2020 AIME II 详解 | LIVE by Po-Shen Loh

1.

求正整数有序对 $(m, n)$ 的个数，使得 $m^2 n = 20^{20}$ 。

Find the number of ordered pairs of positive integers $(m, n)$ such that $m^2 n = 20^{20}.$

答案

知识点：质因数分解基本计数

难度评级：1890

小提示：

写成 $20^{20} = 2^{40} \cdot 5^{20}$ ，并注意一旦选定 $m$ ， $n$ 就完全确定

Write $20^{20} = 2^{40} \cdot 5^{20},$ and note that once $m$ is chosen, $n$ is completely determined

大提示：

需要 $m^2$ 整除 $2^{40} \cdot 5^{20}$ ：分别数出 $m$ 中 $2$ 和 $5$ 的指数可能值

You need $m^2$ to divide $2^{40} \cdot 5^{20}:$ count the possible exponents of $2$ and of $5$ in $m$

解答：

因为 $20^{20} = 2^{40} \cdot 5^{20}$ ，有效的 $m$ 必须形如 $2^a 5^b$ ，此时 $n = 2^{40 - 2a}\,5^{20 - 2b}$ 是正整数，当且仅当 $2a \le 40$ 且 $2b \le 20$ 。反过来，每个这样的选择都可行，并且 $n$ 由 $m$ 唯一确定。

因此 $a$ 可以取 $0, 1, \ldots, 20$ 中任意一个值， $b$ 可以取 $0, 1, \ldots, 10$ 中任意一个值，共有 $21 \cdot 11 = 231$ 个有序对。

Since $20^{20} = 2^{40} \cdot 5^{20},$ a valid $m$ must have the form $2^a 5^b,$ and then $n = 2^{40 - 2a}\,5^{20 - 2b}$ is a positive integer exactly when $2a \le 40$ and $2b \le 20.$ Conversely every such choice works, and $n$ is uniquely determined by $m.$

So $a$ can be any of $0, 1, \ldots, 20$ and $b$ any of $0, 1, \ldots, 10,$ giving $21 \cdot 11 = 231$ ordered pairs.

2.

在顶点为 $(0, 0)$ 、 $(1, 0)$ 、 $(1, 1)$ 、 $(0, 1)$ 的单位正方形内部，均匀随机选取一点 $P$ 。由 $P$ 与点 $\left(\frac{5}{8}, \frac{3}{8}\right)$ 确定的直线斜率大于或等于 $\frac{1}{2}$ 的概率可写成 $\frac{m}{n}$ ，其中 $m$ 与 $n$ 是互质正整数。求 $m + n$ 。

Let $P$ be a point chosen uniformly at random in the interior of the unit square with vertices at $(0, 0),$ $(1, 0),$ $(1, 1),$ and $(0, 1).$ The probability that the slope of the line determined by $P$ and the point $\left(\frac{5}{8}, \frac{3}{8}\right)$ is greater than or equal to $\frac{1}{2}$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

知识点：几何概率斜率梯形

难度评级：2110

小提示：

对 $P = (x, y)$ ，当 $x \gt \frac{5}{8}$ 时，条件是 $y \ge \frac{x}{2} + \frac{1}{16}$ ；当 $x \lt \frac{5}{8}$ 时，不等号方向会反过来

For $P = (x, y)$ with $x \gt \frac{5}{8},$ the condition is $y \ge \frac{x}{2} + \frac{1}{16};$ for $x \lt \frac{5}{8}$ the inequality direction flips

大提示：

有利区域是两个在 $\left(\frac{5}{8}, \frac{3}{8}\right)$ 处相接的梯形；分别求面积后相加

The favorable region is two trapezoids meeting at $\left(\frac{5}{8}, \frac{3}{8}\right);$ find each area and add

解答：

设 $Q = \left(\frac{5}{8}, \frac{3}{8}\right)$ ，且 $P = (x, y)$ 。斜率条件 $\frac{y - \frac{3}{8}}{x - \frac{5}{8}} \ge \frac{1}{2}$ 在 $x \gt \frac{5}{8}$ 时变为 $y \ge \frac{x}{2} + \frac{1}{16}$ ，在 $x \lt \frac{5}{8}$ 时变为 $y \le \frac{x}{2} + \frac{1}{16}$ （乘以负数 $x - \frac{5}{8}$ 会反转不等号）。

当 $x \gt \frac{5}{8}$ 时，正方形内位于直线 $y = \frac{x}{2} + \frac{1}{16}$ 上方的区域是一个梯形，平行的竖边长度分别为 $\frac{5}{8}$ （在 $x = \frac{5}{8}$ 处）和 $\frac{7}{16}$ （在 $x = 1$ 处），宽为 $\frac{3}{8}$ ，面积为 $\frac{3}{8} \cdot \frac{\frac{5}{8} + \frac{7}{16}}{2} = \frac{51}{256}$ 。当 $x \lt \frac{5}{8}$ 时，直线下方的区域是一个梯形，平行边长度分别为 $\frac{1}{16}$ （在 $x = 0$ 处）和 $\frac{3}{8}$ （在 $x = \frac{5}{8}$ 处），宽为 $\frac{5}{8}$ ，面积为 $\frac{5}{8} \cdot \frac{\frac{1}{16} + \frac{3}{8}}{2} = \frac{35}{256}$ 。

概率为 $\frac{51}{256} + \frac{35}{256} = \frac{86}{256} = \frac{43}{128}$ ，所以 $m + n = 43 + 128 = 171$ 。

Let $Q = \left(\frac{5}{8}, \frac{3}{8}\right)$ and $P = (x, y).$ The slope condition $\frac{y - \frac{3}{8}}{x - \frac{5}{8}} \ge \frac{1}{2}$ becomes $y \ge \frac{x}{2} + \frac{1}{16}$ when $x \gt \frac{5}{8},$ and $y \le \frac{x}{2} + \frac{1}{16}$ when $x \lt \frac{5}{8}$ (multiplying by the negative quantity $x - \frac{5}{8}$ reverses the inequality).

For $x \gt \frac{5}{8},$ the region above the line $y = \frac{x}{2} + \frac{1}{16}$ inside the square is a trapezoid with parallel vertical sides of lengths $\frac{5}{8}$ (at $x = \frac{5}{8}$ ) and $\frac{7}{16}$ (at $x = 1$ ) and width $\frac{3}{8},$ with area $\frac{3}{8} \cdot \frac{\frac{5}{8} + \frac{7}{16}}{2} = \frac{51}{256}.$ For $x \lt \frac{5}{8},$ the region below the line is a trapezoid with parallel sides $\frac{1}{16}$ (at $x = 0$ ) and $\frac{3}{8}$ (at $x = \frac{5}{8}$ ) and width $\frac{5}{8},$ with area $\frac{5}{8} \cdot \frac{\frac{1}{16} + \frac{3}{8}}{2} = \frac{35}{256}.$

The probability is $\frac{51}{256} + \frac{35}{256} = \frac{86}{256} = \frac{43}{128},$ so $m + n = 43 + 128 = 171.$

3.

满足 $\log_{2^x} 3^{20} = \log_{2^{x+3}} 3^{2020}$ 的 $x$ 可写成 $\frac{m}{n}$ ，其中 $m$ 与 $n$ 是互质正整数。求 $m + n$ 。

The value of $x$ that satisfies $\log_{2^x} 3^{20} = \log_{2^{x+3}} 3^{2020}$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

知识点：对数分式方程

难度评级：1950

小提示：

换底： $\log_{2^x} 3^{20} = \frac{20 \log 3}{x \log 2}$ ，右边也类似

Change of base: $\log_{2^x} 3^{20} = \frac{20 \log 3}{x \log 2},$ and similarly for the right side

大提示：

方程化为 $\frac{20}{x} = \frac{2020}{x + 3}$ ，这是关于 $x$ 的一次方程

The equation reduces to $\frac{20}{x} = \frac{2020}{x + 3},$ which is linear in $x$

解答：

由换底公式， $\log_{2^x} 3^{20} = \frac{20 \log 3}{x \log 2}$ 且 $\log_{2^{x+3}} 3^{2020} = \frac{2020 \log 3}{(x + 3) \log 2}\text{。}$ 约去公共因子 $\frac{\log 3}{\log 2}$ ，得到 $\frac{20}{x} = \frac{2020}{x + 3}$ 。

交叉相乘得 $20x + 60 = 2020x$ ，所以 $2000x = 60$ ，从而 $x = \frac{3}{100}$ 。因此 $m + n = 3 + 100 = 103$ 。

By the change-of-base formula, $\log_{2^x} 3^{20} = \frac{20 \log 3}{x \log 2}$ and $\log_{2^{x+3}} 3^{2020} = \frac{2020 \log 3}{(x + 3) \log 2}.$ Cancelling the common factor $\frac{\log 3}{\log 2}$ leaves $\frac{20}{x} = \frac{2020}{x + 3}.$

Cross-multiplying gives $20x + 60 = 2020x,$ so $2000x = 60$ and $x = \frac{3}{100}.$ Thus $m + n = 3 + 100 = 103.$

4.

三角形 $\triangle ABC$ 与 $\triangle A'B'C'$ 位于坐标平面内，顶点分别为 $A(0, 0)$ 、 $B(0, 12)$ 、 $C(16, 0)$ 、 $A'(24, 18)$ 、 $B'(36, 18)$ 、 $C'(24, 2)$ 。绕点 $(x, y)$ 顺时针旋转 $m$ 度，其中 $0 \lt m \lt 180$ ，会把 $\triangle ABC$ 变换为 $\triangle A'B'C'$ 。求 $m + x + y$ 。

Triangles $\triangle ABC$ and $\triangle A'B'C'$ lie in the coordinate plane with vertices $A(0, 0),$ $B(0, 12),$ $C(16, 0),$ $A'(24, 18),$ $B'(36, 18),$ $C'(24, 2).$ A rotation of $m$ degrees clockwise around the point $(x, y),$ where $0 \lt m \lt 180,$ will transform $\triangle ABC$ to $\triangle A'B'C'.$ Find $m + x + y.$

答案

知识点：变换坐标几何

难度评级：2300

小提示：

线段 $AB$ 是竖直的，而它的像 $A'B'$ 是水平的，所以旋转角必须是 $90$ 度

Segment $AB$ is vertical while its image $A'B'$ is horizontal, so the rotation must be by $90$ degrees

大提示：

绕 $(a, b)$ 顺时针旋转 $90^\circ$ 会把 $(p, q)$ 送到 $(a + q - b,\ b - p + a)$ ；匹配 $A \mapsto A'$

A $90^\circ$ clockwise rotation about $(a, b)$ sends $(p, q)$ to $(a + q - b,\ b - p + a);$ match $A \mapsto A'$

解答：

向量 $\overrightarrow{AB} = (0, 12)$ 是竖直的，而 $\overrightarrow{A'B'} = (12, 0)$ 是水平的且长度相同，所以旋转把方向顺时针转了 $90^\circ$ ，因此 $m = 90$ 。

绕 $(a, b)$ 顺时针旋转 $90^\circ$ 会把 $(p, q)$ 送到 $(a + q - b,\ b - p + a)$ 。将其用于 $A = (0, 0)$ ，并令像等于 $A' = (24, 18)$ ，得到 $a - b = 24$ 且 $a + b = 18$ ，所以 $a = 21$ ， $b = -3$ 。检查另外两个顶点： $B = (0, 12)$ 映到 $(21 + 12 + 3,\ -3 + 21)$ $= (36, 18) = B'$ ，且 $C = (16, 0)$ 映到 $(21 + 3,\ -3 - 16 + 21)$ $= (24, 2) = C'$ 。

因此 $m + x + y = 90$ $+ 21$ $+ (-3) = 108$ 。

The vector $\overrightarrow{AB} = (0, 12)$ is vertical, while $\overrightarrow{A'B'} = (12, 0)$ is horizontal and of the same length, so the rotation turns directions by $90^\circ$ clockwise, and $m = 90.$

A $90^\circ$ clockwise rotation about $(a, b)$ sends $(p, q)$ to $(a + q - b,\ b - p + a).$ Applying this to $A = (0, 0)$ and setting the image equal to $A' = (24, 18)$ gives $a - b = 24$ and $a + b = 18,$ so $a = 21$ and $b = -3.$ Checking the other vertices: $B = (0, 12)$ maps to $(21 + 12 + 3,\ -3 + 21)$ $= (36, 18) = B',$ and $C = (16, 0)$ maps to $(21 + 3,\ -3 - 16 + 21)$ $= (24, 2) = C'.$

Therefore $m + x + y = 90$ $+ 21$ $+ (-3) = 108.$

5.

对每个正整数 $n$ ，令 $f(n)$ 为 $n$ 以四为底表示时的各位数字之和，令 $g(n)$ 为 $f(n)$ 以八为底表示时的各位数字之和。例如， $f(2020) = f(133210_4) = 10$ $= 12_8$ ，而 $12_8$ 的各位数字之和为 $3$ ，所以 $g(2020) = 3$ 。令 $N$ 为使得 $g(n)$ 以十六为底的表示不能只用数字 $0$ 到 $9$ 表示的最小 $n$ 。求 $N$ 除以 $1000$ 的余数。

For each positive integer $n,$ let $f(n)$ be the sum of the digits in the base-four representation of $n$ and let $g(n)$ be the sum of the digits in the base-eight representation of $f(n).$ For example, $f(2020) = f(133210_4) = 10$ $= 12_8,$ and the digit sum of $12_8$ is $3,$ so $g(2020) = 3.$ Let $N$ be the least value of $n$ such that the base-sixteen representation of $g(n)$ cannot be expressed using only the digits $0$ through $9.$ Find the remainder when $N$ is divided by $1000.$

答案

知识点：进制数字逆推法

难度评级：2450

小提示：

在十六进制表示中需要使用大于 $9$ 的数字的最小 $g(n)$ 是 $10$ ；先找八进制数位和为 $10$ 的最小 $f(n)$

The smallest value of $g(n)$ whose base-sixteen representation uses a digit beyond $9$ is $10;$ first find the smallest possible $f(n)$ whose base-eight digit sum is $10$

大提示：

这样的最小 $f(n)$ 是 $31 = 37_8$ 。以四为底的数字和为 $31$ 的最小 $n$ 是一个 $1$ 后面接十个 $3$ 。

The least such value of $f(n)$ is $31 = 37_8.$ The smallest $n$ whose base-four digits sum to $31$ is a $1$ followed by ten $3$ s.

解答：

在十六进制表示中需要使用大于 $9$ 的数字的最小数是 $10$ 。因此，我们要找使 $f(n)$ 的八进制数位和为 $10$ 的最小 $n$ 。按顺序检查 $f(n)$ 的可能值，所有小于 $31$ 的值，其八进制数位和都至多为 $9$ ，而 $31 = 37_8$ 的数位和为 $10$ 。由于达到给定四进制数位和的最小 $n$ 随该数位和增大而增大，我们要找满足 $f(n) = 31$ 的最小 $n$ 。

一个以四为底的数字至多为 $3$ ，所以数字和为 $31$ 至少需要 $11$ 位；最小的 $11$ 位选择是首位为 $1$ ，后面跟十个 $3$ ： $\begin{aligned} N &= 13333333333_4 \\ &= 4^{10} + (4^{10} - 1) \\ &= 2 \cdot 4^{10} - 1 = 2097151 \end{aligned}\text{。}$

$N = 2097151$ 除以 $1000$ 的余数是 $151$ 。

The smallest value whose base-sixteen representation uses a digit beyond $9$ is $10.$ Thus we seek the least $n$ for which the base-eight digit sum of $f(n)$ is $10.$ Checking possible values of $f(n)$ in order, every value less than $31$ has base-eight digit sum at most $9,$ while $31 = 37_8$ has digit sum $10.$ Since the least $n$ achieving a given base-four digit sum increases with that sum, we want the least $n$ with $f(n) = 31.$

A base-four digit is at most $3,$ so a digit sum of $31$ requires at least $11$ digits, and the smallest $11$ -digit choice is a leading $1$ followed by ten $3$ s: $\begin{aligned} N &= 13333333333_4 \\ &= 4^{10} + (4^{10} - 1) \\ &= 2 \cdot 4^{10} - 1 = 2097151. \end{aligned}$

The remainder when $N = 2097151$ is divided by $1000$ is $151.$

6.

递归定义数列 $t_1 = 20$ 、 $t_2 = 21$ ，并且对所有 $n \ge 3$ ， $t_n = \frac{5t_{n-1} + 1}{25t_{n-2}}\text{。}$ 那么 $t_{2020}$ 可写成 $\frac{p}{q}$ ，其中 $p$ 与 $q$ 是互质正整数。求 $p + q$ 。

Define a sequence recursively by $t_1 = 20,$ $t_2 = 21,$ and $t_n = \frac{5t_{n-1} + 1}{25t_{n-2}}$ for all $n \ge 3.$ Then $t_{2020}$ can be written as $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p + q.$

答案

知识点：递推找规律

难度评级：2340

小提示：

把 $t_3$ 、 $t_4$ 、 $t_5$ 、 $t_6$ 、 $t_7$ 精确计算成分数，会出现一个规律

Compute $t_3,$ $t_4,$ $t_5,$ $t_6,$ and $t_7$ exactly as fractions — a pattern appears

大提示：

该数列以 $5$ 为周期，所以 $t_{2020} = t_5$

The sequence is periodic with period $5,$ so $t_{2020} = t_5$

解答：

精确计算各项： $\begin{aligned} t_3 &= \frac{5 \cdot 21 + 1}{25 \cdot 20} = \frac{53}{250}, \\ t_4 &= \frac{5 \cdot \frac{53}{250} + 1}{25 \cdot 21} = \frac{103}{26250}, \\ t_5 &= \frac{5 \cdot \frac{103}{26250} + 1}{25 \cdot \frac{53}{250}} \\ &= \frac{5353}{5250} \cdot \frac{10}{53} = \frac{101}{525} \end{aligned}\text{，}$ 其中用到 $5353 = 53 \cdot 101$ 。接着 $t_6 = \frac{5 \cdot \frac{101}{525} + 1}{25 \cdot \frac{103}{26250}}$ $= \frac{206}{105} \cdot \frac{1050}{103}$ $= 20 = t_1$ ，且 $t_7 = \frac{101}{\frac{101}{21}} = 21 = t_2$ 。

因为每一项只由前两项决定，数列以 $5$ 为周期。由于 $2020$ 是 $5$ 的倍数，得到 $t_{2020} = t_5 = \frac{101}{525}$ 。因为 $101$ 是质数且不整除 $525$ ，该分数已经约分，所以 $p + q = 101 + 525 = 626$ 。

Computing terms exactly: $\begin{aligned} t_3 &= \frac{5 \cdot 21 + 1}{25 \cdot 20} = \frac{53}{250}, \\ t_4 &= \frac{5 \cdot \frac{53}{250} + 1}{25 \cdot 21} = \frac{103}{26250}, \\ t_5 &= \frac{5 \cdot \frac{103}{26250} + 1}{25 \cdot \frac{53}{250}} \\ &= \frac{5353}{5250} \cdot \frac{10}{53} = \frac{101}{525}, \end{aligned}$ using $5353 = 53 \cdot 101.$ Then $t_6 = \frac{5 \cdot \frac{101}{525} + 1}{25 \cdot \frac{103}{26250}}$ $= \frac{206}{105} \cdot \frac{1050}{103}$ $= 20 = t_1$ and $t_7 = \frac{101}{\frac{101}{21}} = 21 = t_2.$

Since each term depends only on the two preceding terms, the sequence repeats with period $5.$ Because $2020$ is a multiple of $5,$ we get $t_{2020} = t_5 = \frac{101}{525}.$ As $101$ is prime and does not divide $525,$ the fraction is reduced, and $p + q = 101 + 525 = 626.$

7.

两个全等的直圆锥，底面半径均为 $3$ ，高均为 $8$ 。它们的对称轴在两圆锥内部一点垂直相交，且该点到每个圆锥底面的距离都是 $3$ 。一个半径为 $r$ 的球同时位于两个圆锥内部。 $r^2$ 的最大可能值为 $\frac{m}{n}$ ，其中 $m$ 与 $n$ 是互质正整数。求 $m + n$ 。

Two congruent right circular cones each with base radius $3$ and height $8$ have axes of symmetry that intersect at right angles at a point in the interior of the cones a distance $3$ from the base of each cone. A sphere with radius $r$ lies within both cones. The maximum possible value of $r^2$ is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

知识点：圆锥球最优化

难度评级：2560

小提示：

在两条轴所在的平面内，每个圆锥的截面是一个三角形：顶点在交点之外 $5$ 的位置，底边半宽 $3$ ，位于另一侧距离 $3$ 处

In the plane of the two axes, each cone’s cross-section is a triangle with apex $5$ beyond the intersection point and base half-width $3$ at distance $3$ on the other side

大提示：

把球心放在交点；其半径就是该点到经过 $(5, 0)$ 与 $(-3, 3)$ 的斜边直线的距离

Center the sphere at the intersection point; its radius is the distance from that point to the slant line through $(5, 0)$ and $(-3, 3)$

解答：

把两轴交点作为原点，并沿两条轴量取有向坐标 $u_1, u_2$ 。沿某个圆锥的轴，它的顶点距离原点 $8 - 3 = 5$ ，底面平面在另一侧距离 $3$ 处。任何经过轴的平面截该圆锥都会得到一个三角形；用坐标 $(u, w)$ 表示，其中 $w \ge 0$ 为到轴的距离，则其斜边经过 $(5, 0)$ 和 $(-3, 3)$ ，方程为 $3u + 8w = 15$ 。

若半径为 $r$ 的球心对该轴的轴向坐标为 $u$ ，到该轴的距离为 $\rho$ ，那么球的截面圆必须能放入上述三角形，所以 $r \le \frac{15 - 3u - 8\rho}{\sqrt{73}}$ 。由于两条轴互相垂直，球心到轴 $1$ 的距离至少为 $|u_2|$ ，反之亦然。把两个约束相加， $\begin{aligned} 2\sqrt{73}\,r &\le 30 - 3(u_1 + u_2) \\ &\quad {}- 8(|u_1| + |u_2|) \\ &\le 30 \end{aligned}\text{，}$ 因为 $3(u_1 + u_2) \ge -3(|u_1| + |u_2|)$ $\ge -8(|u_1| + |u_2|)$ 。因此 $r \le \frac{15}{\sqrt{73}}$ 。

以原点为球心、半径为 $\frac{15}{\sqrt{73}}$ 的球可以达到这个界：它到每个斜面的距离都是 $\frac{|3 \cdot 0 + 8 \cdot 0 - 15|}{\sqrt{3^2 + 8^2}} = \frac{15}{\sqrt{73}}$ ，且它到每个底面平面的距离 $3$ 更大。所以 $r^2$ 的最大值是 $\frac{225}{73}$ ， $m + n = 225 + 73 = 298$ 。

Put the origin at the point where the axes cross, and measure signed coordinates $u_1, u_2$ along the two axes. Along its axis, each cone has its apex at distance $8 - 3 = 5$ from the origin and its base plane at distance $3$ on the other side. Slicing a cone by any plane through its axis gives a triangle whose slant side, in coordinates $(u, w)$ with $w \ge 0$ the distance from the axis, is the line through $(5, 0)$ and $(-3, 3),$ namely $3u + 8w = 15.$

A sphere of radius $r$ centered at a point with axial coordinate $u$ and distance $\rho$ from the axis fits inside that cone only if its cross-section fits inside the triangle, so $r \le \frac{15 - 3u - 8\rho}{\sqrt{73}}.$ Since the two axes are perpendicular, the distance from the center to axis $1$ is at least $|u_2|,$ and vice versa. Adding the two constraints, $\begin{aligned} 2\sqrt{73}\,r &\le 30 - 3(u_1 + u_2) \\ &\quad {}- 8(|u_1| + |u_2|) \\ &\le 30, \end{aligned}$ because $3(u_1 + u_2) \ge -3(|u_1| + |u_2|)$ $\ge -8(|u_1| + |u_2|).$ Hence $r \le \frac{15}{\sqrt{73}}.$

The sphere of radius $\frac{15}{\sqrt{73}}$ centered at the origin achieves this: its distance to each slant surface is $\frac{|3 \cdot 0 + 8 \cdot 0 - 15|}{\sqrt{3^2 + 8^2}} = \frac{15}{\sqrt{73}},$ and its distance $3$ to each base plane is larger. So the maximum of $r^2$ is $\frac{225}{73},$ and $m + n = 225 + 73 = 298.$

8.

递归定义函数列： $f_1(x) = |x - 1|$ ，且对整数 $n \gt 1$ ， $f_n(x) = f_{n-1}(|x - n|)$ 。求最小的 $n$ ，使得 $f_n$ 的所有零点之和超过 $500{,}000$ 。

Define a sequence recursively by $f_1(x) = |x - 1|$ and $f_n(x) = f_{n-1}(|x - n|)$ for integers $n \gt 1.$ Find the least value of $n$ such that the sum of the zeros of $f_n$ exceeds $500{,}000.$

答案

知识点：绝对值递推三角形数数学归纳法

难度评级：2920

小提示：

$f_n(x) = 0$ 当且仅当 $|x - n|$ 是 $f_{n-1}$ 的一个非负零点。计算前几个 $n$ 的零点集合

$f_n(x) = 0$ exactly when $|x - n|$ is a nonnegative zero of $f_{n-1}.$ Compute the zero sets for the first few $n.$

大提示：

令 $T_k = \frac{k(k+1)}{2}$ ， $f_n$ 的零点是从 $n - T_{n-1}$ 到 $T_n$ 每隔一个取出的整数，所以它们的和是 $n\left(1 + \frac{n(n-1)}{2}\right)$

With $T_k = \frac{k(k+1)}{2},$ the zeros of $f_n$ are every other integer from $n - T_{n-1}$ up to $T_n,$ so their sum is $n\left(1 + \frac{n(n-1)}{2}\right)$

解答：

因为 $f_n(x) = f_{n-1}(|x - n|)$ ，所以 $f_n$ 的零点正好是 $x = n \pm z$ ，其中 $z$ 遍历 $f_{n-1}$ 的非负零点。前几个零点集合为 $\{1\}$ 、 $\{1, 3\}$ 、 $\{0, 2, 4, 6\}$ 、 $\{-2, 0, 2, \ldots, 10\}$ 。记 $T_k = \frac{k(k+1)}{2}$ ，我们断言 $f_n$ 的零点是从 $n - T_{n-1}$ 到 $T_n$ 每隔一个取出的整数。事实上，由归纳假设， $f_{n-1}$ 的非负零点是从 $0$ 或 $1$ 到 $T_{n-1}$ 每隔一个取出的整数；套用 $n \pm z$ 后，就得到从 $n - T_{n-1}$ 到 $n + T_{n-1} = T_n$ 之间奇偶性相应的所有整数。

这个等差数列有 $\frac{T_n + T_{n-1} - n}{2} + 1 = \frac{n(n-1)}{2} + 1$ 项，首末两项之和为 $(n - T_{n-1}) + T_n = 2n$ ，所以零点之和为 $S_n = n\left(\frac{n(n-1)}{2} + 1\right)\text{，}$ 它随 $n$ 增大而增大。

现在 $S_{100} = 100 \cdot 4951$ $= 495{,}100 \le 500{,}000$ ，而 $S_{101} = 101 \cdot 5051$ $= 510{,}151 \gt 500{,}000$ 。最小的这样的 $n$ 是 $101$ 。

Since $f_n(x) = f_{n-1}(|x - n|),$ the zeros of $f_n$ are exactly $x = n \pm z$ as $z$ runs over the nonnegative zeros of $f_{n-1}.$ The first few zero sets are $\{1\},$ $\{1, 3\},$ $\{0, 2, 4, 6\},$ $\{-2, 0, 2, \ldots, 10\}.$ Writing $T_k = \frac{k(k+1)}{2},$ we claim the zeros of $f_n$ are every other integer from $n - T_{n-1}$ through $T_n.$ Indeed, by induction the nonnegative zeros of $f_{n-1}$ are every other integer from $0$ or $1$ up to $T_{n-1},$ and applying $n \pm z$ yields every integer of the appropriate parity from $n - T_{n-1}$ through $n + T_{n-1} = T_n.$

This progression has $\frac{T_n + T_{n-1} - n}{2} + 1 = \frac{n(n-1)}{2} + 1$ terms, and its first and last terms sum to $(n - T_{n-1}) + T_n = 2n,$ so the sum of the zeros is $S_n = n\left(\frac{n(n-1)}{2} + 1\right),$ which is increasing in $n.$

Now $S_{100} = 100 \cdot 4951$ $= 495{,}100 \le 500{,}000,$ while $S_{101} = 101 \cdot 5051$ $= 510{,}151 \gt 500{,}000.$ The least such $n$ is $101.$

9.

看演出时，Ayako、Billy、Carlos、Dahlia、Ehuang 和 Frank 按这个顺序坐在一排六把椅子上。中场休息时，他们去厨房吃点心。回来后，他们坐回这六把椅子，使得如果两个人在休息前相邻，那么休息后他们不相邻。求他们休息后可能选择的座位顺序数。

While watching a show, Ayako, Billy, Carlos, Dahlia, Ehuang, and Frank sat in that order in a row of six chairs. During the break, they went to the kitchen for a snack. When they came back, they sat on those six chairs in such a way that if two of them sat next to each other before the break, then they did not sit next to each other after the break. Find the number of possible seating orders they could have chosen after the break.

答案

知识点：有限制的排列容斥原理排列

难度评级：2840

小提示：

按原来的座位把朋友编号为 $1$ 到 $6$ ；需要数出任意两个连续编号都不相邻的排列

Number the friends $1$ through $6$ by original seat; you need orderings in which no two consecutive numbers stand next to each other

大提示：

对五个数对 $\{i, i+1\}$ 中哪些坐在一起使用容斥；把每一段连续编号粘成一个可正可反的块

Use inclusion-exclusion on which of the five pairs $\{i, i+1\}$ sit together, gluing each run of consecutive numbers into a block that can face two ways

解答：

按原座位顺序把朋友编号为 $1$ 到 $6$ ；我们要数 $1, \ldots, 6$ 的排列，使得没有两个连续整数相邻。对五个数对 $\{i, i+1\}$ 中哪些被迫坐在一起使用容斥。若所选的 $k$ 个数对形成 $r$ 段极大连续整数段，把每段粘成一个块（可按递增或递减两种方向排列），就得到 $2^r (6 - k)!$ 个包含所有所选相邻关系的坐法。

按 $k$ 统计：当 $k = 0$ 时，有 $720$ 。当 $k = 1$ 时，五个集合，每个有 $2 \cdot 120$ ，共 $1200$ 。当 $k = 2$ 时，四个集合形成一段 $(2 \cdot 24)$ ，六个形成两段 $(4 \cdot 24)$ ，共 $768$ 。当 $k = 3$ 时，三个集合形成一段 $(2 \cdot 6)$ ，六个形成两段 $(4 \cdot 6)$ ，一个形成三段 $(8 \cdot 6)$ ，共 $228$ 。当 $k = 4$ 时，两个集合形成一段 $(2 \cdot 2)$ ，三个形成两段 $(4 \cdot 2)$ ，共 $32$ 。当 $k = 5$ 时，一个集合，共 $2$ 。

所求数量为 $720 - 1200 + 768$ $- 228 + 32 - 2 = 90$ 。

Number the friends $1$ through $6$ in original seating order; we count orderings of $1, \ldots, 6$ in which no two consecutive integers are adjacent. Apply inclusion-exclusion over which of the five pairs $\{i, i+1\}$ are forced to sit together. If a chosen set of $k$ pairs forms $r$ maximal runs of consecutive integers, gluing each run into a block (orderable ascending or descending) gives $2^r (6 - k)!$ seatings containing all chosen adjacencies.

Tallying by $k:$ for $k = 0,$ $720.$ For $k = 1,$ five sets, each $2 \cdot 120,$ total $1200.$ For $k = 2,$ four sets form one run $(2 \cdot 24)$ and six form two runs $(4 \cdot 24),$ total $768.$ For $k = 3,$ three sets form one run $(2 \cdot 6),$ six form two runs $(4 \cdot 6),$ one forms three runs $(8 \cdot 6),$ total $228.$ For $k = 4,$ two sets form one run $(2 \cdot 2)$ and three form two runs $(4 \cdot 2),$ total $32.$ For $k = 5,$ one set, total $2.$

The count is $720 - 1200 + 768$ $- 228 + 32 - 2 = 90.$

10.

求所有正整数 $n$ 的和，使得 $1^3 + 2^3 + 3^3 + \cdots + n^3$ 除以 $n + 5$ 时余数为 $17$ 。

Find the sum of all positive integers $n$ such that when $1^3 + 2^3 + 3^3 + \cdots + n^3$ is divided by $n + 5,$ the remainder is $17.$

答案

知识点：前n项立方和模运算整除性

难度评级：2710

小提示：

这个和是 $\left(\frac{n(n+1)}{2}\right)^2$ 。利用 $n \equiv -5$ 对 $n + 5$ 取模化简

The sum is $\left(\frac{n(n+1)}{2}\right)^2.$ Reduce it modulo $n + 5$ using $n \equiv -5.$

大提示：

若 $K = \frac{n(n+1)}{2}$ ，则 $2K \equiv 20 \pmod{n+5}$ ，所以 $n + 5$ 必须整除 $332$ 。乘以 $4$ 可能产生伪解，所以要检查每个候选

If $K = \frac{n(n+1)}{2},$ then $2K \equiv 20 \pmod{n+5},$ so $n + 5$ must divide $332.$ Multiplying by $4$ can create false solutions, so verify each candidate.

解答：

令 $K = \frac{n(n+1)}{2}$ ，则立方和为 $K^2$ 。模 $n + 5$ 时有 $n \equiv -5$ ，因此 $2K = n(n+1)$ $\equiv (-5)(-4) = 20$ ，且 $4K^2 \equiv 400$ 。若 $K^2 \equiv 17 \pmod{n+5}$ ，则 $n + 5$ 整除 $400 - 4 \cdot 17 = 332 = 2^2 \cdot 83$ 。余数为 $17$ 还要求 $n + 5 \gt 17$ ，所以候选为 $n + 5 \in \{83, 166, 332\}$ ，即 $n \in \{78, 161, 327\}$ 。

因为我们乘过 $4$ ，每个候选都必须检查。对 $n = 78$ ： $K = 3081 \equiv 10 \pmod{83}$ ，且 $10^2 = 100 \equiv 17 \pmod{83}$ 。对 $n = 161$ ： $K = 13041 \equiv 93 \pmod{166}$ ，且 $93^2 = 8649 = 52 \cdot 166 + 17$ 。对 $n = 327$ ： $K = 53628 \equiv 176 \pmod{332}$ ，且 $176^2 = 30976 = 93 \cdot 332 + 100$ ，余数 $100 \ne 17$ ，所以此项不符合要求。

有效值为 $n = 78$ 和 $n = 161$ ，它们的和为 $78 + 161 = 239$ 。

Let $K = \frac{n(n+1)}{2},$ so the sum of cubes is $K^2.$ Modulo $n + 5$ we have $n \equiv -5,$ hence $2K = n(n+1)$ $\equiv (-5)(-4) = 20$ and $4K^2 \equiv 400.$ If $K^2 \equiv 17 \pmod{n+5},$ then $n + 5$ divides $400 - 4 \cdot 17 = 332 = 2^2 \cdot 83.$ Since a remainder of $17$ also forces $n + 5 \gt 17,$ the candidates are $n + 5 \in \{83, 166, 332\},$ i.e. $n \in \{78, 161, 327\}.$

Because we multiplied by $4,$ each candidate must be checked. For $n = 78:$ $K = 3081 \equiv 10 \pmod{83},$ and $10^2 = 100 \equiv 17 \pmod{83}.$ For $n = 161:$ $K = 13041 \equiv 93 \pmod{166},$ and $93^2 = 8649 = 52 \cdot 166 + 17.$ For $n = 327:$ $K = 53628 \equiv 176 \pmod{332},$ and $176^2 = 30976 = 93 \cdot 332 + 100,$ remainder $100 \ne 17,$ so this one fails.

The valid values are $n = 78$ and $n = 161,$ with sum $78 + 161 = 239.$

11.

设 $P(x) = x^2 - 3x - 7$ ，且 $Q(x)$ 、 $R(x)$ 是两个二次多项式，它们的 $x^2$ 系数也都为 $1$ 。David 计算三个和 $P + Q$ 、 $P + R$ 、 $Q + R$ ，惊讶地发现这三个和中任意两个都有一个公共根，并且这三个公共根互不相同。若 $Q(0) = 2$ ，则 $R(0) = \frac{m}{n}$ ，其中 $m$ 与 $n$ 是互质正整数。求 $m + n$ 。

Let $P(x) = x^2 - 3x - 7,$ and let $Q(x)$ and $R(x)$ be two quadratic polynomials also with the coefficient of $x^2$ equal to $1.$ David computes each of the three sums $P + Q,$ $P + R,$ and $Q + R$ and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If $Q(0) = 2,$ then $R(0) = \frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

知识点：多项式韦达定理方程组

难度评级：2920

小提示：

每个两两之和的首项系数都是 $2$ ，所以它等于 $2$ 乘以由已知根给出的两个一次因式

Each pairwise sum has leading coefficient $2,$ so it equals $2$ times the product of the two linear factors from its known roots

大提示：

把 $P+Q$ 与 $P+R$ 的根和相加，再减去 $Q+R$ 的根和，就能孤立出前两个和共享的根的两倍

Adding the root sums of $P+Q$ and $P+R$ and subtracting that of $Q+R$ isolates twice the root shared by the first two sums

解答：

令 $a$ 为 $P+Q$ 与 $P+R$ 的公共根， $b$ 为 $P+Q$ 与 $Q+R$ 的公共根， $c$ 为 $P+R$ 与 $Q+R$ 的公共根。每个和都是首项系数为 $2$ 的二次式，所以 $P+Q = 2(x-a)(x-b)$ 、 $P+R = 2(x-a)(x-c)$ ，且 $Q+R = 2(x-b)(x-c)$ 。写 $Q = x^2 + qx + 2$ ， $R = x^2 + rx + s$ 。由韦达定理，根和分别为 $a + b = \frac{3-q}{2}$ 、 $a + c = \frac{3-r}{2}$ 、 $b + c = -\frac{q+r}{2}$ ，所以 $\begin{aligned} 2a &= (a+b) + (a+c) - (b+c) \\ &= \frac{3-q}{2} + \frac{3-r}{2} \\ &\quad {}+ \frac{q+r}{2} = 3 \end{aligned}\text{，}$ 得到 $a = \frac{3}{2}$ 。

$P + Q$ 的常数项为 $-7 + 2 = -5$ ，所以 $2ab = -5$ ，从而 $b = -\frac{5}{3}$ 。另外 $2ac = P(0) + R(0) = s - 7$ ，且 $2bc = Q(0) + R(0) = s + 2$ 。相减得 $2c(b - a) = 9$ ，所以 $c = \frac{9}{2\left(-\frac{5}{3} - \frac{3}{2}\right)} = -\frac{27}{19}$ 。

因此 $s = 2bc - 2$ $= 2 \cdot \left(-\frac{5}{3}\right)\left(-\frac{27}{19}\right) - 2$ $= \frac{90}{19} - \frac{38}{19} = \frac{52}{19}$ ，所以 $R(0) = \frac{52}{19}$ ， $m + n = 52 + 19 = 71$ 。

Let $a$ be the common root of $P+Q$ and $P+R,$ let $b$ be that of $P+Q$ and $Q+R,$ and let $c$ be that of $P+R$ and $Q+R.$ Each sum is quadratic with leading coefficient $2,$ so $P+Q = 2(x-a)(x-b),$ $P+R = 2(x-a)(x-c),$ and $Q+R = 2(x-b)(x-c).$ Write $Q = x^2 + qx + 2$ and $R = x^2 + rx + s.$ By Vieta’s formulas, the root sums are $a + b = \frac{3-q}{2},$ $a + c = \frac{3-r}{2},$ and $b + c = -\frac{q+r}{2},$ so $\begin{aligned} 2a &= (a+b) + (a+c) - (b+c) \\ &= \frac{3-q}{2} + \frac{3-r}{2} \\ &\quad {}+ \frac{q+r}{2} = 3, \end{aligned}$ giving $a = \frac{3}{2}.$

The constant term of $P + Q$ is $-7 + 2 = -5,$ so $2ab = -5$ and $b = -\frac{5}{3}.$ Also $2ac = P(0) + R(0) = s - 7$ and $2bc = Q(0) + R(0) = s + 2.$ Subtracting, $2c(b - a) = 9,$ so $c = \frac{9}{2\left(-\frac{5}{3} - \frac{3}{2}\right)} = -\frac{27}{19}.$

Then $s = 2bc - 2$ $= 2 \cdot \left(-\frac{5}{3}\right)\left(-\frac{27}{19}\right) - 2$ $= \frac{90}{19} - \frac{38}{19} = \frac{52}{19},$ so $R(0) = \frac{52}{19}$ and $m + n = 52 + 19 = 71.$

12.

设 $m$ 与 $n$ 是大于 $1$ 的奇整数。一个 $m \times n$ 矩形由单位正方形组成，最上面一行的方格从左到右编号为 $1$ 到 $n$ ，第二行从左到右编号为 $n + 1$ 到 $2n$ ，依此类推。方格 $200$ 在最上面一行，方格 $2000$ 在最下面一行。求满足以下性质的大于 $1$ 的奇整数有序对 $(m, n)$ 的个数：在该 $m \times n$ 矩形中，经过方格 $200$ 与 $2000$ 中心的直线与方格 $1099$ 的内部相交。

Let $m$ and $n$ be odd integers greater than $1.$ An $m \times n$ rectangle is made up of unit squares where the squares in the top row are numbered left to right with the integers $1$ through $n,$ those in the second row are numbered left to right with the integers $n + 1$ through $2n,$ and so on. Square $200$ is in the top row, and square $2000$ is in the bottom row. Find the number of ordered pairs $(m, n)$ of odd integers greater than $1$ with the property that, in the $m \times n$ rectangle, the line through the centers of squares $200$ and $2000$ intersects the interior of square $1099.$

答案

知识点：坐标几何斜率中点分类讨论

难度评级：3160

小提示：

由于 $m$ 与 $n$ 都是奇数，方格 $200$ 与 $2000$ 中心的中点恰好是方格 $1100$ 的中心，它在方格 $1099$ 右边一格

Since $m$ and $n$ are odd, the midpoint of the centers of squares $200$ and $2000$ is exactly the center of square $1100,$ one square to the right of square $1099$

大提示：

该直线进入方格 $1099$ 当且仅当它的斜率绝对值小于 $1$ ；然后对每个 $m \in \{3, 5, 7, 9\}$ 数出奇数 $n$

The line enters square $1099$ precisely when its slope is less than $1$ in absolute value; then count the odd $n$ for each $m \in \{3, 5, 7, 9\}$

解答：

使用“列、行”坐标。方格 $200$ 在最上面一行，所以 $n \ge 200$ （因 $n$ 为奇数，故 $n \ge 201$ ），其中心为 $(200, 1)$ 。方格 $2000$ 在最下面一行，所以它的列号为 $b = 2000 - (m-1)n$ ，且 $1 \le b \le n$ ，即 $(m-1)n \lt 2000 \le mn$ ，其中心为 $(b, m)$ 。由于 $m$ 和 $n$ 为奇数， $b$ 为偶数，所以这两个中心的中点 $\left(\frac{200 + b}{2}, \frac{m+1}{2}\right)$ 有整数坐标；它的方格编号为 $\frac{(m-1)n + 200 + b}{2} = \frac{2200}{2} = 1100$ 。它的列号 $100 + \frac{b}{2}$ 位于 $101$ 与 $n$ 之间，所以直线经过方格 $1100$ 的中心，而方格 $1099$ 就在同一行的左边一格。

方格 $1099$ 只在那一行中，且该直线穿过这一行横条时得到的线段以方格 $1100$ 的中心为中心（由对称性），向左右各延伸 $\frac{1}{2|s|}$ ，其中 $s = \frac{m-1}{b - 200}$ 是斜率。因此直线与方格 $1099$ 的内部相交，当且仅当 $\frac{1}{2|s|} \gt \frac{1}{2}$ ，即 $|s| \lt 1$ ，也就是 $m - 1 \lt |b - 200|$ $= |1800 - (m-1)n|$ （竖直线 $b = 200$ ，不满足）。

因为 $n \ge 201$ 且 $(m-1)n \lt 2000$ ，需要 $m - 1 \lt 10$ ，所以 $m \in \{3, 5, 7, 9\}$ 。当 $m = 3$ ：奇数 $n \in [667, 999]$ ，共有 $167$ 个，排除 $|n - 900| \le 1$ （奇数情形 $899$ 和 $901$ ）后剩 $165$ 个。当 $m = 5$ ：奇数 $n \in [401, 499]$ ，共有 $50$ 个，排除 $449$ 和 $451$ 后剩 $48$ 个。当 $m = 7$ ：奇数 $n \in [287, 333]$ ，共有 $24$ 个，排除 $299$ 和 $301$ 后剩 $22$ 个。当 $m = 9$ ：奇数 $n \in [223, 249]$ ，共有 $14$ 个，排除 $225$ 后剩 $13$ 个。总数为 $165 + 48 + 22 + 13 = 248$ 。

Use column-row coordinates. Square $200$ is in the top row, so $n \ge 200$ (hence $n \ge 201$ as $n$ is odd) and its center is $(200, 1).$ Square $2000$ is in the bottom row, so its column is $b = 2000 - (m-1)n$ with $1 \le b \le n,$ i.e. $(m-1)n \lt 2000 \le mn,$ and its center is $(b, m).$ Since $m$ and $n$ are odd, $b$ is even, so the midpoint $\left(\frac{200 + b}{2}, \frac{m+1}{2}\right)$ of the two centers has integer coordinates; its square number is $\frac{(m-1)n + 200 + b}{2} = \frac{2200}{2} = 1100.$ Its column $100 + \frac{b}{2}$ lies between $101$ and $n,$ so the line passes through the center of square $1100,$ and square $1099$ sits immediately to its left in the same row.

Square $1099$ lies only in that row, and the line crosses that row’s horizontal strip in a segment centered (by symmetry) at the center of square $1100,$ extending $\frac{1}{2|s|}$ to each side, where $s = \frac{m-1}{b - 200}$ is the slope. So the line meets the interior of square $1099$ exactly when $\frac{1}{2|s|} \gt \frac{1}{2},$ that is $|s| \lt 1,$ i.e. $m - 1 \lt |b - 200|$ $= |1800 - (m-1)n|$ (a vertical line, $b = 200,$ fails).

Since $n \ge 201$ and $(m-1)n \lt 2000,$ we need $m - 1 \lt 10,$ so $m \in \{3, 5, 7, 9\}.$ For $m = 3:$ odd $n \in [667, 999],$ $167$ values, excluding $|n - 900| \le 1$ (the odd cases $899$ and $901$ ) leaves $165.$ For $m = 5:$ odd $n \in [401, 499],$ $50$ values, excluding $449$ and $451$ leaves $48.$ For $m = 7:$ odd $n \in [287, 333],$ $24$ values, excluding $299$ and $301$ leaves $22.$ For $m = 9:$ odd $n \in [223, 249],$ $14$ values, excluding $225$ leaves $13.$ The total is $165 + 48 + 22 + 13 = 248.$

13.

凸五边形 $ABCDE$ 的边长为 $AB = 5$ 、 $BC = CD = DE = 6$ 、 $EA = 7$ 。此外，该五边形有一个内切圆（与五边形每条边都相切的圆）。求 $ABCDE$ 的面积。

Convex pentagon $ABCDE$ has side lengths $AB = 5,$ $BC = CD = DE = 6,$ and $EA = 7.$ Moreover, the pentagon has an inscribed circle (a circle tangent to each side of the pentagon). Find the area of $ABCDE.$

答案

知识点：内切圆、内心与内切圆半径切线三角恒等式

难度评级：3060

小提示：

从同一顶点引出的两条切线段相等：若五个顶点对应的切线段长分别为 $a$ 、 $b$ 、 $c$ 、 $d$ 、 $e$ ，解 $a+b=5$ 、 $b+c=c+d=d+e=6$ 、 $e+a=7$

The two tangent segments from each vertex are equal: if the five corresponding tangent lengths are $a,$ $b,$ $c,$ $d,$ and $e,$ solve $a+b=5,$ $b+c=c+d=d+e=6,$ $e+a=7$

大提示：

若某顶点对应的切线段长为 $t$ ，则其半角满足 $\tan\frac{\theta_V}{2} = \frac{r}{t}$ ；五个半角之和为 $270^\circ$ 。把它转化为关于 $r$ 的多项式方程，再用面积 $= rs$ 。

At a vertex with tangent length $t,$ the half-angle satisfies $\tan\frac{\theta_V}{2} = \frac{r}{t};$ the five half-angles sum to $270^\circ.$ Turn that into a polynomial equation for $r,$ then use area $= rs$

解答：

设从 $A$ 、 $B$ 、 $C$ 、 $D$ 、 $E$ 到内切圆的切线段长分别为 $a$ 、 $b$ 、 $c$ 、 $d$ 、 $e$ 。则 $a + b = 5$ 、 $b + c = 6$ 、 $c + d = 6$ 、 $d + e = 6$ 、 $e + a = 7$ 。中间的方程给出 $d = b$ 、 $e = c$ ，所以 $c + a = 7$ ；结合 $a + b = 5$ 与 $b + c = 6$ ，得到 $a = 3$ 、 $b = d = 2$ 、 $c = e = 4$ 。若 $r$ 为内切圆半径，则切线段长为 $t$ 的顶点处内角满足 $\tan\frac{\theta}{2} = \frac{r}{t}$ ，且半角之和是 $540^\circ$ 的一半： $\begin{aligned} &\arctan\frac{r}{3} + 2\arctan\frac{r}{2} \\ &\quad {}+ 2\arctan\frac{r}{4} = 270^\circ \end{aligned}\text{。}$

令 $\beta = \arctan\frac{r}{2}$ ， $\gamma = \arctan\frac{r}{4}$ 。则 $\arctan\frac{r}{3} = 270^\circ - 2(\beta + \gamma)$ ，并且由于 $\tan(270^\circ - \theta) = \cot\theta$ ，得到 $\frac{r}{3} = \cot 2(\beta + \gamma)$ 。令 $T = \tan(\beta + \gamma) = \frac{\frac{r}{2} + \frac{r}{4}}{1 - \frac{r^2}{8}}$ $= \frac{6r}{8 - r^2}$ ，恒等式 $\frac{r}{3} = \frac{1 - T^2}{2T}$ 变为 $2rT = 3(1 - T^2)$ ，代入 $T$ 并清分母得 $5r^4 - 84r^2 + 64 = 0\text{，}$ 所以 $r^2 = 16$ 或 $r^2 = \frac{4}{5}$ 。当 $r^2 = \frac{4}{5}$ 时，每个半角都远小于 $54^\circ$ ，半角和远不足 $270^\circ$ ；所以这个根是增根。因此 $r = 4$ 。

半周长为 $s = \frac{5 + 6 + 6 + 6 + 7}{2} = 15$ ，所以面积为 $rs = 4 \cdot 15 = 60$ 。

Let the tangent lengths from $A,$ $B,$ $C,$ $D,$ and $E$ to the incircle be $a,$ $b,$ $c,$ $d,$ and $e.$ Then $a + b = 5,$ $b + c = 6,$ $c + d = 6,$ $d + e = 6,$ and $e + a = 7.$ The middle equations give $d = b$ and $e = c,$ so $c + a = 7;$ with $a + b = 5$ and $b + c = 6$ this yields $a = 3,$ $b = d = 2,$ $c = e = 4.$ If $r$ is the inradius, the interior angle at a vertex with tangent length $t$ satisfies $\tan\frac{\theta}{2} = \frac{r}{t},$ and the half-angles sum to half of $540^\circ:$ $\begin{aligned} &\arctan\frac{r}{3} + 2\arctan\frac{r}{2} \\ &\quad {}+ 2\arctan\frac{r}{4} = 270^\circ. \end{aligned}$

Let $\beta = \arctan\frac{r}{2}$ and $\gamma = \arctan\frac{r}{4}.$ Then $\arctan\frac{r}{3} = 270^\circ - 2(\beta + \gamma),$ and since $\tan(270^\circ - \theta) = \cot\theta,$ we get $\frac{r}{3} = \cot 2(\beta + \gamma).$ With $T = \tan(\beta + \gamma) = \frac{\frac{r}{2} + \frac{r}{4}}{1 - \frac{r^2}{8}}$ $= \frac{6r}{8 - r^2},$ the identity $\frac{r}{3} = \frac{1 - T^2}{2T}$ becomes $2rT = 3(1 - T^2),$ and substituting $T$ and clearing denominators gives $5r^4 - 84r^2 + 64 = 0,$ so $r^2 = 16$ or $r^2 = \frac{4}{5}.$ For $r^2 = \frac{4}{5}$ every half-angle is well under $54^\circ,$ so the half-angle sum falls far short of $270^\circ;$ this root is extraneous. Hence $r = 4.$

The semiperimeter is $s = \frac{5 + 6 + 6 + 6 + 7}{2} = 15,$ so the area is $rs = 4 \cdot 15 = 60.$

14.

对实数 $x$ ，令 $\lfloor x \rfloor$ 为小于或等于 $x$ 的最大整数，并定义 $\{x\} = x - \lfloor x \rfloor$ 为 $x$ 的小数部分。例如， $\{3\} = 0$ ，且 $\{4.56\} = 0.56$ 。定义 $f(x) = x\{x\}$ ，令 $N$ 为方程 $f(f(f(x))) = 17$ 在 $0 \le x \le 2020$ 上的实数解个数。求 $N$ 除以 $1000$ 的余数。

For real number $x$ let $\lfloor x \rfloor$ be the greatest integer less than or equal to $x,$ and define $\{x\} = x - \lfloor x \rfloor$ to be the fractional part of $x.$ For example, $\{3\} = 0$ and $\{4.56\} = 0.56.$ Define $f(x) = x\{x\},$ and let $N$ be the number of real-valued solutions to the equation $f(f(f(x))) = 17$ for $0 \le x \le 2020.$ Find the remainder when $N$ is divided by $1000.$

答案

知识点：取整函数函数杨辉三角

难度评级：3160

小提示：

在每个区间 $[k, k+1)$ 上，其中 $k \ge 0$ ，函数 $f$ 从 $0$ 递增并趋近于 $k + 1$ ，恰好取遍 $[0, k+1)$ 中每个值一次

On each interval $[k, k+1)$ with $k \ge 0,$ the function $f$ increases from $0$ toward $k + 1,$ taking every value in $[0, k+1)$ exactly once

大提示：

因此当 $n \lt c \lt n + 1$ 时， $f(w) = c$ 在每个 $[k, k+1)$ 、 $k \ge n$ 中有一个解。数出区间链 $17 \le n \le k \le j$ ，并用冰球杆恒等式求和。

So $f(w) = c$ with $n \lt c \lt n + 1$ has one solution in $[k, k+1)$ for each $k \ge n.$ Count chains of intervals $17 \le n \le k \le j$ and sum with the hockey stick identity.

解答：

在 $[k, k+1)$ 上，其中 $k \ge 0$ 为整数，写 $x = k + t$ ；则 $f(x) = (k + t)t$ 从 $0$ 严格递增并趋近于 $k + 1$ ，所以 $f$ 把 $[k, k+1)$ 双射到 $[0, k+1)$ 。因此对任何满足 $n \lt c \lt n + 1$ 的 $c$ ，对每个整数 $k \ge n$ ，方程 $f(w) = c$ 在 $[k, k+1)$ 中恰有一个解，且没有其他解。

方程 $f(z) = 17$ 对每个 $n \ge 17$ 有一个解 $z_n \in (n, n+1)$ 。接着，对每个 $k \ge n$ ，方程 $f(w) = z_n$ 在 $(k, k+1)$ 中有一个解。最后，对 $x \in [j, j+1)$ ，其中 $0 \le j \le 2019$ ， $f$ 把 $[j, j+1)$ 双射到 $[0, j+1)$ ，所以在该区间内 $f(f(f(x))) = 17$ 的解数等于满足 $w \lt j + 1$ 的这类 $w$ 的个数，也就是满足 $17 \le n \le k \le j$ 的数对 $(n, k)$ 的个数，即 $\binom{j - 15}{2}$ 。（端点 $x = 2020$ 给出 $f(x) = 0$ ，不是解。）

由冰球杆恒等式， $\begin{aligned} N &= \sum_{j=17}^{2019} \binom{j - 15}{2} \\ &= \sum_{a=2}^{2004} \binom{a}{2} \\ &= \binom{2005}{3} \\ &= \frac{2005 \cdot 2004 \cdot 2003}{6} \\ &= 1{,}341{,}349{,}010 \end{aligned}\text{，}$ 所以 $N$ 除以 $1000$ 的余数是 $10$ 。

On $[k, k+1)$ with $k \ge 0$ an integer, write $x = k + t;$ then $f(x) = (k + t)t$ is strictly increasing from $0$ toward $k + 1,$ so $f$ maps $[k, k+1)$ bijectively onto $[0, k+1).$ Hence for any $c$ with $n \lt c \lt n + 1,$ the equation $f(w) = c$ has exactly one solution in $[k, k+1)$ for each integer $k \ge n,$ and no others.

The equation $f(z) = 17$ has one solution $z_n \in (n, n+1)$ for each $n \ge 17.$ In turn, $f(w) = z_n$ has one solution in $(k, k+1)$ for each $k \ge n.$ Finally, for $x \in [j, j+1)$ with $0 \le j \le 2019,$ $f$ maps $[j, j+1)$ bijectively onto $[0, j+1),$ so the number of solutions of $f(f(f(x))) = 17$ there equals the number of such $w$ with $w \lt j + 1,$ namely the number of pairs $(n, k)$ with $17 \le n \le k \le j,$ which is $\binom{j - 15}{2}.$ (The endpoint $x = 2020$ gives $f(x) = 0$ and is not a solution.)

By the hockey stick identity, $\begin{aligned} N &= \sum_{j=17}^{2019} \binom{j - 15}{2} \\ &= \sum_{a=2}^{2004} \binom{a}{2} \\ &= \binom{2005}{3} \\ &= \frac{2005 \cdot 2004 \cdot 2003}{6} \\ &= 1{,}341{,}349{,}010, \end{aligned}$ so the remainder when $N$ is divided by $1000$ is $10.$

15.

设 $\triangle ABC$ 是一个锐角不等边三角形，外接圆为 $\omega$ 。 $\omega$ 在 $B$ 与 $C$ 处的切线相交于 $T$ 。令 $X$ 与 $Y$ 分别为 $T$ 到直线 $AB$ 与 $AC$ 的垂足。已知 $BT = CT = 16$ 、 $BC = 22$ ，且 $TX^2 + TY^2 + XY^2 = 1143$ 。求 $XY^2$ 。

Let $\triangle ABC$ be an acute scalene triangle with circumcircle $\omega.$ The tangents to $\omega$ at $B$ and $C$ intersect at $T.$ Let $X$ and $Y$ be the projections of $T$ onto lines $AB$ and $AC,$ respectively. Suppose $BT = CT = 16,$ $BC = 22,$ and $TX^2 + TY^2 + XY^2 = 1143.$ Find $XY^2.$

答案

知识点：切线圆内接四边形正弦定理坐标几何

难度评级：3370

小提示：

切线-弦定理给出 $\angle ABT = 180^\circ - C$ ，所以 $TX = 16 \sin C$ ；且 $AXTY$ 是以 $AT$ 为直径的圆内接四边形，所以 $XY = AT \sin A$

The tangent-chord angle gives $\angle ABT = 180^\circ - C,$ so $TX = 16 \sin C;$ and $AXTY$ is cyclic with diameter $AT,$ so $XY = AT \sin A$

大提示：

将 $B$ 和 $C$ 放在 $(\mp 11, 0)$ ，于是 $T = (0, -\sqrt{135})$ 。此时 $TX^2 + TY^2$ 与 $XY^2$ 都是 $A$ 的 $y$ 坐标的一次函数。

Place $B$ and $C$ at $(\mp 11, 0),$ so $T = (0, -\sqrt{135}).$ Then $TX^2 + TY^2$ and $XY^2$ are both linear in the $y$ -coordinate of $A.$

解答：

由切线-弦定理， $\angle TBC = A$ ，所以 $\angle ABT = B + A = 180^\circ - C$ ，且 $TX = TB \sin\angle ABT = 16 \sin C$ ；同理 $TY = 16 \sin B$ 。又 $\angle AXT = \angle AYT = 90^\circ$ ，所以 $A$ 、 $X$ 、 $T$ 、 $Y$ 共圆，且直径为 $AT$ ，从而 $XY = AT \sin A$ 。由正弦定理， $\sin A = \frac{11}{R}$ 、 $\sin B = \frac{AC}{2R}$ 、 $\sin C = \frac{AB}{2R}$ ，题设条件变为 $\begin{aligned} &\frac{64\left(AB^2 + AC^2\right) + 121\,AT^2}{R^2} \\ &= 1143 \end{aligned}\text{。}$

取 $B = (-11, 0)$ 、 $C = (11, 0)$ 。因为 $TB = 16$ ，且 $T$ 在 $BC$ 的垂直平分线上，得 $T = (0, -\sqrt{135})$ 。圆心为 $O = (0, k)$ ，且 $OB \perp BT$ ，由此 $121 - k\sqrt{135} = 0$ ，所以 $k = \frac{121}{\sqrt{135}}$ ，并且 $R^2 = 121 + k^2 = \frac{30976}{135}$ 。对圆 $\omega$ 上的 $A = (x, y)$ ，展开 $x^2 + (y - k)^2 = R^2$ ，得到 $x^2 + y^2 = \frac{242}{\sqrt{135}}\,y + 121$ 。因此 $\begin{aligned} AB^2 + AC^2 &= 2(x^2 + y^2) + 242 \\ &= \frac{484}{\sqrt{135}}\,y + 484 \end{aligned}\text{，}$ $\begin{aligned} AT^2 &= x^2 + y^2 \\ &\quad {}+ 2\sqrt{135}\,y + 135 \\ &= \frac{512}{\sqrt{135}}\,y + 256 \end{aligned}\text{。}$

代入后， $64(AB^2 + AC^2) + 121\,AT^2$ $= \frac{92928}{\sqrt{135}}\,y + 61952$ $= 1143 \cdot \frac{30976}{135}$ ，得到 $y = \frac{291}{\sqrt{135}}$ 。于是 $AT^2 = \frac{512 \cdot 291}{135} + 256 = \frac{183552}{135}$ ，且 $\begin{aligned} XY^2 &= AT^2 \sin^2 A \\ &= \frac{121\,AT^2}{R^2} \\ &= \frac{121 \cdot 183552}{30976} \\ &= \frac{183552}{256} = 717 \end{aligned}\text{。}$

By the tangent-chord angle, $\angle TBC = A,$ so $\angle ABT = B + A = 180^\circ - C$ and $TX = TB \sin\angle ABT = 16 \sin C;$ similarly $TY = 16 \sin B.$ Also $\angle AXT = \angle AYT = 90^\circ,$ so $A,$ $X,$ $T,$ and $Y$ lie on a circle with diameter $AT,$ whence $XY = AT \sin A.$ Using the law of sines, $\sin A = \frac{11}{R},$ $\sin B = \frac{AC}{2R},$ and $\sin C = \frac{AB}{2R},$ so the given condition becomes $\begin{aligned} &\frac{64\left(AB^2 + AC^2\right) + 121\,AT^2}{R^2} \\ &= 1143. \end{aligned}$

Place $B = (-11, 0)$ and $C = (11, 0).$ Since $TB = 16$ and $T$ lies on the perpendicular bisector of $BC,$ we get $T = (0, -\sqrt{135}).$ The circumcenter is $O = (0, k)$ with $OB \perp BT,$ which gives $121 - k\sqrt{135} = 0,$ so $k = \frac{121}{\sqrt{135}}$ and $R^2 = 121 + k^2 = \frac{30976}{135}.$ For $A = (x, y)$ on $\omega,$ expanding $x^2 + (y - k)^2 = R^2$ gives $x^2 + y^2 = \frac{242}{\sqrt{135}}\,y + 121.$ Therefore $\begin{aligned} AB^2 + AC^2 &= 2(x^2 + y^2) + 242 \\ &= \frac{484}{\sqrt{135}}\,y + 484, \end{aligned}$ $\begin{aligned} AT^2 &= x^2 + y^2 \\ &\quad {}+ 2\sqrt{135}\,y + 135 \\ &= \frac{512}{\sqrt{135}}\,y + 256. \end{aligned}$

Substituting, $64(AB^2 + AC^2) + 121\,AT^2$ $= \frac{92928}{\sqrt{135}}\,y + 61952$ $= 1143 \cdot \frac{30976}{135}$ yields $y = \frac{291}{\sqrt{135}}.$ Then $AT^2 = \frac{512 \cdot 291}{135} + 256 = \frac{183552}{135},$ and $\begin{aligned} XY^2 &= AT^2 \sin^2 A \\ &= \frac{121\,AT^2}{R^2} \\ &= \frac{121 \cdot 183552}{30976} \\ &= \frac{183552}{256} = 717. \end{aligned}$