2014 AIME II 详解 | LIVE by Po-Shen Loh

1.

Abe 可以在 $15$ 小时内粉刷完一个房间，Bea 的粉刷速度比 Abe 快 $50$ %，Coe 的速度是 Abe 的两倍。Abe 开始粉刷房间，先独自工作一个半小时。然后 Bea 加入 Abe，两人一起工作，直到房间粉刷了一半。接着 Coe 加入 Abe 和 Bea，三人一起工作，直到整个房间粉刷完成。求从 Abe 开始到三人完成粉刷经过了多少分钟。

Abe can paint the room in $15$ hours, Bea can paint $50$ percent faster than Abe, and Coe can paint twice as fast as Abe. Abe begins to paint the room and works alone for the first hour and a half. Then Bea joins Abe, and they work together until half the room is painted. Then Coe joins Abe and Bea, and they work together until the entire room is painted. Find the number of minutes after Abe begins for the three of them to finish painting the room.

答案

知识点：速率分数

难度评级：1890

小提示：

用“每分钟粉刷的房间数”表示速率：Abe 的速率是 $\frac{1}{900}$ ，所以 Bea 是 $\frac{1}{600}$ ，Coe 是 $\frac{1}{450}$

Work in rooms per minute: Abe’s rate is $\frac{1}{900},$ so Bea’s is $\frac{1}{600}$ and Coe’s is $\frac{1}{450}$

大提示：

前 $90$ 分钟 Abe 独自粉刷了房间总量的 $\frac{1}{10}$ 。把之后每段还要完成的房间比例除以当时工作者的合速率。

After $90$ minutes alone, $\frac{1}{10}$ of the room is done. Divide each remaining portion of the room by the combined rate of the painters working then.

解答：

Abe 每分钟粉刷 $\frac{1}{900}$ 个房间，所以 Bea 每分钟粉刷 $\frac{3}{2} \cdot \frac{1}{900} = \frac{1}{600}$ ，Coe 每分钟粉刷 $\frac{2}{900} = \frac{1}{450}$ 。前 $90$ 分钟 Abe 粉刷了 $\frac{90}{900} = \frac{1}{10}$ 个房间。

Abe 和 Bea 合作的速率为 $\frac{1}{900} + \frac{1}{600} = \frac{1}{360}$ ，他们需要把总量从 $\frac{1}{10}$ 提高到 $\frac{1}{2}$ ，所需时间为 $\frac{2}{5} \cdot 360 = 144$ 分钟。三人合速率为 $\frac{1}{360} + \frac{1}{450} = \frac{1}{200}$ ，所以剩下半个房间需要 $\frac{1}{2} \cdot 200 = 100$ 分钟。

总时间为 $90 + 144 + 100 = 334$ 分钟。

Abe paints $\frac{1}{900}$ of the room per minute, so Bea paints $\frac{3}{2} \cdot \frac{1}{900} = \frac{1}{600}$ per minute and Coe paints $\frac{2}{900} = \frac{1}{450}$ per minute. In the first $90$ minutes Abe paints $\frac{90}{900} = \frac{1}{10}$ of the room.

Abe and Bea together paint $\frac{1}{900} + \frac{1}{600} = \frac{1}{360}$ per minute, and they must bring the total from $\frac{1}{10}$ up to $\frac{1}{2},$ which takes $\frac{2}{5} \cdot 360 = 144$ minutes. All three together paint $\frac{1}{360} + \frac{1}{450} = \frac{1}{200}$ per minute, so the remaining half of the room takes $\frac{1}{2} \cdot 200 = 100$ minutes.

The total time is $90 + 144 + 100 = 334$ minutes.

2.

Arnold 正在研究某男性群体中三种健康风险因素的流行情况，分别记为 $A$ 、 $B$ 和 $C$ 。对这三种因素中的每一种，随机选中一名男子只具有这一种风险因素（且不具有另外两种）的概率都是 $0.1$ 。对任意两种风险因素，随机选中一名男子恰好具有这两种风险因素（但不具有第三种）的概率都是 $0.14$ 。已知一名男子具有 $A$ 和 $B$ 的条件下，他同时具有三种风险因素的概率为 $\frac{1}{3}$ 。在一名男子不具有风险因素 $A$ 的条件下，他三种风险因素都不具有的概率为 $\frac{p}{q}$ ，其中 $p$ 和 $q$ 是互质正整数。求 $p + q$ 。

Arnold is studying the prevalence of three health risk factors, denoted by $A,$ $B,$ and $C,$ within a population of men. For each of the three factors, the probability that a randomly selected man in the population has only this risk factor (and none of the others) is $0.1.$ For any two of the three factors, the probability that a randomly selected man has exactly these two risk factors (but not the third) is $0.14.$ The probability that a randomly selected man has all three risk factors, given that he has $A$ and $B,$ is $\frac{1}{3}.$ The probability that a man has none of the three risk factors given that he does not have risk factor $A$ is $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p + q.$

答案

知识点：韦恩图条件概率

难度评级：2110

小提示：

假设总人数为 $100$ ，填入维恩图：每个“恰好一种”的区域有 $10$ 人，每个“恰好两种”的区域有 $14$ 人

Take a population of $100$ and fill in a Venn diagram: each exactly-one region holds $10$ men and each exactly-two region holds $14$

大提示：

若 $x$ 人具有三种因素，则 $\frac{x}{x + 14} = \frac{1}{3}$ 。再数不在 $A$ 中的人和完全没有风险因素的人。

If $x$ men have all three factors, then $\frac{x}{x + 14} = \frac{1}{3}.$ Then count the men outside $A$ and the men with no factor at all.

解答：

假设群体中有 $100$ 名男子并填写维恩图。三个“恰好一种”的区域各有 $10$ 人，三个“恰好两种”的区域各有 $14$ 人。若 $x$ 人具有三种风险因素，则同时具有 $A$ 和 $B$ 的人数为 $x + 14$ ，所以给定的条件概率给出 $\frac{x}{x + 14} = \frac{1}{3}$ ，从而 $x = 7$ 。

三个集合的并集中共有 $3 \cdot 10 + 3 \cdot 14 + 7 = 79$ 人，所以有 $21$ 人没有任何风险因素。具有风险因素 $A$ 的人数为 $10 + 14 + 14 + 7 = 45$ ，所以不具有 $A$ 的人数为 $55$ 。

所求概率为 $\frac{21}{55}$ ，已经是最简分数，因此 $p + q = 21 + 55 = 76$ 。

Take a population of $100$ men and fill in a Venn diagram. Each of the three exactly-one regions contains $10$ men, and each of the three exactly-two regions contains $14.$ If $x$ men have all three factors, then the men with both $A$ and $B$ number $x + 14,$ so the given conditional probability says $\frac{x}{x + 14} = \frac{1}{3},$ giving $x = 7.$

The union of the three sets therefore contains $3 \cdot 10 + 3 \cdot 14 + 7 = 79$ men, leaving $21$ with no risk factor. The men with risk factor $A$ number $10 + 14 + 14 + 7 = 45,$ so $55$ men do not have $A.$

The desired probability is $\frac{21}{55},$ which is in lowest terms, so $p + q = 21 + 55 = 76.$

3.

一个长方形的边长为 $a$ 和 $36$ 。在长方形的每个顶点，以及每条长为 $36$ 的边的中点处都安装铰链。把两条长为 $a$ 的边彼此压近，同时保持这两条边平行，于是长方形变成如图所示的凸六边形。当图形成为一个六边形，且两条长为 $a$ 的边平行并相距 $24$ 时，该六边形的面积与原长方形相同。求 $a^2$ 。

A rectangle has sides of length $a$ and $36.$ A hinge is installed at each vertex of the rectangle and at the midpoint of each side of length $36.$ The sides of length $a$ can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with the sides of length $a$ parallel and separated by a distance of $24,$ the hexagon has the same area as the original rectangle. Find $a^2.$

答案

知识点：勾股定理梯形面积

难度评级：2110

小提示：

每条长为 $36$ 的边的一半是一根长为 $18$ 的铰接杆，在六边形中的竖直跨度为 $12$ ；用勾股定理求它的水平跨度

Each half of a $36$ -side is a hinged bar of length $18$ whose vertical extent in the hexagon is $12;$ the Pythagorean theorem gives its horizontal reach

大提示：

沿两个中点铰链所在的直线把六边形切成两个全等梯形，并令总面积等于 $36a$

Cut the hexagon along the line through the two midpoint hinges into two congruent trapezoids and set the total area equal to $36a$

解答：

在六边形中，每条长为 $36$ 的边都在中点折成两根长为 $18$ 的杆。两条长为 $a$ 的边相距 $24$ ，所以每根杆的竖直跨度为 $12$ ，水平跨度为 $\sqrt{18^2 - 12^2} = \sqrt{180} = 6\sqrt{5}$ 。

过两个中点铰链的直线把六边形分成两个全等梯形，梯形的平行边为 $a$ 和 $a + 12\sqrt{5}$ ，高为 $12$ ，因此六边形面积为 $\begin{aligned} &2 \cdot \frac{a + (a + 12\sqrt{5})}{2} \cdot 12 \\ &= 24a + 144\sqrt{5} \end{aligned}\text{。}$

令它等于长方形面积 $36a$ ，得到 $12a = 144\sqrt{5}$ ，所以 $a = 12\sqrt{5}$ ，且 $a^2 = 720$ 。

In the hexagon, each side of length $36$ has folded at its midpoint into two bars of length $18.$ The two sides of length $a$ are $24$ apart, so each bar spans a vertical distance of $12$ and hence a horizontal distance of $\sqrt{18^2 - 12^2} = \sqrt{180} = 6\sqrt{5}.$

The line through the two midpoint hinges splits the hexagon into two congruent trapezoids with parallel sides $a$ and $a + 12\sqrt{5}$ and height $12,$ so the hexagon has area $\begin{aligned} &2 \cdot \frac{a + (a + 12\sqrt{5})}{2} \cdot 12 \\ &= 24a + 144\sqrt{5}. \end{aligned}$

Setting this equal to the rectangle’s area $36a$ gives $12a = 144\sqrt{5},$ so $a = 12\sqrt{5}$ and $a^2 = 720.$

4.

循环小数 $0.abab\overline{ab}$ 和 $0.abcabc\overline{abc}$ 满足 $0.abab\overline{ab} + 0.abcabc\overline{abc} = \frac{33}{37}\text{，}$ 其中 $a$ 、 $b$ 和 $c$ 是数字，且不一定互不相同。求三位数 $abc$ 。

The repeating decimals $0.abab\overline{ab}$ and $0.abcabc\overline{abc}$ satisfy $0.abab\overline{ab} + 0.abcabc\overline{abc} = \frac{33}{37},$ where $a,$ $b,$ and $c$ are (not necessarily distinct) digits. Find the three-digit number $abc.$

答案

知识点：循环小数数字模运算

难度评级：2230

小提示：

把两个循环小数写成 $\frac{ab}{99}$ 和 $\frac{abc}{999}$ ，再利用 $99 = 9 \cdot 11$ 和 $999 = 27 \cdot 37$ 通分

Write the decimals as $\frac{ab}{99}$ and $\frac{abc}{999},$ then clear denominators using $99 = 9 \cdot 11$ and $999 = 27 \cdot 37$

大提示：

将方程模 $11$ 化简会迫使 $a = b$ ；之后一个简短的一次方程确定数字

Reducing the equation modulo $11$ forces $a = b;$ then a short linear equation pins down the digits

解答：

用 $ab$ 和 $abc$ 表示相应的两位数和三位数，则两个小数分别为 $\frac{ab}{99}$ 和 $\frac{abc}{999}$ 。因为 $99 = 9 \cdot 11$ ，且 $999 = 27 \cdot 37$ ，公分母为 $27 \cdot 37 \cdot 11 = 10989$ ，两边同乘这个数，得到 $\begin{aligned} &111 \cdot ab + 11 \cdot abc \\ &= \frac{33}{37} \cdot 10989 \\ &= 9801 \end{aligned}\text{。}$

模 $11$ 下，因为 $9801 = 11 \cdot 891$ 且 $111 \equiv 1$ ，所以 $ab$ 能被 $11$ 整除，从而 $a = b$ 。于是 $ab = 11a$ ，原方程除以 $11$ 得到 $111a + abc = 891$ 。又 $abc = 110a + c$ ，所以 $221a + c = 891$ ，这要求 $a = 4$ 且 $c = 7$ 。

因此 $a = b = 4$ ， $c = 7$ ，三位数 $abc$ 为 $447$ 。

Writing $ab$ and $abc$ for the two- and three-digit numbers, the decimals equal $\frac{ab}{99}$ and $\frac{abc}{999}.$ Since $99 = 9 \cdot 11$ and $999 = 27 \cdot 37,$ the common denominator is $27 \cdot 37 \cdot 11 = 10989,$ and multiplying the equation by it gives $\begin{aligned} &111 \cdot ab + 11 \cdot abc \\ &= \frac{33}{37} \cdot 10989 \\ &= 9801. \end{aligned}$

Modulo $11,$ since $9801 = 11 \cdot 891$ and $111 \equiv 1,$ this forces $ab$ to be divisible by $11,$ so $a = b.$ Then $ab = 11a,$ and dividing the equation by $11$ gives $111a + abc = 891.$ Since $abc = 110a + c,$ this is $221a + c = 891,$ which requires $a = 4$ and $c = 7.$

Thus $a = b = 4,$ $c = 7,$ and the three-digit number $abc$ is $447.$

5.

实数 $r$ 和 $s$ 是 $p(x) = x^3 + ax + b$ 的根，而 $r + 4$ 和 $s - 3$ 是 $q(x) = x^3 + ax + b + 240$ 的根。求 $|b|$ 的所有可能值之和。

Real numbers $r$ and $s$ are roots of $p(x) = x^3 + ax + b,$ and $r + 4$ and $s - 3$ are roots of $q(x) = x^3 + ax + b + 240.$ Find the sum of all possible values of $|b|.$

答案

知识点：多项式韦达定理方程组

难度评级：2560

小提示：

两个三次多项式都没有 $x^2$ 项，所以第三个根分别是 $t = -r - s$ 和 $t - 1$ 。比较两个三次式中 $x$ 的系数。

Neither cubic has an $x^2$ term, so the third roots are $t = -r - s$ and $t - 1.$ Equate the coefficients of $x$ in the two cubics.

大提示：

比较常数项并代入 $t = 4r - 3s + 13$ ，得到 $(r-s)^2 + 7(r-s) - 8 = 0$ ；再利用 $r + s + t = 0$ 分别完成两个情形。

Comparing constant terms and substituting $t = 4r - 3s + 13$ yields $(r-s)^2 + 7(r-s) - 8 = 0;$ finish each case using $r + s + t = 0.$

解答：

两个三次多项式的 $x^2$ 系数都为 $0$ ，所以它们的根之和为零： $p$ 的第三个根是 $t = -r - s$ ，而 $q$ 的第三个根是 $-(r+4) - (s-3) = t - 1$ 。两个多项式中 $x$ 的系数同为 $a$ ，所以 $\begin{aligned} &rs + st + tr \\ &= (r+4)(s-3) \\ &\quad {}+ (s-3)(t-1) \\ &\quad {}+ (t-1)(r+4) \end{aligned}\text{，}$ 化简得 $t = 4r - 3s + 13$ 。

常数项给出 $b = -rst$ 和 $b + 240 =$ $-(r+4)(s-3)(t-1)$ ，所以 $240 =$ $rst - (r+4)(s-3)(t-1)$ ，即 $rs - 4st + 3tr - 3r$ $+ 4s + 12t - 252 = 0$ 。代入 $t = 4r - 3s + 13$ ，化为 $12\left[(r-s)^2 + 7(r-s) - 8\right] = 0$ ，所以 $r - s = 1$ 或 $r - s = -8$ 。

若 $r - s = 1$ ，则 $t = 4r - 3s + 13 = r + 16$ ，且 $t = -r - s = -2r + 1$ ，所以 $r = -5$ ：根为 $-5$ 、 $-6$ 、 $11$ ，且 $b = -rst = -330$ 。若 $r - s = -8$ ，则 $t = r - 11 = -2r - 8$ ，所以 $r = 1$ ：根为 $1$ 、 $9$ 、 $-10$ ，且 $b = 90$ 。所求和为 $330 + 90 = 420$ 。

Both cubics have zero $x^2$ coefficient, so their roots sum to $0:$ the third root of $p$ is $t = -r - s,$ and the third root of $q$ is $-(r+4) - (s-3) = t - 1.$ The coefficient of $x$ is $a$ in both, so $\begin{aligned} &rs + st + tr \\ &= (r+4)(s-3) \\ &\quad {}+ (s-3)(t-1) \\ &\quad {}+ (t-1)(r+4), \end{aligned}$ which simplifies to $t = 4r - 3s + 13.$

The constant terms give $b = -rst$ and $b + 240 =$ $-(r+4)(s-3)(t-1),$ so $240 =$ $rst - (r+4)(s-3)(t-1),$ i.e. $rs - 4st + 3tr - 3r$ $+ 4s + 12t - 252 = 0.$ Substituting $t = 4r - 3s + 13$ reduces this to $12\left[(r-s)^2 + 7(r-s) - 8\right] = 0,$ so $r - s = 1$ or $r - s = -8.$

If $r - s = 1,$ then $t = 4r - 3s + 13 = r + 16$ and $t = -r - s = -2r + 1,$ so $r = -5:$ the roots are $-5,$ $-6,$ $11,$ and $b = -rst = -330.$ If $r - s = -8,$ then $t = r - 11 = -2r - 8,$ so $r = 1:$ the roots are $1,$ $9,$ $-10,$ and $b = 90.$ The requested sum is $330 + 90 = 420.$

6.

Charles 有两个六面骰子。其中一个是公平骰子，另一个有偏骰子掷出六点的概率为 $\frac{2}{3}$ ，其余五个面各自出现的概率为 $\frac{1}{15}$ 。Charles 随机选择这两个骰子中的一个并掷三次。已知前两次都掷出六点，第三次也掷出六点的概率为 $\frac{p}{q}$ ，其中 $p$ 和 $q$ 是互质正整数。求 $p + q$ 。

Charles has two six-sided dice. One of the dice is fair, and the other die is biased so that it comes up six with probability $\frac{2}{3},$ and each of the other five sides has probability $\frac{1}{15}.$ Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixes, the probability that the third roll will also be a six is $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p + q.$

答案

知识点：条件概率贝叶斯定理骰子（概率）

难度评级：2390

小提示：

所求概率等于“三次都是六点”的概率除以“前两次都是六点”的概率；这两个概率都要对两个等可能骰子取平均

The answer is the probability of three sixes divided by the probability of two sixes, each averaged over the two equally likely dice

大提示：

连续两次六点会让有偏骰子的可能性大大增加：先按每个骰子产生这两次六点的概率加权，再预测第三次

Two sixes in a row make the biased die much more likely: weight each die by its chance of producing the observed two sixes before predicting the third roll

解答：

所求条件概率为 $\begin{aligned} &\frac{\Pr(\text{三次都是六})}{\Pr(\text{前两次是六})} \\ &= \frac{\frac{1}{2}\left(\frac{2}{3}\right)^3 + \frac{1}{2}\left(\frac{1}{6}\right)^3} {\frac{1}{2}\left(\frac{2}{3}\right)^2 + \frac{1}{2}\left(\frac{1}{6}\right)^2} \end{aligned}\text{，}$ 因为每个骰子被选中的概率都是 $\frac{1}{2}$ ，公平骰子掷出六点的概率为 $\frac{1}{6}$ 。

分子为 $\frac{1}{2}\left(\frac{8}{27} + \frac{1}{216}\right) = \frac{65}{432}$ ，分母为 $\frac{1}{2}\left(\frac{4}{9} + \frac{1}{36}\right) = \frac{17}{72}$ ，所以概率为 $\frac{65}{432} \cdot \frac{72}{17} = \frac{65}{102}$ 。

因为 $65 = 5 \cdot 13$ ，而 $102 = 2 \cdot 3 \cdot 17$ ，二者没有公因数，所以 $p + q = 65 + 102 = 167$ 。

The desired conditional probability is $\begin{aligned} &\frac{\Pr(\text{three sixes})}{\Pr(\text{first two are sixes})} \\ &= \frac{\frac{1}{2}\left(\frac{2}{3}\right)^3 + \frac{1}{2}\left(\frac{1}{6}\right)^3} {\frac{1}{2}\left(\frac{2}{3}\right)^2 + \frac{1}{2}\left(\frac{1}{6}\right)^2}, \end{aligned}$ since each die is chosen with probability $\frac{1}{2}$ and the fair die shows a six with probability $\frac{1}{6}.$

The numerator is $\frac{1}{2}\left(\frac{8}{27} + \frac{1}{216}\right) = \frac{65}{432}$ and the denominator is $\frac{1}{2}\left(\frac{4}{9} + \frac{1}{36}\right) = \frac{17}{72},$ so the probability is $\frac{65}{432} \cdot \frac{72}{17} = \frac{65}{102}.$

Since $65 = 5 \cdot 13$ and $102 = 2 \cdot 3 \cdot 17$ share no factor, $p + q = 65 + 102 = 167.$

7.

令 $f(x) = \left(x^2 + 3x + 2\right)^{\cos(\pi x)}$ 。求所有满足下式的正整数 $n$ 之和： $\left|\sum_{k=1}^{n} \log_{10} f(k)\right| = 1$

Let $f(x) = \left(x^2 + 3x + 2\right)^{\cos(\pi x)}.$ Find the sum of all positive integers $n$ for which $\left|\sum_{k=1}^{n} \log_{10} f(k)\right| = 1.$

答案

知识点：裂项相消对数奇偶性

难度评级：2450

小提示：

因为 $\cos(\pi k) = (-1)^k$ ，乘积中的 $f(k)$ 会裂项相消：偶数 $k$ 贡献 $(k+1)(k+2)$ ，奇数 $k$ 贡献其倒数

Since $\cos(\pi k) = (-1)^k,$ the product of the $f(k)$ telescopes: even $k$ contribute $(k+1)(k+2)$ and odd $k$ contribute its reciprocal

大提示：

当 $n$ 为偶数时乘积为 $\frac{n+2}{2}$ ；当 $n$ 为奇数时乘积为 $\frac{1}{2(n+2)}$ 。对数和为 $\pm 1$ 当且仅当乘积为 $10$ 或 $\frac{1}{10}$ 。

For even $n$ the product is $\frac{n+2}{2};$ for odd $n$ it is $\frac{1}{2(n+2)}.$ The sum of logs is $\pm 1$ when the product is $10$ or $\frac{1}{10}.$

解答：

因为 $\cos(\pi k) = (-1)^k$ ，且 $k^2 + 3k + 2 = (k+1)(k+2)$ ，所以 $f(k) = \left[(k+1)(k+2)\right]^{(-1)^k}$ 。这些对数之和等于乘积 $\prod_{k=1}^n f(k)$ 的对数，而乘积会裂项相消：相邻因子 $\frac{1}{(k+1)(k+2)}$ 和 $(k+2)(k+3)$ 只留下边界项。

当 $n$ 为偶数时，乘积为 $\frac{3 \cdot 4 \cdots (n+2)}{2 \cdot 3 \cdots (n+1)} = \frac{n+2}{2}$ ，当 $n$ 为奇数时，乘积为 $\frac{3 \cdot 4 \cdots (n+1)}{2 \cdot 3 \cdots (n+2)} = \frac{1}{2(n+2)}$ 。对数的绝对值等于 $1$ ，当且仅当乘积为 $10$ 或 $\frac{1}{10}$ 。

偶数 $n$ 时， $\frac{n+2}{2} = 10$ 给出 $n = 18$ ；奇数 $n$ 时， $2(n+2) = 10$ 给出 $n = 3$ 。所求和为 $18 + 3 = 21$ 。

Since $\cos(\pi k) = (-1)^k$ and $k^2 + 3k + 2 = (k+1)(k+2),$ we have $f(k) = \left[(k+1)(k+2)\right]^{(-1)^k}.$ The sum of the logarithms is the log of the product $\prod_{k=1}^n f(k),$ which telescopes: consecutive factors $\frac{1}{(k+1)(k+2)}$ and $(k+2)(k+3)$ leave only boundary terms.

For even $n$ the product is $\frac{3 \cdot 4 \cdots (n+2)}{2 \cdot 3 \cdots (n+1)} = \frac{n+2}{2},$ and for odd $n$ it is $\frac{3 \cdot 4 \cdots (n+1)}{2 \cdot 3 \cdots (n+2)} = \frac{1}{2(n+2)}.$ The absolute value of the log equals $1$ exactly when the product is $10$ or $\frac{1}{10}.$

For even $n,$ $\frac{n+2}{2} = 10$ gives $n = 18;$ for odd $n,$ $2(n+2) = 10$ gives $n = 3.$ The requested sum is $18 + 3 = 21.$

8.

半径为 $2$ 的圆 $C$ 有直径 $\overline{AB}$ 。圆 $D$ 在点 $A$ 与圆 $C$ 内切。圆 $E$ 与圆 $C$ 内切、与圆 $D$ 外切，并与 $\overline{AB}$ 相切。圆 $D$ 的半径是圆 $E$ 半径的三倍，并可写成 $\sqrt{m} - n$ ，其中 $m$ 和 $n$ 是正整数。求 $m + n$ 。

Circle $C$ with radius $2$ has diameter $\overline{AB}.$ Circle $D$ is internally tangent to circle $C$ at $A.$ Circle $E$ is internally tangent to circle $C,$ externally tangent to circle $D,$ and tangent to $\overline{AB}.$ The radius of circle $D$ is three times the radius of circle $E$ and can be written in the form $\sqrt{m} - n,$ where $m$ and $n$ are positive integers. Find $m + n.$

答案

知识点：相切圆勾股定理根式

难度评级：2710

小提示：

设圆 $E$ 的半径为 $s$ ，并从其圆心向 $\overline{AB}$ 作垂线，垂足为 $F$ ；用勾股定理表示 $CF$ 和 $DF$ 。

Let $s$ be the radius of circle $E$ and drop its center to a foot $F$ on $\overline{AB};$ express $CF$ and $DF$ with the Pythagorean theorem

大提示：

$DF = DC + CF$ 给出 $s\sqrt{15} = (2 - 3s) + \sqrt{4 - 4s}$ ；孤立一个根式并平方两次，会得到关于 $s$ 的二次方程

$DF = DC + CF$ gives $s\sqrt{15} = (2 - 3s) + \sqrt{4 - 4s};$ isolating one radical and squaring twice leaves a quadratic in $s$

解答：

也用 $C$ 、 $D$ 、 $E$ 表示这些圆的圆心。设圆 $E$ 的半径为 $s$ ，则圆 $D$ 的半径为 $3s$ ，并设 $F$ 是 $E$ 到 $\overline{AB}$ 的垂足。相切关系给出 $CE = 2 - s$ ， $DE = 3s + s = 4s$ ， $EF = s$ ，而 $D$ 在 $\overline{AB}$ 上且 $DC = 2 - 3s$ 。

直角三角形 $CEF$ 和 $DEF$ 给出 $CF = \sqrt{(2-s)^2 - s^2}$ $= \sqrt{4 - 4s}$ ，以及 $DF = \sqrt{(4s)^2 - s^2} = s\sqrt{15}$ 。因为 $F$ 与 $A$ 位于 $C$ 的相反侧，所以 $DF = DC + CF$ ，即 $s\sqrt{15} = (2 - 3s) + \sqrt{4 - 4s}\text{。}$

将 $2 - 3s$ 移到左边并平方，得到 $24s^2 - 8s = 2\sqrt{15}\,s\,(2 - 3s)$ ，也就是 $12s - 4 = \sqrt{15}\,(2 - 3s)$ ；再次平方得 $9s^2 + 84s - 44 = 0$ ，所以 $s = \frac{-14 + 4\sqrt{15}}{3}$ 。圆 $D$ 的半径为 $3s = 4\sqrt{15} - 14 = \sqrt{240} - 14$ ，因此 $m + n = 240 + 14 = 254$ 。

Let $C,$ $D,$ $E$ also name the circles’ centers, let $s$ be the radius of circle $E,$ so circle $D$ has radius $3s,$ and let $F$ be the foot of $E$ on $\overline{AB}.$ Tangency gives $CE = 2 - s,$ $DE = 3s + s = 4s,$ and $EF = s,$ while $D$ lies on $\overline{AB}$ with $DC = 2 - 3s.$

Right triangles $CEF$ and $DEF$ give $CF = \sqrt{(2-s)^2 - s^2}$ $= \sqrt{4 - 4s}$ and $DF = \sqrt{(4s)^2 - s^2} = s\sqrt{15}.$ Since $F$ is on the opposite side of $C$ from $A,$ we have $DF = DC + CF,$ so $s\sqrt{15} = (2 - 3s) + \sqrt{4 - 4s}.$

Moving $2 - 3s$ to the left and squaring gives $24s^2 - 8s = 2\sqrt{15}\,s\,(2 - 3s),$ i.e. $12s - 4 = \sqrt{15}\,(2 - 3s);$ squaring again yields $9s^2 + 84s - 44 = 0,$ so $s = \frac{-14 + 4\sqrt{15}}{3}.$ The radius of circle $D$ is $3s = 4\sqrt{15} - 14 = \sqrt{240} - 14,$ and $m + n = 240 + 14 = 254.$

9.

十把椅子围成一圈。求这些椅子的子集中，包含至少三把相邻椅子的子集个数。

Ten chairs are arranged in a circle. Find the number of subsets of this set of chairs that contain at least three adjacent chairs.

答案

知识点：环形排列子集容斥原理

难度评级：2760

小提示：

除全集外，按顺时针找到一把空椅子，后面紧接着三把属于该子集的椅子

Count subsets other than the full set by locating an empty chair followed clockwise by three chairs of the subset

大提示：

每个这样的空-选-选-选模式让剩下六把椅子任意选择，共有 $2^6$ 种；再减去含有两段相互分离的长度至少为三的连续段而被重复计算的子集。

Each such empty-full-full-full pattern leaves $2^6$ choices for the remaining chairs; subtract the double-counted subsets containing two separate runs of three or more

解答：

全部 $10$ 把椅子的子集符合条件；先数其他子集。对每个长度至少为三的极大连续选中段，定位其顺时针起点。这样的子集必含一个连续四椅块，形如空-选-选-选。这个块有 $10$ 个位置，其余 $6$ 把椅子任意选择，给出 $10 \cdot 2^6 = 640$ 次计数。

每个子集被计数的次数等于其长度至少为 $3$ 的极大连续段数。两个这样的连续段至少需要 $3 + 3$ 把选中椅子和两个空隙，所以不可能有三段；恰有两段的子集被计数两次。要有两段，放置两个不相交的空-选-选-选块：有 $\frac{10 \cdot 3}{2} = 15$ 种方法（第二个块在剩余 $6$ 把椅子中有 $3$ 个位置），最后 $2$ 把椅子任意选择，因此有 $15 \cdot 2^2 = 60$ 个这样的子集。

总数为 $1 + 640 - 60 = 581$ 。

The full set of $10$ chairs qualifies; count the others by locating each maximal run of at least three adjacent chosen chairs at its clockwise start. Any such subset contains a block of four consecutive chairs that is empty-chosen-chosen-chosen. There are $10$ positions for this block, and the remaining $6$ chairs are free, giving $10 \cdot 2^6 = 640.$

This counts once for each maximal run of length at least $3.$ Two such runs require at least $3 + 3$ chosen chairs plus two gaps, so three runs are impossible, and subsets with exactly two runs are counted twice. To have two runs, place two disjoint empty-chosen-chosen-chosen blocks: $\frac{10 \cdot 3}{2} = 15$ ways (the second block fits in $3$ positions among the remaining $6$ chairs), with the last $2$ chairs free, for $15 \cdot 2^2 = 60$ subsets.

The total is $1 + 640 - 60 = 581.$

10.

设 $z$ 是满足 $|z| = 2014$ 的复数。令 $P$ 为复平面中的多边形，其顶点包括 $z$ 以及所有满足 $\frac{1}{z+w} = \frac{1}{z} + \frac{1}{w}$ 的 $w$ 。则 $P$ 所围成的面积可写成 $n\sqrt{3}$ ，其中 $n$ 是整数。求 $n$ 除以 $1000$ 的余数。

Let $z$ be a complex number with $|z| = 2014.$ Let $P$ be the polygon in the complex plane whose vertices are $z$ and every $w$ such that $\frac{1}{z+w} = \frac{1}{z} + \frac{1}{w}.$ Then the area enclosed by $P$ can be written in the form $n\sqrt{3},$ where $n$ is an integer. Find the remainder when $n$ is divided by $1000.$

答案

知识点：复数单位根等边三角形

难度评级：2560

小提示：

将 $\frac{1}{z+w} = \frac{1}{z} + \frac{1}{w}$ 清分母，得到 $z^2 + zw + w^2 = 0$

Clear denominators in $\frac{1}{z+w} = \frac{1}{z} + \frac{1}{w}$ to get $z^2 + zw + w^2 = 0$

大提示：

乘以 $z - w$ 得到 $w^3 = z^3$ ，所以顶点是 $z$ 乘以三次单位根：这是外接圆半径为 $2014$ 的等边三角形

Multiplying by $z - w$ gives $w^3 = z^3,$ so the vertices are $z$ times the cube roots of unity: an equilateral triangle with circumradius $2014$

解答：

将 $\frac{1}{z+w} = \frac{1}{z} + \frac{1}{w}$ 两边同乘 $zw(z+w)$ ，得到 $zw = (z+w)^2$ ，即 $z^2 + zw + w^2 = 0$ 。再乘以 $z - w$ ，得到 $z^3 - w^3 = 0$ ，所以 $w = \omega z$ 或 $w = \omega^2 z$ ，其中 $\omega$ 是本原三次单位根（并且二者确实满足原方程）。

因此 $P$ 是顶点为 $z$ 、 $\omega z$ 、 $\omega^2 z$ 的等边三角形，内接于半径为 $2014$ 的圆。其面积为 $\frac{3\sqrt{3}}{4}\,(2014)^2 = 3 \cdot 1007^2 \sqrt{3}\text{，}$ 所以 $n = 3 \cdot 1007^2 = 3042147$ 。

$n$ 除以 $1000$ 的余数为 $147$ 。

Multiplying $\frac{1}{z+w} = \frac{1}{z} + \frac{1}{w}$ by $zw(z+w)$ gives $zw = (z+w)^2,$ i.e. $z^2 + zw + w^2 = 0.$ Multiplying by $z - w$ yields $z^3 - w^3 = 0,$ so $w = \omega z$ or $w = \omega^2 z,$ where $\omega$ is a primitive cube root of unity (and both indeed satisfy the original equation).

Thus $P$ is the equilateral triangle with vertices $z,$ $\omega z,$ $\omega^2 z,$ inscribed in the circle of radius $2014.$ Its area is $\frac{3\sqrt{3}}{4}\,(2014)^2 = 3 \cdot 1007^2 \sqrt{3},$ so $n = 3 \cdot 1007^2 = 3042147.$

The remainder when $n$ is divided by $1000$ is $147.$

11.

在 $\triangle RED$ 中， $RD = 1$ 、 $\angle DRE = 75^\circ$ 、 $\angle RED = 45^\circ$ 。令 $M$ 为线段 $\overline{RD}$ 的中点。点 $C$ 在边 $\overline{ED}$ 上，且 $\overline{RC} \perp \overline{EM}$ 。将线段 $\overline{DE}$ 沿 $E$ 的方向延长至点 $A$ ，使 $CA = AR$ 。于是 $AE = \frac{a - \sqrt{b}}{c}$ ，其中 $a$ 和 $c$ 是互质正整数， $b$ 是正整数。求 $a + b + c$ 。

In $\triangle RED,$ $RD = 1,$ $\angle DRE = 75^\circ$ and $\angle RED = 45^\circ.$ Let $M$ be the midpoint of segment $\overline{RD}.$ Point $C$ lies on side $\overline{ED}$ such that $\overline{RC} \perp \overline{EM}.$ Extend segment $\overline{DE}$ through $E$ to point $A$ such that $CA = AR.$ Then $AE = \frac{a - \sqrt{b}}{c},$ where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer. Find $a + b + c.$

答案

知识点：坐标几何正弦定理斜率

难度评级：3060

小提示：

令 $D$ 在原点， $E$ 在正 $x$ 轴上；则 $R = (\cos 60^\circ, \sin 60^\circ)$ ，并由正弦定理得到 $DE = \frac{\sqrt{3}+1}{2}$

Place $D$ at the origin and $E$ on the positive $x$ -axis; then $R = (\cos 60^\circ, \sin 60^\circ)$ and the law of sines gives $DE = \frac{\sqrt{3}+1}{2}$

大提示：

用垂直斜率确定 $x$ 轴上的 $C$ ，而 $CA = AR$ 会给出关于 $A$ 的 $x$ 坐标的一次方程

Perpendicular slopes locate $C$ on the $x$ -axis, and $CA = AR$ is a linear equation in the $x$ -coordinate of $A$

解答：

因为 $\angle RDE = 180^\circ - 75^\circ - 45^\circ$ $= 60^\circ$ ，令 $D = (0,0)$ ，且 $E$ 在正 $x$ 轴上，则 $R = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ 。正弦定理给出 $DE = \frac{\sin 75^\circ}{\sin 45^\circ} = \frac{\sqrt{3}+1}{2}$ ，并且 $M = \left(\frac{1}{4}, \frac{\sqrt{3}}{4}\right)$ 。

直线 $EM$ 的斜率为 $\frac{\frac{\sqrt{3}}{4}}{\frac{1}{4} - \frac{\sqrt{3}+1}{2}} = \frac{-\sqrt{3}}{1 + 2\sqrt{3}}$ ，所以直线 $RC$ 的斜率为 $\frac{1 + 2\sqrt{3}}{\sqrt{3}}$ 。从 $R$ 下降 $\frac{\sqrt{3}}{2}$ 到 $x$ 轴，水平向左移动 $\frac{\frac{3}{2}}{1 + 2\sqrt{3}} = \frac{6\sqrt{3} - 3}{22}$ ，所以 $C = (c, 0)$ ，其中 $c = \frac{1}{2} - \frac{6\sqrt{3} - 3}{22} = \frac{7 - 3\sqrt{3}}{11}$ 。

对 $A = (t, 0)$ ，条件 $CA = AR$ 写作 $(t - c)^2 = \left(t - \frac{1}{2}\right)^2 + \frac{3}{4}$ ，这是关于 $t$ 的一次方程： $t = \frac{1 - c^2}{1 - 2c} = \frac{9 + 4\sqrt{3}}{11}$ 。因此 $\begin{aligned} AE &= t - \frac{\sqrt{3}+1}{2} \\ &= \frac{18 + 8\sqrt{3} - 11\sqrt{3} - 11}{22} \\ &= \frac{7 - 3\sqrt{3}}{22} = \frac{7 - \sqrt{27}}{22} \end{aligned}\text{，}$ 所以 $a + b + c = 7 + 27 + 22 = 56$ 。

Since $\angle RDE = 180^\circ - 75^\circ - 45^\circ$ $= 60^\circ,$ place $D = (0,0)$ with $E$ on the positive $x$ -axis, so $R = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right).$ The law of sines gives $DE = \frac{\sin 75^\circ}{\sin 45^\circ} = \frac{\sqrt{3}+1}{2},$ and $M = \left(\frac{1}{4}, \frac{\sqrt{3}}{4}\right).$

The slope of $EM$ is $\frac{\frac{\sqrt{3}}{4}}{\frac{1}{4} - \frac{\sqrt{3}+1}{2}} = \frac{-\sqrt{3}}{1 + 2\sqrt{3}},$ so line $RC$ has slope $\frac{1 + 2\sqrt{3}}{\sqrt{3}}.$ Descending from $R$ by $\frac{\sqrt{3}}{2}$ to the $x$ -axis moves us left by $\frac{\frac{3}{2}}{1 + 2\sqrt{3}} = \frac{6\sqrt{3} - 3}{22},$ so $C = (c, 0)$ with $c = \frac{1}{2} - \frac{6\sqrt{3} - 3}{22} = \frac{7 - 3\sqrt{3}}{11}.$

For $A = (t, 0),$ the condition $CA = AR$ reads $(t - c)^2 = \left(t - \frac{1}{2}\right)^2 + \frac{3}{4},$ which is linear in $t:$ $t = \frac{1 - c^2}{1 - 2c} = \frac{9 + 4\sqrt{3}}{11}.$ Then $\begin{aligned} AE &= t - \frac{\sqrt{3}+1}{2} \\ &= \frac{18 + 8\sqrt{3} - 11\sqrt{3} - 11}{22} \\ &= \frac{7 - 3\sqrt{3}}{22} = \frac{7 - \sqrt{27}}{22}, \end{aligned}$ so $a + b + c = 7 + 27 + 22 = 56.$

12.

假设 $\triangle ABC$ 的角满足 $\cos(3A) + \cos(3B)$ $+ \cos(3C) = 1$ 。该三角形的两条边长为 $10$ 和 $13$ 。存在正整数 $m$ ，使得 $\triangle ABC$ 剩余一边的最大可能长度为 $\sqrt{m}$ 。求 $m$ 。

Suppose that the angles of $\triangle ABC$ satisfy $\cos(3A) + \cos(3B)$ $+ \cos(3C) = 1.$ Two sides of the triangle have lengths $10$ and $13.$ There is a positive integer $m$ so that the maximum possible length for the remaining side of $\triangle ABC$ is $\sqrt{m}.$ Find $m.$

答案

知识点：三角恒等式因式分解余弦定理

难度评级：2990

小提示：

写成 $1 - \cos 3A = 2\sin^2 \frac{3A}{2}$ ，再用和差化积处理 $\cos 3B + \cos 3C$ ；整个条件会分解

Write $1 - \cos 3A = 2\sin^2 \frac{3A}{2}$ and combine $\cos 3B + \cos 3C$ by sum-to-product; the whole condition factors

大提示：

必有一个角等于 $120^\circ$ 。当这个角夹在长为 $10$ 和 $13$ 的两边之间时，剩余边最长。

One angle must equal $120^\circ.$ The remaining side is longest when that angle lies between the sides of lengths $10$ and $13.$

解答：

使用 $1 - \cos 3A = 2\sin^2\frac{3A}{2}$ 和 $\cos 3B + \cos 3C =$ $2\cos\frac{3(B+C)}{2}\cos\frac{3(B-C)}{2}$ ，又因为 $\frac{3(B+C)}{2} = 270^\circ - \frac{3A}{2}$ ，所以 $\cos\frac{3(B+C)}{2} = -\sin\frac{3A}{2}$ ，于是条件变为 $\begin{aligned} 0 &= 2\sin\tfrac{3A}{2} \\ &\quad {}\cdot \left(\sin\tfrac{3A}{2} + \cos\tfrac{3(B-C)}{2}\right) \\ &= 2\sin\tfrac{3A}{2} \\ &\quad {}\cdot \left(\cos\tfrac{3(B-C)}{2} - \cos\tfrac{3(B+C)}{2}\right) \\ &= 4\sin\tfrac{3A}{2}\sin\tfrac{3B}{2}\sin\tfrac{3C}{2} \end{aligned}\text{。}$

对三角形的任意角 $X$ ， $\frac{3X}{2}$ 严格位于 $0^\circ$ 和 $270^\circ$ 之间，所以 $\sin\frac{3X}{2} = 0$ 当且仅当 $X = 120^\circ$ 。因此该三角形有一个角为 $120^\circ$ 。

当 $120^\circ$ 角夹在长为 $10$ 和 $13$ 的两边之间时，剩余边最长（如果 $120^\circ$ 角对着其中一条已知边，则剩余边会比那条边短）。由余弦定理，该边长为 $\sqrt{10^2 + 13^2 + 10 \cdot 13} = \sqrt{399}$ ，所以 $m = 399$ 。

Using $1 - \cos 3A = 2\sin^2\frac{3A}{2}$ and $\cos 3B + \cos 3C =$ $2\cos\frac{3(B+C)}{2}\cos\frac{3(B-C)}{2},$ together with $\frac{3(B+C)}{2} = 270^\circ - \frac{3A}{2}$ so that $\cos\frac{3(B+C)}{2} = -\sin\frac{3A}{2},$ the condition becomes $\begin{aligned} 0 &= 2\sin\tfrac{3A}{2} \\ &\quad {}\cdot \left(\sin\tfrac{3A}{2} + \cos\tfrac{3(B-C)}{2}\right) \\ &= 2\sin\tfrac{3A}{2} \\ &\quad {}\cdot \left(\cos\tfrac{3(B-C)}{2} - \cos\tfrac{3(B+C)}{2}\right) \\ &= 4\sin\tfrac{3A}{2}\sin\tfrac{3B}{2}\sin\tfrac{3C}{2}. \end{aligned}$

For an angle $X$ of a triangle, $\frac{3X}{2}$ lies strictly between $0^\circ$ and $270^\circ,$ so $\sin\frac{3X}{2} = 0$ exactly when $X = 120^\circ.$ Hence one angle of the triangle is $120^\circ.$

The remaining side is longest when the $120^\circ$ angle sits between the sides of lengths $10$ and $13$ (if $120^\circ$ were opposite one of them, the remaining side would be shorter than that side). By the law of cosines its length is $\sqrt{10^2 + 13^2 + 10 \cdot 13} = \sqrt{399},$ so $m = 399.$

13.

十个成年人进入一个房间，脱下鞋子并把鞋子扔成一堆。之后，一个孩子随机地把每只左鞋与一只右鞋配成一双，不考虑它们是否原本属于同一个人。对于每个满足 $k \lt 5$ 的正整数，任意由孩子配出的 $k$ 双鞋都不会恰好涉及 $k$ 个成年人。这个事件的概率为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

Ten adults enter a room, remove their shoes, and toss their shoes into a pile. Later, a child randomly pairs each left shoe with a right shoe without regard to which shoes belong together. The probability that for every positive integer $k \lt 5,$ no collection of $k$ pairs made by the child contains the shoes from exactly $k$ of the adults is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

知识点：排列基本概率

难度评级：3060

小提示：

将左鞋 $j$ 与右鞋 $\pi(j)$ 配对定义了一个随机排列；由 $k$ 双鞋组成并恰好涉及 $k$ 个成年人的集合就是若干个循环的并

Pairing left shoe $j$ with right shoe $\pi(j)$ defines a random permutation; a collection of $k$ pairs using exactly $k$ adults’ shoes is a union of cycles

大提示：

条件等价于 $\pi$ 的每个循环长度都至少为 $5$ ：数一个 $10$ -循环和两个 $5$ -循环的乘积

The condition says every cycle of $\pi$ has length at least $5:$ count $10$ -cycles and products of two $5$ -cycles

解答：

孩子的配对可以看作把左鞋 $j$ 配给右鞋 $\pi(j)$ ，其中 $\pi$ 是 $\{1, \ldots, 10\}$ 上的均匀随机排列。由 $k$ 双鞋组成的集合含有 $k$ 只左鞋和 $k$ 只右鞋，所以它恰好涉及 $k$ 个成年人，当且仅当这些成年人的编号在 $\pi$ 下封闭；也就是说，该集合是 $\pi$ 的若干个循环的并。因此条件等价于 $\pi$ 没有长度小于 $5$ 的循环。

循环长度必须把 $10$ 分拆成每部分至少为 $5$ ：要么是一个 $10$ -循环，要么是两个 $5$ -循环。共有 $9!$ 个十循环，而两个 $5$ -循环的排列数为 $\frac{1}{2}\binom{10}{5}(4!)^2 = \frac{9!}{5}$ 。

概率为 $\frac{9! + \frac{1}{5} \cdot 9!}{10!} = \frac{1 + \frac{1}{5}}{10} = \frac{3}{25}\text{，}$ 所以 $m + n = 3 + 25 = 28$ 。

The child’s pairing matches left shoe $j$ with right shoe $\pi(j)$ for a uniformly random permutation $\pi$ of $\{1, \ldots, 10\}.$ A collection of $k$ pairs uses $k$ left and $k$ right shoes, so it involves exactly $k$ adults precisely when those adults’ indices are closed under $\pi$ — that is, when the collection is a union of cycles of $\pi.$ The condition therefore says $\pi$ has no cycle of length less than $5.$

The cycle lengths must partition $10$ into parts of size at least $5:$ either one $10$ -cycle or two $5$ -cycles. There are $9!$ ten-cycles, and $\frac{1}{2}\binom{10}{5}(4!)^2 = \frac{9!}{5}$ permutations that are products of two $5$ -cycles.

The probability is $\frac{9! + \frac{1}{5} \cdot 9!}{10!} = \frac{1 + \frac{1}{5}}{10} = \frac{3}{25},$ so $m + n = 3 + 25 = 28.$

14.

在 $\triangle ABC$ 中， $AB = 10$ 、 $\angle A = 30^\circ$ 、 $\angle C = 45^\circ$ 。设 $H$ 、 $D$ 和 $M$ 是直线 $\overline{BC}$ 上的点，满足 $\overline{AH} \perp \overline{BC}$ 、 $\angle BAD = \angle CAD$ ，且 $BM = CM$ 。点 $N$ 是线段 $\overline{HM}$ 的中点，点 $P$ 在射线 $AD$ 上且 $\overline{PN} \perp \overline{BC}$ 。于是 $AP^2 = \frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

In $\triangle ABC,$ $AB = 10,$ $\angle A = 30^\circ,$ and $\angle C = 45^\circ.$ Let $H,$ $D,$ and $M$ be points on line $\overline{BC}$ such that $\overline{AH} \perp \overline{BC},$ $\angle BAD = \angle CAD,$ and $BM = CM.$ Point $N$ is the midpoint of segment $\overline{HM},$ and point $P$ is on ray $AD$ such that $\overline{PN} \perp \overline{BC}.$ Then $AP^2 = \frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

知识点：外接圆、外心与外接圆半径角平分线正弦定理中点

难度评级：3160

小提示：

将射线 $AD$ 延长到外接圆：它在弧 $BC$ 的中点 $E$ 处再次相交，而该点到直线 $BC$ 的投影正好是 $M$ 。

Extend ray $AD$ to the circumcircle: it meets it at the midpoint $E$ of arc $BC,$ whose projection onto line $BC$ is exactly $M$

大提示：

因为 $N$ 是 $\overline{HM}$ 的中点，所以 $P$ 是 $\overline{AE}$ 的中点。用正弦定理在三角形 $ABE$ 中求 $AE$ 。

Since $N$ is the midpoint of $\overline{HM},$ $P$ is the midpoint of $\overline{AE}.$ Find $AE$ from triangle $ABE$ with the law of sines.

解答：

设射线 $AD$ 与 $\triangle ABC$ 的外接圆再次交于 $E$ 。因为 $AD$ 平分角 $A$ ，点 $E$ 是弧 $BC$ 的中点，所以 $E$ 在 $\overline{BC}$ 的垂直平分线上，并投影到直线 $BC$ 上的点 $M$ 。共线点 $A$ 、 $P$ 、 $E$ 到直线 $BC$ 的投影分别为 $H$ 、 $N$ 、 $M$ ，而投影保持同一直线上的比例；由于 $N$ 是 $\overline{HM}$ 的中点，点 $P$ 是 $\overline{AE}$ 的中点。

这里 $\angle B = 105^\circ$ ，且 $\angle CBE = \angle CAE = 15^\circ$ （都对着弧 $CE$ ），所以 $\angle ABE = 120^\circ$ 。又 $\angle AEB = \angle ACB = 45^\circ$ （都对着弧 $AB$ ）。在 $\triangle ABE$ 中由正弦定理得到 $\begin{aligned} AE &= AB \cdot \frac{\sin \angle ABE}{\sin \angle AEB} \\ &= 10 \cdot \frac{\sin 120^\circ}{\sin 45^\circ} \\ &= 5\sqrt{6} \end{aligned}\text{。}$

因此 $AP = \frac{1}{2} AE = \frac{5\sqrt{6}}{2}$ ，所以 $AP^2 = \frac{75}{2}$ ，且 $m + n = 75 + 2 = 77$ 。

Let ray $AD$ meet the circumcircle of $\triangle ABC$ again at $E.$ Since $AD$ bisects angle $A,$ the point $E$ is the midpoint of arc $BC,$ so $E$ lies on the perpendicular bisector of $\overline{BC}$ and projects onto line $BC$ at $M.$ The projections of the collinear points $A,$ $P,$ $E$ onto line $BC$ are $H,$ $N,$ $M,$ and projection preserves ratios along a line; since $N$ is the midpoint of $\overline{HM},$ point $P$ is the midpoint of $\overline{AE}.$

Here $\angle B = 105^\circ,$ and $\angle CBE = \angle CAE = 15^\circ$ (both subtend arc $CE$ ), so $\angle ABE = 120^\circ.$ Also $\angle AEB = \angle ACB = 45^\circ$ (both subtend arc $AB$ ). The law of sines in $\triangle ABE$ gives $\begin{aligned} AE &= AB \cdot \frac{\sin \angle ABE}{\sin \angle AEB} \\ &= 10 \cdot \frac{\sin 120^\circ}{\sin 45^\circ} \\ &= 5\sqrt{6}. \end{aligned}$

Therefore $AP = \frac{1}{2} AE = \frac{5\sqrt{6}}{2},$ so $AP^2 = \frac{75}{2}$ and $m + n = 75 + 2 = 77.$

15.

对任意整数 $k \ge 1$ ，令 $p(k)$ 为不整除 $k$ 的最小素数。定义整数函数 $X(k)$ ：若 $p(k) \gt 2$ ，则该函数值为所有小于 $p(k)$ 的素数的乘积；若 $p(k) = 2$ ，则 $X(k) = 1$ 。设序列 $\{x_n\}$ 由 $x_0 = 1$ 和 $x_{n+1} X(x_n) = x_n p(x_n)$ （ $n \ge 0$ ）定义。求满足 $x_t = 2090$ 的最小正整数 $t$ 。

For any integer $k \ge 1,$ let $p(k)$ be the smallest prime which does not divide $k.$ Define the integer function $X(k)$ to be the product of all primes less than $p(k)$ if $p(k) \gt 2,$ and $X(k) = 1$ if $p(k) = 2.$ Let $\{x_n\}$ be the sequence defined by $x_0 = 1,$ and $x_{n+1} X(x_n) = x_n p(x_n)$ for $n \ge 0.$ Find the smallest positive integer $t$ such that $x_t = 2090.$

答案

知识点：质数进制递推找规律

难度评级：3270

小提示：

计算 $x_1$ 到 $x_8$ 并分解质因数；比较 $x_n$ 中出现的素数集合和 $n$ 的二进制数位。

Compute $x_1$ through $x_8$ and factor each; compare the set of primes appearing in $x_n$ with the binary digits of $n$

大提示：

这个递推就是二进制计数：除以 $X(x_n)$ 会清掉末尾为一的若干位，乘以 $p(x_n)$ 则把 $1$ 进到下一位。

The recursion is binary counting: dividing by $X(x_n)$ clears the trailing prime “ones” and multiplying by $p(x_n)$ carries a $1$ into the next bit

解答：

按顺序列出素数为 $\rho_0 = 2$ 、 $\rho_1 = 3$ 、 $\rho_2 = 5, \ldots$ 。每个 $x_n$ 都是无平方因子的，所以它由整除它的素数集合决定。我们断言这个集合编码了 $n$ 的二进制表示：若 $n = \sum_i d_i 2^i$ ，其中 $d_i \in \{0, 1\}$ ，则 $x_n = \prod_i \rho_i^{d_i}$ 。

的确，假设 $x_n = \prod_i \rho_i^{d_i}$ ，并令 $j$ 为使 $d_j = 0$ 的最小下标。则 $p(x_n) = \rho_j$ ，而 $X(x_n) = \rho_0 \rho_1 \cdots \rho_{j-1}$ ，这正是对应末尾 $1$ 位的素数乘积（当 $j = 0$ 时 $X(x_n) = 1$ ）。所以 $x_{n+1} = \frac{x_n \, \rho_j}{\rho_0 \rho_1 \cdots \rho_{j-1}}$ 会去掉末尾的一串 $1$ ，并插入 $\rho_j$ ，这正是二进制加一。因为 $x_0 = 1$ 对应 $0$ ，归纳证明了断言。

现在 $2090 = 2 \cdot 5 \cdot 11 \cdot 19 = \rho_0 \rho_2 \rho_4 \rho_7$ ，对应二进制表示中位置 $0$ 、 $2$ 、 $4$ 、 $7$ 上的数字为一。因此 $t = 2^0 + 2^2 + 2^4 + 2^7 = 149$ 。

List the primes in order as $\rho_0 = 2,$ $\rho_1 = 3,$ $\rho_2 = 5, \ldots.$ Every $x_n$ is squarefree, so it is described by the set of primes dividing it, and we claim this set encodes $n$ in binary: if $n = \sum_i d_i 2^i$ with $d_i \in \{0, 1\},$ then $x_n = \prod_i \rho_i^{d_i}.$

Indeed, suppose $x_n = \prod_i \rho_i^{d_i}$ and let $j$ be the smallest index with $d_j = 0.$ Then $p(x_n) = \rho_j,$ and $X(x_n) = \rho_0 \rho_1 \cdots \rho_{j-1}$ is exactly the product of the primes for the trailing $1$ -bits (with $X(x_n) = 1$ when $j = 0$ ). So $x_{n+1} = \frac{x_n \, \rho_j}{\rho_0 \rho_1 \cdots \rho_{j-1}}$ removes the trailing ones and inserts $\rho_j$ — precisely adding $1$ in binary. Since $x_0 = 1$ corresponds to $0,$ induction proves the claim.

Now $2090 = 2 \cdot 5 \cdot 11 \cdot 19 = \rho_0 \rho_2 \rho_4 \rho_7,$ which corresponds to binary digits at positions $0,$ $2,$ $4,$ $7.$ Hence $t = 2^0 + 2^2 + 2^4 + 2^7 = 149.$