种情况中， $|x-y|=x-y$ 且 $|x+y|=x+y$ ，所以

$\begin{gathered} x^2+y^2=6x \\ \quad\Longrightarrow\quad (x-3)^2+y^2=9 \end{gathered}\text{。}$

其余三种情况类似，得到半径为 $3$ 、圆心分别在 $(0,3)$ 、 $(-3,0)$ 和 $(0,-3)$ 的圆。相应的四段全等圆弧就构成了图像的边界。

围成区域由边长 $6$ 的中心正方形和四个半径为 $3$ 的半圆组成。正方形面积为 $36$ ，四个半圆的总面积等于两个半径为 $3$ 的圆，即 $18\pi$ 。

因此面积为 $36+18\pi$ ，所以 $m+n=36+18=54$ 。

所以正确答案是 E。

Consider the four sign cases for $x-y$ and $x+y.$ In one case, for example, $|x-y|=x-y$ and $|x+y|=x+y,$ so

$\begin{gathered} x^2+y^2=6x \\ \quad\Longrightarrow\quad (x-3)^2+y^2=9. \end{gathered}$

The other three cases similarly give circles of radius $3$ centered at $(0,3),$ $(-3,0),$ and $(0,-3).$ The relevant arcs form the boundary shown by these four congruent circle pieces.

The region consists of a central square of side length $6,$ together with four semicircles of radius $3.$ The square contributes area $36,$ and the four semicircles have the area of two full radius- $3$ circles, namely $18\pi.$

Therefore the area is $36+18\pi,$ so $m+n=36+18=54.$

Thus, E is the correct answer.

20.

将数列 $1$ ， $2$ ， $3$ ， $4$ ， $5$ 重新排列，有多少种排列使得不存在连续三项递增，也不存在连续三项递减？

In how many ways can the sequence $1,$ $2,$ $3,$ $4,$ $5$ be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?

$10$

$18$

$24$

$32$

$44$

答案：D

知识点：有限制的排列双射

难度评级：1950

小提示：

有效排列中，相邻项之间的大小比较符号必须交替。

A valid permutation must have comparison signs that alternate

大提示：

先数“升降升降”型排列，再用对称性得到相反型。

Count the up-down-up-down permutations and use symmetry for the reverse pattern

视频讲解：

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文字解答：

一个排列有效，当且仅当相邻两项之间的四个比较符号交替出现。因此这些符号只能是升降升降或降升降升。

对升降升降型，最大的数 $5$ 必须位于第 $2$ 位或第 $4$ 位。若它在第 $2$ 位，设第 $4$ 位上的数为 $r$ 。它的两个相邻数必须是小于 $r$ 的两个不同的数，共有 $(r-1)(r-2)$ 种排法。对 $r=1,2,3,4$ 求和，得到 $0+0+2+6=8$ 个排列。由对称性，当 $5$ 在第 $4$ 位时另有 $8$ 个，所以这种比较模式共有 $16$ 个排列。

把每个数 $x$ 替换为 $6-x$ ，可与降升降升型一一对应，所以另有 $16$ 个。

总共有 $16+16=32$ 个有效排列。

所以正确答案是 D。

A permutation is valid exactly when the four comparison signs between consecutive terms alternate. Thus the signs must be either up-down-up-down or down-up-down-up.

For the up-down-up-down pattern, the largest entry $5$ must be in position $2$ or position $4.$ If it is in position $2,$ let the entry in position $4$ be $r.$ Its two neighbors must be distinct numbers less than $r,$ which can be ordered in $(r-1)(r-2)$ ways. Summing over $r=1,2,3,4$ gives $0+0+2+6=8$ permutations. By symmetry there are another $8$ when $5$ is in position $4,$ for a total of $16$ with this comparison pattern.

Replacing every entry $x$ by $6-x$ gives a bijection to the down-up-down-up permutations, so there are another $16.$

The total number of valid rearrangements is $16+16=32.$

Thus, D is the correct answer.

21.

设 $ABCDEF$ 是一个等角六边形。直线 $AB$ ， $CD$ 和 $EF$ 确定一个面积为 $192\sqrt{3}$ 的三角形，直线 $BC$ ， $DE$ 和 $FA$ 确定一个面积为 $324\sqrt{3}$ 的三角形。六边形 $ABCDEF$ 的周长可表示为 $m +n\sqrt{p}$ ，其中 $m$ ， $n$ 和 $p$ 为正整数，且 $p$ 不被任何质数的平方整除。 $m + n + p$ 是多少？

Let $ABCDEF$ be an equiangular hexagon. The lines $AB,$ $CD,$ and $EF$ determine a triangle with area $192\sqrt{3},$ and the lines $BC,$ $DE,$ and $FA$ determine a triangle with area $324\sqrt{3}.$ The perimeter of hexagon $ABCDEF$ can be expressed as $m +n\sqrt{p},$ where $m,$ $n,$ and $p$ are positive integers and $p$ is not divisible by the square of any prime. What is $m + n + p?$

$47$

$52$

$55$

$58$

$63$

答案：C

知识点：等角多边形等边三角形三角形面积

难度评级：2150

小提示：

三条相隔的边所在直线形成等边三角形。

The three alternating side lines form equilateral triangles

大提示：

把两个给定三角形面积转化为边长。

Convert the two given triangle areas into side lengths

解答：

设直线 $AB,CD,EF$ 的交点形成三角形 $PQR$ ，直线 $BC,DE,FA$ 的交点形成三角形 $XYZ$ 。因为六边形等角，这些外侧三角形都是等边三角形。

若等边三角形边长为 $s$ ，面积为 $\frac{\sqrt3}{4}s^2$ 。两个面积条件可写成：

$\begin{aligned} \frac{\sqrt3}{4}PQ^2 &=192\sqrt3, \\ \frac{\sqrt3}{4}YZ^2 &=324\sqrt3 \end{aligned}\text{。}$

所以 $PQ=16\sqrt3$ ， $YZ=36$ 。为了说明这个周长关系，把六边形依次的边长记为 $a,b,c,d,e,f$ 。两条交替延长线截出的三角形的边长分别为 $b+c+d$ 和 $c+d+e$ ，而六边形封闭又给出 $a+f=c+d$ 。因此它们的边长之和为 $\begin{aligned} &(b+c+d)\\ &\quad+(c+d+e)\\ &=b+e+2(c+d)\\ &=(a+b+c)+(d+e+f) \end{aligned}\text{，}$ 这正是六边形的周长。所以周长为

$PQ+YZ=16\sqrt3+36\text{。}$

因此 $m+n+p=36+16+3=55$ 。

所以正确答案是 C。

Let the intersections of lines $AB,CD,EF$ form triangle $PQR,$ and let the intersections of lines $BC,DE,FA$ form triangle $XYZ.$ Because the hexagon is equiangular, all these outer triangles are equilateral.

For an equilateral triangle with side length $s,$ the area is $\frac{\sqrt3}{4}s^2.$ Hence

$\begin{aligned} \frac{\sqrt3}{4}PQ^2 &=192\sqrt3, \\ \frac{\sqrt3}{4}YZ^2 &=324\sqrt3. \end{aligned}$

So $PQ=16\sqrt3$ and $YZ=36.$ To justify the perimeter relation, write the consecutive hexagon side lengths as $a,b,c,d,e,f.$ The two alternating-line triangles have side lengths $b+c+d$ and $c+d+e,$ while closure of the hexagon gives $a+f=c+d.$ Hence their side-length sum is $\begin{aligned} &(b+c+d)\\ &\quad+(c+d+e)\\ &=b+e+2(c+d)\\ &=(a+b+c)+(d+e+f), \end{aligned}$ the hexagon’s perimeter. Therefore the perimeter is

$PQ+YZ=16\sqrt3+36.$

Thus $m+n+p=36+16+3=55.$

Thus, C is the correct answer.

22.

Hiram 的代数笔记共有 $50$ 页，印在 $25$ 张纸上；第一张纸含第 $1$ 页和第 $2$ 页，第二张纸含第 $3$ 页和第 $4$ 页，依此类推。有一天他午饭前把笔记留在桌上，室友决定从笔记中间借走一些页。Hiram 回来时发现，室友拿走的是连续若干张纸，并且所有剩余纸张上的页码平均数恰好为 $19$ 。室友借走了多少张纸？

Hiram’s algebra notes are $50$ pages long and are printed on $25$ sheets of paper; the first sheet contains pages $1$ and $2,$ the second sheet contains pages $3$ and $4,$ and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly $19.$ How many sheets were borrowed?

$10$

$13$

$15$

$17$

$20$

答案：B

知识点：平均数丢番图方程因式分解

难度评级：1820

小提示：

设被借走的是第 $a$ 张到第 $b$ 张纸。

Let the borrowed sheets run from sheet $a$ through sheet $b$

大提示：

用总页码和与剩余页码平均数，化简出一个可因式分解的方程。

Use the total page sum and the mean of the remaining pages to factor an equation

视频讲解：

Click to load, then click again to play

文字解答：

设借走的是第 $a$ 张到第 $b$ 张纸，并设 $s=b-a+1$ 。被借走的页码从 $2a-1$ 到 $2b$ ，所以共有 $2s$ 页，页码之和为 $s(2a+2b-1)$ 。

所有页码之和为 $\frac{50\cdot51}{2}=1275$ 。若剩余页码的平均数为 $19$ ，则 $\begin{aligned} 1275&-s(2a+2b-1)\\ &=19(50-2s),\\ s(2a+2b-39)&=325 \end{aligned}\text{。}$

因为 $b-a+1$ 是 $325$ 的正因数，且不超过 $25$ ，所以它只能是 $1,5,13,25$ 这四种可能。前两种会迫使 $b>25$ ，而 $25$ 会借走所有纸张。因此唯一有效的可能是

$\begin{aligned} 2a+2b-39 &=25, \\ b-a+1 &=13 \end{aligned}\text{。}$

所以 $a+b=32$ ，且 $b-a=12$ ，得到 $a=10$ 、 $b=22$ 。因此借走了 $13$ 张纸。

所以正确答案是 B。

Suppose the borrowed sheets are sheets $a$ through $b,$ and let $s=b-a+1.$ The borrowed pages run from $2a-1$ through $2b,$ so there are $2s$ borrowed pages and their sum is $s(2a+2b-1).$

The total sum of all page numbers is $\frac{50\cdot51}{2}=1275.$ If the remaining pages have mean $19,$ then $\begin{aligned} 1275&-s(2a+2b-1)\\ &=19(50-2s),\\ s(2a+2b-39)&=325. \end{aligned}$

Because $b-a+1$ is a positive divisor of $325$ and is at most $25,$ its only possibilities are $1,5,13,25.$ The first two would force $b>25,$ and $25$ would remove every sheet. Thus the only valid possibility is

$\begin{aligned} 2a+2b-39 &=25, \\ b-a+1 &=13. \end{aligned}$

Thus $a+b=32$ and $b-a=12,$ so $a=10$ and $b=22.$ Therefore $13$ sheets were borrowed.

Thus, B is the correct answer.

23.

青蛙 Frieda 在一个 $3 \times 3$ 方格网中开始一串跳跃，每次跳一格，并随机选择跳跃方向：上、下、左、右。她不斜着跳。如果某次跳跃方向会让 Frieda 跳出方格网，她会“绕回”并跳到相对的边。例如，如果 Frieda 从中心格开始并连续向上跳两次，第一次会到达上排中间格，第二次会让她跳到相对边，落在下排中间格。

假设 Frieda 从中心格开始，最多随机跳四次，并且一旦落在角格就停止。她在四次跳跃中的某一次到达角格的概率是多少？

Frieda the frog begins a sequence of hops on a $3 \times 3$ grid of squares, moving one square on each hop and choosing at random the direction of each hop—up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she “wraps around” and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops “up”, the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square.

Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?

$\dfrac{9}{16}$

$\dfrac{5}{8}$

$\dfrac{3}{4}$

$\dfrac{25}{32}$

$\dfrac{13}{16}$

答案：D

知识点：随机游走分类讨论

难度评级：1720

小提示：

把位置分为中心格、边格和角格。

Classify positions as center, edge, or corner

大提示：

计算第 $2$ 、 $3$ 、 $4$ 次跳跃首次到达角格的方式。

Count the ways to first hit a corner by hop $2,$ $3,$ or $4$

解答：

用 $M$ 表示中心格， $E$ 表示非角的边格， $C$ 表示角格。Frieda 从 $M$ 出发，第一次跳一定到 $E$ 。

从一个边格出发，到 $C,E,M$ 的概率分别为 $\frac12,\frac14,\frac14$ 。从 $M$ 出发下一步一定到 $E$ 。

现在列出四次以内首次到达角格的状态模式及其概率。

$EC:\quad 1\cdot\frac12=\frac12\text{，}$

$EEC:\quad 1\cdot\frac14\cdot\frac12=\frac18\text{，}$

$EEEC:\quad 1\cdot\frac14\cdot\frac14\cdot\frac12=\frac1{32}\text{，}$

$EMEC:\quad 1\cdot\frac14\cdot1\cdot\frac12=\frac18\text{。}$

把这些概率相加：

$\frac12+\frac18+\frac1{32}+\frac18=\frac{25}{32}\text{。}$

所以正确答案是 D。

Classify a square as $M$ for the center, $E$ for a non-corner edge square, and $C$ for a corner. Frieda starts at $M,$ and the first hop always takes her to an $E.$

From an edge square, the probabilities of moving to $C,E,M$ are $\frac12,\frac14,\frac14,$ respectively. From $M,$ the next hop always goes to an $E.$

Now count the possible first-hit patterns within four hops:

$EC:\quad 1\cdot\frac12=\frac12,$

$EEC:\quad 1\cdot\frac14\cdot\frac12=\frac18,$

$EEEC:\quad 1\cdot\frac14\cdot\frac14\cdot\frac12=\frac1{32},$

$EMEC:\quad 1\cdot\frac14\cdot1\cdot\frac12=\frac18.$

Adding gives

$\frac12+\frac18+\frac1{32}+\frac18=\frac{25}{32}.$

Thus, D is the correct answer.

24.

一个四边形的内部由图像 $(x+ay)^2=4a^2$ 和 $(ax-y)^2=a^2$ 围成，其中 $a$ 是正实数。对所有 $a > 0$ ，这个区域的面积用 $a$ 表示是多少？

The interior of a quadrilateral is bounded by the graphs of $(x+ay)^2=4a^2$ and $(ax-y)^2=a^2,$ where $a$ is a positive real number. What is the area of this region in terms of $a,$ valid for all $a > 0?$

$\dfrac{8a^2}{(a+1)^2}$

$\dfrac{4a}{a+1}$

$\dfrac{8a}{a+1}$

$\dfrac{8a^2}{a^2+1}$

$\dfrac{8a}{a^2+1}$

答案：D

知识点：平行线距离公式矩形

难度评级：1720

小提示：

每个平方方程都表示一对平行直线。

Each squared equation represents a pair of parallel lines

大提示：

两组平行线互相垂直，所以面积是两组平行线距离的乘积。

The two pairs of lines are perpendicular, so the area is the product of the two distances between parallel lines

解答：

注意，每个方程都会给出两条平行直线。

$(x + ay)^2 = 4a^2$ 给出两条直线 $x + ay - 2a = 0$ 和 $x + ay + 2a = 0\text{。}$ 这两条直线的斜率都是 $-\dfrac{1}{a}$ 。

类似地， $(ax-y)^2 = a^2$ 给出直线 $ax - y - a = 0$ 和 $ax - y + a = 0\text{。}$ 这些直线的斜率为 $a$ 。

两组直线互相垂直，因此围成一个矩形。

回忆两条平行线 $\begin{cases} Ax+By+C_1=0 \\ Ax+By+C_2=0 \end{cases}$ 之间的距离 $d$ 为 $d = \dfrac{\mid C_2 - C_1 \mid}{\sqrt{A^2 + B^2}}\text{。}$

用这个公式，第一组平行线之间的距离为 $\dfrac{4a}{\sqrt{a^2 + 1}}\text{。}$ 类似地，第二组平行线之间的距离为 $\dfrac{2a}{\sqrt{a^2 + 1}}\text{。}$

这两个距离就是矩形的边长，相乘得面积 $\dfrac{8a^2}{a^2 + 1}\text{。}$

所以正确答案是 D。

Note that each of the equations yields two parallel lines.

$(x + ay)^2 = 4a^2$ results in the two lines $x + ay - 2a = 0$ and $x + ay + 2a = 0.$ Both of these lines have a slope of $-\dfrac{1}{a}.$

Similarly, $(ax-y)^2 = a^2$ results in the lines $ax - y - a = 0$ and $ax - y + a = 0.$ These lines have slope $a.$

Note that each pair of lines is perpendicular to the other pair of lines. This shows that the equations form a rectangle.

Recall that the formula for the distance $d$ between two parallel lines $\begin{cases} Ax+By+C_1=0 \\ Ax+By+C_2=0 \end{cases}$ is $d = \dfrac{\mid C_2 - C_1 \mid}{\sqrt{A^2 + B^2}}.$

Using this formula, we get that the distance between the first pair of lines is $\dfrac{4a}{\sqrt{a^2 + 1}}.$ Similarly, the distance between the second pair of lines is $\dfrac{2a}{\sqrt{a^2 + 1}}.$

These are the side lengths of the rectangle. Multiplying yields the area $\dfrac{8a^2}{a^2 + 1}.$

Thus, D is the correct answer.

25.

有多少种方法把 $3$ 枚不可区分的红色筹码、 $3$ 枚不可区分的蓝色筹码和 $3$ 枚不可区分的绿色筹码放入一个 $3 \times 3$ 方格的格子中，使得任意两个同色筹码都不在上下或左右方向相邻？

How many ways are there to place $3$ indistinguishable red chips, $3$ indistinguishable blue chips, and $3$ indistinguishable green chips in the squares of a $3 \times 3$ grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally?

$12$

$18$

$24$

$30$

$36$

答案：E

知识点：有限制的排列分类讨论

难度评级：1820

小提示：

先选定中心格的颜色，以及被这种颜色占据的两个角格。

First choose the color in the center and the two corners occupied by that color

大提示：

这三枚筹码放好后，另外两种颜色的位置就被确定，至多只差一次互换。

Once those three chips are placed, the other two colors are forced up to interchange

解答：

中心格的颜色有 $3$ 种选法。这种颜色的另外两枚筹码都不能放在边中格，因为这些格子与中心格相邻。因此它们只能占据四个角格中的两个，共有 $\binom42=6$ 种选法。

无论这两个角格是哪一种情形——处于对角，还是处于同一边的两端——相邻条件都会把剩下两种颜色的位置唯一确定，至多相差二者的互换。因此对中心颜色及其两个角格的每一种选择，都恰有 $2$ 种补全方式。总数为 $3\binom42\cdot2=36\text{。}$

所以正确答案是 E。

Choose the center color in $3$ ways. Its other two chips cannot occupy any edge-middle square, because those squares are adjacent to the center. Thus they must occupy two of the four corners, which can be chosen in $\binom42=6$ ways.

For either possible corner pattern—two opposite corners or two corners on the same side—the adjacency conditions force the remaining two colors up to interchanging them. Hence there are $2$ completions for each choice of the center color and its two corners. The total is $3\binom42\cdot2=36.$

Thus, E is the correct answer.