2021 AMC 10B Spring 详解与视频讲解 | LIVE by Po-Shen Loh

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2021 AMC 10B Spring 详解

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所有题目均经美国数学协会（MAA）官方合法授权使用。

1.

有多少个整数 $x$ 满足 $|x| < 3\pi$ ？

How many integer values of $x$ satisfy $|x| < 3\pi?$

$9$

$10$

$18$

$19$

$20$

知识点：绝对值区间内整数计数

难度评级：560

解答：

从 $-9$ 到 $9$ 的每个整数都满足条件，共有 $9-(-9)+1 = 19$ 个。

所以正确答案是 D。

Every integer from $-9$ to $9,$ inclusive, works. This yields $9-(-9)+1 = 19$ solutions.

Thus, the correct answer is D .

2.

下式的值是多少？ $\begin{aligned} &\sqrt{\left(3-2\sqrt{3}\right)^2} \\ &{}+\sqrt{\left(3+2\sqrt{3}\right)^2} \end{aligned}$

What is the value of $\begin{aligned} &\sqrt{\left(3-2\sqrt{3}\right)^2} \\ &{}+\sqrt{\left(3+2\sqrt{3}\right)^2}? \end{aligned}$

$0$

$4\sqrt{3}-6$

$6$

$4\sqrt{3}$

$4\sqrt{3}+6$

知识点：根式绝对值

难度评级：770

解答：

先利用平方根与绝对值的关系： $\sqrt{\left(3-2\sqrt{3}\right)^2}+\sqrt{\left(3+2\sqrt{3}\right)^2 }$ $= |3-2\sqrt{3}| + |3+2\sqrt{3}|.$ 因为 $3 < 2 \sqrt 3$ ，所以 $3-2\sqrt{3} < 0$ 。

因此原式为 $-3+2\sqrt{3} + 3+2\sqrt{3}$ $= 4 \sqrt 3.$

所以正确答案是 D。

We know $\sqrt{\left(3-2\sqrt{3}\right)^2}+\sqrt{\left(3+2\sqrt{3}\right)^2 }$ $= |3-2\sqrt{3}| + |3+2\sqrt{3}|.$ Since $3 < 2 \sqrt 3,$ we know that $3-2\sqrt{3} < 0.$

Therefore, our desired equation expression is equal to $-3+2\sqrt{3} + 3+2\sqrt{3}$ $= 4 \sqrt 3.$

Thus, the correct answer is D .

3.

在一个面向低年级和高年级学生的课后项目中，有一支辩论队，队中来自两个年级的学生人数相同。这个项目共有 $28$ 名学生，其中 $25\%$ 的低年级学生和 $10\%$ 的高年级学生在辩论队中。这个项目中有多少名低年级学生？

In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the $28$ students in the program, $25\%$ of the juniors and $10\%$ of the seniors are on the debate team. How many juniors are in the program?

$5$

$6$

$8$

$11$

$20$

知识点：百分数方程组

难度评级：870

视频讲解：

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文字解答：

设低年级学生有 $j$ 人，高年级学生有 $s$ 人。那么 $j+s = 28$ ，且 $0.25j = 0.1s$ 。这说明 $2.5j = s$ ，所以 $3.5j = 28$ 。因此 $j=8$ 。

所以正确答案是 C。

Let the number of juniors be $j$ and the number of seniors be $s.$ Then, $j+s = 28$ and $0.25j = 0.1s.$ This means $2.5j = s,$ so $3.5j = 28.$ This makes $j=8.$

Thus, the correct answer is C .

4.

在一场数学竞赛中， $57$ 名学生穿蓝色衬衫，另外 $75$ 名学生穿黄色衬衫。这 $132$ 名学生被分成 $66$ 对。其中恰有 $23$ 对中两名学生都穿蓝色衬衫。有多少对中两名学生都穿黄色衬衫？

At a math contest, $57$ students are wearing blue shirts, and another $75$ students are wearing yellow shirts. The $132$ students are assigned into $66$ pairs. In exactly $23$ of these pairs, both students are wearing blue shirts. In how many pairs are both students wearing yellow shirts?

$23$

$32$

$37$

$41$

$64$

知识点：数对计数

难度评级：960

视频讲解：

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文字解答：

蓝蓝配对用掉 $2\cdot 23 = 46$ 名穿蓝色衬衫的学生。因此还有 $57-46=11$ 名穿蓝色衬衫的学生在混合配对中，也就恰有 $11$ 名穿黄色衬衫的学生在混合配对中。

剩下只有 $64$ 名穿黄色衬衫的学生与另一名穿黄色衬衫的学生配对，所以黄黄配对数为 $\dfrac{64}2 = 32$ 。

所以答案是 B。

There are $2\cdot 23 = 46$ students with blue shirts that are in a pair with just blue shirts. This means there are $57-46=11$ students in blue shirts who are paired with someone wearing a yellow shirt, meaning exactly $11$ people wearing yellow shirts are paired with someone wearing a blue shirt.

This leaves just $64$ students wearing a yellow shirt who are paired with someone else wearing a yellow shirt. This yields $\dfrac{64}2 = 32$ pairs.

Thus, the answer is B .

5.

Jonie 的四个表亲年龄互不相同，且都是一位正整数。其中两个表亲的年龄相乘为 $24$ ，另外两个相乘为 $30$ 。Jonie 的四个表亲年龄之和是多少？

The ages of Jonie's four cousins are distinct single-digit positive integers. Two of the cousins' ages multiplied together give $24,$ while the other two multiply to $30.$ What is the sum of the ages of Jonie's four cousins?

$21$

$22$

$23$

$24$

$25$

知识点：因数年龄问题

难度评级：900

视频讲解：

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文字解答：

另外两个年龄相乘为 $30$ ，且都是一位数，所以其中一个必须是 $5,$ 另一个是 $6$ 。前两个年龄相乘为 $24$ ，需要找乘积为 $24$ 的一位数因数对。不能再使用 $4$ 或 $6$ ，所以只有 $3,8.$ 因此四个年龄为 $3,5,6,8,$ 和为 $22.$

所以答案是 B。

Since the last two are multiplied to $30$ and both are single-digit numbers, one of them must be $5,$ making the other person $6.$ The first two are of ages that multiply to $24.$ The only pair of single-digit numbers whose product is $24$ and none of them are $4$ or $6$ is the pair $3,8.$ Thus, the ages are $3,5,6,8,$ making their sum $22.$

Thus, the answer is B .

6.

Blackwell 老师给两个班考试。上午班学生的平均分是 $84$ ，下午班学生的平均分是 $70$ 。上午班人数与下午班人数之比为 $\frac{3}{4}$ 。所有学生的平均分是多少？

Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is $84,$ and the afternoon class's mean score is $70.$ The ratio of the number of students in the morning class to the number of students in the afternoon class is $\frac{3}{4}.$ What is the mean of the scores of all the students?

$74$

$75$

$76$

$77$

$78$

知识点：平均数比与比例

难度评级：900

视频讲解：

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文字解答：

设上午班有 $3x$ 人，那么下午班有 $4x$ 人。

上午班总分为 $84\cdot 3x = 252x$ ，下午班总分为 $70\cdot 4x = 280x$ 。因此总分为 $532x,$ 总人数为 $7x$ 。

所以所有学生的平均分为 $\dfrac{532x}{7x} = 76$ 。

所以正确答案是 C。

Let the number of people in the first class be $3x.$ This means the number of people in the second class is $4x.$

Thus, the sum of the scores of the first class is $84\cdot 3x = 252x$ and the sum of the scores for the people in the second class is $70\cdot 4x = 280x.$ This means the total sum is $532x,$ with $7x$ people.

Therefore, the average of all the students is $\dfrac{532x}{7x} = 76.$

Thus, the correct answer is C .

7.

在平面中，半径分别为 $1,3,5$ 、 $7$ 的四个圆都与直线 $\ell$ 在同一点 $A$ 相切，但它们可以位于 $\ell$ 的任一侧。区域 $S$ 由恰好位于这四个圆中一个圆内部的所有点组成。区域 $S$ 的最大可能面积是多少？

In a plane, four circles with radii $1,3,5,$ and $7$ are tangent to line $\ell$ at the same point $A,$ but they may be on either side of $\ell.$ Region $S$ consists of all the points that lie inside exactly one of the four circles. What is the maximum possible area of region $S?$

$24\pi$

$32\pi$

$64\pi$

$65\pi$

$84\pi$

知识点：相切圆圆面积最优化

难度评级：1240

视频讲解：

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文字解答：

在 $\ell$ 的同一侧，所有在 $A$ 处相切的圆互相包含。若套在一起的圆半径为 $r_1>r_2>\cdots$ ，则恰好在一个圆内的面积为 $\pi(r_1^2-r_2^2)$ ；更小圆内的点同时位于至少三个圆内，不符合“恰好一个”的条件。

要使面积最大，把半径 $7$ 的圆单独放在一侧，把半径 $5,3,1$ 的圆放在另一侧。这给出

$\begin{aligned} &7^2\pi+(5^2-3^2)\pi \\ &=49\pi+16\pi=65\pi. \end{aligned}$

所以答案是 D。

On one side of $\ell$ , circles tangent at $A$ are nested. For nested circles with radii $r_1>r_2>\cdots$ , the points inside exactly one of those circles have area $\pi(r_1^2-r_2^2)$ if there are at least two circles; a third smaller nested circle does not count because its points are inside three circles, not exactly one.

To maximize the area, put the circle of radius $7$ alone on one side, and put the circles of radii $5,3,1$ on the other side. This gives

$\begin{aligned} &7^2\pi+(5^2-3^2)\pi \\ &=49\pi+16\pi=65\pi. \end{aligned}$

Thus, the answer is D .

8.

Zhou 先生把从 $1$ 到 $225$ 的所有整数放入一个 $15$ 行 $15$ 列的方格中。他把 $1$ 放在正中间的方格（第八行第八列），然后如图所示按顺时针方向逐个放入其他数。在从上往下数第二行中出现的最大数与最小数之和是多少？

Mr. Zhou places all the integers from $1$ to $225$ into a $15$ by $15$ grid. He places $1$ in the middle square (eighth row and eighth column) and places other numbers one by one clockwise, as shown in part in the diagram below. What is the sum of the greatest number and the least number that appear in the second row from the top?

$367$

$368$

$369$

$379$

$380$

知识点：找规律完全平方数

难度评级：1420

视频讲解：

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文字解答：

在外层 $15\times15$ 的一圈中，最上面一行包含 $211,212,\ldots,225$ ，因此第二行中正好在 $211$ 下方的数是 $210$ 。

这是第二行的最大数。内层 $13\times13$ 螺旋的右上角是 $13^2=169$ 。在整个方格第二行中，内层这一行的数从 $157$ 到 $169$ ，因此最小数是 $157$ 。

所求和为 $210+157=367$ 。

所以答案是 A。

In the outer $15\times15$ ring, the top row contains $211,212,\ldots,225$ , so the number just below $211$ in the second row is $210$ . This is the greatest number in the second row.

The inner $13\times13$ spiral has $13^2=169$ in its upper-right corner. In the second row of the full grid, the inner-ring entries run from $157$ to $169$ . Thus the least entry in that row is $157$ .

The required sum is $210+157=367$ .

Thus, the answer is A .

9.

$xy$ 平面中的点 $P(a,b)$ 先绕点 $(1,5)$ 逆时针旋转 $90^\circ$ ，再关于直线 $y = -x$ 反射。经过这两个变换后， $P$ 的像为 $(-6,3)$ 。 $b - a$ 等于多少？

The point $P(a,b)$ in the $xy$ -plane is first rotated counterclockwise by $90^\circ$ around the point $(1,5)$ and then reflected about the line $y = -x.$ The image of $P$ after these two transformations is at $(-6,3).$ What is $b - a ?$

$1$

$3$

$5$

$7$

$9$

知识点：变换坐标几何逆推法

难度评级：1220

视频讲解：

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文字解答：

倒推变换。点 $(-6,3)$ 关于 $y=-x$ 反射后得到 $(-3,6)$ 。

接着把 $(-3,6)$ 绕 $(1,5)$ 顺时针旋转 $90^\circ$ ，以撤销原来的逆时针旋转。相对于 $(1,5)$ ，该点为 $(-4,1)$ 。顺时针旋转四分之一圈得到 $(1,4)$ ，再平移回去得到 $(2,9)$ 。

因此 $a=2$ 、 $b=9$ ，所以 $b-a=7$ 。

所以答案是 D。

Work backward. Reflecting $(-6,3)$ across $y=-x$ gives $(-3,6)$ .

Now undo the $90^\circ$ counterclockwise rotation by rotating $(-3,6)$ clockwise about $(1,5)$ . Relative to $(1,5)$ , the point is $(-4,1)$ . A clockwise quarter-turn sends this to $(1,4)$ , and translating back gives $(2,9)$ .

Thus $a=2$ , $b=9$ , and $b-a=7$ .

Thus, the answer is D .

10.

一个倒置的圆锥底面半径为 $12 \mathrm{ cm}$ ，高为 $18 \mathrm{ cm}$ ，里面装满了水。把水倒入一个高圆柱中，圆柱的水平底面半径为 $24 \mathrm{ cm}$ 。圆柱中水的高度是多少厘米？

An inverted cone with base radius $12 \mathrm{ cm}$ and height $18 \mathrm{ cm}$ is full of water. The water is poured into a tall cylinder whose horizontal base has radius of $24 \mathrm{ cm}.$ What is the height in centimeters of the water in the cylinder?

$1.5$

$3$

$4$

$4.5$

$6$

知识点：体积圆锥圆柱

难度评级：1020

视频讲解：

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文字解答：

水的体积不变。 $\frac{r^2h \pi}3 = \frac{12^2\cdot 18 \pi}3 = 864 \pi$ 。

圆柱中水的体积为 $r^2h \pi =24^2h \pi = 576h \pi$ 。于是 $576 h \pi = 864 \pi,$ 所以 $h = 1.5$ 。

所以答案是 A。

The volumes must be the same since the water is poured from one to another. The volume of the cone is $\frac{r^2h \pi}3 = \frac{12^2\cdot 18 \pi}3 = 864 \pi.$

The volume of the cylinder is $r^2h \pi =24^2h \pi = 576h \pi.$ This makes $576 h \pi = 864 \pi,$ so $h = 1.5.$

Thus, the answer is A .

11.

奶奶刚烤好一大盘长方形布朗尼。她打算沿着盘子的边平行地直线切开，做成大小和形状相同的长方形小块。每一刀都必须从盘子的一边完全切到另一边。奶奶希望内部小块的数量与位于周边的小块数量相同。她最多可以做出多少块布朗尼？

Grandma has just finished baking a large rectangular pan of brownies. She is planning to make rectangular pieces of equal size and shape, with straight cuts parallel to the sides of the pan. Each cut must be made entirely across the pan. Grandma wants to make the same number of interior pieces as pieces along the perimeter of the pan. What is the greatest possible number of brownies she can produce?

$24$

$30$

$48$

$60$

$64$

知识点：西蒙最爱的因式分解技巧最优化

难度评级：1420

视频讲解：

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文字解答：

假设切成 $l\times w$ 的网格。内部小块数为 $(l-2)(w-2)$ ，总块数为 $lw$ 。

$(l-2)(w-2)=\frac{lw}{2}.$

展开得到 $lw-4l-4w+8=0$ ，整理为：

$(l-4)(w-4)=8.$

$8$ 的正因数对给出 $(l,w)=(5,12)$ 或 $(6,8)$ ，顺序可交换。它们分别产生 $60$ 或 $48$ 块，所以最大可能数量是 $60$ 。

所以答案是 D。

Suppose the cuts make an $l\times w$ grid of pieces. The number of interior pieces is $(l-2)(w-2)$ , and the total number of pieces is $lw$ . Since the number of interior pieces equals the number of perimeter pieces, the interior pieces make up half the total:

$(l-2)(w-2)=\frac{lw}{2}.$

Multiplying out gives $lw-4l-4w+8=0$ , or

$(l-4)(w-4)=8.$

The positive factor pairs of $8$ give $(l,w)=(5,12)$ or $(6,8)$ , up to order. These produce $60$ or $48$ pieces, respectively, so the greatest possible number is $60$ .

Thus, the answer is D .

12.

设 $N = 34 \cdot 34 \cdot 63 \cdot 270$ 。 $N$ 的奇因数之和与 $N$ 的偶因数之和的比是多少？

Let $N = 34 \cdot 34 \cdot 63 \cdot 270.$ What is the ratio of the sum of the odd divisors of $N$ to the sum of the even divisors of $N?$

$1 : 16$

$1 : 15$

$1 : 14$

$1 : 8$

$1 : 3$

知识点：因数之和质因数分解奇偶性

难度评级：1140

解答：

分解质因数得 $N =2^3\cdot 3^5\cdot 5\cdot 7\cdot 17^2.$

若 $x$ 是 $N$ 的一个奇因数，则 $2x,4x,8x$ 都是 $N$ 的偶因数，它们的和为 $14x$ 。把所有奇因数都这样配对，偶因数之和就是奇因数之和的 $14$ 倍。因此所求比为 $1:14$ 。

所以答案是 C。

Using prime factorization, we get $N =2^3\cdot 3^5\cdot 5\cdot 7\cdot 17^2.$

If we have an odd divisor $x$ of $N,$ then $2x,4x,8x$ are divisors of $N,$ which has a combined sum of $14x.$ If we take the sum of every odd divisor, then the even divisors must have a sum which is $14$ times the sum of the odd divisors. Therefore, the requested ratio is $1:14$ .

Thus, the answer is C .

13.

设 $n$ 是正整数， $d$ 是一个数字。以 $n$ 为底的数 $\underline{32d}$ 的值等于 $263$ ，并且以 $n$ 为底的数 $\underline{324}$ 的值等于以六为底的数 $\underline{11d1}$ 的值。 $n + d$ 等于多少？

Let $n$ be a positive integer and $d$ be a digit such that the value of the numeral $\underline{32d}$ in base $n$ equals $263,$ and the value of the numeral $\underline{324}$ in base $n$ equals the value of the numeral $\underline{11d1}$ in base six. What is $n + d ?$

$10$

$11$

$13$

$15$

$16$

知识点：进制方程组

难度评级：1280

视频讲解：

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文字解答：

第一个条件表示 $3n^2 + 2n+d = 263.$

类似地，第二个条件表示 $3n^2 +2n +4$ $= 6^3 +6^2 +6d+1$ $= 253 + 6d.$

两式相减得 $4-d = 6d-10$ $7d = 14$ $d=2.$ 因此 $3n^2 + 2n + 2 = 263$ ，所以 $n(3n+2) = 261$ 。

这给出 $n=9$ 。因此 $n+d = 11$ 。

所以答案是 B。

The first statement means $3n^2 + 2n+d = 263.$

Similarly, the second statement means $3n^2 +2n +4$ $= 6^3 +6^2 +6d+1$ $= 253 + 6d.$

Subtracting these shows us that $4-d = 6d-10$ $7d = 14$ $d=2.$ Therefore, $3n^2 + 2n + 2 = 263,$ so $n(3n+2) = 261.$

This implies $n=9.$ Therefore, $n+d = 11.$

Thus, the answer is B .

14.

三条等距平行线与一个圆相交，形成三条长度分别为 $38,38$ , $34$ 的弦。相邻两条平行线之间的距离是多少？

Three equally spaced parallel lines intersect a circle, creating three chords of lengths $38,38,$ and $34.$ What is the distance between two adjacent parallel lines?

$5\frac12$

$6$

$6\frac12$

$7$

$7\frac12$

知识点：弦圆勾股定理

难度评级：1540

视频讲解：

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文字解答：

两条长度为 $38$ 的弦到圆心距离相等。设圆心到其中一条 $38$ 弦的距离为 $d$ 。那么另一条相等的弦也距离圆心 $d$ ，而 $34$ 弦距离圆心 $3d$ 。

由于三条平行线等距，这两条相等的弦必须位于相邻的直线上，圆心在它们中间。若圆半径为 $r$ ，则

$r^2=19^2+d^2=17^2+(3d)^2.$

因此 $19^2-17^2=8d^2$ ，所以 $72=8d^2$ ，得到 $d=3$ 。相邻平行线之间的距离为 $2d=6$ 。

所以答案是 B。

The two chords of length $38$ are equally far from the center of the circle. Because the three parallel lines are equally spaced, those two equal chords must lie on adjacent lines, with the center halfway between them. Let that half-distance be $d$ . Then each $38$ -chord is distance $d$ from the center, and the $34$ -chord is distance $3d$ from the center.

If the circle has radius $r$ , then

$r^2=19^2+d^2=17^2+(3d)^2.$

Thus $19^2-17^2=8d^2$ , so $72=8d^2$ , and $d=3$ . The distance between adjacent parallel lines is $2d=6$ .

Thus, the answer is B .

15.

实数 $x$ 满足方程 $x+\frac{1}{x} = \sqrt{5}.$ 求 $x^{11}-7x^{7}+x^3$ 的值。

The real number $x$ satisfies the equation $x+\frac{1}{x} = \sqrt{5}.$ What is the value of $x^{11}-7x^{7}+x^3?$

$-1$

$0$

$1$

$2$

$\sqrt{5}$

知识点：代数变形对称性（代数）

难度评级：1340

视频讲解：

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文字解答：

由 $x+\frac{1}{x} = \sqrt{5}$ ，平方得 $x^2 + 2 + \frac 1{x^2} = 5$ $x^2 + \frac 1{x^2} = 3.$ 再平方一次： $x^4 + 2 + \frac 1{x^4} = 9$ 即 $x^4 -7 + \frac 1{x^4} = 0.$ 两边乘以 $x^7$ ，得到 $x^{11}-7x^{7}+x^3=0.$

所以答案是 B。

Since $x+\frac{1}{x} = \sqrt{5},$ squaring yields $x^2 + 2 + \frac 1{x^2} = 5$ $x^2 + \frac 1{x^2} = 3.$ Squaring again yields $x^4 + 2 + \frac 1{x^4} = 9$ $x^4 -7 + \frac 1{x^4} = 0.$ Multiplying by $x^7$ yields $x^{11}-7x^{7}+x^3=0.$

Thus, the answer is B .

16.

如果一个正整数的每一位数字都严格大于前一位数字，则称它为上坡整数。例如 $1357, 89,$ 和 $5$ 都是上坡整数，但 $32, 1240,$ 和 $466$ 不是。多少个上坡整数能被 $15$ 整除？

Call a positive integer an uphill integer if every digit is strictly greater than the previous digit. For example, $1357, 89,$ and $5$ are all uphill integers, but $32, 1240,$ and $466$ are not. How many uphill integers are divisible by $15?$

$4$

$5$

$6$

$7$

$8$

知识点：子集整除性

难度评级：1480

视频讲解：

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文字解答：

能被 $15$ 整除的数，个位必须是 $0$ 或 $5$ 。若上坡整数的个位是 $0$ ，则这个数只能是 $0$ ，不是正整数。因此只需考虑个位为 $5$ 的数。

接着要求这个上坡整数是 $3$ 的倍数。其余数字只能从 $\{1,2,3,4\}$ 中选，所选集合的数字和除以 $3$ 必须余 $1$ 。数字 $3$ 取或不取不影响余数，所以先在不含 $3$ 的情况下计数再乘以 $2$ 。满足条件的子集只有 $3$ 个，即 $\{1\}, \{4\},$ 和 $\{1,2,4\}$ 。因此总共有 $6$ 个子集。

所以正确答案是 C。

If a number is divisible by $15,$ it has a units digit of $0$ or $5.$ If the units digit is $0$ and the digits are strictly increasing, then the number is $0,$ which isn't positive. Therefore, we can just look at numbers with a units digit of $5.$

Next, we need to find uphill integers that are a multiple of $3.$ This means the other digits are a subset of $\{1,2,3,4\}.$ Taking the sum of the set must have a remainder of $1$ when divided by $3.$ Also, having or taking out $3$ wouldn't affect the remainder, so we can take the number of subsets without a $3$ and multiply it by $2.$ There are only $3$ such subsets, namely $\{1\}, \{4\},$ and $\{1,2,4\}.$ Thus, there are $6$ total subsets.

Thus, the correct answer is C .

17.

Ravon、Oscar、Aditi、Tyrone 和 Kim 玩一个纸牌游戏。每人从编号 $1,2,3, \dots,10$ 的 $10$ 张牌中拿到 $2$ 张。玩家的得分是自己两张牌上数字之和。五人的得分如下：Ravon 为 $11$ ，Oscar 为 $4$ ，Aditi 为 $7$ ，Tyrone 为 $16$ ，Kim 为 $17$ 。下列哪项陈述正确？

Ravon, Oscar, Aditi, Tyrone, and Kim play a card game. Each person is given $2$ cards out of a set of $10$ cards numbered $1,2,3, \dots,10.$ The score of a player is the sum of the numbers of their cards. The scores of the players are as follows: Ravon-- $11,$ Oscar-- $4,$ Aditi-- $7,$ Tyrone-- $16,$ Kim-- $17.$ Which of the following statements is true?

Ravon 拿到了 3 号牌。

Ravon was given card 3.

Aditi 拿到了 3 号牌。

Aditi was given card 3.

Ravon 拿到了 4 号牌。

Ravon was given card 4.

Aditi 拿到了 4 号牌。

Aditi was given card 4.

Tyrone 拿到了 7 号牌。

Tyrone was given card 7.

知识点：逻辑推理

难度评级：1420

视频讲解：

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文字解答：

Oscar 有 $2$ 张牌，和为 $4$ ，只能是 $1$ 和 $3$ 。

因此 $3$ 号牌已被 Oscar 拿走，选项 A 和 B 都不可能。Aditi 有 $2$ 张牌，和为 $7$ ，不能用 Oscar 的牌，所以只能是 $2$ 和 $5$ ，从而她没有 $1$ 号牌。

如果某人拿到 $4$ 号牌，则另一张牌至多为 $10$ ，所以其得分至多为 $14$ 。因此 Ravon 必须拿到 $4$ 和 $7$ 。

所以答案是 C。

If there are $2$ cards for Oscar that add up to $4,$ he must have both $1$ and $3.$ This eliminates choices A and B.

If there are $2$ cards for Aditi that add up to $7,$ he must have both $2$ and $5$ so she doesn't have $1$ and $3.$

If someone has $4,$ their sum must be equal to or under $14$ since the other number must be under or equal to $10.$ Thus, Ravon must have the $4$ and $7,$ making C true and D and E false.

Thus, the answer is C .

18.

反复掷一枚公平的 $6$ 面骰，直到第一次出现奇数为止。在第一次出现奇数之前，每个偶数都至少出现一次的概率是多少？

A fair $6$ -sided die is repeatedly rolled until an odd number appears. What is the probability that every even number appears at least once before the first occurrence of an odd number?

$\dfrac{1}{120}$

$\dfrac{1}{32}$

$\dfrac{1}{20}$

$\dfrac{3}{20}$

$\dfrac{1}{6}$

知识点：骰子（概率）基本概率

难度评级：1220

视频讲解：

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文字解答：

第一个不同的点数是偶数的概率为 $\frac 36$ 。

在已经出现一个偶数后，第二个不同点数仍为偶数的概率为 $\frac 25$ 。

在已经出现两个不同偶数后，第三个不同点数仍为偶数的概率为 $\frac 14$ 。

因此所求概率为 $\dfrac{3\cdot 2\cdot 1}{6\cdot 5\cdot 4} = \dfrac 1{20}.$

所以答案是 C。

The probability that the first number is even is $\frac 36.$

The probability that the second distinct number is even is $\frac 25.$

The probability that the third distinct number is even is $\frac 14.$

The combined probability is $\dfrac{3\cdot 2\cdot 1}{6\cdot 5\cdot 4} = \dfrac 1{20}.$

Thus, the answer is C .

19.

设 $S$ 是一个有限的正整数集合。

如果从 $S$ 中移除 $S$ 中最大的整数，剩余整数的平均值为 $32$ 。如果再从 $S$ 中移除最小的整数，剩余整数的平均值为 $35$ 。如果随后把最大的整数放回 $S$ ，平均值升高到 $40$ 。原集合 $S$ 中最大的整数比最小的整数大 $72$ 。

集合 $S$ 中所有整数的平均值是多少？

Suppose that $S$ is a finite set of positive integers.

If the greatest integer in $S$ is removed from $S,$ then the average value (arithmetic mean) of the integers remaining is $32.$ If the least integer in $S$ is also removed, then the average value of the integers remaining is $35.$ If the greatest integer is then returned to the set, the average value of the integers rises to $40.$ The greatest integer in the original set $S$ is $72$ greater than the least integer in $S.$

What is the average value of all the integers in the set $S?$

$36.2$

$36.4$

$36.6$

$36.8$

$37$

知识点：平均数方程组

难度评级：1540

视频讲解：

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文字解答：

设所有整数之和为 $s$ ，最大数为 $g$ ，最小数为 $l$ ，集合 $S$ 的大小为 $n$ 。

条件给出 $\dfrac{s-l}{n-1} = 40$ $\dfrac{s-g}{n-1} = 32.$ 相减得 $\dfrac{g-l}{n-1} = 8.$ 由于 $g-l=72$ ，所以 $\frac{72}{n-1} =8$ ，即 $n=10$ 。

又有 $\dfrac{s-g-l}{n-2} = 35,$ 所以 $s-g-l = 8\cdot 35$ $=280.$ 由 $\dfrac{s-l}{n-1} = 40,$ 得 $s-l = 360$ 。于是 $g=80$ 。再由 $g=l+72$ ，得 $l=8$ ，所以 $s=368$ 。

平均值为 $\dfrac sn = \dfrac{368}{10}$ $= 36.8.$

所以答案是 D。

Let the sum of all the integers be $s,$ the greatest number be $g,$ the least number be $l,$ and the size of $S$ be $n.$

From the info given, we know $\dfrac{s-l}{n-1} = 40$ $\dfrac{s-g}{n-1} = 32.$ Subtracting these yields $\dfrac{g-l}{n-1} = 8.$ Since we know $g-l=72,$ we know $\frac{72}{n-1} =8,$ so $n=10.$

We also know $\dfrac{s-g-l}{n-2} = 35,$ so $s-g-l = 8\cdot 35$ $=280.$ Since $\dfrac{s-l}{n-1} = 40,$ we know $s-l = 360.$ This makes $g=80.$ Using $g=l+72,$ we get $l=8.$ Thus, $s=368.$

The average is $\dfrac sn = \dfrac{368}{10}$ $= 36.8.$

Thus, the answer is D .

20.

下图由 $11$ 条线段构成，每条线段长度都为 $2$ 。五边形 $ABCDE$ 的面积可写成 $\sqrt{m} + \sqrt{n}$ ，其中 $m$ 和 $n$ 为正整数。 $m + n$ 等于多少？

The figure is constructed from $11$ line segments, each of which has length $2.$ The area of pentagon $ABCDE$ can be written as $\sqrt{m} + \sqrt{n},$ where $m$ and $n$ are positive integers. What is $m + n ?$

$20$

$21$

$22$

$23$

$24$

知识点：面积分割等边三角形特殊直角三角形

难度评级：1950

解答：

相等的线段表明 $B$ 和 $E$ 附近的两侧部分来自边长为 $2$ 的等边三角形的一半。边长为 $2$ 的等边三角形高为 $\sqrt3$ ，面积为 $\sqrt3$ ，所以两侧部分合起来面积为 $2\sqrt3=\sqrt{12}$ 。

剩下的中间三角形是 $\triangle ACD$ 。由同样的等边三角形高可知 $AC=AD=2\sqrt3=\sqrt{12}$ ，且 $CD=2$ 。它到底边 $CD$ 的高为

$\sqrt{(\sqrt{12})^2-1^2}=\sqrt{11}.$

因此 $\triangle ACD$ 的面积为 $\frac12\cdot2\cdot\sqrt{11}=\sqrt{11}$ 。五边形总面积为

$\sqrt{12}+\sqrt{11},$

所以 $m+n=12+11=23$ 。

所以答案是 D。

The equal length segments show that the side pieces near $B$ and $E$ are made from halves of equilateral triangles of side length $2$ . An equilateral triangle of side length $2$ has altitude $\sqrt3$ and area $\sqrt3$ , so the two side pieces together contribute area $2\sqrt3=\sqrt{12}$ .

The remaining central triangle is $\triangle ACD$ . From the same equilateral-triangle altitudes, $AC=AD=2\sqrt3=\sqrt{12}$ , and $CD=2$ . Its altitude to $CD$ is

$\sqrt{(\sqrt{12})^2-1^2}=\sqrt{11}.$

Therefore the area of $\triangle ACD$ is $\frac12\cdot2\cdot\sqrt{11}=\sqrt{11}$ . The pentagon's total area is

$\sqrt{12}+\sqrt{11},$

so $m+n=12+11=23$ .

Thus, the answer is D .

21.

一张正方形纸片边长为 $1$ ，顶点按顺序为 $A,B,C,$ 和 $D$ 。如图所示，将纸片折叠，使顶点 $C$ 落在边 $\overline{AD}$ 上的点 $C'$ ，且边 $\overline{BC}$ 与边 $\overline{AB}$ 相交于点 $E$ 。若 $C'D = \frac{1}{3}$ ，三角形 $\bigtriangleup AEC'$ 的周长是多少？

A square piece of paper has side length $1$ and vertices $A,B,C,$ and $D$ in that order. As shown in the figure, the paper is folded so that vertex $C$ meets edge $\overline{AD}$ at point $C',$ and edge $\overline{BC}$ intersects edge $\overline{AB}$ at point $E.$ Suppose that $C'D = \frac{1}{3}.$ What is the perimeter of triangle $\bigtriangleup AEC' ?$

$2$

$1+\dfrac{2}{3}\sqrt{3}$

$\dfrac{13}{6}$

$1 + \dfrac{3}{4}\sqrt{3}$

$\dfrac{7}{3}$

知识点：折纸坐标几何

难度评级：2230

视频讲解：

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文字解答：

取坐标 $A=(0,1)$ 、 $B=(0,0)$ 、 $C=(1,0)$ 、 $D=(1,1)$ 。因为 $C'D=\frac13$ ，所以 $C'=(\frac23,1)$ ，从而 $AC'=\frac23$ 。

折叠把 $C$ 反射到 $C'$ ，所以边 $BC$ 的像是过 $C'$ 和 $E$ 的直线。把 $B=(0,0)$ 关于 $CC'$ 的垂直平分线反射，得到 $(-\frac{2}{15},\frac25)$ 。过该点与 $C'$ 的直线与 $AB$ 相交于 $E=(0,\frac12)$ 。

因此 $AE=\frac12$ ，并且

$EC'=\sqrt{\left(\frac23\right)^2+\left(\frac12\right)^2}=\frac56.$

$\triangle AEC'$ 的周长为

$\frac12+\frac23+\frac56=2.$

所以答案是 A。

Use coordinates with $A=(0,1)$ , $B=(0,0)$ , $C=(1,0)$ , and $D=(1,1)$ . Since $C'D=\frac13$ , we have $C'=(\frac23,1)$ , so $AC'=\frac23$ .

The fold reflects $C$ to $C'$ , so the image of side $BC$ is the line through $C'$ and $E$ . Reflecting $B=(0,0)$ across the perpendicular bisector of $CC'$ gives $(-\frac{2}{15},\frac25)$ . The line through this point and $C'$ meets $AB$ at $E=(0,\frac12)$ .

Thus $AE=\frac12$ , and

$EC'=\sqrt{\left(\frac23\right)^2+\left(\frac12\right)^2}=\frac56.$

The perimeter of $\triangle AEC'$ is

$\frac12+\frac23+\frac56=2.$

Thus, the answer is A .

22.

Ang、Ben 和 Jasmin 每人都有 $5$ 块积木，颜色分别为红、蓝、黄、白、绿；另有 $5$ 个空盒子。三个人各自随机且彼此独立地把自己的每种颜色积木各放入一个盒子中。至少有一个盒子收到 $3$ 块同色积木的概率为 $\frac{m}{n}$ ，其中 $m$ 与 $n$ 是互质正整数。 $m + n$ 等于多少？

Ang, Ben, and Jasmin each have $5$ blocks, colored red, blue, yellow, white, and green; and there are $5$ empty boxes. Each of the people randomly and independently of the other two people places one of their blocks into each box. The probability that at least one box receives $3$ blocks all of the same color is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. What is $m + n ?$

$47$

$94$

$227$

$471$

$542$

知识点：容斥原理排列基本概率

难度评级：2150

视频讲解：

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文字解答：

固定 Ang 的摆放方式，并用 Ang 放入的颜色标记每个盒子。Ben 和 Jasmin 各自选择五种颜色的一个排列，因此共有 $(5!)^2$ 个等可能的摆放对。

若指定 $k$ 个盒子都收到三块同色积木，则 Ben 和 Jasmin 在这 $k$ 个盒子中都必须与 Ang 的颜色相同。这可以用 $((5-k)!)^2$ 种方式完成。由容斥原理，成功的摆放对数为

$\begin{aligned} &\binom51(4!)^2-\binom52(3!)^2 \\ &\quad {}+\binom53(2!)^2-\binom54(1!)^2 \\ &\quad {}+\binom55(0!)^2. \end{aligned}$

上式等于

$2880-360+40-5+1=2556.$

这等于

$\frac{2556}{(5!)^2}=\frac{2556}{14400}=\frac{71}{400}.$

因此 $m+n=71+400=471$ 。

所以答案是 D。

Fix Ang's placement and label each box by the color Ang put in it. Ben and Jasmin each choose a permutation of the five colors, so there are $(5!)^2$ equally likely pairs of placements.

For a specified set of $k$ boxes to receive three blocks of the same color, both Ben and Jasmin must match Ang in those $k$ boxes. This can happen in $((5-k)!)^2$ ways. By inclusion-exclusion, the number of successful placement pairs is

$\begin{aligned} &\binom51(4!)^2-\binom52(3!)^2 \\ &\quad {}+\binom53(2!)^2-\binom54(1!)^2 \\ &\quad {}+\binom55(0!)^2. \end{aligned}$

This equals

$2880-360+40-5+1=2556.$

Therefore the probability is

$\frac{2556}{(5!)^2}=\frac{2556}{14400}=\frac{71}{400}.$

Thus $m+n=71+400=471$ .

Thus, the answer is D .

23.

一个边长为 $8$ 的正方形中，除了 $4$ 个角上腿长为 $2$ 的阴影等腰直角三角形，以及正方形中心一个边长为 $2\sqrt{2}$ 的阴影菱形外，其余部分都不涂阴影，如图所示。

一枚直径为 $1$ 的圆形硬币落到正方形上，并随机落在一个使硬币完全包含在正方形内的位置。硬币覆盖到正方形阴影区域一部分的概率可写成 $\frac{1}{196}\left(a+b\sqrt{2}+\pi\right)$ ，其中 $a$ 与 $b$ 是正整数。 $a+b$ 等于多少？

A square with side length $8$ is unshaded except for $4$ shaded isosceles right triangular regions with legs of length $2$ in each corner of the square and a shaded diamond with side length $2\sqrt{2}$ in the center of the square, as shown in the diagram.

A circular coin with diameter $1$ is dropped onto the square and lands in a random location where the coin is completely contained within the square. The probability that the coin will cover part of the shaded region of the square can be written as $\frac{1}{196}\left(a+b\sqrt{2}+\pi\right),$ where $a$ and $b$ are positive integers. What is $a+b?$

$64$

$66$

$68$

$70$

$72$

知识点：几何概率面积分割

难度评级：2390

解答：

硬币半径为 $\frac12$ ，所以它的圆心均匀分布在一个 $7\times7$ 的正方形内，该区域面积为 $49$ 。

对每个角上的阴影三角形，能覆盖到它的圆心位置是在允许区域内距离该三角形不超过 $\frac12$ 的点集。每个角对应一个高为 $\frac{1+\sqrt2}{2}$ 的等腰直角三角形，所以面积为

$\left(\frac{1+\sqrt2}{2}\right)^2=\frac{3+2\sqrt2}{4}.$

四个角合计贡献 $3+2\sqrt2$ 。

中心阴影菱形是一个边长为 $2\sqrt2$ 的正方形。把它向外扩张距离 $\frac12$ ，除了菱形本身面积 $8$ ，还增加总面积 $4\sqrt2$ 的四个矩形，以及合起来面积为 $\frac\pi4$ 的四个四分之一圆。因此中心部分贡献

$8+4\sqrt2+\frac\pi4.$

有利面积为

$\begin{aligned} &3+2\sqrt2+8+4\sqrt2+\frac\pi4 \\ &=11+6\sqrt2+\frac\pi4. \end{aligned}$

概率为

$\begin{aligned} &\frac{11+6\sqrt2+\frac\pi4}{49} \\ &=\frac{44+24\sqrt2+\pi}{196}. \end{aligned}$

所以 $a+b=44+24=68$ 。

所以答案是 C。

The coin has radius $\frac12$ , so its center is uniformly distributed over a $7\times7$ square of area $49$ .

A shaded corner triangle contributes the set of center positions within distance $\frac12$ of that triangle, inside the allowed center square. For each corner this is a right isosceles triangle whose altitude is $\frac{1+\sqrt2}{2}$ , so its area is

$\left(\frac{1+\sqrt2}{2}\right)^2=\frac{3+2\sqrt2}{4}.$

All four corners contribute $3+2\sqrt2$ .

The center shaded diamond is a square of side $2\sqrt2$ . Expanding it by distance $\frac12$ adds four rectangles of total area $4\sqrt2$ and four quarter-circles of total area $\frac\pi4$ , in addition to the diamond's area $8$ . Thus the center contribution is

$8+4\sqrt2+\frac\pi4.$

The favorable area is

$\begin{aligned} &3+2\sqrt2+8+4\sqrt2+\frac\pi4 \\ &=11+6\sqrt2+\frac\pi4. \end{aligned}$

The probability is

$\begin{aligned} &\frac{11+6\sqrt2+\frac\pi4}{49} \\ &=\frac{44+24\sqrt2+\pi}{196}. \end{aligned}$

So $a+b=44+24=68$ .

Thus, the answer is C .

24.

Arjun 和 Beth 玩一个游戏：他们轮流从若干堵砖墙中的一堵移除一块砖，或移除相邻的两块砖；移除后产生的空隙可能把一堵墙分成新的墙。每堵墙都只有一块砖高。例如，大小为 $4$ 和 $2$ 的一组墙，经过一步可以变成以下任意一种： $(3,2),(2,1,2),(4),(4,1),(2,2)$ ，或 $(1,1,2)$ 。

Arjun 先手，移除最后一块砖的玩家获胜。对于哪一种初始配置，Beth 有必胜策略？

Arjun and Beth play a game in which they take turns removing one brick or two adjacent bricks from one "wall" among a set of several walls of bricks, with gaps possibly creating new walls. The walls are one brick tall. For example, a set of walls of sizes $4$ and $2$ can be changed into any of the following by one move: $(3,2),(2,1,2),(4),(4,1),(2,2),$ or $(1,1,2).$

Arjun plays first, and the player who removes the last brick wins. For which starting configuration is there a strategy that guarantees a win for Beth?

$(6,1,1)$

$(6,2,1)$

$(6,2,2)$

$(6,3,1)$

$(6,3,2)$

知识点：组合游戏

难度评级：2390

解答：

对长度为 $n$ 的单独一堵墙，根据所有可能操作计算它的 Sprague-Grundy 值。

$\begin{aligned} &g(1)=1,\quad g(2)=2, \\ &g(3)=3,\quad g(4)=1, \\ &g(5)=4,\quad g(6)=3. \end{aligned}$

多堵墙的局面在这些值按位异或为 $0$ 时，正好是轮到行动者的必败局面。逐一计算选项：

$(6,1,1): 3\oplus1\oplus1=3,$

$(6,2,1): 3\oplus2\oplus1=0,$

$(6,2,2): 3\oplus2\oplus2=3,$

$(6,3,1): 3\oplus3\oplus1=1,$

$(6,3,2): 3\oplus3\oplus2=2.$

只有 $(6,2,1)$ 是轮到行动者的必败局面，所以 Beth 恰好在这个初始配置下有必胜策略。

所以答案是 B。

For a single wall of length $n$ , compute its Sprague-Grundy value from the possible moves. For the wall lengths needed here, the values are

$\begin{aligned} &g(1)=1,\quad g(2)=2, \\ &g(3)=3,\quad g(4)=1, \\ &g(5)=4,\quad g(6)=3. \end{aligned}$

For several walls, the position is losing for the player to move exactly when the xor of the wall values is $0$ . Evaluating the choices gives

$(6,1,1): 3\oplus1\oplus1=3,$

$(6,2,1): 3\oplus2\oplus1=0,$

$(6,2,2): 3\oplus2\oplus2=3,$

$(6,3,1): 3\oplus3\oplus1=1,$

$(6,3,2): 3\oplus3\oplus2=2.$

Only $(6,2,1)$ is losing for the player to move, so Beth has a guaranteed win exactly for that starting configuration.

Thus, the answer is B .

25.

设 $S$ 是坐标平面中的格点集合，其中两个坐标都是从 $1$ 到 $30$ 的整数。恰有 $300$ 个 $S$ 中的点位于直线 $y=mx$ 上或其下方。所有可能的 $m$ 值构成一个长度为 $\frac ab$ 的区间，其中 $a$ 与 $b$ 是互质正整数。 $a+b$ 等于多少？

Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positive integers. What is $a+b?$

$31$

$47$

$62$

$72$

$85$

知识点：格点取整函数

难度评级：2390

解答：

在答案附近的斜率范围内，对固定斜率 $m$ ，集合 $S$ 中位于 $y=mx$ 上或下方的点数为

$\sum_{x=1}^{30}\lfloor mx\rfloor,$

在所考虑的斜率范围内，计数不会受到纵坐标上限的截断。

当 $m=\frac23$ 时，把 $x=3k+1,3k+2,3k+3$ 分组，其中 $k=0,1,\ldots,9$ 。于是

$\lfloor 2x/3\rfloor=2k,\ 2k+1, \ 2k+2,$

总和为 $6k+3$

$\sum_{k=0}^9(6k+3)=270+30=300.$

若 $m<\frac23$ ，十个比值 $y/x=2/3$ 的点不再被计入，所以点数小于 $300$ 。因此区间下端为 $\frac23$ 。

下一个大于 $\frac23$ 的可能比值 $y/x$ ，在 $1\le x,y\le30$ 下可按 $x$ 模 $3$ 检查。最佳候选为

$\frac{19}{28},\qquad \frac{20}{29},\qquad \frac{21}{30}=\frac{7}{10},$

其中最小的是 $\frac{19}{28}$ 。因此区间长度为

$\frac{19}{28}-\frac23=\frac1{84}.$

所以 $a+b=1+84=85$ 。

所以答案是 E。

For a fixed slope $m$ , the number of points in $S$ on or below $y=mx$ is

$\sum_{x=1}^{30}\lfloor mx\rfloor,$

for the slopes near the answer.

At $m=\frac23$ , grouping $x=3k+1,3k+2,3k+3$ for $k=0,1,\ldots,9$ gives

$\lfloor 2x/3\rfloor=2k,\ 2k+1, \ 2k+2,$

whose sum over each block is $6k+3$ . Thus the total is

$\sum_{k=0}^9(6k+3)=270+30=300.$

If $m<\frac23$ , the ten points with ratios $y/x=2/3$ are no longer counted, so the count is less than $300$ . Therefore the lower end is $\frac23$ .

The next possible ratio $y/x$ greater than $\frac23$ , with $1\le x,y\le30$ , is minimized by checking $x$ modulo $3$ . The best candidates are

$\frac{19}{28},\qquad \frac{20}{29},\qquad \frac{21}{30}=\frac{7}{10},$

and the smallest is $\frac{19}{28}$ . Hence the interval length is

$\frac{19}{28}-\frac23=\frac1{84}.$

Thus $a+b=1+84=85$ .

Thus, the answer is E .