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2021 AMC 10B Spring 真题

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1.

有多少个整数 xx 满足 ∣x∣<3π|x| < 3\pi？
How many integer values of xx satisfy ∣x∣<3π?|x| < 3\pi?
99
1010
1818
1919
2020
答案：D
知识点：绝对值区间内整数计数
难度评级：560
小提示：
把 3π3\pi 看成略大于 99 的数。
Use 3π3\pi as a little more than 99
大提示：
数出严格位于 −3π-3\pi 与 3π3\pi 之间的整数。
Count the integers strictly between −3π-3\pi and 3π3\pi
解答：
从 −9-9 到 99 的每个整数都满足条件，共有 9−(−9)+1=199-(-9)+1 = 19 个。
所以正确答案是 D。
Every integer from −9-9 to 9,9, inclusive, works. This yields 9−(−9)+1=199-(-9)+1 = 19 solutions.
Thus, the correct answer is D .

2.

下式的值是多少？(3−23)2+(3+23)2？ \begin{aligned} &\sqrt{\left(3-2\sqrt{3}\right)^2} \\ &{}+\sqrt{\left(3+2\sqrt{3}\right)^2} \end{aligned}\text{？}
What is the value of (3−23)2+(3+23)2? \begin{aligned} &\sqrt{\left(3-2\sqrt{3}\right)^2} \\ &{}+\sqrt{\left(3+2\sqrt{3}\right)^2}? \end{aligned}
00
43−64\sqrt{3}-6
66
434\sqrt{3}
43+64\sqrt{3}+6
答案：D
知识点：根式绝对值
难度评级：770
小提示：
记住 u2=∣u∣\sqrt{u^2}=|u|。
Remember that u2=∣u∣\sqrt{u^2}=|u|
大提示：
分别判断 3−233-2\sqrt3 和 3+233+2\sqrt3 的正负。
Decide the signs of 3−233-2\sqrt3 and 3+233+2\sqrt3 separately
解答：
因为 u2=∣u∣\sqrt{u^2}=|u|，所以原式等于 ∣3−23∣+∣3+23∣。|3-2\sqrt3|+|3+2\sqrt3|\text{。} 又因为 23>32\sqrt3>3，所以它化为 (23−3)+(3+23)=43。(2\sqrt3-3)+(3+2\sqrt3)=4\sqrt3\text{。}
所以正确答案是 D。
Because u2=∣u∣,\sqrt{u^2}=|u|, the expression equals ∣3−23∣+∣3+23∣.|3-2\sqrt3|+|3+2\sqrt3|. Since 23>3,2\sqrt3>3, this becomes (23−3)+(3+23)=43.(2\sqrt3-3)+(3+2\sqrt3)=4\sqrt3.
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 2828 名学生，其中 25%25\% 的低年级学生和 10%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 2828 students in the program, 25%25\% of the juniors and 10%10\% of the seniors are on the debate team. How many juniors are in the program?
55
66
88
1111
2020
答案：C
知识点：百分数方程组
难度评级：870
小提示：
设低年级和高年级学生人数分别为变量。
Let the numbers of juniors and seniors be variables
大提示：
辩论队中两个年级人数相同，这给出第二个方程。
The equal debate-team counts give a second equation
视频讲解：
解答视频缩略图
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文字解答：
设低年级学生有 jj 人，高年级学生有 ss 人。那么 j+s=28j+s = 28，且 0.25j=0.1s0.25j = 0.1s。这说明 2.5j=s2.5j = s，所以 3.5j=283.5j = 28。因此 j=8j=8。
所以正确答案是 C。
Let the number of juniors be jj and the number of seniors be s.s. Then, j+s=28j+s = 28 and 0.25j=0.1s.0.25j = 0.1s. This means 2.5j=s,2.5j = s, so 3.5j=28.3.5j = 28. This makes j=8.j=8.
Thus, the correct answer is C .

4.

在一场数学竞赛中，5757 名学生穿蓝色衬衫，另外 7575 名学生穿黄色衬衫。这 132132 名学生被分成 6666 对。其中恰有 2323 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, 5757 students are wearing blue shirts, and another 7575 students are wearing yellow shirts. The 132132 students are assigned into 6666 pairs. In exactly 2323 of these pairs, both students are wearing blue shirts. In how many pairs are both students wearing yellow shirts?
2323
3232
3737
4141
6464
答案：B
知识点：数对计数
难度评级：960
小提示：
先用蓝蓝配对的数量求出混合配对的数量。
Use the blue-blue pairs to find how many mixed pairs there are
大提示：
每一对混合配对恰好用掉一名穿黄色衬衫的学生。
Every mixed pair uses exactly one yellow-shirted student
视频讲解：
解答视频缩略图
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文字解答：
蓝蓝配对用掉 2⋅23=462\cdot 23 = 46 名穿蓝色衬衫的学生。因此还有 57−46=1157-46=11 名穿蓝色衬衫的学生在混合配对中，也就恰有 1111 名穿黄色衬衫的学生在混合配对中。
剩下只有 6464 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 642=32\dfrac{64}2 = 32。
所以答案是 B。
There are 2⋅23=462\cdot 23 = 46 students with blue shirts that are in a pair with just blue shirts. This means there are 57−46=1157-46=11 students in blue shirts who are paired with someone wearing a yellow shirt, meaning exactly 1111 people wearing yellow shirts are paired with someone wearing a blue shirt.
This leaves just 6464 students wearing a yellow shirt who are paired with someone else wearing a yellow shirt. This yields 642=32\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 2424，另外两个相乘为 3030。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 24,24, while the other two multiply to 30.30. What is the sum of the ages of Jonie’s four cousins?
2121
2222
2323
2424
2525
答案：B
知识点：因数年龄问题
难度评级：900
小提示：
先列出 3030 的一位数因数对。
List the single-digit factor pairs of 3030 first
大提示：
年龄互不相同，所以排除会重复使用某个年龄的因数对。
The ages are distinct, so exclude any factor pair that reuses an age
视频讲解：
解答视频缩略图
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文字解答：
乘积为 3030 的一位正整数对只有 5,65,6。乘积为 2424 的一位数因数对是 3,83,8 和 4,64,6。因为四个年龄互不相同，若取 4,64,6 就会重复年龄 66，所以另一对必须是 3,83,8。因此四个年龄是 3,5,6,83,5,6,8，它们的和为 2222。
所以答案是 B。
The only pair of single-digit positive integers with product 3030 is 5,6.5,6. The single-digit factor pairs of 2424 are 3,83,8 and 4,6.4,6. Because all four ages are distinct, the pair 4,64,6 would repeat the age 6,6, so the other pair must be 3,8.3,8. Thus the ages are 3,5,6,8,3,5,6,8, whose sum is 22.22.
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 8484，下午班学生的平均分是 7070。上午班人数与下午班人数之比为 34\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 84,84, and the afternoon class’s mean score is 70.70. The ratio of the number of students in the morning class to the number of students in the afternoon class is 34.\frac{3}{4}. What is the mean of the scores of all the students?
7474
7575
7676
7777
7878
答案：C
知识点：平均数比与比例
难度评级：900
小提示：
设两个班的人数分别为 3x3x 和 4x4x。
Use 3x3x and 4x4x students for the two classes
大提示：
用两个班的总分计算加权平均数。
Compute the weighted average from the two total score sums
视频讲解：
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文字解答：
设上午班有 3x3x 人，那么下午班有 4x4x 人。
上午班总分为 84⋅3x=252x84\cdot 3x = 252x，下午班总分为 70⋅4x=280x70\cdot 4x = 280x。因此总分为 532x532x，总人数为 7x7x。
所以所有学生的平均分为 532x7x=76\dfrac{532x}{7x} = 76。
所以正确答案是 C。
Let the number of people in the first class be 3x.3x. This means the number of people in the second class is 4x.4x.
Thus, the sum of the scores of the first class is 84⋅3x=252x84\cdot 3x = 252x and the sum of the scores for the people in the second class is 70⋅4x=280x.70\cdot 4x = 280x. This means the total sum is 532x,532x, with 7x7x people.
Therefore, the average of all the students is 532x7x=76. \dfrac{532x}{7x} = 76.
Thus, the correct answer is C .

7.

在平面中，半径分别为 11、33、55 和 77 的四个圆都与直线 ℓ\ell 在同一点 AA 相切，但它们可以位于 ℓ\ell 的任一侧。区域 SS 由恰好位于这四个圆中一个圆内部的所有点组成。区域 SS 的最大可能面积是多少？
In a plane, four circles with radii 1,1, 3,3, 5,5, and 77 are tangent to line ℓ\ell at the same point A,A, but they may be on either side of ℓ.\ell. Region SS consists of all the points that lie inside exactly one of the four circles. What is the maximum possible area of region S?S?
24π24\pi
32π32\pi
64π64\pi
65π65\pi
84π84\pi
答案：D
知识点：相切圆圆面积最优化
难度评级：1240
小提示：
在直线同一侧且同点相切的圆是相互套在一起的。
Circles on the same side of the line are nested
大提示：
对套在一起的圆，恰好在一个圆内的部分是大圆内但小一号圆外的部分。
Inside exactly one nested circle means outside the next smaller circle
视频讲解：
解答视频缩略图
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文字解答：
在 ℓ\ell 的同一侧，所有在 AA 处相切的圆互相包含。若套在一起的圆半径为 r1>r2>⋯r_1>r_2>\cdots，则恰好在一个圆内的面积为 π(r12−r22)\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 77 的圆单独放在一侧，把半径 5,3,15,3,1 的圆放在另一侧。这给出
72π+(52−32)π=49π+16π=65π。 \begin{aligned} &7^2\pi+(5^2-3^2)\pi \\ &=49\pi+16\pi=65\pi \end{aligned}\text{。}
所以答案是 D。
On one side of ℓ,\ell, circles tangent at AA are nested. For nested circles with radii r1>r2>⋯ ,r_1>r_2>\cdots, the points inside exactly one of those circles have area π(r12−r22)\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 77 alone on one side, and put the circles of radii 5,3,15,3,1 on the other side. This gives
72π+(52−32)π=49π+16π=65π. \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 11 到 225225 的所有整数放入一个 1515 行 1515 列的方格中。他把 11 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 11 to 225225 into a 1515 by 1515 grid. He places 11 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?
367367
368368
369369
379379
380380
答案：A
知识点：找规律完全平方数
难度评级：1420
小提示：
外层 15×1515\times15 的一圈以右上角的 225225 结束。
The outer 15×1515\times15 ring ends with 225225 in the upper right
大提示：
第二行包含内层 13×1313\times13 螺旋的顶行。
The second row includes the top row of the inner 13×1313\times13 spiral
视频讲解：
解答视频缩略图
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文字解答：
在外层 15×1515\times15 的一圈中，最上面一行包含 211,212,…,225211,212,\ldots,225，因此第二行中正好在 211211 下方的数是 210210。
这是第二行的最大数。内层 13×1313\times13 螺旋的右上角是 132=16913^2=169。在整个方格第二行中，内层这一行的数从 157157 到 169169，因此最小数是 157157。
所求和为 210+157=367210+157=367。
所以答案是 A。
In the outer 15×1515\times15 ring, the top row contains 211,212,…,225,211,212,\ldots,225, so the number just below 211211 in the second row is 210.210. This is the greatest number in the second row.
The inner 13×1313\times13 spiral has 132=16913^2=169 in its upper-right corner. In the second row of the full grid, the inner-ring entries run from 157157 to 169.169. Thus the least entry in that row is 157.157.
The required sum is 210+157=367.210+157=367.
Thus, the answer is A .

9.

xyxy 平面中的点 P(a,b)P(a,b) 先绕点 (1,5)(1,5) 逆时针旋转 90∘90^\circ，再关于直线 y=−xy = -x 反射。经过这两个变换后，PP 的像为 (−6,3)(-6,3)。b−ab - a 等于多少？
The point P(a,b)P(a,b) in the xyxy-plane is first rotated counterclockwise by 90∘90^\circ around the point (1,5)(1,5) and then reflected about the line y=−x.y = -x. The image of PP after these two transformations is at (−6,3).(-6,3). What is b−a?b - a ?
11
33
55
77
99
答案：D
知识点：变换坐标几何逆推法
难度评级：1220
小提示：
先倒着做，撤销反射。
Undo the reflection first
大提示：
撤销旋转时，先把旋转中心平移到原点。
Undo the rotation by translating the center of rotation to the origin
视频讲解：
解答视频缩略图
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文字解答：
倒推变换。点 (−6,3)(-6,3) 关于 y=−xy=-x 反射后得到 (−3,6)(-3,6)。
接着把 (−3,6)(-3,6) 绕 (1,5)(1,5) 顺时针旋转 90∘90^\circ，以撤销原来的逆时针旋转。相对于 (1,5)(1,5)，该点为 (−4,1)(-4,1)。顺时针旋转四分之一圈得到 (1,4)(1,4)，再平移回去得到 (2,9)(2,9)。
因此 a=2a=2、b=9b=9，所以 b−a=7b-a=7。
所以答案是 D。
Work backward. Reflecting (−6,3)(-6,3) across y=−xy=-x gives (−3,6).(-3,6).
Now undo the 90∘90^\circ counterclockwise rotation by rotating (−3,6)(-3,6) clockwise about (1,5).(1,5). Relative to (1,5),(1,5), the point is (−4,1).(-4,1). A clockwise quarter-turn sends this to (1,4),(1,4), and translating back gives (2,9).(2,9).
Thus a=2,a=2, b=9,b=9, and b−a=7.b-a=7.
Thus, the answer is D .

10.

一个倒置的圆锥底面半径为 12cm12 \mathrm{ cm}，高为 18cm18 \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 24cm24 \mathrm{ cm}。圆柱中水的高度是多少厘米？
An inverted cone with base radius 12cm12 \mathrm{ cm} and height 18cm18 \mathrm{ cm} is full of water. The water is poured into a tall cylinder whose horizontal base has a radius of 24cm.24 \mathrm{ cm}. What is the height in centimeters of the water in the cylinder?
1.51.5
33
44
4.54.5
66
答案：A
知识点：体积圆锥圆柱
难度评级：1020
小提示：
令圆锥体积等于圆柱中水的体积。
Set the cone volume equal to the cylinder volume
大提示：
圆柱半径是圆锥半径的两倍。
The cylinder radius is twice the cone radius
视频讲解：
解答视频缩略图
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文字解答：
因为水是从一个容器倒入另一个容器，所以体积保持不变。圆锥中水的体积为 r2hπ3=122⋅18π3=864π\frac{r^2h \pi}3 = \frac{12^2\cdot 18 \pi}3 = 864 \pi。
圆柱中水的体积为 r2hπ=242hπ=576hπr^2h \pi =24^2h \pi = 576h \pi。于是 576hπ=864π576 h \pi = 864 \pi，所以 h=1.5h = 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 r2hπ3=122⋅18π3=864π.\frac{r^2h \pi}3 = \frac{12^2\cdot 18 \pi}3 = 864 \pi.
The volume of the cylinder is r2hπ=242hπ=576hπ.r^2h \pi =24^2h \pi = 576h \pi. This makes 576hπ=864π,576 h \pi = 864 \pi, so h=1.5.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?
2424
3030
4848
6060
6464
答案：D
知识点：西蒙最爱的因式分解技巧最优化
难度评级：1420
小提示：
设盘子被切成一个 l×wl\times w 的小块网格。
Let the pan be cut into an l×wl\times w grid of pieces
大提示：
令内部小块数量等于周边小块数量。
Set the number of interior pieces equal to the number of perimeter pieces
视频讲解：
解答视频缩略图
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文字解答：
假设切成 l×wl\times w 的网格。内部小块数为 (l−2)(w−2)(l-2)(w-2)，总块数为 lwlw。因为内部小块数等于周边小块数，所以内部小块占总块数的一半：
(l−2)(w−2)=lw2。(l-2)(w-2)=\frac{lw}{2}\text{。}
展开得到 lw−4l−4w+8=0lw-4l-4w+8=0，整理为：
(l−4)(w−4)=8。(l-4)(w-4)=8\text{。}
88 的正因数对给出 (l,w)=(5,12)(l,w)=(5,12) 或 (6,8)(6,8)，顺序可交换。它们分别产生 6060 或 4848 块，所以最大可能数量是 6060。
所以答案是 D。
Suppose the cuts make an l×wl\times w grid of pieces. The number of interior pieces is (l−2)(w−2),(l-2)(w-2), and the total number of pieces is lw.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)=lw2.(l-2)(w-2)=\frac{lw}{2}.
Multiplying out gives lw−4l−4w+8=0,lw-4l-4w+8=0, or
(l−4)(w−4)=8.(l-4)(w-4)=8.
The positive factor pairs of 88 give (l,w)=(5,12)(l,w)=(5,12) or (6,8),(6,8), up to order. These produce 6060 or 4848 pieces, respectively, so the greatest possible number is 60.60.
Thus, the answer is D .

12.

设 N=34⋅34⋅63⋅270N = 34 \cdot 34 \cdot 63 \cdot 270。NN 的奇因数之和与 NN 的偶因数之和的比是多少？
Let N=34⋅34⋅63⋅270.N = 34 \cdot 34 \cdot 63 \cdot 270. What is the ratio of the sum of the odd divisors of NN to the sum of the even divisors of N?N?
1:161 : 16
1:151 : 15
1:141 : 14
1:81 : 8
1:31 : 3
答案：C
知识点：因数之和质因数分解奇偶性
难度评级：1140
小提示：
按因数中 22 的幂次把因数分开。
Separate divisors by their power of 22
大提示：
每个奇因数都对应偶因数 2d,4d,8d2d,4d,8d。
Each odd divisor corresponds to even divisors 2d,4d,8d2d,4d,8d
解答：
分解质因数得 N=23⋅35⋅5⋅7⋅172。N =2^3\cdot 3^5\cdot 5\cdot 7\cdot 17^2\text{。}
若 xx 是 NN 的一个奇因数，则 2x,4x,8x2x,4x,8x 都是 NN 的偶因数，它们的和为 14x14x。把所有奇因数都这样配对，偶因数之和就是奇因数之和的 1414 倍。因此所求比为 1:141:14。
所以答案是 C。
Using prime factorization, we get N=23⋅35⋅5⋅7⋅172.N =2^3\cdot 3^5\cdot 5\cdot 7\cdot 17^2.
If we have an odd divisor xx of N,N, then 2x,4x,8x2x,4x,8x are divisors of N,N, which has a combined sum of 14x.14x. If we take the sum of every odd divisor, then the even divisors must have a sum which is 1414 times the sum of the odd divisors. Therefore, the requested ratio is 1:14.1:14.
Thus, the answer is C .

13.

设 nn 是正整数，dd 是一个数字。以 nn 为底的数 32d‾\underline{32d} 的值等于 263263，并且以 nn 为底的数 324‾\underline{324} 的值等于以六为底的数 11d1‾\underline{11d1} 的值。n+dn + d 等于多少？
Let nn be a positive integer and dd be a digit such that the value of the numeral 32d‾\underline{32d} in base nn equals 263,263, and the value of the numeral 324‾\underline{324} in base nn equals the value of the numeral 11d1‾\underline{11d1} in base six. What is n+d?n + d ?
1010
1111
1313
1515
1616
答案：B
知识点：进制方程组
难度评级：1280
小提示：
把每个进制表示的数都转换成十进制。
Translate each base numeral into base ten
大提示：
将两个方程相减，先求出这个数字。
Subtract the two equations to solve for the digit first
视频讲解：
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文字解答：
第一个条件表示 3n2+2n+d=263。3n^2 + 2n+d = 263\text{。}
类似地，第二个条件表示 3n2+2n+43n^2 +2n +4 =63+62+6d+1= 6^3 +6^2 +6d+1 =253+6d。= 253 + 6d\text{。}
两式相减得 4−d=6d−104-d = 6d-10 7d=147d = 14 d=2。d=2\text{。} 因此 3n2+2n+2=2633n^2 + 2n + 2 = 263，所以 n(3n+2)=261n(3n+2) = 261。
这给出 n=9n=9。因此 n+d=11n+d = 11。
所以答案是 B。
The first statement means 3n2+2n+d=263.3n^2 + 2n+d = 263.
Similarly, the second statement means 3n2+2n+43n^2 +2n +4 =63+62+6d+1= 6^3 +6^2 +6d+1 =253+6d.= 253 + 6d.
Subtracting these shows us that 4−d=6d−104-d = 6d-10 7d=147d = 14 d=2.d=2. Therefore, 3n2+2n+2=263,3n^2 + 2n + 2 = 263, so n(3n+2)=261.n(3n+2) = 261.
This implies n=9.n=9. Therefore, n+d=11.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 3838、3838 和 3434 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, 38,38, and 34.34. What is the distance between two adjacent parallel lines?
5125\frac12
66
6126\frac12
77
7127\frac12
答案：B
知识点：弦圆勾股定理
难度评级：1540
小提示：
长度相等的弦到圆心的距离相等。
Equal chord lengths are equally far from the circle center
大提示：
若两条相等的弦在相邻直线上，则第三条直线到圆心的距离是半间距的三倍。
If the equal chords are adjacent lines, the third line is three half-spacings from the center
视频讲解：
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文字解答：
两条长度为 3838 的弦到圆心的距离相等。它们不能位于两条最外侧的直线上：否则圆心会在中间直线上，该线所截的弦会是直径，长度大于 3838，而不是给定的 3434。因此两条相等弦位于相邻直线上，圆心在两线中间。设这个半间距为 dd。则每条长 3838 的弦到圆心的距离是 dd，长 3434 的弦到圆心的距离是 3d3d。
若圆的半径为 rr，则
r2=192+d2=172+(3d)2。r^2=19^2+d^2=17^2+(3d)^2\text{。}
因此 192−172=8d219^2-17^2=8d^2，所以 72=8d272=8d^2，得到 d=3d=3。相邻平行线的距离是 2d=62d=6。
所以答案是 B。
The two chords of length 3838 are equally far from the center of the circle. They cannot lie on the two outer lines: then the center would lie on the middle line, whose chord would be a diameter longer than 38,38, not the given 34.34. Therefore, the equal chords lie on adjacent lines, with the center halfway between them. Let that half-distance be d.d. Then each 3838-chord is distance dd from the center, and the 3434-chord is distance 3d3d from the center.
If the circle has radius r,r, then
r2=192+d2=172+(3d)2.r^2=19^2+d^2=17^2+(3d)^2.
Thus 192−172=8d2,19^2-17^2=8d^2, so 72=8d2,72=8d^2, and d=3.d=3. The distance between adjacent parallel lines is 2d=6.2d=6.
Thus, the answer is B .

15.

实数 xx 满足方程 x+1x=5。x+\frac{1}{x} = \sqrt{5}\text{。} 求 x11−7x7+x3x^{11}-7x^{7}+x^3 的值。
The real number xx satisfies the equation x+1x=5.x+\frac{1}{x} = \sqrt{5}. What is the value of x11−7x7+x3?x^{11}-7x^{7}+x^3?
−1-1
00
11
22
5\sqrt{5}
答案：B
知识点：代数变形对称性（代数）
难度评级：1340
小提示：
将已知方程平方，求出 x2+x−2x^2+x^{-2}。
Square the given equation to find x2+x−2x^2+x^{-2}
大提示：
再平方一次，得到含有 x4+x−4x^4+x^{-4} 的关系。
Square again to find a relation involving x4+x−4x^4+x^{-4}
视频讲解：
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文字解答：
由 x+1x=5x+\frac{1}{x} = \sqrt{5}，两边平方得 x2+2+1x2=5x^2 + 2 + \frac 1{x^2} = 5 x2+1x2=3。x^2 + \frac 1{x^2} = 3\text{。} 再平方一次，得 x4+2+1x4=9x^4 + 2 + \frac 1{x^4} = 9 x4−7+1x4=0。x^4 -7 + \frac 1{x^4} = 0\text{。} 两边乘以 x7x^7，得到 x11−7x7+x3=0。x^{11}-7x^{7}+x^3=0\text{。}
所以答案是 B。
Since x+1x=5,x+\frac{1}{x} = \sqrt{5}, squaring yields x2+2+1x2=5x^2 + 2 + \frac 1{x^2} = 5 x2+1x2=3.x^2 + \frac 1{x^2} = 3. Squaring again yields x4+2+1x4=9x^4 + 2 + \frac 1{x^4} = 9 x4−7+1x4=0.x^4 -7 + \frac 1{x^4} = 0. Multiplying by x7x^7 yields x11−7x7+x3=0.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 13571357，8989 和 55 都是上坡整数，但 3232，12401240 和 466466 不是。多少个上坡整数能被 1515 整除？
Call a positive integer an uphill integer if every digit is strictly greater than the previous digit. For example, 1357,1357, 89,89, and 55 are all uphill integers, but 32,32, 1240,1240, and 466466 are not. How many uphill integers are divisible by 15?15?
44
55
66
77
88
答案：C
知识点：子集整除性
难度评级：1480
小提示：
个位数字必须是 55。
The last digit must be 55
大提示：
从 1,2,3,4{1,2,3,4} 中选择一个子集，使数字和满足被 33 整除的条件。
Choose a subset of 1,2,3,4{1,2,3,4} whose digit sum gives divisibility by 33
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文字解答：
能被 1515 整除的数，个位必须是 00 或 55。若上坡整数的个位是 00，则这个数只能是 00，不是正整数。因此只需考虑个位为 55 的数。
接着要求这个上坡整数是 33 的倍数。其余数字只能从 {1,2,3,4}\{1,2,3,4\} 中选，所选集合的数字和除以 33 必须余 11。数字 33 取或不取不影响余数，所以先在不含 33 的情况下计数再乘以 22。满足条件的子集只有 33 个，即 {1},{4}\{1\}, \{4\} 和 {1,2,4}\{1,2,4\}。因此总共有 66 个子集。
所以正确答案是 C。
If a number is divisible by 15,15, it has a units digit of 00 or 5.5. If the units digit is 00 and the digits are strictly increasing, then the number is 0,0, which isn’t positive. Therefore, we can just look at numbers with a units digit of 5.5.
Next, we need to find uphill integers that are a multiple of 3.3. This means the other digits are a subset of {1,2,3,4}.\{1,2,3,4\}. Taking the sum of the set must have a remainder of 11 when divided by 3.3. Also, having or taking out 33 wouldn’t affect the remainder, so we can take the number of subsets without a 33 and multiply it by 2.2. There are only 33 such subsets, namely {1},{4},\{1\}, \{4\}, and {1,2,4}.\{1,2,4\}. Thus, there are 66 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 11，22，33，…\dots，1010 的 1010 张牌中拿到 22 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 1111，Oscar 为 44，Aditi 为 77，Tyrone 为 1616，Kim 为 1717。下列哪项陈述正确？
Ravon, Oscar, Aditi, Tyrone, and Kim play a card game. Each person is given 22 cards out of a set of 1010 cards numbered 1,1, 2,2, 3,3, …,\dots, 10.10. The score of a player is the sum of the numbers of their cards. The scores of the players are as follows: Ravon--11,11, Oscar--4,4, Aditi--7,7, Tyrone--16,16, Kim--17.17. Which of the following statements is true?
Ravon 拿到了 33 号牌。
Ravon was given card 3.3.
Aditi 拿到了 33 号牌。
Aditi was given card 3.3.
Ravon 拿到了 44 号牌。
Ravon was given card 4.4.
Aditi 拿到了 44 号牌。
Aditi was given card 4.4.
Tyrone 拿到了 77 号牌。
Tyrone was given card 7.7.
答案：C
知识点：逻辑推理
难度评级：1420
小提示：
Oscar 的得分会强制确定他的两张牌。
Oscar’s cards are forced by his score
大提示：
确定 Oscar 和 Aditi 后，再看剩余的可能和。
After Oscar and Aditi are determined, use the remaining possible sums
视频讲解：
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文字解答：
Oscar 的得分为 44，所以他的牌必须是 11 和 33。在剩余牌中，和为 77 的唯一一对是 22 和 55，所以它们是 Aditi 的牌。
未使用的牌是 4,6,7,8,9,104,6,7,8,9,10。Tyrone 的得分 1616 只能来自 6+106+10 或 7+97+9。若 Tyrone 拿到 77 和 99，剩余的 4,6,8,104,6,8,10 无法配成 Ravon 的得分 1111。因此 Tyrone 拿到 66 和 1010。
剩余的牌是 4,7,8,94,7,8,9。Ravon 必须拿 44 和 77，得分才为 1111，留下 88 和 99 给 Kim，得分为 1717。所以 Ravon 拿到了 44 号牌。
所以答案是 C。
Oscar’s score of 44 forces his cards to be 11 and 3.3. From the remaining cards, the only pair with sum 77 is 22 and 5,5, so those are Aditi’s cards.
The unused cards are now 4,6,7,8,9,10.4,6,7,8,9,10. Tyrone’s score of 1616 can come only from 6+106+10 or 7+9.7+9. If Tyrone had 77 and 9,9, then the remaining cards 4,6,8,104,6,8,10 could not be paired to give Ravon’s score 11.11. Thus Tyrone has 66 and 10.10.
The remaining cards are 4,7,8,9.4,7,8,9. Ravon must have 44 and 77 to score 11,11, leaving 88 and 99 for Kim’s score of 17.17. Therefore, Ravon was given card 4.4.
Thus, the answer is C .

18.

反复掷一枚公平的 66 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 66-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?
1120\dfrac{1}{120}
132\dfrac{1}{32}
120\dfrac{1}{20}
320\dfrac{3}{20}
16\dfrac{1}{6}
答案：C
知识点：骰子（概率）基本概率
难度评级：1220
小提示：
暂时忽略重复掷出的结果，只看第一次出现的新点数类型。
Ignore repeated rolls until a new face type appears
大提示：
前三个不同的结果必须先是三个偶数，然后才出现任何奇数。
The first three distinct outcomes must be the three even numbers before any odd number appears
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 36\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 25\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 14\frac 14。
因此所求概率为 3⋅2⋅16⋅5⋅4=120。\dfrac{3\cdot 2\cdot 1}{6\cdot 5\cdot 4} = \dfrac 1{20}\text{。}
所以答案是 C。
The probability that the first number is even is 36.\frac 36.
The probability that the second distinct number is even is 25.\frac 25.
The probability that the third distinct number is even is 14.\frac 14.
The combined probability is 3⋅2⋅16⋅5⋅4=120.\dfrac{3\cdot 2\cdot 1}{6\cdot 5\cdot 4} = \dfrac 1{20}.
Thus, the answer is C .

19.

设 SS 是一个有限的正整数集合。
如果从 SS 中移除 SS 中最大的整数，剩余整数的平均值为 3232。如果再从 SS 中移除最小的整数，剩余整数的平均值为 3535。如果随后把最大的整数放回 SS，平均值升高到 4040。原集合 SS 中最大的整数比最小的整数大 7272。
集合 SS 中所有整数的平均值是多少？
Suppose that SS is a finite set of positive integers.
If the greatest integer in SS is removed from S,S, then the average value (arithmetic mean) of the integers remaining is 32.32. If the least integer in SS is also removed, then the average value of the integers remaining is 35.35. If the greatest integer is then returned to the set, the average value of the integers rises to 40.40. The greatest integer in the original set SS is 7272 greater than the least integer in S.S.
What is the average value of all the integers in the set S?S?
36.236.2
36.436.4
36.636.6
36.836.8
3737
答案：D
知识点：平均数方程组
难度评级：1540
小提示：
用总和、最小值、最大值和集合大小写出三个平均数方程。
Write equations for the three averages using the total sum, least value, greatest value, and set size
大提示：
将两个分母为 n−1n-1 的平均数方程相减，利用差为 7272。
Subtract the two averages with denominator n−1n-1 to use the difference 7272
视频讲解：
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文字解答：
设所有整数之和为 ss，最大数为 gg，最小数为 ll，集合 SS 的大小为 nn。
两个含 n−1n-1 个元素的集合的平均数给出 s−l=40(n−1),s−g=32(n−1)。 \begin{aligned} s-l&=40(n-1),\\ s-g&=32(n-1) \end{aligned}\text{。} 两式相减并利用 g−l=72g-l=72，得 72=8(n−1)72=8(n-1)，所以 n=10n=10。
同时去掉最大数和最小数后，还剩 88 个数，其和为 8⋅35=2808\cdot35=280。另一方面，只去掉最小数时剩下的和为 9⋅40=3609\cdot40=360，所以 g=360−280=80g=360-280=80。于是 l=80−72=8l=80-72=8，原来的总和为 s=280+80+8=368s=280+80+8=368。
平均值为 sn=36810=36.8\frac{s}{n}=\frac{368}{10}=36.8。
所以答案是 D。
Let the sum of all the integers be s,s, the greatest number be g,g, the least number be l,l, and the size of SS be n.n.
The two averages of sets with n−1n-1 elements give s−l=40(n−1),s−g=32(n−1). \begin{aligned} s-l&=40(n-1),\\ s-g&=32(n-1). \end{aligned} Subtracting and using g−l=72g-l=72 gives 72=8(n−1),72=8(n-1), so n=10.n=10.
Removing both extremes leaves 88 numbers with sum 8⋅35=280.8\cdot35=280. On the other hand, removing only the least number leaves a sum of 9⋅40=360,9\cdot40=360, so g=360−280=80.g=360-280=80. Hence l=80−72=8,l=80-72=8, and the original sum is s=280+80+8=368.s=280+80+8=368.
The required average is sn=36810=36.8.\frac{s}{n}=\frac{368}{10}=36.8.
Thus, the answer is D .

20.

下图由 1111 条线段构成，每条线段长度都为 22。五边形 ABCDEABCDE 的面积可写成 m+n\sqrt{m} + \sqrt{n}，其中 mm 和 nn 为正整数。m+nm + n 等于多少？
The figure below is constructed from 1111 line segments, each of which has length 2.2. The area of pentagon ABCDEABCDE can be written as m+n,\sqrt{m} + \sqrt{n}, where mm and nn are positive integers. What is m+n?m + n ?
2020
2121
2222
2323
2424
答案：D
知识点：面积分割等边三角形特殊直角三角形
难度评级：1950
小提示：
利用相等线段识别等边三角形的高。
Use the equal length segments to identify equilateral-triangle altitudes
大提示：
把五边形分成两侧部分和中间的等腰三角形。
Split the pentagon into two side pieces and the central isosceles triangle
解答：
设 FF 为与 A,BA,B 和 CC 都相连但未标字母的点。因为所有画出的线段长度都是 22，三角形 ABFABF 和 CBFCBF 是位于 BFBF 两侧的等边三角形。因此 ∠ABC=120∘\angle ABC=120^\circ，所以
[ABC]=12⋅2⋅2sin⁡120∘=3。[ABC]=\frac12\cdot2\cdot2\sin120^\circ=\sqrt3\text{。}
另一侧同理可得 [ADE]=3[ADE]=\sqrt3。此外，在 △ABC\triangle ABC 中应用余弦定理，得 AC2=12AC^2=12，同理 AD2=12AD^2=12。因此等腰三角形 ACDACD 到底边 CD=2CD=2 的高为
(12)2−12=11。\sqrt{(\sqrt{12})^2-1^2}=\sqrt{11}\text{。}
所以 [ACD]=12⋅2⋅11=11[ACD]=\frac12\cdot2\cdot\sqrt{11}=\sqrt{11}。五边形的总面积为
12+11，\sqrt{12}+\sqrt{11}\text{，}
所以 m+n=12+11=23m+n=12+11=23。
所以答案是 D。
Let FF be the unlabeled point joined to A,B,A,B, and C.C. Because all the drawn segments have length 2,2, triangles ABFABF and CBFCBF are equilateral and lie on opposite sides of BF.BF. Hence ∠ABC=120∘,\angle ABC=120^\circ, so
[ABC]=12⋅2⋅2sin⁡120∘=3.[ABC]=\frac12\cdot2\cdot2\sin120^\circ=\sqrt3.
The same reasoning on the other side gives [ADE]=3.[ADE]=\sqrt3. Also, the Law of Cosines in △ABC\triangle ABC gives AC2=12,AC^2=12, and similarly AD2=12.AD^2=12. Thus the altitude of isosceles triangle ACDACD to its base CD=2CD=2 is
(12)2−12=11.\sqrt{(\sqrt{12})^2-1^2}=\sqrt{11}.
Therefore [ACD]=12⋅2⋅11=11.[ACD]=\frac12\cdot2\cdot\sqrt{11}=\sqrt{11}. The pentagon’s total area is
12+11,\sqrt{12}+\sqrt{11},
so m+n=12+11=23.m+n=12+11=23.
Thus, the answer is D .

21.

一张正方形纸片边长为 11，顶点按顺序为 AA，BB，CC 和 DD。如图所示，将纸片折叠，使顶点 CC 落在边 AD‾\overline{AD} 上的点 C′C'，且边 BC‾\overline{BC} 与边 AB‾\overline{AB} 相交于点 EE。若 C′D=13C'D = \frac{1}{3}，三角形 △AEC′\triangle AEC' 的周长是多少？
A square piece of paper has side length 11 and vertices A,A, B,B, C,C, and DD in that order. As shown in the figure, the paper is folded so that vertex CC meets edge AD‾\overline{AD} at point C′,C', and edge BC‾\overline{BC} intersects edge AB‾\overline{AB} at point E.E. Suppose that C′D=13.C'D = \frac{1}{3}. What is the perimeter of △AEC′?\triangle AEC'?
22
1+2331+\dfrac{2}{3}\sqrt{3}
136\dfrac{13}{6}
1+3431 + \dfrac{3}{4}\sqrt{3}
73\dfrac{7}{3}
答案：A
知识点：折纸坐标几何
难度评级：2230
小提示：
折叠是反射，所以折痕是 CC′CC' 的垂直平分线。
A fold is a reflection, so the crease is the perpendicular bisector of CC′CC'
大提示：
从折叠后的直角三角形中求出 AEAE、AC′AC' 和 EC′EC'。
Find AE,AE, AC′,AC', and EC′EC' from the folded right triangle
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文字解答：
取坐标 A=(0,1)A=(0,1)、B=(0,0)B=(0,0)、C=(1,0)C=(1,0)、D=(1,1)D=(1,1)。因为 C′D=13C'D=\frac13，所以 C′=(23,1)C'=(\frac23,1)，从而 AC′=23AC'=\frac23。
折叠把 CC 反射到 C′C'，所以边 BCBC 的像是过 C′C' 和 EE 的直线。把 B=(0,0)B=(0,0) 关于 CC′CC' 的垂直平分线反射，得到 (−215,25)(-\frac{2}{15},\frac25)。过该点与 C′C' 的直线与 ABAB 相交于 E=(0,12)E=(0,\frac12)。
因此 AE=12AE=\frac12，并且
EC′=(23)2+(12)2=56。EC'=\sqrt{\left(\frac23\right)^2+\left(\frac12\right)^2}=\frac56\text{。}
△AEC′\triangle AEC' 的周长为
12+23+56=2。\frac12+\frac23+\frac56=2\text{。}
所以答案是 A。
Use coordinates with A=(0,1),A=(0,1), B=(0,0),B=(0,0), C=(1,0),C=(1,0), and D=(1,1).D=(1,1). Since C′D=13,C'D=\frac13, we have C′=(23,1),C'=(\frac23,1), so AC′=23.AC'=\frac23.
The fold reflects CC to C′,C', so the image of side BCBC is the line through C′C' and E.E. Reflecting B=(0,0)B=(0,0) across the perpendicular bisector of CC′CC' gives (−215,25).(-\frac{2}{15},\frac25). The line through this point and C′C' meets ABAB at E=(0,12).E=(0,\frac12).
Thus AE=12,AE=\frac12, and
EC′=(23)2+(12)2=56.EC'=\sqrt{\left(\frac23\right)^2+\left(\frac12\right)^2}=\frac56.
The perimeter of △AEC′\triangle AEC' is
12+23+56=2.\frac12+\frac23+\frac56=2.
Thus, the answer is A .

22.

Ang、Ben 和 Jasmin 每人都有 55 块积木，颜色分别为红、蓝、黄、白、绿；另有 55 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 33 块同色积木的概率为 mn\frac{m}{n}，其中 mm 与 nn 是互质正整数。m+nm + n 等于多少？
Ang, Ben, and Jasmin each have 55 blocks, colored red, blue, yellow, white, and green; and there are 55 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 33 blocks all of the same color is mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. What is m+n?m + n ?
4747
9494
227227
471471
542542
答案：D
知识点：容斥原理排列基本概率
难度评级：2150
小提示：
固定 Ang 的摆放方式，然后数 Ben 和 Jasmin 的排列。
Fix Ang’s placement and count Ben’s and Jasmin’s permutations
大提示：
对 Ben 和 Jasmin 都与 Ang 颜色匹配的盒子使用容斥原理。
Use inclusion-exclusion on the boxes where both other placements match Ang’s color
视频讲解：
解答视频缩略图
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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(5!)^2 个等可能的摆放对。
若指定 kk 个盒子都收到三块同色积木，则 Ben 和 Jasmin 在这 kk 个盒子中都必须与 Ang 的颜色相同。这可以用 ((5−k)!)2((5-k)!)^2 种方式完成。由容斥原理，成功的摆放对数为
(51)(4!)2−(52)(3!)2+(53)(2!)2−(54)(1!)2+(55)(0!)2。 \begin{aligned} &\binom51(4!)^2-\binom52(3!)^2 \\ &\quad {}+\binom53(2!)^2-\binom54(1!)^2 \\ &\quad {}+\binom55(0!)^2 \end{aligned}\text{。}
上式等于
2880−360+40−5+1=2556。2880-360+40-5+1=2556\text{。}
这等于
2556(5!)2=255614400=71400。\frac{2556}{(5!)^2}=\frac{2556}{14400}=\frac{71}{400}\text{。}
因此 m+n=71+400=471m+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(5!)^2 equally likely pairs of placements.
For a specified set of kk boxes to receive three blocks of the same color, both Ben and Jasmin must match Ang in those kk boxes. This can happen in ((5−k)!)2((5-k)!)^2 ways. By inclusion-exclusion, the number of successful placement pairs is
(51)(4!)2−(52)(3!)2+(53)(2!)2−(54)(1!)2+(55)(0!)2. \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=2556.2880-360+40-5+1=2556.
Therefore the probability is
2556(5!)2=255614400=71400.\frac{2556}{(5!)^2}=\frac{2556}{14400}=\frac{71}{400}.
Thus m+n=71+400=471.m+n=71+400=471.
Thus, the answer is D .

23.

一个边长为 88 的正方形中，除了 44 个角上腿长为 22 的阴影等腰直角三角形，以及正方形中心一个边长为 222\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 11 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 1196(a+b2+π)\frac{1}{196}\left(a+b\sqrt{2}+\pi\right)，其中 aa 与 bb 是正整数。a+ba+b 等于多少？
A square with side length 88 is colored white except for 44 black isosceles right triangular regions with legs of length 22 in each corner of the square and a black diamond with side length 222\sqrt{2} in the center of the square, as shown in the diagram.
A circular coin with diameter 11 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 black region of the square can be written as 1196(a+b2+π),\frac{1}{196}\left(a+b\sqrt{2}+\pi\right), where aa and bb are positive integers. What is a+b?a+b?
6464
6666
6868
7070
7272
答案：C
知识点：几何概率面积分割
难度评级：2390
小提示：
硬币圆心的可选区域是一个 7×77\times7 的正方形。
The coin center ranges over a 7×77\times7 square
大提示：
计算距离阴影角三角形和中心阴影菱形不超过 12\frac{1}{2} 的面积。
Count the area within distance 12\frac{1}{2} of the shaded corner triangles and the shaded center diamond
解答：
硬币半径为 12\frac12，所以它的圆心均匀分布在一个 7×77\times7 的正方形内，该区域面积为 4949。
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 12\frac12 的点集。每个角对应一个高为 1+22\frac{1+\sqrt2}{2} 的等腰直角三角形，所以面积为
(1+22)2=3+224。\left(\frac{1+\sqrt2}{2}\right)^2=\frac{3+2\sqrt2}{4}\text{。}
四个角合计贡献 3+223+2\sqrt2。
中心阴影菱形是一个边长为 222\sqrt2 的正方形。把它向外扩张距离 12\frac12，除了菱形本身面积 88，还增加总面积 424\sqrt2 的四个矩形，以及合起来面积为 π4\frac\pi4 的四个四分之一圆。因此中心部分贡献
8+42+π4。8+4\sqrt2+\frac\pi4\text{。}
有利面积为
3+22+8+42+π4=11+62+π4。 \begin{aligned} &3+2\sqrt2+8+4\sqrt2+\frac\pi4 \\ &=11+6\sqrt2+\frac\pi4 \end{aligned}\text{。}
概率为
11+62+π449=44+242+π196。 \begin{aligned} &\frac{11+6\sqrt2+\frac\pi4}{49} \\ &=\frac{44+24\sqrt2+\pi}{196} \end{aligned}\text{。}
所以 a+b=44+24=68a+b=44+24=68。
所以答案是 C。
The coin has radius 12,\frac12, so its center is uniformly distributed over a 7×77\times7 square of area 49.49.
A shaded corner triangle contributes the set of center positions within distance 12\frac12 of that triangle, inside the allowed center square. For each corner this is a right isosceles triangle whose altitude is 1+22,\frac{1+\sqrt2}{2}, so its area is
(1+22)2=3+224.\left(\frac{1+\sqrt2}{2}\right)^2=\frac{3+2\sqrt2}{4}.
All four corners contribute 3+22.3+2\sqrt2.
The center shaded diamond is a square of side 22.2\sqrt2. Expanding it by distance 12\frac12 adds four rectangles of total area 424\sqrt2 and four quarter-circles of total area π4,\frac\pi4, in addition to the diamond’s area 8.8. Thus the center contribution is
8+42+π4.8+4\sqrt2+\frac\pi4.
The favorable area is
3+22+8+42+π4=11+62+π4. \begin{aligned} &3+2\sqrt2+8+4\sqrt2+\frac\pi4 \\ &=11+6\sqrt2+\frac\pi4. \end{aligned}
The probability is
11+62+π449=44+242+π196. \begin{aligned} &\frac{11+6\sqrt2+\frac\pi4}{49} \\ &=\frac{44+24\sqrt2+\pi}{196}. \end{aligned}
So a+b=44+24=68.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 44 和 22 的一组墙，经过一步可以变成以下任意一种：(3,2)(3,2)，(2,1,2)(2,1,2)，(4)(4)，(4,1)(4,1)，(2,2)(2,2)，或 (1,1,2)(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 44 and 22 can be changed into any of the following by one move: (3,2),(3,2), (2,1,2),(2,1,2), (4),(4), (4,1),(4,1), (2,2),(2,2), or (1,1,2).(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)(6,1,1)
(6,2,1)(6,2,1)
(6,2,2)(6,2,2)
(6,3,1)(6,3,1)
(6,3,2)(6,3,2)
答案：B
知识点：组合游戏
难度评级：2390
小提示：
计算每个相关单独墙长的游戏值。
Compute the game value for a single wall of each relevant size
大提示：
当所有墙的游戏值按位异或为 00 时，轮到行动的玩家处于必败状态。
A position is losing for the player to move exactly when the xor of the wall values is 00
解答：
对长度为 nn 的单独一堵墙，根据所有可能操作计算它的 Sprague-Grundy 值。这里用到的各个墙长的值为
g(1)=1,g(2)=2,g(3)=3,g(4)=1,g(5)=4,g(6)=3。 \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}\text{。}
多堵墙的局面在这些值按位异或为 00 时，正好是轮到行动者的必败局面。逐一计算选项：
(6,1,1):3⊕1⊕1=3，(6,1,1): 3\oplus1\oplus1=3\text{，}
(6,2,1):3⊕2⊕1=0，(6,2,1): 3\oplus2\oplus1=0\text{，}
(6,2,2):3⊕2⊕2=3，(6,2,2): 3\oplus2\oplus2=3\text{，}
(6,3,1):3⊕3⊕1=1，(6,3,1): 3\oplus3\oplus1=1\text{，}
(6,3,2):3⊕3⊕2=2。(6,3,2): 3\oplus3\oplus2=2\text{。}
只有 (6,2,1)(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,n, compute its Sprague-Grundy value from the possible moves. For the wall lengths needed here, the values are
g(1)=1,g(2)=2,g(3)=3,g(4)=1,g(5)=4,g(6)=3. \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 0.0. Evaluating the choices gives
(6,1,1):3⊕1⊕1=3,(6,1,1): 3\oplus1\oplus1=3,
(6,2,1):3⊕2⊕1=0,(6,2,1): 3\oplus2\oplus1=0,
(6,2,2):3⊕2⊕2=3,(6,2,2): 3\oplus2\oplus2=3,
(6,3,1):3⊕3⊕1=1,(6,3,1): 3\oplus3\oplus1=1,
(6,3,2):3⊕3⊕2=2.(6,3,2): 3\oplus3\oplus2=2.
Only (6,2,1)(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 SS 是坐标平面中的格点集合，其中两个坐标都是从 11 到 3030 的整数。恰有 300300 个 SS 中的点位于直线 y=mxy=mx 上或其下方。所有可能的 mm 值构成一个长度为 ab\frac ab 的区间，其中 aa 与 bb 是互质正整数。a+ba+b 等于多少？
Let SS be the set of lattice points in the coordinate plane, both of whose coordinates are integers between 11 and 30,30, inclusive. Exactly 300300 points in SS lie on or below a line with equation y=mx.y=mx. The possible values of mm lie in an interval of length ab,\frac ab, where aa and bb are relatively prime positive integers. What is a+b?a+b?
3131
4747
6262
7272
8585
答案：E
知识点：格点取整函数
难度评级：2390
小提示：
在相关的斜率附近，点数为 ∑x=130⌊mx⌋\sum_{x=1}^{30}\lfloor mx\rfloor。
Near the relevant slopes, the number of points is ∑x=130⌊mx⌋\sum_{x=1}^{30}\lfloor mx\rfloor
大提示：
找到第 300300 个格点比值对应的斜率，以及下一个更大的可能比值。
Find the slope where the 300300th lattice-point ratio occurs and the next larger possible ratio
解答：
对固定斜率 mm，集合 SS 中位于 y=mxy=mx 上或下方的点数为
∑x=130⌊mx⌋，\sum_{x=1}^{30}\lfloor mx\rfloor\text{，}
这适用于答案附近的斜率。
当 m=23m=\frac23 时，把 x=3k+1,3k+2,3k+3x=3k+1,3k+2,3k+3 分组，其中 k=0,1,…,9k=0,1,\ldots,9。于是
⌊2x3⌋=2k, 2k+1, 2k+2，\lfloor \frac{2x}{3}\rfloor=2k,\ 2k+1, \ 2k+2\text{，}
每个区块上的和为 6k+36k+3。因此总数为
∑k=09(6k+3)=270+30=300。\sum_{k=0}^9(6k+3)=270+30=300\text{。}
若 m<23m<\frac23，十个比值 yx=23\frac{y}{x}=\frac{2}{3} 的点不再被计入，所以点数小于 300300。因此区间下端为 23\frac23。
下一个大于 23\frac23 的可能比值 yx\frac{y}{x}，在 1≤x,y≤301\le x,y\le30 下可按 xx 模 33 检查。最佳候选为
1928,2029,2130=710，\frac{19}{28},\qquad \frac{20}{29},\qquad \frac{21}{30}=\frac{7}{10}\text{，}
其中最小的是 1928\frac{19}{28}。因此区间长度为
1928−23=184。\frac{19}{28}-\frac23=\frac1{84}\text{。}
所以 a+b=1+84=85a+b=1+84=85。
所以答案是 E。
For a fixed slope m,m, the number of points in SS on or below y=mxy=mx is
∑x=130⌊mx⌋,\sum_{x=1}^{30}\lfloor mx\rfloor,
for the slopes near the answer.
At m=23,m=\frac23, grouping x=3k+1,3k+2,3k+3x=3k+1,3k+2,3k+3 for k=0,1,…,9k=0,1,\ldots,9 gives
⌊2x3⌋=2k, 2k+1, 2k+2,\lfloor \frac{2x}{3}\rfloor=2k,\ 2k+1, \ 2k+2,
whose sum over each block is 6k+3.6k+3. Thus the total is
∑k=09(6k+3)=270+30=300.\sum_{k=0}^9(6k+3)=270+30=300.
If m<23,m<\frac23, the ten points with ratios yx=23\frac{y}{x}=\frac{2}{3} are no longer counted, so the count is less than 300.300. Therefore the lower end is 23.\frac23.
The next possible ratio yx\frac{y}{x} greater than 23,\frac23, with 1≤x,y≤30,1\le x,y\le30, is minimized by checking xx modulo 3.3. The best candidates are
1928,2029,2130=710,\frac{19}{28},\qquad \frac{20}{29},\qquad \frac{21}{30}=\frac{7}{10},
and the smallest is 1928.\frac{19}{28}. Hence the interval length is
1928−23=184.\frac{19}{28}-\frac23=\frac1{84}.
Thus a+b=1+84=85.a+b=1+84=85.
Thus, the answer is E .