2017 AMC 12B 详解 | LIVE by Po-Shen Loh

1.

Kymbrea 的漫画书收藏目前有 $30$ 本漫画书，并且她每月增加 $2$ 本。LaShawn 的收藏目前有 $10$ 本漫画书，并且他每月增加 $6$ 本。多少个月后，LaShawn 收藏的漫画书数量会是 Kymbrea 的两倍？

Kymbrea’s comic book collection currently has $30$ comic books in it, and she is adding to her collection at the rate of $2$ comic books per month. LaShawn’s collection currently has $10$ comic books in it, and he is adding to his collection at the rate of $6$ comic books per month. After how many months will LaShawn’s collection have twice as many comic books as Kymbrea’s?

$1$

$4$

$5$

$20$

$25$

知识点：一次方程

难度评级：920

小提示：

$m$ 个月后，两人的收藏分别有 $30 + 2m$ 本和 $10 + 6m$ 本漫画书

After $m$ months the collections hold $30 + 2m$ and $10 + 6m$ comic books

大提示：

列方程 $10 + 6m = 2(30 + 2m)$ ，并解出 $m$ 。

Set $10 + 6m = 2(30 + 2m)$ and solve for $m$

解答：

$m$ 个月后，Kymbrea 有 $30 + 2m$ 本漫画书，LaShawn 有 $10 + 6m$ 本。令 $10 + 6m = 2(30 + 2m)$ ，得到 $10 + 6m = 60 + 4m$ ，因此 $2m = 50$ ， $m = 25$ 。

所以正确答案是 E。

After $m$ months, Kymbrea has $30 + 2m$ comic books and LaShawn has $10 + 6m.$ Setting $10 + 6m = 2(30 + 2m)$ gives $10 + 6m = 60 + 4m,$ so $2m = 50$ and $m = 25.$

Thus, the correct answer is E.

2.

实数 $x$ 、 $y$ 和 $z$ 满足不等式

$0 \lt x \lt 1, \quad -1 \lt y \lt 0\text{，}$ 以及 $1 \lt z \lt 2\text{。}$

下列哪个数一定为正？

Real numbers $x,$ $y,$ and $z$ satisfy the inequalities

$0 \lt x \lt 1, \quad -1 \lt y \lt 0,$ and $1 \lt z \lt 2.$

Which of the following numbers is necessarily positive?

$y + x^2$

$y + xz$

$y + y^2$

$y + 2y^2$

$y + z$

知识点：不等式反例

难度评级：1020

小提示：

因为 $y \gt -1$ 且 $z \gt 1$ ，你能推出 $y + z$ 有什么性质？

Since $y \gt -1$ and $z \gt 1,$ what can you say about $y + z?$

大提示：

对其他四个选项，试用 $x = \tfrac18$ ， $y = -\tfrac14$ ， $z = \tfrac32$ 使其为负

For the other four, try $x = \tfrac18,$ $y = -\tfrac14,$ $z = \tfrac32$ to force a negative value

解答：

将 $y \gt -1$ 与 $z \gt 1$ 相加，得到 $y + z \gt 0$ ，所以 $y + z$ 总是正数。其他四个选项都可以取负值：当 $x = \tfrac18$ ， $y = -\tfrac14$ ， $z = \tfrac32$ 时， $y + x^2$ ， $y + xz$ ， $y + y^2$ ，和 $y + 2y^2$ 都为负。

所以正确答案是 E。

Adding $y \gt -1$ and $z \gt 1$ gives $y + z \gt 0,$ so $y + z$ is always positive. Each of the other four choices can be made negative: with $x = \tfrac18,$ $y = -\tfrac14,$ $z = \tfrac32,$ every one of $y + x^2,$ $y + xz,$ $y + y^2,$ and $y + 2y^2$ is negative.

Thus, the correct answer is E.

3.

假设 $x$ 和 $y$ 是非零实数，且满足

$\frac{3x + y}{x - 3y} = -2\text{。}$

下式的值是多少 $\frac{x + 3y}{3x - y}\text{？}$

Suppose that $x$ and $y$ are nonzero real numbers such that

$\frac{3x + y}{x - 3y} = -2.$

What is the value of $\frac{x + 3y}{3x - y}?$

$-3$

$-1$

$1$

$2$

$3$

知识点：代数变形换元法

难度评级：1130

小提示：

清除分母： $3x + y = -2(x - 3y)$

Clear the denominator: $3x + y = -2(x - 3y)$

大提示：

这会化简为 $x = y$ ；再代入目标表达式

This simplifies to $x = y;$ substitute into the target expression

解答：

原方程给出 $3x + y = -2(x - 3y)$ $= -2x + 6y$ ，所以 $5x = 5y$ ，即 $x = y$ 。因此 $\frac{x + 3y}{3x - y} = \frac{y + 3y}{3y - y} = \frac{4y}{2y} = 2\text{。}$

所以正确答案是 D。

The equation gives $3x + y = -2(x - 3y)$ $= -2x + 6y,$ so $5x = 5y,$ meaning $x = y.$ Then $\frac{x + 3y}{3x - y} = \frac{y + 3y}{3y - y} = \frac{4y}{2y} = 2.$

Thus, the correct answer is D.

4.

Samia 骑自行车出发去拜访朋友，平均速度为每小时 $17$ 千米。当她骑完到朋友家一半的距离时，车胎漏气了，于是她以每小时 $5$ 千米的速度步行剩下的路程。她总共花了 $44$ 分钟到达朋友家。以千米为单位并四舍五入到小数点后一位，Samia 步行了多远？

Samia set off on her bicycle to visit her friend, traveling at an average speed of $17$ kilometers per hour. When she had gone half the distance to her friend’s house, a tire went flat, and she walked the rest of the way at $5$ kilometers per hour. In all it took her $44$ minutes to reach her friend’s house. In kilometers rounded to the nearest tenth, how far did Samia walk?

$2.0$

$2.2$

$2.8$

$3.4$

$4.4$

知识点：路程、速度与时间一次方程

难度评级：1270

小提示：

设她单程骑行的距离（也就是步行的距离）为 $d$ ；使用时间 $= \dfrac{\text{距离}}{\text{速度}}$ 。

Let $d$ be the one-way distance she biked (and also walked); use time $= \dfrac{\text{distance}}{\text{rate}}$

大提示：

解方程 $\dfrac{d}{17} + \dfrac{d}{5} = \dfrac{44}{60}$ ，求出 $d$ 。

Solve $\dfrac{d}{17} + \dfrac{d}{5} = \dfrac{44}{60}$ for $d$

解答：

设总距离为 $2d$ ，因此她以 $17$ 千米/小时骑了 $d$ 千米，又以 $5$ 千米/小时走了 $d$ 千米。总时间（小时）满足 $\frac{d}{17} + \frac{d}{5} = \frac{44}{60}\text{。}$ 左边合并得到 $\dfrac{22d}{85} = \dfrac{11}{15}$ ，所以 $d = \dfrac{11}{15} \cdot \dfrac{85}{22} = \dfrac{17}{6} = 2.833\ldots$ 。她步行了约 $2.8$ 千米。

所以正确答案是 C。

Let $2d$ be the total distance, so she biked $d$ at $17$ km/h and walked $d$ at $5$ km/h. The total time in hours is $\frac{d}{17} + \frac{d}{5} = \frac{44}{60}.$ Combining the left side gives $\dfrac{22d}{85} = \dfrac{11}{15},$ so $d = \dfrac{11}{15} \cdot \dfrac{85}{22} = \dfrac{17}{6} = 2.833\ldots$ She walked about $2.8$ kilometers.

Thus, the correct answer is C.

5.

数据集 $[6, 19, 33, 33, 39,$ $41,$ $41,$ $43,$ $51,$ $57]$ 的中位数为 $Q_2 = 40$ ，第一四分位数为 $Q_1 = 33$ ，第三四分位数为 $Q_3 = 43$ 。数据集中的异常值是指比第一四分位数 $(Q_1)$ 低超过四分位距的 $1.5$ 倍，或比第三四分位数 $(Q_3)$ 高超过四分位距的 $1.5$ 倍的值，其中四分位距定义为 $Q_3 - Q_1$ 。这个数据集中有多少个异常值？

The data set $[6, 19, 33, 33, 39,$ $41,$ $41,$ $43,$ $51,$ $57]$ has median $Q_2 = 40,$ first quartile $Q_1 = 33,$ and third quartile $Q_3 = 43.$ An outlier in a data set is a value that is more than $1.5$ times the interquartile range below the first quartile $(Q_1)$ or more than $1.5$ times the interquartile range above the third quartile $(Q_3),$ where the interquartile range is defined as $Q_3 - Q_1.$ How many outliers does this data set have?

$0$

$1$

$2$

$3$

$4$

知识点：中位数（数据）

难度评级：1130

小提示：

四分位距是 $Q_3 - Q_1 = 43 - 33 = 10$ 。

The interquartile range is $Q_3 - Q_1 = 43 - 33 = 10$

大提示：

统计小于 $33 - 1.5(10)$ 或大于 $43 + 1.5(10)$ 的值

Count values below $33 - 1.5(10)$ or above $43 + 1.5(10)$

解答：

四分位距为 $43 - 33 = 10$ ，它的 $1.5$ 倍是 $15$ 。异常值是小于 $33 - 15 = 18$ 或大于 $43 + 15 = 58$ 的值。只有 $6$ 小于 $18$ ，且没有数大于 $58$ ，所以恰好有 $1$ 个异常值。

所以正确答案是 B。

The interquartile range is $43 - 33 = 10,$ so $1.5$ times it is $15.$ Outliers are values less than $33 - 15 = 18$ or greater than $43 + 15 = 58.$ Only $6$ falls below $18,$ and nothing exceeds $58,$ so there is exactly $1$ outlier.

Thus, the correct answer is B.

6.

以 $(0, 0)$ 和 $(8, 6)$ 为直径端点的圆与 $x$ -轴相交于另一个点。这个点的 $x$ -坐标是多少？

The circle having $(0, 0)$ and $(8, 6)$ as the endpoints of a diameter intersects the $x$ -axis at a second point. What is the $x$ -coordinate of this point?

$4\sqrt{2}$

$6$

$5\sqrt{2}$

$8$

$6\sqrt{2}$

知识点：坐标几何圆中点

难度评级：1350

小提示：

圆心是中点 $(4, 3)$ ，半径是 $\sqrt{4^2 + 3^2} = 5$ 。

The center is the midpoint $(4, 3)$ and the radius is $\sqrt{4^2 + 3^2} = 5$

大提示：

在 $(x - 4)^2 + (y - 3)^2 = 25$ 中令 $y = 0$ 。

Set $y = 0$ in $(x - 4)^2 + (y - 3)^2 = 25$

解答：

圆心是直径的中点 $(4, 3)$ ，半径为 $\sqrt{4^2 + 3^2} = 5$ ，圆的方程为 $(x - 4)^2 + (y - 3)^2 = 25$ 。令 $y = 0$ ，得到 $(x - 4)^2 = 16$ ，所以 $x = 0$ 或 $x = 8$ 。与 $x$ -轴的另一个交点位于 $x = 8$ 。

所以正确答案是 D。

The center is the midpoint of the diameter, $(4, 3),$ and the radius is $\sqrt{4^2 + 3^2} = 5.$ The circle is $(x - 4)^2 + (y - 3)^2 = 25.$ Setting $y = 0$ gives $(x - 4)^2 = 16,$ so $x = 0$ or $x = 8.$ The second intersection with the $x$ -axis is at $x = 8.$

Thus, the correct answer is D.

7.

函数 $\sin(x)$ 和 $\cos(x)$ 都是周期函数，最小正周期为 $2\pi$ 。函数 $\cos(\sin(x))$ 的最小正周期是多少？

The functions $\sin(x)$ and $\cos(x)$ are periodic with least period $2\pi.$ What is the least period of the function $\cos(\sin(x))?$

$\dfrac{\pi}{2}$

$\pi$

$2\pi$

$4\pi$

它不是周期函数。

It’s not periodic.

知识点：三角恒等式函数

难度评级：1500

小提示：

使用 $\sin(x + \pi) = -\sin(x)$ ，以及 $\cos$ 是偶函数

Use that $\sin(x + \pi) = -\sin(x)$ and $\cos$ is even

大提示：

要排除更小的周期，注意 $\cos(\sin(x)) = 1$ 只在 $\sin(x) = 0$ 时成立，也就是在 $\pi$ 的整数倍处

To rule out a smaller period, note $\cos(\sin(x)) = 1$ only when $\sin(x) = 0,$ i.e. at multiples of $\pi$

解答：

因为 $\cos(\sin(x + \pi))$ $= \cos(-\sin(x))$ $= \cos(\sin(x))$ ，所以该函数有周期 $\pi$ 。它不可能有更小的周期： $\cos(\sin(x)) = 1$ 当且仅当 $\sin(x) = 0$ ，这只发生在 $\pi$ 的整数倍处，所以最大值之间相隔 $\pi$ 。最小正周期为 $\pi$ 。

所以正确答案是 B。

Since $\cos(\sin(x + \pi))$ $= \cos(-\sin(x))$ $= \cos(\sin(x)),$ the function has period $\pi.$ It cannot be smaller: $\cos(\sin(x)) = 1$ exactly when $\sin(x) = 0,$ which happens only at integer multiples of $\pi,$ so the maxima are spaced $\pi$ apart. The least period is $\pi.$

Thus, the correct answer is B.

8.

某个长方形的短边与长边之比，等于长边与对角线之比。这个长方形的短边与长边之比的平方是多少？

The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side to the diagonal. What is the square of the ratio of the short side to the long side of this rectangle?

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

$\dfrac{1}{2}$

$\dfrac{\sqrt{5} - 1}{2}$

$\dfrac{\sqrt{2}}{2}$

$\dfrac{\sqrt{6} - 1}{2}$

知识点：二次方程勾股定理比与比例

难度评级：1440

小提示：

设短边为 $x$ ，长边为 $y$ ，则对角线为 $\sqrt{x^2 + y^2}$ ；写出 $\dfrac{x}{y} = \dfrac{y}{\sqrt{x^2+y^2}}$ 。

With short side $x$ and long side $y,$ the diagonal is $\sqrt{x^2 + y^2};$ write $\dfrac{x}{y} = \dfrac{y}{\sqrt{x^2+y^2}}$

大提示：

令 $r = \dfrac{x^2}{y^2}$ 。将比例方程平方，得到 $\dfrac{1}{r} = r + 1$ 。

Let $r = \dfrac{x^2}{y^2}.$ Squaring the ratio equation gives $\dfrac{1}{r} = r + 1$

解答：

设 $x$ 和 $y$ 分别为短边和长边，则对角线为 $\sqrt{x^2 + y^2}$ ，并且 $\dfrac{x^2}{y^2} = \dfrac{y^2}{x^2 + y^2}$ 。令 $r = \dfrac{x^2}{y^2}$ ，右边为 $\dfrac{y^2}{x^2 + y^2} = \dfrac{1}{r + 1}$ ，因此 $r = \dfrac{1}{r+1}$ ，得 $r^2 + r - 1 = 0$ 。正根为 $r = \dfrac{\sqrt{5} - 1}{2}$ 。

所以正确答案是 C。

Let $x$ and $y$ be the short and long sides, so the diagonal is $\sqrt{x^2 + y^2}$ and $\dfrac{x^2}{y^2} = \dfrac{y^2}{x^2 + y^2}.$ Writing $r = \dfrac{x^2}{y^2},$ the right side is $\dfrac{y^2}{x^2 + y^2} = \dfrac{1}{r + 1},$ so $r = \dfrac{1}{r+1},$ giving $r^2 + r - 1 = 0.$ The positive root is $r = \dfrac{\sqrt{5} - 1}{2}.$

Thus, the correct answer is C.

9.

一个圆的圆心为 $(-10, -4)$ ，半径为 $13$ 。另一个圆的圆心为 $(3, 9)$ ，半径为 $\sqrt{65}$ 。经过这两个圆交点的直线方程为 $x + y = c$ 。 $c$ 是多少？

A circle has center $(-10, -4)$ and radius $13.$ Another circle has center $(3, 9)$ and radius $\sqrt{65}.$ The line passing through the two points of intersection of the two circles has equation $x + y = c.$ What is $c?$

$3$

$3\sqrt{3}$

$4\sqrt{2}$

$6$

$\dfrac{13}{2}$

知识点：根轴圆坐标几何

难度评级：1410

小提示：

将两个圆写为 $(x+10)^2 + (y+4)^2 = 169$ 和 $(x-3)^2 + (y-9)^2 = 65$

Write both circles as $(x+10)^2 + (y+4)^2 = 169$ and $(x-3)^2 + (y-9)^2 = 65$

大提示：

展开后用一个方程减去另一个方程；二次项会相消

Subtract one expanded equation from the other; the quadratic terms cancel

解答：

两个圆为 $(x+10)^2 + (y+4)^2 = 169$ 和 $(x-3)^2 + (y-9)^2 = 65$ 。展开并用第一个方程减去第二个方程， $x^2$ 与 $y^2$ 项相消，化简为 $x + y = 3$ 。任一交点都满足该方程，因此这就是经过两个交点的直线，且 $c = 3$ 。

所以正确答案是 A。

The circles are $(x+10)^2 + (y+4)^2 = 169$ and $(x-3)^2 + (y-9)^2 = 65.$ Expanding and subtracting the second from the first cancels the $x^2$ and $y^2$ terms and simplifies to $x + y = 3.$ Any intersection point satisfies this, so it is the line through both, and $c = 3.$

Thus, the correct answer is A.

10.

在 Typico 高中， $60\%$ 的学生喜欢跳舞，其余学生不喜欢。在喜欢跳舞的学生中， $80\%$ 说自己喜欢跳舞，其余说自己不喜欢。在不喜欢跳舞的学生中， $90\%$ 说自己不喜欢跳舞，其余说自己喜欢。在所有说自己不喜欢跳舞的学生中，实际喜欢跳舞的学生占多少比例？

At Typico High School, $60\%$ of the students like dancing, and the rest dislike it. Of those who like dancing, $80\%$ say that they like it, and the rest say that they dislike it. Of those who dislike dancing, $90\%$ say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?

$10\%$

$12\%$

$20\%$

$25\%$

$33\tfrac{1}{3}\%$

知识点：条件概率贝叶斯定理

难度评级：1440

小提示：

喜欢跳舞但说自己不喜欢的学生占全体学生的 $60\% \cdot 20\%$ 。

Students who like dancing but say they dislike it are $60\% \cdot 20\%$ of all students

大提示：

用这一组人数除以所有说自己不喜欢跳舞的人数

Divide that group by the total who say they dislike dancing

解答：

喜欢跳舞但说自己不喜欢的学生占全体学生的 $60\% \cdot 20\% = 12\%$ 。不喜欢跳舞且如实这么说的学生占 $40\% \cdot 90\% = 36\%$ 。在所有说自己不喜欢跳舞的人中，实际喜欢跳舞的比例为 $\frac{12}{12 + 36} = \frac{12}{48} = \frac{1}{4} = 25\%\text{。}$

所以正确答案是 D。

Students who like dancing but say they dislike it make up $60\% \cdot 20\% = 12\%$ of all students. Students who dislike dancing and say so make up $40\% \cdot 90\% = 36\%.$ Among everyone who says they dislike dancing, the fraction who actually like it is $\frac{12}{12 + 36} = \frac{12}{48} = \frac{1}{4} = 25\%.$

Thus, the correct answer is D.

11.

若一个正整数是一位数，或者从左到右读它的数字时，其数字构成严格递增或严格递减的序列，就称它为单调数。例如， $3$ 、 $23578$ 和 $987620$ 是单调数，但 $88$ 、 $7434$ 和 $23557$ 不是。共有多少个单调正整数？

Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3,$ $23578,$ and $987620$ are monotonous, but $88,$ $7434,$ and $23557$ are not. How many monotonous positive integers are there?

$1024$

$1524$

$1533$

$1536$

$2048$

知识点：子集双射容斥原理

难度评级：1590

小提示：

$\{1, 2, \ldots, 9\}$ 的每个非空子集都唯一对应一个严格递增数

Each nonempty subset of $\{1, 2, \ldots, 9\}$ gives exactly one strictly increasing number

大提示：

递减数对应 $\{0, 1, \ldots, 9\}$ 的子集，但排除 $\varnothing$ 和 $\{0\}$ ；一位数被重复计算

Decreasing numbers correspond to subsets of $\{0, 1, \ldots, 9\}$ except $\varnothing$ and $\{0\};$ single digits are counted twice

解答：

严格递增的单调数对应 $\{1, \ldots, 9\}$ 的非空子集，得到 $2^9 - 1 = 511$ 。严格递减的数对应 $\{0, 1, \ldots, 9\}$ 的子集，但要排除 $\varnothing$ 和 $\{0\}$ （不允许前导 $0$ ），得到 $2^{10} - 2 = 1022$ 。九个一位数在两类中被重复计算，所以总数为 $511 + 1022 - 9 = 1524$ 。

所以正确答案是 B。

Strictly increasing monotonous numbers correspond to nonempty subsets of $\{1, \ldots, 9\},$ giving $2^9 - 1 = 511.$ Strictly decreasing ones correspond to subsets of $\{0, 1, \ldots, 9\}$ other than $\varnothing$ and $\{0\}$ (a leading $0$ is not allowed), giving $2^{10} - 2 = 1022.$ The nine single-digit numbers are counted in both, so the total is $511 + 1022 - 9 = 1524.$

Thus, the correct answer is B.

12.

$z^{12} = 64$ 的所有根中，实部为正的根之和是多少？

What is the sum of the roots of $z^{12} = 64$ that have a positive real part?

$2$

$4$

$2 + 2\sqrt{3}$

$2\sqrt{2} + \sqrt{6}$

$(1 + \sqrt{3}) + (1 + \sqrt{3})i$

知识点：单位根复数对称性

难度评级：1630

小提示：

十二个根位于半径为 $\sqrt{2}$ 的圆上，相邻根相隔 $30^\circ$ 。

The twelve roots lie on a circle of radius $\sqrt{2},$ spaced $30^\circ$ apart

大提示：

实部为正的根对应角度 $0, \pm 30^\circ, \pm 60^\circ$ ；虚部会成对抵消

The roots with positive real part are at angles $0, \pm 30^\circ, \pm 60^\circ;$ imaginary parts cancel in pairs

解答：

$z^{12} = 64$ 的根位于半径 $64^{\frac{1}{12}} = \sqrt{2}$ 的圆上，角度为 $30^\circ$ 的整数倍。实部为正的根对应角度 $0, \pm 30^\circ, \pm 60^\circ$ 。它们的虚部相互抵消，所以和为 $\begin{aligned} &\sqrt{2} + 2\sqrt{2}\cos 30^\circ \\ &\quad {}+ 2\sqrt{2}\cos 60^\circ \\ &\quad {}= \sqrt{2}\bigl(1 + \sqrt{3} + 1\bigr) \\ &\quad {}= 2\sqrt{2} + \sqrt{6} \end{aligned}\text{。}$

所以正确答案是 D。

The roots of $z^{12} = 64$ lie on the circle of radius $64^{\frac{1}{12}} = \sqrt{2},$ at angles that are multiples of $30^\circ.$ Those with positive real part are at angles $0, \pm 30^\circ, \pm 60^\circ.$ Their imaginary parts cancel, so the sum is $\begin{aligned} &\sqrt{2} + 2\sqrt{2}\cos 30^\circ \\ &\quad {}+ 2\sqrt{2}\cos 60^\circ \\ &\quad {}= \sqrt{2}\bigl(1 + \sqrt{3} + 1\bigr) \\ &\quad {}= 2\sqrt{2} + \sqrt{6}. \end{aligned}$

Thus, the correct answer is D.

13.

如下图所示，要把 $6$ 个圆盘中的 $3$ 个涂成蓝色， $2$ 个涂成红色， $1$ 个涂成绿色。如果两种涂色可以通过整个图形的旋转或翻折互相得到，则认为它们相同。共有多少种不同的涂色？

In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?

$6$

$8$

$9$

$12$

$15$

知识点：分类讨论对称性

难度评级：1660

小提示：

先数带标号的涂色种数，再对三角形的 $6$ 个对称使用 Burnside 引理

Count the labeled colorings first, then apply Burnside’s Lemma to the $6$ symmetries of the triangle

大提示：

非恒等旋转不固定任何涂色；在反射下，个数为奇数的绿色和蓝色圆盘必须占据两个不动的圆盘

A nonidentity rotation fixes none; under a reflection, the odd green and blue counts must use the two fixed disks

解答：

在考虑对称性之前，共有 $\dfrac{6!}{3!2!}=60$ 种涂色。两个非恒等旋转把这些圆盘分成两个 $3$ -循环，所以它们都不能固定颜色数量为 $3,2,1$ 的涂色。

$3$ 个反射中的每一个都固定 $2$ 个圆盘，并把其余 $4$ 个圆盘两两交换成 $2$ 对。要使一种涂色保持不变，唯一的绿色圆盘和 $3$ 个蓝色圆盘之一必须占据这两个不动位置，共有 $2$ 种顺序；两组交换对中任意一组都可以是红色对，因此每个反射固定 $2\cdot2=4$ 种涂色。于是由 Burnside 引理得到 $\dfrac{60+3(4)}{6}=12$ 种不同的涂色。

所以正确答案是 D。

Before accounting for symmetry, there are $\dfrac{6!}{3!2!}=60$ paintings. The two nonidentity rotations partition the disks into two $3$ -cycles, so neither can fix a painting having color counts $3,2,1.$

Each of the $3$ reflections fixes $2$ disks and swaps the other $4$ in $2$ pairs. For a painting to be fixed, the lone green disk and one of the $3$ blue disks must occupy the two fixed positions, in $2$ orders. Of the two swapped pairs, either one can be the red pair, giving $2\cdot2=4$ fixed paintings per reflection. Burnside’s Lemma therefore gives $\dfrac{60+3(4)}{6}=12$ distinct paintings.

Thus, the correct answer is D.

14.

一种冰淇淋新品由一个杯子和冰淇淋组成。杯子形状是一个高 $4$ 英寸的直圆锥台，底部底面直径为 $2$ 英寸，顶部底面直径为 $4$ 英寸，杯内装满实心冰淇淋；此外上面还有一个高 $4$ 英寸的实心冰淇淋圆锥，其底面在下方，正好是圆锥台的上底面。冰淇淋的总体积是多少立方英寸？

An ice-cream novelty item consists of a cup in the shape of a $4$ -inch-tall frustum of a right circular cone, with a $2$ -inch-diameter base at the bottom and a $4$ -inch-diameter base at the top, packed solid with ice cream, together with a solid cone of ice cream of height $4$ inches, whose base, at the bottom, is the top base of the frustum. What is the total volume of the ice cream, in cubic inches?

$8\pi$

$\dfrac{28\pi}{3}$

$12\pi$

$14\pi$

$\dfrac{44\pi}{3}$

知识点：圆锥体积相似

难度评级：1530

小提示：

圆锥台是一个完整圆锥（半径 $2$ ，高 $8$ ）减去一个小圆锥（半径 $1$ ，高 $4$ ）

The frustum is a full cone (radius $2,$ height $8$ ) minus a small cone (radius $1,$ height $4$ )

大提示：

再加上顶部半径为 $2$ 、高为 $4$ 的实心圆锥；使用 $V = \tfrac13 \pi r^2 h$ 。

Add the top solid cone of radius $2$ and height $4;$ use $V = \tfrac13 \pi r^2 h$

解答：

将圆锥台的侧边延长至一点，由相似三角形可知，该圆锥台等于一个半径为 $2$ 、高为 $8$ 的圆锥减去一个半径为 $1$ 、高为 $4$ 的圆锥。因此圆锥台的体积为 $\begin{aligned} &\tfrac13 \pi (2^2)(8) \\ &\quad {}- \tfrac13 \pi (1^2)(4) \\ &\quad {}= \tfrac{32}{3}\pi - \tfrac{4}{3}\pi \\ &\quad {}= \tfrac{28}{3}\pi\text{。} \end{aligned}$ 顶部半径为 $2$ 、高为 $4$ 的圆锥体积为 $\tfrac13 \pi (2^2)(4) = \tfrac{16}{3}\pi$ 。总体积为 $\tfrac{28}{3}\pi + \tfrac{16}{3}\pi = \tfrac{44}{3}\pi$ 。

所以正确答案是 E。

Extending the frustum’s sides to a point, similar triangles show the frustum equals a cone of radius $2$ and height $8$ minus a cone of radius $1$ and height $4:$ $\begin{aligned} &\tfrac13 \pi (2^2)(8) \\ &\quad {}- \tfrac13 \pi (1^2)(4) \\ &\quad {}= \tfrac{32}{3}\pi - \tfrac{4}{3}\pi \\ &\quad {}= \tfrac{28}{3}\pi. \end{aligned}$ The top cone of radius $2$ and height $4$ adds $\tfrac13 \pi (2^2)(4) = \tfrac{16}{3}\pi.$ The total is $\tfrac{28}{3}\pi + \tfrac{16}{3}\pi = \tfrac{44}{3}\pi.$

Thus, the correct answer is E.

15.

设 $ABC$ 是等边三角形。将边 $\overline{AB}$ 从 $B$ 向外延长到点 $B'$ ，使 $BB' = 3 \cdot AB$ 。同样地，将边 $\overline{BC}$ 从 $C$ 向外延长到点 $C'$ ，使 $CC' = 3 \cdot BC$ ，并将边 $\overline{CA}$ 从 $A$ 向外延长到点 $A'$ ，使 $AA' = 3 \cdot CA$ 。 $\triangle A'B'C'$ 的面积与 $\triangle ABC$ 的面积之比是多少？

Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB' = 3 \cdot AB.$ Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC' = 3 \cdot BC,$ and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA' = 3 \cdot CA.$ What is the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC?$

$9 : 1$

$16 : 1$

$25 : 1$

$36 : 1$

$37 : 1$

知识点：面积比面积分割等边三角形

难度评级：1660

小提示：

设 $\triangle ABC$ 的面积为 $X$ 。作 $CB'$ ， $AC'$ ， $BA'$ ，把 $\triangle A'B'C'$ 分成若干部分

Let $X$ be the area of $\triangle ABC.$ Draw $CB',$ $AC',$ $BA'$ to split $\triangle A'B'C'$ into pieces

大提示：

底边为 $k$ 倍且高相同的三角形面积为 $k$ 倍；把每一部分表示为 $X$ 的倍数

A triangle with $k$ times the base and the same height has $k$ times the area; find each piece as a multiple of $X$

解答：

令 $X = [\triangle ABC]$ ，并作线段 $CB'$ 、 $AC'$ 和 $BA'$ 。三角形 $BB'C$ 的底边 $BB' = 3 \cdot AB$ ，且从 $C$ 到直线 $AB$ 的高与 $\triangle ABC$ 相同，所以面积为 $3X$ ；同理， $\triangle CC'A$ 和 $\triangle AA'B$ 的面积也各为 $3X$ 。接着， $\triangle AA'C'$ 与 $\triangle ACC'$ 高相同，而前者的底边为后者的 $3$ 倍，所以面积为 $9X$ ；同理， $\triangle CC'B'$ 和 $\triangle BB'A'$ 的面积也各为 $9X$ 。因此 $\begin{aligned} &[\triangle A'B'C'] = X \\ &\quad {}+ 3(3X) + 3(9X) \\ &\quad {}= 37X\text{，} \end{aligned}$ 所以面积比为 $37 : 1$ 。

所以正确答案是 E。

Let $X = [\triangle ABC],$ and draw segments $CB',$ $AC',$ and $BA'.$ Triangle $BB'C$ has base $BB' = 3 \cdot AB$ and the same altitude as $\triangle ABC$ from $C$ to line $AB,$ so its area is $3X;$ likewise $\triangle CC'A$ and $\triangle AA'B$ each have area $3X.$ Next, $\triangle AA'C'$ has $3$ times the base and the same height as $\triangle ACC',$ so its area is $9X;$ similarly $\triangle CC'B'$ and $\triangle BB'A'$ each have area $9X.$ Thus $\begin{aligned} &[\triangle A'B'C'] = X \\ &\quad {}+ 3(3X) + 3(9X) \\ &\quad {}= 37X, \end{aligned}$ so the ratio is $37 : 1.$

Thus, the correct answer is E.

16.

数 $21!$ $= 51{,}090{,}942{,}171{,}709{,}440{,}000$ 有超过 $60{,}000$ 个正整数因数。从这些因数中随机选一个。它是奇数的概率是多少？

The number $21!$ $= 51{,}090{,}942{,}171{,}709{,}440{,}000$ has over $60{,}000$ positive integer divisors. One of them is chosen at random. What is the probability that it is odd?

$\dfrac{1}{21}$

$\dfrac{1}{19}$

$\dfrac{1}{18}$

$\dfrac{1}{2}$

$\dfrac{11}{21}$

知识点：勒让德公式因数个数质因数分解

难度评级：1730

小提示：

用 $\lfloor \frac{21}{2} \rfloor + \lfloor \frac{21}{4} \rfloor + \cdots$ 求 $21!$ 中 $2$ 的指数

Find the exponent of $2$ in $21!$ using $\lfloor \frac{21}{2} \rfloor + \lfloor \frac{21}{4} \rfloor + \cdots$

大提示：

一个因数是奇数当且仅当它使用 $2^0$ 。若指数为 $e$ ，则奇因数占全部因数的 $\dfrac{1}{e+1}$ 。

A divisor is odd iff it uses $2^0.$ With exponent $e,$ the odd divisors are a fraction $\dfrac{1}{e+1}$ of all divisors

解答：

$21!$ 中 $2$ 的指数为 $\lfloor \frac{21}{2} \rfloor$ $+ \lfloor \frac{21}{4} \rfloor$ $+ \lfloor \frac{21}{8} \rfloor$ $+ \lfloor \frac{21}{16} \rfloor$ $= 10 + 5 + 2 + 1$ $= 18$ 。每个因数都形如 $2^i b$ ，其中 $0 \le i \le 18$ 且 $b$ 为奇数；它为奇数正好在 $i = 0$ 时。因此奇因数的比例为 $\dfrac{1}{18 + 1} = \dfrac{1}{19}$ 。

所以正确答案是 B。

The exponent of $2$ in $21!$ is $\lfloor \frac{21}{2} \rfloor$ $+ \lfloor \frac{21}{4} \rfloor$ $+ \lfloor \frac{21}{8} \rfloor$ $+ \lfloor \frac{21}{16} \rfloor$ $= 10 + 5 + 2 + 1$ $= 18.$ Every divisor has the form $2^i b$ with $0 \le i \le 18$ and $b$ odd; it is odd exactly when $i = 0.$ So the fraction of odd divisors is $\dfrac{1}{18 + 1} = \dfrac{1}{19}.$

Thus, the correct answer is B.

17.

一枚硬币有偏，且每次抛掷出现正面的概率为 $\dfrac{2}{3}$ ，出现反面的概率为 $\dfrac{1}{3}$ 。各次抛掷结果相互独立。一名玩家可以选择玩游戏 A 或游戏 B。在游戏 A 中，她抛掷硬币三次，若三次结果全相同则获胜。在游戏 B 中，她抛掷硬币四次，若第一次和第二次的结果相同且第三次和第四次的结果相同，则获胜。游戏 A 的获胜机会与游戏 B 的获胜机会相比如何？

A coin is biased in such a way that on each toss the probability of heads is $\dfrac{2}{3}$ and the probability of tails is $\dfrac{1}{3}.$ The outcomes of the tosses are independent. A player has the choice of playing Game A or Game B. In Game A she tosses the coin three times and wins if all three outcomes are the same. In Game B she tosses the coin four times and wins if both the outcomes of the first and second tosses are the same and the outcomes of the third and fourth tosses are the same. How do the chances of winning Game A compare to the chances of winning Game B?

游戏 A 的获胜概率比游戏 B 的获胜概率小 $\dfrac{4}{81}$ 。

The probability of winning Game A is $\dfrac{4}{81}$ less than the probability of winning Game B.

游戏 A 的获胜概率比游戏 B 的获胜概率小 $\dfrac{2}{81}$ 。

The probability of winning Game A is $\dfrac{2}{81}$ less than the probability of winning Game B.

两个概率相同。

The probabilities are the same.

游戏 A 的获胜概率比游戏 B 的获胜概率大 $\dfrac{2}{81}$ 。

The probability of winning Game A is $\dfrac{2}{81}$ greater than the probability of winning Game B.

游戏 A 的获胜概率比游戏 B 的获胜概率大 $\dfrac{4}{81}$ 。

The probability of winning Game A is $\dfrac{4}{81}$ greater than the probability of winning Game B.

知识点：独立事件基本概率

难度评级：1800

小提示：

游戏 A 的获胜概率为 $p^3 + (1-p)^3$ ；游戏 B 的获胜概率为 $\bigl(p^2 + (1-p)^2\bigr)^2$

Game A wins with probability $p^3 + (1-p)^3;$ Game B with $\bigl(p^2 + (1-p)^2\bigr)^2$

大提示：

代入 $p = \dfrac{2}{3}$ ，并计算两个概率之差

Substitute $p = \dfrac{2}{3}$ and subtract the two probabilities

解答：

令 $p = \dfrac23$ 。游戏 A 在三次抛掷结果全部相同时获胜，概率为 $p^3 + (1-p)^3$ 。游戏 B 要求第一对相同且第二对相同，每对相同的概率为 $p^2 + (1-p)^2$ ，所以获胜概率为 $\bigl(p^2 + (1-p)^2\bigr)^2$ 。当 $p = \tfrac23$ 时，游戏 A 的概率为 $\left(\tfrac23\right)^3 + \left(\tfrac13\right)^3 = \tfrac{9}{27} = \tfrac13$ ，游戏 B 的概率为 $\left(\tfrac49 + \tfrac19\right)^2 = \left(\tfrac59\right)^2 = \tfrac{25}{81}$ 。差为 $\tfrac{27}{81} - \tfrac{25}{81} = \tfrac{2}{81}$ ，所以游戏 A 的胜率高 $\tfrac{2}{81}$ 。

所以正确答案是 D。

Let $p = \dfrac23.$ Game A is won when all three tosses match: $p^3 + (1-p)^3.$ Game B needs the first pair to match and the second pair to match, each with probability $p^2 + (1-p)^2,$ so the win probability is $\bigl(p^2 + (1-p)^2\bigr)^2.$ With $p = \tfrac23,$ Game A gives $\left(\tfrac23\right)^3 + \left(\tfrac13\right)^3 = \tfrac{9}{27} = \tfrac13,$ and Game B gives $\left(\tfrac49 + \tfrac19\right)^2 = \left(\tfrac59\right)^2 = \tfrac{25}{81}.$ The difference is $\tfrac{27}{81} - \tfrac{25}{81} = \tfrac{2}{81},$ so Game A is $\tfrac{2}{81}$ more likely.

Thus, the correct answer is D.

18.

将半径为 $2$ 的圆的直径 $\overline{AB}$ 延长到圆外一点 $D$ ，使 $BD = 3$ 。点 $E$ 满足 $ED = 5$ ，且直线 $ED$ 垂直于直线 $AD$ 。线段 $\overline{AE}$ 与该圆相交于位于 $A$ 和 $E$ 之间的一点 $C$ 。 $\triangle ABC$ 的面积是多少？

The diameter $\overline{AB}$ of a circle of radius $2$ is extended to a point $D$ outside the circle so that $BD = 3.$ Point $E$ is chosen so that $ED = 5$ and line $ED$ is perpendicular to line $AD.$ Segment $\overline{AE}$ intersects the circle at a point $C$ between $A$ and $E.$ What is the area of $\triangle ABC?$

$\dfrac{120}{37}$

$\dfrac{140}{39}$

$\dfrac{145}{39}$

$\dfrac{140}{37}$

$\dfrac{120}{31}$

知识点：圆周角相似面积比

难度评级：1860

小提示：

$\angle ACB$ 是半圆上的圆周角，所以是直角；因此 $\triangle ABC \sim \triangle AED$ 。

$\angle ACB$ is inscribed in a semicircle, so it is a right angle; hence $\triangle ABC \sim \triangle AED$

大提示：

它们的面积比为 $AB^2 : AE^2$ ；计算 $AE^2 = AD^2 + ED^2$ 。

Their areas are in ratio $AB^2 : AE^2;$ compute $AE^2 = AD^2 + ED^2$

解答：

因为 $\angle ACB$ 是半圆上的圆周角，所以它是直角，因此 $\triangle ABC \sim \triangle AED$ （都是直角三角形且共享角 $A$ ）。它们的面积比为 $AB^2 : AE^2$ 。这里 $AB = 4$ ，所以 $AB^2 = 16$ ，且 $AD = AB + BD = 7$ ，因此 $AE^2 = AD^2 + ED^2$ $= 49 + 25$ $= 74$ 。 $\triangle AED$ 的面积为 $\tfrac12 \cdot 7 \cdot 5 = \tfrac{35}{2}$ 。所以 $[\triangle ABC] = \frac{16}{74} \cdot \frac{35}{2} = \frac{140}{37}\text{。}$

所以正确答案是 D。

Since $\angle ACB$ is inscribed in a semicircle, it is a right angle, so $\triangle ABC \sim \triangle AED$ (both right-angled and sharing angle $A$ ). Their areas are in ratio $AB^2 : AE^2.$ Here $AB = 4,$ so $AB^2 = 16,$ and $AD = AB + BD = 7,$ so $AE^2 = AD^2 + ED^2$ $= 49 + 25$ $= 74.$ The area of $\triangle AED$ is $\tfrac12 \cdot 7 \cdot 5 = \tfrac{35}{2}.$ Thus $[\triangle ABC] = \frac{16}{74} \cdot \frac{35}{2} = \frac{140}{37}.$

Thus, the correct answer is D.

19.

设 $N = 123456789101112\ldots4344$ 是把整数 $1$ 到 $44$ 按顺序一个接一个写下形成的 $79$ 位数。 $N$ 除以 $45$ 的余数是多少？

Let $N = 123456789101112\ldots4344$ be the $79$ -digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45?$

$1$

$4$

$9$

$18$

$44$

知识点：模运算整除性数字

难度评级：1910

小提示：

因为 $45 = 9 \cdot 5$ ，分别求 $N \bmod 5$ （最后一位）和 $N \bmod 9$ （数字和）

Since $45 = 9 \cdot 5,$ find $N \bmod 5$ (the last digit) and $N \bmod 9$ (the digit sum)

大提示：

数字和是 $9$ 的倍数，所以 $N \equiv 0 \pmod 9$ ；再结合模 $5$ 的余数

The digit sum is a multiple of $9,$ so $N \equiv 0 \pmod 9;$ combine with the residue mod $5$

解答：

$N$ 的最后一位是 $4$ ，所以 $N \equiv 4 \pmod 5$ 。对模 $9$ ，求数字和：数字 $1$ 到 $9$ 贡献它们自身的数字， $10$ 到 $44$ 的十位数字与个位数字合计为 $270$ ，这是 $9$ 的倍数，所以 $N \equiv 0 \pmod 9$ 。因此 $N - 9$ 是 $9$ 的倍数，且它的最后一位是 $5$ ，所以也是 $5$ 的倍数；于是 $N - 9$ 是 $45$ 的倍数。因此 $N \equiv 9 \pmod{45}$ 。

所以正确答案是 C。

The last digit of $N$ is $4,$ so $N \equiv 4 \pmod 5.$ For mod $9,$ sum the digits: the numbers $1$ – $9$ contribute their digits, the tens digits of $10$ – $44$ and the units digits together sum to $270,$ which is a multiple of $9,$ so $N \equiv 0 \pmod 9.$ The number $N - 9$ is then a multiple of $9,$ and its last digit is $5,$ so it is a multiple of $5;$ hence $N - 9$ is a multiple of $45.$ Therefore $N \equiv 9 \pmod{45}.$

Thus, the correct answer is C.

20.

实数 $x$ 和 $y$ 独立地从区间 $(0, 1)$ 中均匀随机选取。若 $\lfloor r \rfloor$ 表示小于或等于实数 $r$ 的最大整数，那么 $\lfloor \log_2 x \rfloor = \lfloor \log_2 y \rfloor$ 的概率是多少？

Real numbers $x$ and $y$ are chosen independently and uniformly at random from the interval $(0, 1).$ What is the probability that $\lfloor \log_2 x \rfloor = \lfloor \log_2 y \rfloor,$ where $\lfloor r \rfloor$ denotes the greatest integer less than or equal to the real number $r?$

$\dfrac{1}{8}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

知识点：几何概率取整函数等比数列

难度评级：1990

小提示：

$\lfloor \log_2 x \rfloor = -n$ 当且仅当 $\dfrac{1}{2^n} \le x \lt \dfrac{1}{2^{n-1}}$ 。

$\lfloor \log_2 x \rfloor = -n$ exactly when $\dfrac{1}{2^n} \le x \lt \dfrac{1}{2^{n-1}}$

大提示：

对每个 $n$ ， $x$ 和 $y$ 都必须落在这个长度为 $\dfrac{1}{2^n}$ 的区间内；把对应面积相加。

For each $n,$ both $x$ and $y$ must land in that interval of length $\dfrac{1}{2^n};$ sum the resulting areas

解答：

对每个正整数 $n$ ， $\lfloor \log_2 x \rfloor = -n$ 当且仅当 $\dfrac{1}{2^n} \le x \lt \dfrac{1}{2^{n-1}}$ ，这是长度为 $\dfrac{1}{2^n}$ 的区间。两个取整值都等于 $-n$ 的事件是面积为 $\dfrac{1}{4^n}$ 的正方形。对所有 $n$ 求和，概率为 $\sum_{n=1}^{\infty} \frac{1}{4^n} = \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{1}{3}\text{。}$

所以正确答案是 D。

For each positive integer $n,$ $\lfloor \log_2 x \rfloor = -n$ exactly when $\dfrac{1}{2^n} \le x \lt \dfrac{1}{2^{n-1}},$ an interval of length $\dfrac{1}{2^n}.$ The event that both floors equal $-n$ is a square of area $\dfrac{1}{4^n}.$ Summing over all $n,$ the probability is $\sum_{n=1}^{\infty} \frac{1}{4^n} = \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{1}{3}.$

Thus, the correct answer is D.

21.

去年 Isabella 参加了 $7$ 次数学测验，得到 $7$ 个互不相同的分数，每个分数都是 $91$ 到 $100$ 之间（含端点）的整数。每次测验后，她都注意到到目前为止所有测验分数的平均数是整数。她第七次测验的分数是 $95$ 。她第六次测验的分数是多少？

Last year Isabella took $7$ math tests and received $7$ different scores, each an integer between $91$ and $100,$ inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was $95.$ What was her score on the sixth test?

$92$

$94$

$96$

$98$

$100$

知识点：整除性平均数极限情形界定

难度评级：2040

小提示：

所有 $7$ 个分数之和 $S$ 是 $7$ 的倍数，并且介于 $658$ 和 $679$ 之间

The sum $S$ of all $7$ scores is a multiple of $7$ and lies between $658$ and $679$

大提示：

还要有 $S - 95$ 是 $6$ 的倍数；然后利用前 $5$ 个分数之和是 $5$ 的倍数，限制第六个分数的余数。

Also $S - 95$ is a multiple of $6;$ then the first $5$ scores sum to a multiple of $5,$ forcing the sixth score’s residue

解答：

令 $S$ 为七个分数之和。则 $S$ 是 $7$ 的倍数且 $658 \le S \le 679$ ，所以 $S \in \{658, 665, 672, 679\}$ 。因为前六次测验后的平均数是整数， $S - 95$ 必须是 $6$ 的倍数，这迫使 $S = 665$ 。因此前六个分数之和为 $570$ ，它是 $5$ 的倍数。前五次测验后的平均数也是整数，所以前五个分数之和是 $5$ 的倍数，于是第六个分数也是 $5$ 的倍数。由于所有分数互不相同且第七个分数为 $95$ ，第六个分数必须是 $100$ 。

所以正确答案是 E。

Let $S$ be the sum of all seven scores. Then $S$ is a multiple of $7$ with $658 \le S \le 679,$ so $S \in \{658, 665, 672, 679\}.$ Since the average after six tests is an integer, $S - 95$ is a multiple of $6,$ which forces $S = 665.$ Then the first six scores sum to $570,$ a multiple of $5;$ the average after five tests is an integer, so the first five scores also sum to a multiple of $5,$ making the sixth score a multiple of $5.$ Since all scores differ and the seventh is $95,$ the sixth must be $100.$

Thus, the correct answer is E.

22.

Abby、Bernardo、Carl 和 Debra 玩一个游戏，每个人开始时都有四枚硬币。游戏共四轮。每一轮，把四个球放入一个瓮中：一个绿色、一个红色、两个白色。玩家们依次随机抽球且不放回。抽到绿球的人给抽到红球的人一枚硬币。第四轮结束时，每位玩家都有四枚硬币的概率是多少？

Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn—one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?

$\dfrac{7}{576}$

$\dfrac{5}{192}$

$\dfrac{1}{36}$

$\dfrac{5}{144}$

$\dfrac{7}{48}$

知识点：基本概率分类讨论

难度评级：2330

小提示：

每轮有 $4 \cdot 3 = 12$ 个等可能的（给出者，接收者）结果，所以共有 $12^4$ 个结果序列

Each round has $4 \cdot 3 = 12$ equally likely (giver, receiver) outcomes, for $12^4$ total

大提示：

统计能让每个人回到 $4$ 枚硬币的结果序列：每个人的净转移必须抵消；按交换模式分类

Count outcome sequences that return everyone to $4$ coins: each player’s net transfers must cancel; split into cases by the pattern of exchanges

解答：

每轮有 $4 \cdot 3 = 12$ 个等可能的（给出者，接收者）有序对，所以共有 $12^4$ 个结果序列。每个人最后都有四枚硬币，当且仅当四次转移相互抵消。有利模式为：一次 $4$ -循环赠送（ $24 \cdot 6 = 144$ 种），两组不相交的相互交换（ $24 \cdot 3 = 72$ 种），同一对玩家互相交换两次（ $6 \cdot 6 = 36$ 种），以及某个玩家既给另外两人各一枚又从这两人各收一枚（ $4 \cdot 3 \cdot 24 = 288$ 种）。总数为 $144 + 72 + 36 + 288 = 540$ ，概率为 $\dfrac{540}{12^4} = \dfrac{540}{20736} = \dfrac{5}{192}$ 。

所以正确答案是 B。

Each round has $4 \cdot 3 = 12$ equally likely (giver, receiver) pairs, so there are $12^4$ outcome sequences. Everyone ends with four coins exactly when the four transfers cancel. The favorable patterns are: a $4$ -cycle of gifts ( $24 \cdot 6 = 144$ ways), two disjoint mutual exchanges ( $24 \cdot 3 = 72$ ), one pair exchanging twice ( $6 \cdot 6 = 36$ ), and one player both giving to and receiving from each of two others ( $4 \cdot 3 \cdot 24 = 288$ ). These total $144 + 72 + 36 + 288 = 540.$ The probability is $\dfrac{540}{12^4} = \dfrac{540}{20736} = \dfrac{5}{192}.$

Thus, the correct answer is B.

23.

在 $y = f(x)$ 的图像中， $f(x)$ 是一个 $3$ 次多项式，且图像包含点 $A(2, 4)$ 、 $B(3, 9)$ 和 $C(4, 16)$ 。直线 $AB$ 、 $AC$ 和 $BC$ 分别再次与图像相交于点 $D$ 、 $E$ 和 $F$ 。若 $D$ 、 $E$ 和 $F$ 的 $x$ -坐标之和为 $24$ ，那么 $f(0)$ 是多少？

The graph of $y = f(x),$ where $f(x)$ is a polynomial of degree $3,$ contains points $A(2, 4),$ $B(3, 9),$ and $C(4, 16).$ Lines $AB,$ $AC,$ and $BC$ intersect the graph again at points $D,$ $E,$ and $F,$ respectively, and the sum of the $x$ -coordinates of $D,$ $E,$ and $F$ is $24.$ What is $f(0)?$

$-2$

$0$

$2$

$\dfrac{24}{5}$

$8$

知识点：韦达定理多项式

难度评级：2370

小提示：

因为 $A, B, C$ 在 $y = x^2$ 上，令 $g(x) = f(x) - x^2$ ，则 $g(x) = a(x-2)(x-3)(x-4)$

Since $A, B, C$ lie on $y = x^2,$ let $g(x) = f(x) - x^2,$ so $g(x) = a(x-2)(x-3)(x-4)$

大提示：

一条直线与三次曲线相交时满足 $f(x) - L(x) = 0$ ；由韦达定理，每组三个 $x$ -坐标之和为 $9 - \tfrac1a$ 。

A line meets the cubic where $f(x) - L(x) = 0;$ by Vieta the three $x$ -coordinates of each such triple sum to $9 - \tfrac1a$

解答：

点 $A, B, C$ 都在 $y = x^2$ 上，所以 $g(x) = f(x) - x^2$ 以 $2, 3, 4$ 为根，即 $g(x) = a(x-2)(x-3)(x-4)$ ，其中 $a \ne 0$ 。 $x^3$ 和 $x^2$ 在 $f$ 中的系数分别为 $a$ 和 $1 - 9a$ ，因此由韦达定理， $f(x) - L(x)$ （对任意一次函数 $L$ ）的三个根之和为 $9 - \tfrac1a$ 。直线 $AB, AC, BC$ 与三次曲线相交的三个横坐标集合分别为 $\{2, 3, x_D\}$ 、 $\{2, 4, x_E\}$ 、 $\{3, 4, x_F\}$ ，所以 $\begin{aligned} &x_D + x_E + x_F \\ &\quad {}= 3\left(9 - \tfrac1a\right) \\ &\quad {}- 2(2 + 3 + 4) \\ &\quad {}= 9 - \tfrac3a = 24\text{，} \end{aligned}$ 解得 $a = -\tfrac15$ 。因此 $f(x) = x^2$ $- \tfrac15(x-2)(x-3)(x-4)$ ，所以 $f(0) = 0 - \tfrac15(-2)(-3)(-4)$ $= \tfrac{24}{5}$ 。

所以正确答案是 D。

The points $A, B, C$ lie on $y = x^2,$ so $g(x) = f(x) - x^2$ has roots $2, 3, 4:$ $g(x) = a(x-2)(x-3)(x-4)$ for some $a \ne 0.$ The coefficients of $x^3$ and $x^2$ in $f$ are $a$ and $1 - 9a,$ so by Vieta the three roots of $f(x) - L(x)$ (for any linear $L$ ) sum to $9 - \tfrac1a.$ The lines $AB, AC, BC$ meet the cubic in triples $\{2, 3, x_D\},$ $\{2, 4, x_E\},$ $\{3, 4, x_F\},$ so $\begin{aligned} &x_D + x_E + x_F \\ &\quad {}= 3\left(9 - \tfrac1a\right) \\ &\quad {}- 2(2 + 3 + 4) \\ &\quad {}= 9 - \tfrac3a = 24, \end{aligned}$ giving $a = -\tfrac15.$ Then $f(x) = x^2$ $- \tfrac15(x-2)(x-3)(x-4),$ so $f(0) = 0 - \tfrac15(-2)(-3)(-4)$ $= \tfrac{24}{5}.$

Thus, the correct answer is D.

24.

四边形 $ABCD$ 在 $B$ 和 $C$ 处有直角， $\triangle ABC \sim \triangle BCD$ ，且 $AB \gt BC$ 。在 $ABCD$ 内部有一点 $E$ ，使得 $\triangle ABC \sim \triangle CEB$ ，并且 $\triangle AED$ 的面积是 $\triangle CEB$ 面积的 $17$ 倍。 $\dfrac{AB}{BC}$ 是多少？

Quadrilateral $ABCD$ has right angles at $B$ and $C,$ $\triangle ABC \sim \triangle BCD,$ and $AB \gt BC.$ There is a point $E$ in the interior of $ABCD$ such that $\triangle ABC \sim \triangle CEB$ and the area of $\triangle AED$ is $17$ times the area of $\triangle CEB.$ What is $\dfrac{AB}{BC}?$

$1 + \sqrt{2}$

$2 + \sqrt{2}$

$\sqrt{17}$

$2 + \sqrt{5}$

$1 + 2\sqrt{3}$

知识点：相似坐标几何鞋带公式

难度评级：2550

小提示：

令 $BC = 1$ ， $AB = r$ ，并取 $C = (0,0)$ ， $B = (0,1)$ ， $A = (r,1)$ ， $D = (\tfrac1r, 0)$

Set $BC = 1,$ $AB = r,$ and place $C = (0,0),$ $B = (0,1),$ $A = (r,1),$ $D = (\tfrac1r, 0)$

大提示：

由 $\triangle ABC \sim \triangle CEB$ ，求出 $E$ ，计算两个面积，并令 $[\triangle AED] = 17[\triangle CEB]$ 。

Find $E$ from $\triangle ABC \sim \triangle CEB,$ compute both areas, and set $[\triangle AED] = 17[\triangle CEB]$

解答：

令 $BC = 1$ ， $AB = r \gt 1$ 。由 $\triangle ABC \sim \triangle BCD$ 以及直角位置，可取 $C = (0,0)$ ， $B = (0,1)$ ， $A = (r,1)$ ， $D = \bigl(\tfrac1r, 0\bigr)$ 。设 $E = (x, y)$ ，其中 $x, y \gt 0$ 。由 $\triangle ABC \sim \triangle CEB$ 得 $\dfrac{x}{y} = \tan(\angle ECB)$ $= \tan(\angle BAC)$ $= \dfrac1r$ 且 $x^2 + y^2 = \dfrac{r^2}{1 + r^2}$ ，所以 $x = \dfrac{r}{1+r^2}$ ， $y = \dfrac{r^2}{1+r^2}$ 。这两个相关的面积为 $\begin{aligned} [\triangle CEB]&=\dfrac{r}{2(1+r^2)},\\ [\triangle AED]&=\dfrac{r^4-r^2+1}{2r(1+r^2)} \end{aligned}\text{。}$ 令后者等于前者的 $17$ 倍，得到 $r^4 - 18r^2 + 1 = 0$ 。于是 $r^2 = 9 + 4\sqrt5 = (2 + \sqrt5)^2$ ，所以 $r = 2 + \sqrt5$ 。

所以正确答案是 D。

Set $BC = 1$ and $AB = r \gt 1.$ The similarity $\triangle ABC \sim \triangle BCD$ with the right angles places the figure at $C = (0,0),$ $B = (0,1),$ $A = (r,1),$ $D = \bigl(\tfrac1r, 0\bigr).$ Let $E = (x, y)$ with $x, y \gt 0.$ From $\triangle ABC \sim \triangle CEB$ we get $\dfrac{x}{y} = \tan(\angle ECB)$ $= \tan(\angle BAC)$ $= \dfrac1r$ and $x^2 + y^2 = \dfrac{r^2}{1 + r^2},$ so $x = \dfrac{r}{1+r^2},$ $y = \dfrac{r^2}{1+r^2}.$ The two relevant areas are $\begin{aligned} [\triangle CEB]&=\dfrac{r}{2(1+r^2)},\\ [\triangle AED]&=\dfrac{r^4-r^2+1}{2r(1+r^2)}. \end{aligned}$ Setting the second equal to $17$ times the first gives $r^4 - 18r^2 + 1 = 0.$ Then $r^2 = 9 + 4\sqrt5 = (2 + \sqrt5)^2,$ so $r = 2 + \sqrt5.$

Thus, the correct answer is D.

25.

有 $n$ 个人参加一个在线视频篮球锦标赛。每个人可以属于任意数量的 $5$ 人球队，但任意两支球队不能有完全相同的 $5$ 名成员。网站统计显示一个有趣的事实：在所有由这 $n$ 名参赛者组成的 $9$ 人子集中，这 $9$ 人内完整球队数量的平均值，等于在所有由 $n$ 名参赛者组成的 $8$ 人子集中，这 $8$ 人内完整球队数量平均值的倒数。满足 $9 \le n \le 2017$ 的参赛人数 $n$ 可以有多少个值？

A set of $n$ people participate in an online video basketball tournament. Each person may be a member of any number of $5$ -player teams, but no two teams may have exactly the same $5$ members. The site statistics show a curious fact: The average, over all subsets of size $9$ of the set of $n$ participants, of the number of complete teams whose members are among those $9$ people is equal to the reciprocal of the average, over all subsets of size $8$ of the set of $n$ participants, of the number of complete teams whose members are among those $8$ people. How many values $n,$ $9 \le n \le 2017,$ can be the number of participants?

$477$

$482$

$487$

$557$

$562$

知识点：双重计数整除性中国剩余定理

难度评级：2650

小提示：

设球队数为 $T$ 。每支球队在 $9$ 人子集求和中被计数 $\binom{n-5}{4}$ 次，在 $8$ 人子集求和中被计数 $\binom{n-5}{3}$ 次

Let $T$ be the number of teams. Each team is counted $\binom{n-5}{4}$ times in the size- $9$ sum and $\binom{n-5}{3}$ times in the size- $8$ sum

大提示：

条件化为 $T =$ $\dfrac{n(n-1)(n-2)(n-3)(n-4)}{2^5 \cdot 3^2 \cdot 5 \cdot 7}$ ；统计使它为整数的 $n$ 。

The condition becomes $T =$ $\dfrac{n(n-1)(n-2)(n-3)(n-4)}{2^5 \cdot 3^2 \cdot 5 \cdot 7};$ count $n$ making this an integer

解答：

设球队数为 $T$ 。对 $9$ 人子集求和时，每支球队被计数 $\binom{n-5}{4}$ 次；对 $8$ 人子集求和时，每支球队被计数 $\binom{n-5}{3}$ 次。两个平均值分别为 $\dfrac{\binom{n-5}{4}T}{\binom n9}$ 和 $\dfrac{\binom{n-5}{3}T}{\binom n8}$ 。令第一个等于第二个的倒数并化简，得到 $\begin{aligned} &T \\ &\quad {}= \scriptsize \frac{n(n-1)(n-2)(n-3)(n-4)}{2^5 \cdot 3^2 \cdot 5 \cdot 7} \end{aligned}\text{。}$ 我们需要它在 $n \ge 9$ 时为正整数。令 $N =$ $n(n-1)(n-2)(n-3)(n-4)$ ；作为五个连续整数的乘积， $N$ 总能被 $5$ 整除。 $N$ 是 $7$ 的倍数这一条件对模 $7$ 的 $5$ 个余数成立；是 $9$ 的倍数对模 $9$ 的 $7$ 个余数成立；是 $32$ 的倍数对模 $16$ 的 $8$ 个余数成立。因此由中国剩余定理，模 $\operatorname{lcm}(7,9,16)=1008$ 共有 $5\cdot7\cdot8=280$ 个解。于是 $1 \le n \le 2016$ 中有 $560$ 个值；去掉 $n = 1, 2, 3, 4$ （它们小于 $9$ ），再加上 $n = 2017$ （因为 $2017 \equiv 1 \pmod{1008}$ ），得到 $560 - 4 + 1 = 557$ 个有效值。

所以正确答案是 D。

Let $T$ be the number of teams. Summing over size- $9$ subsets counts each team $\binom{n-5}{4}$ times and over size- $8$ subsets $\binom{n-5}{3}$ times. The averages are $\dfrac{\binom{n-5}{4}T}{\binom n9}$ and $\dfrac{\binom{n-5}{3}T}{\binom n8};$ setting the first equal to the reciprocal of the second and simplifying gives $\begin{aligned} &T \\ &\quad {}= \scriptsize \frac{n(n-1)(n-2)(n-3)(n-4)}{2^5 \cdot 3^2 \cdot 5 \cdot 7}. \end{aligned}$ We need this to be a positive integer with $n \ge 9.$ Let $N =$ $n(n-1)(n-2)(n-3)(n-4);$ as a product of five consecutive integers, $N$ is always divisible by $5.$ The condition that $N$ is divisible by $7$ holds for $5$ residues modulo $7;$ divisibility by $9$ holds for $7$ residues modulo $9;$ and divisibility by $32$ holds for $8$ residues modulo $16.$ The Chinese Remainder Theorem therefore gives $5\cdot7\cdot8=280$ solutions modulo $\operatorname{lcm}(7,9,16)=1008.$ So there are $560$ values in $1 \le n \le 2016;$ removing $n = 1, 2, 3, 4$ (which are below $9$ ) and adding $n = 2017$ (since $2017 \equiv 1 \pmod{1008}$ ) gives $560 - 4 + 1 = 557$ valid values.

Thus, the correct answer is D.