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2018 AMC 8 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.

一个游乐园收藏了全国各地建筑和景观的比例模型，比例为 1:201: 20。美国国会大厦高 289289 英尺。这个公园中它的复制模型高多少英尺？答案四舍五入到最接近的整数。
An amusement park has a collection of scale models, with a ratio of 1:20, 1: 20, of buildings and other sights from around the country. The height of the United States Capitol is 289289 feet. What is the height in feet of its replica at this park, rounded to the nearest whole number?
1414
1515
1616
1818
2020
知识点：比与比例估算
难度评级：370
小提示：
1:201:20 的比例表示模型高度是真实高度除以 2020
A 1:201:20 scale means the replica height is the real height divided by 2020.
大提示：
计算 289289 除以 2020，再四舍五入到最接近的整数英尺
After dividing 289289 by 2020, round to the nearest whole foot.
视频讲解：
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文字解答：
复制模型的高度是 28920=14.45\dfrac{289}{20}=14.45 英尺，四舍五入得到 1414 英尺。
所以正确答案是 A。
The replica is 28920=14.45\dfrac{289}{20}=14.45 feet tall, which rounds to 1414 feet.
Thus, the correct answer is A.

2.

下列乘积的值是多少？(1+11)⋅(1+12)⋅(1+13)⋅(1+14)⋅(1+15)⋅(1+16)？\begin{align*} &\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\\ &\quad{}\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\\ &\quad{}\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right) \end{align*}\text{？}
What is the value of the product (1+11)⋅(1+12)⋅(1+13)⋅(1+14)⋅(1+15)⋅(1+16)?\begin{align*} &\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\\ &\quad{}\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\\ &\quad{}\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)? \end{align*}
76\dfrac{7}{6}
43\dfrac{4}{3}
72\dfrac{7}{2}
77
88
知识点：裂项相消分数
难度评级：660
小提示：
把每个因子 1+1n1+\frac1n 改写成 n+1n\frac{n+1}{n}
Rewrite each factor 1+1n1+\frac1n as n+1n\frac{n+1}{n}.
大提示：
改写后相邻分子和分母会相消
The product telescopes after the fractions are rewritten.
视频讲解：
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文字解答：
先注意，形如 1+1n1 + \frac{1}{n} 的因子可以改写为 nn+1n=n+1n。\frac{n}{n} + \frac{1}{n} = \frac{n+1}{n}\text{。}因此题中的乘积可以写成 (1+11)⋅(1+12)⋅(1+13)⋅(1+14)⋅(1+15)⋅(1+16)=21⋅32⋅43⋅54⋅65⋅76=7\begin{align*} &\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\\ &\quad{}\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\\ &\quad{}\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)\\ &=\dfrac{2}{1} \cdot \dfrac{3}{2} \cdot \dfrac{4}{3} \cdot\dfrac{5}{4} \cdot\dfrac{6}{5} \cdot\dfrac{7}{6}\\ &=7 \end{align*} 所以正确答案是 D。
Let’s first note that if we are given an expression of the form 1+1n,1 + \frac{1}{n}, we can rewrite this as nn+1n=n+1n.\frac{n}{n} + \frac{1}{n} = \frac{n+1}{n}. With that in mind, we can rewrite the expression given to us in the problem, as shown below: (1+11)⋅(1+12)⋅(1+13)⋅(1+14)⋅(1+15)⋅(1+16)=21⋅32⋅43⋅54⋅65⋅76=7\begin{align*} &\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\\ &\quad{}\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\\ &\quad{}\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)\\ &=\dfrac{2}{1} \cdot \dfrac{3}{2} \cdot \dfrac{4}{3} \cdot\dfrac{5}{4} \cdot\dfrac{6}{5} \cdot\dfrac{7}{6}\\ &=7 \end{align*} Thus, the correct answer is D.

3.

学生 Arn、Bob、Cyd、Dan、Eve 和 Fon 按这个顺序围成一圈。他们开始报数：Arn 先报，然后 Bob，以此类推。当一个数含有数字 77（例如 4747）或者是 77 的倍数时，报到这个数的人离开圆圈，报数继续。最后留在圆圈里的是谁？
Students Arn, Bob, Cyd, Dan, Eve, and Fon are arranged in that order in a circle. They start counting: Arn first, then Bob, and so forth. When the number contains a 77 as a digit (such as 4747) or is a multiple of 77 that person leaves the circle and the counting continues. Who is the last one present in the circle?
Arn\text{Arn}
Bob\text{Bob}
Cyd\text{Cyd}
Dan\text{Dan}
Eve\text{Eve}
知识点：整除性数字过程模拟
难度评级：1070
小提示：
只模拟会让人离开的那些数
Simulate only the numbers that cause someone to leave.
大提示：
每次有人离开后仍按原来的顺序继续报数，下一个数由圆圈中下一位还在的人报出
Keep the counting order after each person leaves; the next number goes to the next person still in the circle.
视频讲解：
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文字解答：
最先会让人离开的五个数是 7,14,17,21,277,14,17,21,27。只跟踪这些轮次即可。
Arn 报到 77，Cyd 报到 1414，Fon 报到 1717，Bob 报到 2121，Eve 报到 2727。
Dan 是唯一留在圆圈里的学生，所以正确答案是 D。
The first five removal numbers are 7,14,17,21,27.7,14,17,21,27. Tracking only those turns gives:
Arn says 77, Cyd says 1414, Fon says 1717, Bob says 2121, and Eve says 2727.
Dan is the only student remaining, so D is the correct answer.

4.

图中的十二边形画在 1 cm×1 cm1 \text{ cm}\times 1 \text{ cm} 的方格纸上。该图形的面积是多少 cm2\text{cm}^2？
The twelve-sided figure shown has been drawn on 1 cm×1 cm1 \text{ cm}\times 1 \text{ cm} graph paper. What is the area of the figure in cm2?\text{cm}^2?
1212
12.512.5
1313
13.513.5
1414
知识点：面积分割三角形面积
难度评级：770
小提示：
可以把图形放进一个便于计算的长方形中，或者把它分成中间的正方形和周围的小三角形
Enclose the figure in an easy rectangle or split it into a central square and small triangles.
大提示：
斜边部分形成四个全等的直角三角形，每个面积为 11
The slanted parts form four congruent right triangles of area 11.
视频讲解：
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文字解答：
如图所示，可以把图形分解：
现在可以看出，中间是一个 3×33 \times 3 正方形，周围有 44 个较小的阴影三角形。
正方形面积为 3⋅3=93 \cdot 3 = 9。每个小三角形的底为 22，高为 11，所以面积为 bh2=2⋅12=1。\dfrac{bh}{2} = \dfrac{2\cdot 1}{2} =1\text{。}这样的三角形有 44 个，所以它们的总面积为 1⋅4=41\cdot 4 = 4。
因此总面积是 9+4=139+4 = 13。
所以正确答案是 C。
To solve for the area of the figure, we separate the compound shape into parts that are easier to work with, as such:
As is now clear, there is the center 3×33 \times 3 square, with 44 smaller shaded triangles surrounding it.
The area of the square is 3⋅3=9.3 \cdot 3 = 9. The other triangles each have a base of 22 and a height of 1,1, so their area is equal to bh2=2⋅12=1.\dfrac{bh}{2} = \dfrac{2\cdot 1}{2} =1 . There are 44 of these triangles, so their total area is 1⋅4=4.1\cdot 4 = 4.
Therefore, the total area is 9+4=13.9+4 = 13.
Thus, the correct answer is C.

5.

下列表达式的值是多少？1+3+5+⋯+2017+2019−2−4−6−⋯−2016−2018？\begin{align*} &1+3+5+\cdots+2017+2019 \\ -&2-4-6-\cdots-2016-2018 \end{align*}\text{？}
What is the value of 1+3+5+⋯+2017+2019−2−4−6−⋯−2016−2018?\begin{align*} &1+3+5+\cdots+2017+2019 \\ -&2-4-6-\cdots-2016-2018? \end{align*}
−1010-1010
−1009-1009
10081008
10091009
10101010
知识点：等差数列配对与分组
难度评级：870
小提示：
把每个 11 之后的正奇数与它前面的偶数配对
Pair each positive odd number after 11 with the even number just before it.
大提示：
配对之后，数一数还剩下多少个 11
Count how many 11’s remain after pairing.
视频讲解：
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文字解答：
重新分组可得 1+(3−2)+(5−4)+⋯+(2017−2016)+(2019−2018)。\begin{align*}&1 + (3-2) + (5-4) + \cdots +\\ &(2017-2016) + (2019-2018) \end{align*}\text{。}每一项都等于 11，而这样的项共有 2019−12+1=1010\frac{2019-1}2+1 = 1010 项，所以总和为 1010⋅1=10101010\cdot1 = 1010。
所以正确答案是 E。
Rearranging the terms, notice that the expression in the question is equal to: 1+(3−2)+(5−4)+⋯+(2017−2016)+(2019−2018).\begin{align*}&1 + (3-2) + (5-4) + \cdots +\\ &(2017-2016) + (2019-2018). \end{align*} Each term is equal to 1,1, and there are 2019−12+1=1010\frac{2019-1}2+1 = 1010 terms, so the total sum is 1010⋅1=1010.1010\cdot1 = 1010.
Thus, E is the correct answer.

6.

Anh 去海滩旅行时，在高速公路上行驶了 5050 英里，在海边通道上行驶了 1010 英里。他在高速公路上的速度是海边通道上的三倍。如果 Anh 在海边通道上开了 3030 分钟，那么他的整趟行程用了多少分钟？
On a trip to the beach, Anh traveled 5050 miles on the highway and 1010 miles on a coastal access road. He drove three times as fast on the highway as on the coastal road. If Anh spent 3030 minutes driving on the coastal road, how many minutes did his entire trip take?
5050
7070
8080
9090
100100
知识点：路程、速度与时间单位换算
难度评级：900
小提示：
用海边通道的距离和时间先求出海边通道上的速度
Use the coastal-road information to find Anh’s coastal-road speed.
大提示：
高速公路上的速度是海边通道上速度的三倍，再由高速公路上的 5050 英里求出所用时间
The highway speed is three times the coastal-road speed, so the highway time follows from the 5050 highway miles.
视频讲解：
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文字解答：
Anh 在海边通道上开了 1010 英里，用了 3030 分钟。因此海边通道速度（记为 vcv_c）是 vc=1030=13，\begin{align*}v_c&=\dfrac{10}{30}\\ &=\dfrac13\end{align*}\text{，}也就是每分钟 13\dfrac13 英里。因为他在高速公路上的速度是 33 倍，也就是 vh=3vcv_h=3v_c，所以高速公路速度为 3⋅13=13\cdot \dfrac13 = 1 英里每分钟。他在海边通道上开了 3030 分钟，在高速公路上开了 5050 英里，速度为每分钟 11 英里，所以这 5050 英里需要 5050 分钟。
因此总行程时间为 50+30=8050+30=80 分钟。
所以正确答案是 C。
Anh drove 1010 miles on the coastal road in 3030 minutes. Therefore, his speed on the coastal road (notated as vcv_c) is vc=1030=13.\begin{align*}v_c&=\dfrac{10}{30}\\ &=\dfrac13.\end{align*} This is 13\dfrac13 mile per minute. Since he drives 33 times as fast on the highway (i.e. vh=3vcv_h=3v_c), his highway speed is 3⋅13=13\cdot \dfrac13 = 1 mile per minute. Armed with these two facts, we know that Anh drove for 3030 minutes on the coastal road, and he drove 5050 miles at 11 mile per minute. This means it takes 5050 minutes to drive the 5050 miles on the highway.
As such, the total travel time is 50+30=8050+30=80 minutes.
Thus, the correct answer is C.

7.

55 位数 2‾\underline{2} 0‾\underline{0} 1‾\underline{1} 8‾\underline{8} U‾\underline{U} 能被 99 整除。这个数除以 88 的余数是多少？
The 55-digit number 2‾\underline{2} 0‾\underline{0} 1‾\underline{1} 8‾\underline{8} U‾\underline{U} is divisible by 9.9. What is the remainder when this number is divided by 8?8?
11
33
55
66
77
知识点：整除性模运算
难度评级：960
小提示：
先用 99 的整除规则确定 UU
Use the divisibility-by-99 rule to determine UU.
大提示：
UU 确定后，求模 88 的余数只需要看最后三位
Once UU is known, only the last three digits matter for the remainder modulo 88.
视频讲解：
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文字解答：
注意，一个数能被 99 整除，当且仅当它的各位数字之和也能被 99 整除。
题中这个 55 位数的数字和为 2+0+1+8+U=11+U。2+0+1+8+U= 11+U\text{。}因为 2‾\underline{2} 0‾\underline{0} 1‾\underline{1} 8‾\underline{8} U‾\underline{U} 能被 99 整除，11+U11+U 也必须被 99 整除。又因为 UU 是一位数字，0≤U≤90\le U\le 9，所以 UU 只能是 77。
现在可知这个 55 位数是 2018720187，而我们要求 2018720187 除以 88 的余数。长除法给出 20187=2523⋅8+320187=2523\cdot 8 + 3，所以余数是 33。
所以正确答案是 B。
Notice that a number is divisible by 99 if and only if the sum of its digits is also divisible by 9.9.
The sum of the digits of the 55-digit number in the problem is: 2+0+1+8+U=11+U.2+0+1+8+U= 11+U. As 2‾\underline{2} 0‾\underline{0} 1‾\underline{1} 8‾\underline{8} U‾\underline{U} is divisible by 9,9, 11+U11+U must also be divisible by 9.9. Also, as UU is a digit, we know that 0≤U≤9.0\le U\le 9. This means that UU can only be 7.7.
Now we know that the 55-digit number in question is 20187,20187, and we want to find the remainder when we divide 2018720187 by 8.8. To solve this, simply use long division to see that 20187=2523⋅8+3.20187=2523\cdot 8 + 3. Therefore, the remainder is 3.3.
Thus, the correct answer is B.

8.

Garcia 老师询问健康课的学生上周有多少天至少锻炼了 3030 分钟。结果汇总在下面的条形图中，柱子的高度表示学生人数。
Garcia 老师班上学生报告的上周锻炼天数的平均数是多少？答案四舍五入到最接近的百分之一。
Mr. Garcia asked the members of his health class how many days last week they exercised for at least 3030 minutes. The results are summarized in the following bar graph, where the heights of the bars represent the number of students.
What was the mean number of days of exercise last week, rounded to the nearest hundredth, reported by the students in Mr. Garcia’s class?
3.503.50
3.573.57
4.364.36
4.504.50
5.005.00
知识点：平均数数据与图表解读
难度评级：960
小提示：
根据条形图高度计算加权平均数
Compute the weighted average from the bar heights.
大提示：
先求学生总数，再求报告的锻炼天数总和
First find the total number of students, then find the total number of reported exercise days.
视频讲解：
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文字解答：
从条形图可知，锻炼 1,2,3,4,5,6,71,2,3,4,5,6,7 天的学生人数依次为 1,3,2,6,8,3,21,3,2,6,8,3,2，学生总数为 2525。
学生报告的锻炼天数总和为 1⋅1+2⋅3+3⋅2+4⋅6+5⋅8+6⋅3+7⋅2=109。\begin{aligned} &1\cdot1+2\cdot3+3\cdot2+4\cdot6\\ &\qquad+5\cdot8+6\cdot3+7\cdot2=109 \end{aligned}\text{。}
平均数为 10925=4.36\frac{109}{25}=4.36 天。
所以正确答案是 C。
The bar heights for 1,2,3,4,5,6,71,2,3,4,5,6,7 days are 1,3,2,6,8,3,21,3,2,6,8,3,2, for a total of 2525 students.
The total number of reported exercise days is 1⋅1+2⋅3+3⋅2+4⋅6+5⋅8+6⋅3+7⋅2=109.\begin{aligned} &1\cdot1+2\cdot3+3\cdot2+4\cdot6\\ &\qquad+5\cdot8+6\cdot3+7\cdot2=109. \end{aligned}
The mean is 10925=4.36\frac{109}{25}=4.36 days.
Thus, C is the correct answer.

9.

Tyler 正在给他 1212 英尺乘 1616 英尺的客厅铺地砖。他计划沿房间边缘用一英尺乘一英尺的正方形地砖铺一圈边框，其余部分用二英尺乘二英尺的正方形地砖铺满。他一共会用多少块地砖？
Tyler is tiling the floor of his 1212 foot by 1616 foot living room. He plans to place one-foot by one-foot square tiles to form a border along the edges of the room and to fill in the rest of the floor with two-foot by two-foot square tiles. How many tiles will he use?
4848
8787
9191
9696
120120
知识点：铺砖面积
难度评级：1100
小提示：
先数 11 英尺宽边框上的地砖，注意角落不要重复计算
Count the 11-foot border tiles first, being careful not to double-count the corners.
大提示：
去掉一圈 11 英尺边框后，剩余矩形是 1010 英尺乘 1414 英尺
After removing a 11-foot border, the remaining rectangle is 1010 feet by 1414 feet.
视频讲解：
解答视频缩略图
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文字解答：
注意边框每平方英尺需要一块地砖，所以四边相加会得到 16+12+16+12=5616+12+16+12=56 块。不过这样四个角各重复计算了一次，所以要减去 44。因此边框共需要 56−4=5256-4=52 块 1×11\times 1 正方形地砖。
由于每边都去掉了一英尺宽的边框，剩下的内部矩形为 1010 英尺乘 1414 英尺。它要完全铺上 2×22 \times 2 的地砖，所以这部分需要 10⋅142⋅2=35 \dfrac{10\cdot14}{2\cdot2} = 35 块地砖。
边框需要 5252 块 1×11\times 1 地砖，内部需要 3535 块 2×22\times 2 地砖，因此一共需要 52+35=8752+35=87 块地砖。
所以正确答案是 B。
Note that each square foot of the border would require one tile, meaning that the border will take 16+12+16+12=5616+12+16+12=56 tiles. However, notice that this will cause overlapping tiles in each of the four corners, so to fix this, we subtract 4.4. Therefore, the border will take 56−4=5256-4=52 1×11\times 1 square tiles to completely tile.
Since we have removed one foot from each side due to the border, the remaining rectangle is 1010 feet by 1414 feet. This must be tiled completely by 2×22 \times 2 tiles, so it will take 10⋅142⋅2=35 \dfrac{10\cdot14}{2\cdot2} = 35 tiles in total to tile this area.
As it takes 5252 1×11\times 1 square tiles to tile the border, and 3535 2×22\times 2 square tiles to tile the remaining area, it will take 52+35=8752+35=87 tiles in total to fill in Tyler’s entire living room floor.
Thus, the correct answer is B.

10.

一组非零数的调和平均数定义为这些数的倒数的平均数再取倒数。11、22 和 44 的调和平均数是多少？
The harmonic mean of a set of non-zero numbers is the reciprocal of the average of the reciprocals of the numbers. What is the harmonic mean of 1,1, 2,2, and 4?4?
37\dfrac{3}{7}
712\dfrac{7}{12}
127\dfrac{12}{7}
74\dfrac{7}{4}
73\dfrac{7}{3}
知识点：调和平均数分数
难度评级：900
小提示：
先求 1,21,2 和 44 的倒数的平均数
Average the reciprocals of 1,2,1,2, and 44.
大提示：
调和平均数就是这个平均数的倒数
The harmonic mean is the reciprocal of that average.
视频讲解：
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文字解答：
11、22 和 44 的倒数分别是 11\dfrac11、12\dfrac12 和 14\dfrac14。它们的平均数为 (1+12+14)3=(74)3=712。\begin{align*}\dfrac{\left(1 + \dfrac12 + \dfrac14\right)}{3} &= \dfrac{\left(\dfrac{7}{4}\right)}{3} \\&= \dfrac{7}{12} \end{align*}\text{。}
调和平均数是这些倒数的平均数的倒数；刚才算出的平均数是 712\dfrac{7}{12}，所以调和平均数为 127\dfrac{12}{7}。
正确答案是 C。
The reciprocals of 11, 22, and 44 are 11\dfrac11, 12\dfrac12, and 14\dfrac14, respectively. The average of these reciprocals is (1+12+14)3=(74)3=712.\begin{align*}\dfrac{\left(1 + \dfrac12 + \dfrac14\right)}{3} &= \dfrac{\left(\dfrac{7}{4}\right)}{3} \\&= \dfrac{7}{12}. \end{align*}
As the harmonic mean is the reciprocal of the average of the reciprocals of the numbers (which we just calculated to be 712\dfrac{7}{12}), we conclude that the harmonic mean is 127.\dfrac{12}{7}.
Thus, the correct answer is C.

11.

Abby、Bridget 和另外四位同学将坐成两排、每排三人来拍集体照，如下图所示。XXXXXX\begin{array}{ccc} \text{\text{X}}&\text{X}&\text{X} \\ \text{X}&\text{X}&\text{X} \end{array}如果座位随机分配，那么 Abby 和 Bridget 在同一行或同一列相邻的概率是多少？
Abby, Bridget, and four of their classmates will be seated in two rows of three for a group picture, as shown. XXXXXX\begin{array}{ccc} \text{\text{X}}&\text{X}&\text{X} \\ \text{X}&\text{X}&\text{X} \end{array} If the seating positions are assigned randomly, what is the probability that Abby and Bridget are adjacent to each other in the same row or the same column?
13\dfrac{1}{3}
25\dfrac{2}{5}
715\dfrac{7}{15}
12\dfrac{1}{2}
23\dfrac{2}{3}
知识点：基本概率分类讨论
难度评级：1210
小提示：
先固定 Abby 的座位，再数与它相邻的座位数
First choose Abby’s seat, then count how many seats are adjacent to it.
大提示：
中间列的座位和角落座位拥有的相邻座位数不同
Middle seats and corner seats have different numbers of adjacent seats.
视频讲解：
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文字解答：
把情况分成两类：第 11 类是 Abby 坐在中间列的两个座位之一，第 22 类是她坐在外侧的 44 个座位之一。
Abby 坐在中间列的概率是 26=13\dfrac26 = \dfrac13，这就是第 11 类。此时 Bridget 要与 Abby 相邻，可以坐在 Abby 左右两个座位中的一个，或者坐在同一列的座位，共有 33 个可选位置；剩下共有 55 个空位，所以条件概率是 35\frac35。因此这一类的概率为 13⋅35=315。 \dfrac13 \cdot \dfrac35 = \dfrac{3}{15}\text{。}
Abby 坐在外侧的概率是 46=23\dfrac46 = \dfrac23，这就是第 22 类。此时 Bridget 要与 Abby 相邻，只能坐在同一行相邻的那个座位，或同一列的座位，共有 22 个可选位置；剩下仍有 55 个空位，所以条件概率是 25\frac25。因此这一类的概率为 23⋅25=415。 \frac23 \cdot \frac25 = \frac{4}{15}\text{。}
因此任一类发生的总概率为 315+415=715\dfrac{3}{15} + \dfrac{4}{15} = \dfrac{7}{15} 。
所以正确答案是 C。
We can split the problem into two cases. In case 1,1, Abby is in one of the middle two seats, and in case 2,2, she is in one of the outer 44 seats.
Firstly notice that there is a 26=13 \dfrac26 = \dfrac13 probability of case 11 being true (i.e. Abby is in the middle two seats). For Bridget to be adjacent to Abby in this case, she must be in either of the two seats beside Abby in the same row, or she is in the same column as her. There are 33 ways to make this happen out of a possible 55 open seats, so there is a 35 \frac35 chance of this happening. Therefore, the total probability of this case is 13⋅35=315. \dfrac13 \cdot \dfrac35 = \dfrac{3}{15} .
Next, notice that there is a 46=23 \dfrac46 = \dfrac23 probability of case 22 being true (i.e. Abby is in the outer four seats). For Bridget to be adjacent to Abby in this case, she must either be in the single seat next to Abby in the same row, or she is in the same column as Abby. There are 22 ways to make this happen out of a possible 55 open seats, so there is a 25 \frac25 chance of this happening. Therefore, the total probability of this case is 23⋅25=415. \frac23 \cdot \frac25 = \frac{4}{15} .
Therefore, the final probability of either of these cases happening is 315+415=715. \dfrac{3}{15} + \dfrac{4}{15} = \dfrac{7}{15} .
Thus, C is the correct answer.

12.

Sri 车里的钟不准，并且以恒定速度走快。某天他开始购物时，车钟和他准确的手表都显示中午 12:0012{:}00 整。购物结束时，他的手表显示 12:3012{:}30，车钟显示 12:3512{:}35。当天晚些时候，Sri 弄丢了手表。他看车钟，显示 7:007{:}00 整。实际时间是多少？
The clock in Sri’s car, which is not accurate, gains time at a constant rate. One day as he begins shopping he notes that his car clock and his watch (which is accurate) both say 12:0012{:}00 noon. When he is done shopping, his watch says 12:3012{:}30 and his car clock says 12:35.12{:}35. Later that day, Sri loses his watch. He looks at his car clock and it says 7:00.7{:}00. What is the actual time?
5:505:50
6:006:00
6:306:30
6:556:55
8:108:10
知识点：时钟比与比例
难度评级：1140
小提示：
比较车钟多走的时间和真实经过的时间
Compare how much time the car clock gains to how much real time passes.
大提示：
车钟的 3535 分钟对应真实的 3030 分钟
The car clock’s 3535 minutes correspond to 3030 actual minutes.
视频讲解：
解答视频缩略图
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文字解答：
从 12:0012:00 中午开始，真实经过 3030 分钟时，车钟走快到经过了 3535 分钟。
因此车钟每走一分钟，真实时间经过 3035=67\frac {30}{35} = \frac 67 分钟。从 12:0012:00 到 7:007:00，车钟走了 7⋅60=4207\cdot 60 = 420 分钟，所以真实时间经过了 420⋅67=360420 \cdot \frac67 = 360 分钟，也就是 66 小时。从 12:0012:00 开始经过 66 小时，实际时间是 6:006:00 。
所以正确答案是 B。
Starting from 12:0012:00 noon, after 3030 minutes of time elapsed, the car clock went 3535 minutes ahead.
Therefore, for every minute the car clock goes ahead, 3035=67 \frac {30}{35} = \frac 67 minutes of actual time pass by. From the time 12:0012:00 to 7:00,7:00, the car clock goes ahead 7⋅60=4207\cdot 60 = 420 minutes, and therefore, 420⋅67=360420 \cdot \frac67 = 360 minutes, or 66 hours, of actual time have passed by. If we start at 12:0012:00 and 66 hours pass by, the time is 6:00.6:00 .
Thus, B is the correct answer.

13.

Laila 参加了五次数学测试，每次满分 100100 分。每次成绩都是 00 到 100100 之间的整数。她前四次测试得了相同分数，最后一次分数更高。五次测试的平均分是 8282。Laila 最后一次测试的分数可能有多少个不同的值？
Laila took five math tests, each worth a maximum of 100100 points. Laila’s score on each test was an integer between 00 and 100,100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82.82. How many values are possible for Laila’s score on the last test?
44
55
99
1010
1818
知识点：平均数模运算
难度评级：1250
小提示：
设前四次相同分数为 ff，最后一次分数为 ll
Let the repeated score be ff and the last test score be ll.
大提示：
使用 4f+l=4104f+l=410，并注意 ll 大于平均分
Use 4f+l=4104f+l=410 and the fact that ll is larger than the average.
视频讲解：
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文字解答：
因为五次测试平均分是 8282，所以五次总分为 5⋅82=4105\cdot82 = 410 。
设 ff 为前 44 次测试的分数，ll 为最后一次测试的分数。
我们知道 f<l≤100f < l \leq 100 且 4f+l=410。4f + l = 410\text{。}又因为 410=4f+l<5l410 = 4f + l < 5l ，所以 4105=82<l\frac{410}{5} = 82 < l 。
另外，由于 4f+l=4104f + l = 410 ，而 410410 除以 44 的余数是 22，所以 ll 除以 44 也必须余 22，因为 4f4f 除以 44 没有余数。等价地，l≡2mod  4。l \equiv 2 \mod 4\text{。}因为 82<l≤10082 < l \leq 100 且 l≡2mod  4l \equiv 2 \mod 4，所以 ll 只能是 86,90,94,9886,90,94,98。这给出四组不同的解：(f,l)=(81,86);(80,90);(79,94);(78,98)\begin{align*} (f,l) =& (81,86);\\ &(80,90);\\ &(79,94);\\ &(78,98) \end{align*} 因此共有 44 种可能，正确答案是 A。
Since the average score on the five tests is 82,82, the total score of those five tests must be 5⋅82=410.5\cdot82 = 410 .
Now, let ff be the score on the first 44 tests and let ll be the score for the last test.
We know that f<l≤100f < l \leq 100 and 4f+l=410.4f + l = 410. And as 410=4f+l<5l,410 = 4f + l < 5l , we know 4105=82<l.\frac{410}{5} = 82 < l .
Also, since 4f+l=410,4f + l = 410 , and dividing 410410 by 44 gives us a remainder of 2,2, we know that dividing ll by 44 must leave a remainder of 22 as 4f4f will leave no remainder when divided by 4.4. Equivalently: l≡2mod  4.l \equiv 2 \mod 4 . Since 82<l≤10082 < l \leq 100 and l≡2mod  4,l \equiv 2 \mod 4, the only options for ll are 86,90,94,98.86,90,94,98. This yields four distinct solutions as follows: (f,l)=(81,86);(80,90);(79,94);(78,98)\begin{align*} (f,l) =& (81,86);\\ &(80,90);\\ &(79,94);\\ &(78,98) \end{align*} Therefore, there are 44 solutions, and A is the correct answer.

14.

设 NN 是数字乘积为 120120 的最大五位数。NN 的各位数字之和是多少？
Let NN be the greatest five-digit number whose digits have a product of 120.120. What is the sum of the digits of N?N?
1515
1616
1717
1818
2020
知识点：数字最优化
难度评级：1140
小提示：
为了让五位数最大，从左到右尽量让数字最大
To maximize the five-digit number, maximize the digits from left to right.
大提示：
每次选择的数字都必须整除剩余需要得到的乘积
Each chosen digit must divide the remaining required product.
视频讲解：
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文字解答：
为了使 55 位数最大，必须先让第一位，也就是万位，尽可能大。
小于 1010 且能整除 120120 的最大数字是 88，所以第一位必须是 88。剩下各位数字的乘积为 1515。
同样地，现在要让第二位尽可能大。
小于 1010 且能整除 1515 的最大数字是 55，所以第二位是 55。剩下各位数字的乘积为 33。
接着让第三位尽可能大。
小于 1010 且能整除 33 的最大数字是 33，所以第三位是 33。剩下的乘积是 11，这说明第 44 位和第 55 位都是 11。
于是 N=85311N = 85311，各位数字之和为 8+5+3+1+1=188+5+3+1+1=18
所以正确答案是 D。
To make the largest possible 55 digit number, we must maximize the first digit (the digit in the ten-thousands place).
The largest number that is strictly less than 1010 and divides 120120 is 8,8, so the first digit must be 8.8. Therefore, the product of the remaining number is 15.15.
Similarly, we must now maximize the second digit.
The largest number that is less than 1010 and divides 1515 is 5,5, so the second digit is 5.5. Therefore, the product of the remaining number is 3.3.
We must then maximize the third digit.
The largest number that is less than 1010 and divides 33 is 3,3, so the third digit is 3.3. Therefore, the product of the remaining number is 1.1. This means the 44th and 55th digits are 1.1.
This makes N=85311,N = 85311, so the sum of the digits is 8+5+3+1+1=188+5+3+1+1=18
Thus, D is the correct answer.

15.

如下图，在两个较小圆中，每个较小圆的一条直径都是大圆的一条半径。如果两个较小圆的总面积为 11 平方单位，那么阴影区域的面积是多少平方单位？
In the diagram below, a diameter of each of the two smaller circles is a radius of the larger circle. If the two smaller circles have a combined area of 11 square unit, then what is the area of the shaded region, in square units?
14\dfrac{1}{4}
13\dfrac{1}{3}
12\dfrac{1}{2}
11
π2\dfrac{\pi}{2}
知识点：圆面积长度、面积与体积的缩放关系
难度评级：1070
小提示：
比较小圆半径与大圆半径
Compare the radius of a smaller circle to the radius of the larger circle.
大提示：
面积按长度比例的平方缩放
Area scales by the square of the scale factor.
视频讲解：
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文字解答：
设大圆面积为 AA。
每个小圆的直径等于大圆半径，所以每个小圆的半径是大圆半径的一半。
用符号表示，若大圆半径为 rlr_l，每个小圆半径为 rsr_s，则 rs=12rlr_s = \dfrac12 r_l 因为大圆面积为 A=πrl2A=\pi r_l^2，所以每个小圆面积为 πrs2=π(12rl)2=14(πrl2)=14A。\begin{align*}\pi r_s^2 &= \pi \left(\dfrac12 r_l\right)^2 \\&= \dfrac14 (\pi r_l^2)\\&=\dfrac14 A\end{align*}\text{。}两个小圆总面积为 11 平方单位，于是 2⋅14A=12\cdot \dfrac14 A=1 平方单位，所以 A=2A=2 平方单位。
阴影面积等于大圆面积 (A)(A) 减去两个小圆的总面积 (1)(1)，所以阴影面积为 A−1=2−1=1A - 1=2-1=1 平方单位。
所以正确答案是 D。
Let AA be the area of the large circle.
Since the diameter of each of the two smaller circles is itself the radius of the larger circle, the radius of each smaller circle is half that of the larger circle.
Symbolically, if we allow rlr_l to be the radius of the large circle and rsr_s to be the radius of each of the smaller circles: rs=12rlr_s = \dfrac12 r_l As the area of the larger circle is equal to A=πrl2,A=\pi r_l^2, the area of the smaller circles are equal to πrs2=π(12rl)2=14(πrl2)=14A.\begin{align*}\pi r_s^2 &= \pi \left(\dfrac12 r_l\right)^2 \\&= \dfrac14 (\pi r_l^2)\\&=\dfrac14 A.\end{align*} As the area of two of these smaller circles combined is equal to 11 square unit, then it follows that 2⋅14A=12\cdot \dfrac14 A=1 square unit, implying that A=2A=2 square units.
As the area of the shaded region is equal to the area of the larger circle (A)(A) minus the combined area of the two smaller circles (1),(1), the area of the shaded region is A−1=2−1=1 A - 1=2-1=1 square unit.
Thus, the correct answer is D

16.

Chang 教授有九本不同的语言书排在书架上：两本阿拉伯语书、三本德语书和四本西班牙语书。若要求阿拉伯语书放在一起，西班牙语书也放在一起，那么这九本书有多少种排列方式？
Professor Chang has nine different language books lined up on a bookshelf: two Arabic, three German, and four Spanish. How many ways are there to arrange the nine books on the shelf keeping the Arabic books together and keeping the Spanish books together?
14401440
28802880
57605760
182,440182{,}440
362,880362{,}880
知识点：排列有限制的排列
难度评级：1100
小提示：
把两本阿拉伯语书看成一个整体，把四本西班牙语书看成一个整体
Treat the two Arabic books as one block and the four Spanish books as one block.
大提示：
先排列五个对象，再排列阿拉伯语块和西班牙语块内部的书
Arrange the five objects, then arrange the books within the Arabic and Spanish blocks.
视频讲解：
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文字解答：
把两本阿拉伯语书作为一个块，四本西班牙语书作为一个块。
此时共有 55 个对象：三本德语书、一个阿拉伯语书块、一个西班牙语书块。可用 5!5! 种方式排列这 55 个对象。阿拉伯语书块内部有 2!2! 种排列，西班牙语书块内部有 4!4! 种排列，所以总排列数为 5!⋅4!⋅2!=120⋅24⋅2=5760\begin{align*}5! \cdot 4! \cdot 2! &= 120 \cdot 24 \cdot 2 \\&= 5760 \end{align*}
所以正确答案是 C。
Since we are keeping the Arabic books together and the Spanish books together, we can look at each group as a single block.
As such, there are 55 objects on the bookshelf: three German books, one collection of Arabic books, and one collection of Spanish books. There are 5!5! ways to order the 55 objects. As we already have the books together, there are 2!2! ways of ordering the Arabic books and 4!4! ways of ordering the Spanish books. Therefore, the total ways to order the books is 5!⋅4!⋅2!=120⋅24⋅2=5760\begin{align*}5! \cdot 4! \cdot 2! &= 120 \cdot 24 \cdot 2 \\&= 5760 \end{align*}
Thus, the correct answer is C.

17.

Bella 从自己家出发，朝朋友 Ella 家走去。与此同时，Ella 从自己家骑自行车朝 Bella 家出发。她们都保持恒定速度，Ella 骑车速度是 Bella 步行速度的 55 倍。两家相距 22 英里，即 10,56010{,}560 英尺，Bella 每一步走 2122 \tfrac{1}{2} 英尺。到两人相遇时，Bella 会走多少步？
Bella begins to walk from her house toward her friend Ella’s house. At the same time, Ella begins to ride her bicycle toward Bella’s house. They each maintain a constant speed, and Ella rides 55 times as fast as Bella walks. The distance between their houses is 22 miles, which is 10,56010{,}560 feet, and Bella covers 2122 \tfrac{1}{2} feet with each step. How many steps will Bella take by the time she meets Ella?
704704
845845
10561056
17601760
35203520
知识点：路程、速度与时间比与比例单位换算
难度评级：1210
小提示：
Ella 的速度是 Bella 的 55 倍，所以把总距离按 1:51:5 分配
Since Ella is 55 times as fast as Bella, split the distance in the ratio 1:51:5.
大提示：
找到 Bella 走的英尺数后，除以每步 2122\frac12 英尺
After finding Bella’s walking distance in feet, divide by 2122\frac12 feet per step.
视频讲解：
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文字解答：
Ella 每骑 55 英尺，Bella 走一英尺，所以相遇时 Bella 走了两家距离的 16\frac16。因此 Bella 走了 16⋅10560=1760\dfrac16 \cdot 10560=1760 英尺。因为她每步走 2.52.5 英尺，到相遇时她走了 17602.5=704\dfrac{1760}{2.5} = 704 步。
所以正确答案是 A。
Since for every foot Bella walks, Ella rides 55 feet, we know that Bella will walk 16\frac16 of the distance between the two houses, and so she walks 16⋅10560=1760\dfrac16 \cdot 10560=1760 feet. Since she walks 2.52.5 feet per step, she takes 17602.5=704\dfrac{1760}{2.5} = 704 steps by the time she meets Ella.
Thus, A is the correct answer.

18.

23,23223{,}232 有多少个正因数？
How many positive factors does 23,23223{,}232 have?
99
1212
2828
3636
4242
知识点：质因数分解因数个数
难度评级：1170
小提示：
先把 23,23223,232 分解质因数
Prime-factorize 23,23223,232.
大提示：
若 n=paqbrcn=p^a q^b r^c，则正因数个数为 (a+1)(b+1)(c+1)(a+1)(b+1)(c+1)
If n=paqbrcn=p^a q^b r^c, then its number of positive factors is (a+1)(b+1)(c+1)(a+1)(b+1)(c+1).
视频讲解：
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文字解答：
质因数分解为 23,232=26⋅3⋅11223{,}232=2^6\cdot3\cdot11^2。构造正因数时，指数可以从 00 到 66（对应质因数 22）、从 00 到 11（对应 33）、从 00 到 22（对应 1111）中选择。因此正因数的个数为 (6+1)(1+1)(2+1)=42(6+1)(1+1)(2+1)=42。
所以正确答案是 E。
The prime factorization is 23,232=26⋅3⋅112.23{,}232=2^6\cdot3\cdot11^2. A divisor may use any exponent from 00 through 66 on 22, from 00 through 11 on 33, and from 00 through 22 on 1111. Therefore, the number of positive divisors is (6+1)(1+1)(2+1)=42.(6+1)(1+1)(2+1)=42.
Thus, E is the correct answer.

19.

在一个符号金字塔中，如果某格下面两个格子的符号相同，则该格填 “+”；如果下面两个格子的符号不同，则该格填 “-”。下图展示了一个四层符号金字塔。有多少种方法可以填最底行的四个格子，使得金字塔顶端为 “+”？
In a sign pyramid a cell gets a “+” if the two cells below it have the same sign, and it gets a “-” if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a “+” at the top of the pyramid?
22
44
88
1212
1616
知识点：奇偶性乘法原理
难度评级：1310
小提示：
如果知道下方一格和它们上方那格，另一个下方格子就被确定
If you know one lower cell and the cell above a pair, the other lower cell is forced.
大提示：
从顶端向下数，每新添一行有多少选择
Work downward from the top, counting how many choices appear at each new row.
视频讲解：
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文字解答：
设有两个相邻的下方格子和它们上方的格子。如果已知左下格和上方格，就总能确定右下格：
如果上方格是 ++，则右下格必须与左下格相同；如果上方格是 −-，则右下格必须与左下格相反。
现在假设已知某一行。只要选择这一行最左格左下方的那个格子的符号，就能依次确定下一整行。
因此，顶行已知为 ++ 时，下一行有 22 种选择；同理，第三行有 22 种选择，第四行也有 22 种选择。底行共有 2⋅2⋅2=82\cdot2\cdot2=8 种可能。
所以正确答案是 C。
Suppose we have two cells and the cell above them. If we are given the bottom left cell and the top cell, we can always find the bottom right cell as follows:
If the top cell is +,+, then the bottom right cell must be the same as the bottom left cell, and if the top cell is −,-, the bottom right cell must be the opposite of the bottom left cell.
Now, suppose we are given a row. Then, suppose we choose a value for the cell below and to the left of the leftmost cell in our given row. We then can inductively determine the entire row below our given by first finding the bottom-right cell of the leftmost cell in our row, and using that newly found cell as the bottom-left reference for the second to the left cell in the given row to find its bottom-right counterpart. The process continues on until the row below the given row is fully solved.
Therefore, since we know that the top row has a cell labelled +,+, we have 22 choices for the row below, depending on our choice of the bottom-left cell. Similarly, we have 22 choices for the third row, and thus 22 choices for the fourth row. This makes 2⋅2⋅2=82\cdot2\cdot2=8 total choices for the bottom row of the sign pyramid.
Thus, the correct answer is C.

20.

在 △ABC\triangle ABC 中，点 EE 在 AB‾\overline{AB} 上，且 AE=1AE=1、EB=2EB=2。点 DD 在 AC‾\overline{AC} 上，使得 DE‾∥BC‾\overline{DE} \parallel \overline{BC}；点 FF 在 BC‾\overline{BC} 上，使得 EF‾∥AC‾\overline{EF} \parallel \overline{AC}。CDEFCDEF 的面积与 △ABC\triangle ABC 的面积之比是多少？
In △ABC,\triangle ABC, a point EE is on AB‾\overline{AB} with AE=1AE=1 and EB=2.EB=2. Point DD is on AC‾\overline{AC} so that DE‾∥BC‾\overline{DE} \parallel \overline{BC} and point FF is on BC‾\overline{BC} so that EF‾∥AC‾.\overline{EF} \parallel \overline{AC}. What is the ratio of the area of CDEFCDEF to the area of △ABC?\triangle ABC?
49\dfrac{4}{9}
12\dfrac{1}{2}
59\dfrac{5}{9}
35\dfrac{3}{5}
23\dfrac{2}{3}
知识点：相似面积比
难度评级：1340
小提示：
利用两组平行线得到相似三角形
Use similarity from the two parallel-line conditions.
大提示：
把两个角上的小三角形面积与 △ABC\triangle ABC 面积比较
Compare the areas of the two small corner triangles to the area of △ABC\triangle ABC.
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文字解答：
设 △ABC\triangle ABC 的面积为 tt。由于 DE∥BCDE \parallel BC 且 FE∥CAFE \parallel CA ，可得 ADE∼ABCADE \sim ABC 以及 EFB∼ABC。EFB \sim ABC\text{。}因为 AE=AB3AE = \dfrac{AB}3，所以 ADEADE 的面积为 (13)2t=t9\left(\dfrac13\right)^2 t = \dfrac{t}{9} 。因为 EB=2AB3EB = \dfrac{2AB}3，所以 EFBEFB 的面积为 (23)2t=49t\left(\dfrac23\right)^2 t = \dfrac{4}{9}t 。最后，要找 CDEFCDEF 的面积，就从 ABC=tABC =t 中减去 ADEADE 和 EFBEFB 的面积，即 t−t9−4t9=4t9。t- \frac{t}{9} - \frac{4t}{9} = \frac{4t}{9}\text{。}因此 CDEFCDEF 与 ABCABC 的面积比为 (4t9)t=49\dfrac{\left(\dfrac{4t}{9}\right)}{t} = \dfrac{4}{9} 。
所以正确答案是 A。
Let the area of △ABC\triangle ABC be equal to t.t. Since DE∥BCDE \parallel BC and FE∥CA,FE \parallel CA , we can deduce that ADE∼ABCADE \sim ABC and EFB∼ABC.EFB \sim ABC. Since AE=AB3,AE = \dfrac{AB}3, the area of ADEADE is equal to (13)2t=t9.\left(\dfrac13\right)^2 t = \dfrac{t}{9} . Since EB=2AB3,EB = \dfrac{2AB}3, the area of EFBEFB is equal to (23)2t=49t.\left(\dfrac23\right)^2 t = \dfrac{4}{9}t . Finally, to find the area of CDEF,CDEF, we take the area of ABC=tABC =t and subtract the areas of ADEADE and EFB.EFB. This is equivalent to the expression t−t9−4t9=4t9.t- \frac{t}{9} - \frac{4t}{9} = \frac{4t}{9} . Therefore, the ratio of the area of CDEFCDEF and ABCABC is (4t9)t=49.\dfrac{\left(\dfrac{4t}{9}\right)}{t} = \dfrac{4}{9} .
Thus, A is the correct answer.

21.

有多少个正的三位整数，除以 66 余 22，除以 99 余 55，且除以 1111 余 77？
How many positive three-digit integers have a remainder of 22 when divided by 6,6, a remainder of 55 when divided by 9,9, and a remainder of 77 when divided by 11?11?
11
22
33
44
55
知识点：中国剩余定理最小公倍数区间内整数计数
难度评级：1490
小提示：
每个余数条件都说明这个数比相应除数的某个倍数小 44
Each remainder condition says the number is 44 less than a multiple of the divisor.
大提示：
数形如 198k−4198k-4 的三位数有多少个
Count three-digit numbers of the form 198k−4198k-4.
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文字解答：
每个所给余数都比相应除数小 44，所以 x+4x+4 必须同时被 6,96,9 和 1111 整除。因此 x+4x+4 是 lcm⁡(6,9,11)=198\operatorname{lcm}(6,9,11)=198 的倍数。
因为 xx 是三位数，所以 104≤x+4≤1003104\le x+4\le1003。这个区间内 198198 的倍数是 198,396,594,792,990198,396,594,792,990，因此共有 55 个符合条件的整数。
正确答案是 E。
Each required remainder is 44 less than its divisor, so x+4x+4 must be divisible by 6,9,6,9, and 1111. Hence x+4x+4 is a multiple of lcm⁡(6,9,11)=198.\operatorname{lcm}(6,9,11)=198.
For a three-digit xx, we have 104≤x+4≤1003104\le x+4\le1003. The multiples of 198198 in this interval are 198,396,594,792,990198,396,594,792,990, giving 55 possible integers.
Thus, E is the correct answer.

22.

在正方形 ABCDABCD 中，点 EE 是边 CD‾\overline{CD} 的中点。BE‾\overline{BE} 与对角线 AC‾\overline{AC} 相交于 FF。四边形 AFEDAFED 的面积为 4545。正方形 ABCDABCD 的面积是多少？
Point EE is the midpoint of side CD‾\overline{CD} in square ABCD,ABCD, and BE‾\overline{BE} meets diagonal AC‾\overline{AC} at F.F. The area of quadrilateral AFEDAFED is 45.45. What is the area of ABCD?ABCD?
100100
108108
120120
135135
144144
知识点：相似正方形（几何）面积比
难度评级：1770
小提示：
设正方形边长为 ss
Let the square have side length ss.
大提示：
求出从 △ACD\triangle ACD 中切掉的小三角形面积，再从半个正方形中减去它
Find the small triangle cut from △ACD\triangle ACD, then subtract it from half the square.
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文字解答：
设正方形边长为 ss，并设 HH 是从 FF 向 BC‾\overline{BC} 作垂线的垂足。直角三角形 CABCAB 与 CFHCFH 相似，所以 CH=FHCH=FH。另外，三角形 BFHBFH 与 BECBEC 相似，因此 FHEC=BHBC。 \frac{FH}{EC}=\frac{BH}{BC}\text{。}因为 EC=s2EC=\frac{s}{2}、BC=sBC=s，且 BH=s−CH=s−FHBH=s-CH=s-FH，所以 2FHs=1−FHs\frac{2FH}{s}=1-\frac{FH}{s}，从而 FH=CH=s3FH=CH=\frac{s}{3}。
因此，三角形 EFCEFC 的底为 EC=s2EC=\frac{s}{2}，高为 CH=s3CH=\frac{s}{3}，所以面积为 s212\frac{s^2}{12}。于是 [AFED]=[△ACD]−[△EFC]=s22−s212=5s212。\begin{aligned} [AFED]&=[\triangle ACD]-[\triangle EFC]\\ &=\dfrac{s^2}{2}-\dfrac{s^2}{12}\\ &=\dfrac{5s^2}{12} \end{aligned}\text{。}因为这个面积是 4545，所以 s2=108s^2=108，这就是正方形的面积。
Let the square have side length s,s, and let HH be the foot of the perpendicular from FF to BC‾.\overline{BC}. The right triangles CABCAB and CFHCFH are similar, so CH=FH.CH=FH. Also, triangles BFHBFH and BECBEC are similar, giving FHEC=BHBC. \frac{FH}{EC}=\frac{BH}{BC}. Since EC=s2,EC=\frac{s}{2}, BC=s,BC=s, and BH=s−CH=s−FH,BH=s-CH=s-FH, this becomes 2FHs=1−FHs,\frac{2FH}{s}=1-\frac{FH}{s}, so FH=CH=s3.FH=CH=\frac{s}{3}.
Thus triangle EFCEFC has base EC=s2EC=\frac{s}{2} and height CH=s3,CH=\frac{s}{3}, so its area is s212.\frac{s^2}{12}. Therefore [AFED]=[△ACD]−[△EFC]=s22−s212=5s212.\begin{aligned} [AFED]&=[\triangle ACD]-[\triangle EFC]\\ &=\dfrac{s^2}{2}-\dfrac{s^2}{12}\\ &=\dfrac{5s^2}{12}. \end{aligned} Since this area is 4545, we get s2=108s^2=108, the area of the square.

23.

从一个正八边形中随机选择三个顶点并连接成一个三角形。这个三角形至少有一条边也是八边形的一条边的概率是多少？
From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?
27\dfrac{2}{7}
542\dfrac{5}{42}
1114\dfrac{11}{14}
57\dfrac{5}{7}
67\dfrac{6}{7}
知识点：补集计数隔板法
难度评级：1650
小提示：
使用补集：数没有相邻顶点被选中的三角形
Use complementary counting: count triangles with no adjacent chosen vertices.
大提示：
固定一个顶点后，数三个选中顶点之间的正间隔
After fixing one vertex, count the positive gaps around the octagon between the chosen vertices.
视频讲解：
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文字解答：
用补集计数：计算任意两个所选顶点都不相邻的三角形。从一个所选顶点出发，设三个间隔中未选顶点的正整数个数为 x,y,zx,y,z。那么 x+y+z=5x+y+z=5，共有 (42)=6\binom42=6 个正整数解。
起点有 88 种选择，而每个三角形会从它的 33 个顶点各被计算一次，所以补集中有 8⋅63=16\frac{8\cdot6}{3}=16 个三角形。全部三角形共有 (83)=56\binom83=56 个，因此所求概率是 1−1656=571-\frac{16}{56}=\frac{5}{7}。
所以正确答案是 D。
Count the complement: triangles with no two chosen vertices adjacent. Starting at a chosen vertex, let x,y,zx,y,z be the positive numbers of unchosen vertices in the three gaps. Then x+y+z=5x+y+z=5, which has (42)=6\binom42=6 positive solutions.
There are 88 choices for the starting vertex, and each triangle is counted from each of its 33 vertices, so the complement contains 8⋅63=16\frac{8\cdot6}{3}=16 triangles. Out of (83)=56\binom83=56 total triangles, the desired probability is 1−1656=57.1-\frac{16}{56}=\frac{5}{7}.
Thus, D is the correct answer.

24.

在立方体 ABCDEFGHABCDEFGH 中，CC 与 EE 是相对顶点，JJ 和 II 分别是棱 FB‾\overline{FB} 和 HD‾\overline{HD} 的中点。设 RR 为截面 EJCIEJCI 的面积与立方体一个面的面积之比。R2R^2 是多少？
In the cube ABCDEFGHABCDEFGH with opposite vertices CC and E,E, JJ and II are the midpoints of edges FB‾\overline{FB} and HD‾,\overline{HD}, respectively. Let RR be the ratio of the area of the cross-section EJCIEJCI to the area of one of the faces of the cube. What is R2?R^2?
54\dfrac{5}{4}
43\dfrac{4}{3}
32\dfrac{3}{2}
2516\dfrac{25}{16}
94\dfrac{9}{4}
知识点：正方体菱形勾股定理
难度评级：1910
小提示：
四边形 EJCIEJCI 是菱形，可以用对角线求面积
The quadrilateral EJCIEJCI is a rhombus, so use its diagonals.
大提示：
用立方体边长表示对角线 IJIJ 和 CECE
Express the diagonals IJIJ and CECE in terms of the cube side length.
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文字解答：
设立方体边长为 ss。注意截面四边形每条边长度相同，所以 EJCIEJCI 是菱形。菱形面积等于两条对角线乘积的一半，因此它的面积为 12IJ⋅CE\frac12 IJ\cdot CE。由勾股定理，IJ=FH=s2。IJ=FH=s\sqrt{2}\text{。}再次使用勾股定理可得 CE=AC2+AE2=(s2)2+s2=2s2+s2=s3\begin{align*}CE&=\sqrt{AC^2+AE^2}\\&=\sqrt{(s\sqrt{2})^2+s^2}\\&=\sqrt{2s^2+s^2}\\&=s\sqrt{3}\end{align*} 因此 R=12IJ⋅CEs2=12s223s2=32\begin{align*}R&=\dfrac{\frac12 IJ\cdot CE}{s^2}\\&=\dfrac{\frac12 s^2\sqrt{2}\sqrt{3}}{s^2}\\&=\sqrt{\dfrac32}\end{align*} 所以 R2=32R^2=\dfrac32，正确答案是 C。
Allow ss to represent the length of an edge of the cube. Noting that each side of the cross section is equal in length, we conclude that EJCIEJCI is a rhombus. The area of this rhombus can be calculated as 12IJ⋅CE,\frac12 IJ\cdot CE, as the area of a rhombus is equal to half the product of its diagonals. Using the Pythagorean Theorem: IJ=FH=s2.IJ=FH=s\sqrt{2}. Similarly, using the Pythagorean Theorem again lets us see that: CE=AC2+AE2=(s2)2+s2=2s2+s2=s3\begin{align*}CE&=\sqrt{AC^2+AE^2}\\&=\sqrt{(s\sqrt{2})^2+s^2}\\&=\sqrt{2s^2+s^2}\\&=s\sqrt{3}\end{align*} Therefore, R=12IJ⋅CEs2=12s223s2=32\begin{align*}R&=\dfrac{\frac12 IJ\cdot CE}{s^2}\\&=\dfrac{\frac12 s^2\sqrt{2}\sqrt{3}}{s^2}\\&=\sqrt{\dfrac32}\end{align*} Thus, R2=32,R^2=\dfrac32, and the correct answer is C.

25.

在 28+12^8+1 和 218+12^{18}+1 之间（含端点）有多少个完全立方数？
How many perfect cubes lie between 28+12^8+1 and 218+1,2^{18}+1, inclusive?
44
99
1010
5757
5858
知识点：完全幂区间内整数计数
难度评级：1280
小提示：
区间中的完全立方数形如 n3n^3
A perfect cube in the interval has the form n3n^3.
大提示：
把上下界与 63,73,6436^3,7^3,64^3 和 65365^3 比较
Compare the bounds to 63,73,643,6^3,7^3,64^3, and 65365^3.
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文字解答：
因为 63=216<2576^3=216<257，且 257<343=73257<343=7^3，区间内最小的完全立方数是 737^3。又因为 643=218<218+1<65364^3=2^{18}<2^{18}+1<65^3，最大的完全立方数是 64364^3。
所以整数立方根为 7,8,…,647,8,\ldots,64，共有 64−7+1=5864-7+1=58 个。
所以正确答案是 E。
Because 63=216<2576^3=216<257 and 257<343=73257<343=7^3, the smallest cube in the interval is 737^3. Also, 643=218<218+1<65364^3=2^{18}<2^{18}+1<65^3, so the largest cube is 64364^3.
The integer cube roots are therefore 7,8,…,647,8,\ldots,64, a total of 64−7+1=5864-7+1=58.
Thus, E is the correct answer.