2014 AIME I 详解 | LIVE by Po-Shen Loh

1.

一只运动鞋鞋带穿过的 $8$ 个鞋眼都在一个长方形上，较长的两边各有四个等距的鞋眼。这个长方形宽 $50$ 毫米，长 $80$ 毫米，并且每个顶点都有一个鞋眼。鞋带必须先在长方形一条宽边的两个顶点鞋眼之间穿过，然后在相邻鞋眼之间交叉穿插，直到到达另一条宽边上的两个鞋眼，如图所示。穿过最后这两个鞋眼之后，鞋带两端还必须各至少再伸出 $200$ 毫米，以便打结。求这根鞋带的最小长度，单位为毫米。

The $8$ eyelets for the lace of a sneaker all lie on a rectangle, four equally spaced on each of the longer sides. The rectangle has a width of $50$ mm and a length of $80$ mm. There is one eyelet at each vertex of the rectangle. The lace itself must pass between the vertex eyelets along a width side of the rectangle and then crisscross between successive eyelets until it reaches the two eyelets at the other width side of the rectangle as shown. After passing through these final eyelets, each of the ends of the lace must extend at least $200$ mm farther to allow a knot to be tied. Find the minimum length of the lace in millimeters.

答案

知识点：勾股定理周长

难度评级：1890

小提示：

这些鞋眼把每条 $80$ 毫米的边分成三个相等的间隔；最短鞋带只使用直线段

The eyelets divide each $80$ mm side into three equal gaps, and the shortest lace uses straight segments only

大提示：

六段交叉线段各长 $\sqrt{50^2 + (\frac{80}{3})^2}$ ；再加上一条宽边和两段 $200$ 毫米的末端

Six crisscross segments each measure $\sqrt{50^2 + (\frac{80}{3})^2};$ add one width and two $200$ mm ends

解答：

每条 $80$ 毫米边上的四个鞋眼等距排列，且两端在顶点，所以同一边上相邻鞋眼相距 $\frac{80}{3}$ 毫米。鞋带由一段横跨 $50$ 毫米宽度的线段、六段交叉线段（经过宽边连接后，两股鞋带各再交叉三次到达上端）以及两段至少 $200$ 毫米的自由末端组成。要使总长最短，每一段都应是直线段。

每段交叉线段横跨整个宽度，并沿长边方向上升一个间隔，所以长度为 $\begin{aligned} &\sqrt{50^2 + \left(\tfrac{80}{3}\right)^2} \\ &= \sqrt{\tfrac{28900}{9}} = \frac{170}{3} \end{aligned}\text{。}$

最小长度为 $50 + 6 \cdot \frac{170}{3}$ $+ 2 \cdot 200$ $= 50 + 340 + 400$ $= 790$ 。

The four eyelets on each $80$ mm side are equally spaced with one at each vertex, so consecutive eyelets on a side are $\frac{80}{3}$ mm apart. The lace consists of one segment across the $50$ mm width, six crisscross pieces (after the width crossing, each of the two strands makes three crossings to reach the top), and two free ends of at least $200$ mm each. The lace is shortest when every piece is a straight segment.

Each crisscross piece spans the full width and rises one gap, so its length is $\begin{aligned} &\sqrt{50^2 + \left(\tfrac{80}{3}\right)^2} \\ &= \sqrt{\tfrac{28900}{9}} = \frac{170}{3}. \end{aligned}$

The minimum length is $50 + 6 \cdot \frac{170}{3}$ $+ 2 \cdot 200$ $= 50 + 340 + 400$ $= 790.$

2.

一个罐子中有 $4$ 个绿球和 $6$ 个蓝球。第二个罐子中有 $16$ 个绿球和 $N$ 个蓝球。从每个罐子中各随机取出一个球。两个球颜色相同的概率为 $0.58$ 。求 $N$ 。

An urn contains $4$ green balls and $6$ blue balls. A second urn contains $16$ green balls and $N$ blue balls. A single ball is drawn at random from each urn. The probability that both balls are of the same color is $0.58.$ Find $N.$

答案

知识点：基本概率独立事件一次方程

难度评级：1750

小提示：

把取到两个绿球的概率和取到两个蓝球的概率相加

Add the probabilities of green-green and blue-blue draws

大提示：

解方程 $\frac{4}{10} \cdot \frac{16}{16+N} + \frac{6}{10} \cdot \frac{N}{16+N} = \frac{29}{50}$ ，即可求得 $N$

Solve $\frac{4}{10} \cdot \frac{16}{16+N} + \frac{6}{10} \cdot \frac{N}{16+N} = \frac{29}{50}$ for $N$

解答：

两个球都是绿色的概率为 $\frac{4}{10} \cdot \frac{16}{16+N}$ ，两个球都是蓝色的概率为 $\frac{6}{10} \cdot \frac{N}{16+N}$ 。条件给出 $\frac{64 + 6N}{10(16 + N)} = \frac{29}{50}\text{。}$

清除分母，得到 $5(64 + 6N) = 29(16 + N)$ ，所以 $320 + 30N = 464 + 29N$ ，从而 $N = 144$ 。

Both balls are green with probability $\frac{4}{10} \cdot \frac{16}{16+N},$ and both are blue with probability $\frac{6}{10} \cdot \frac{N}{16+N}.$ The condition is $\frac{64 + 6N}{10(16 + N)} = \frac{29}{50}.$

Clearing denominators, $5(64 + 6N) = 29(16 + N),$ so $320 + 30N = 464 + 29N,$ giving $N = 144.$

3.

求有多少个有理数 $r$ ，满足 $0 \lt r \lt 1$ ，并且当 $r$ 写成最简分数时，分子和分母之和为 $1000$ 。

Find the number of rational numbers $r,$ $0 \lt r \lt 1,$ such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of $1000.$

答案

知识点：欧拉函数最大公约数分数

难度评级：2110

小提示：

将 $r = \frac{a}{b}$ 写成最简分数；则 $a + b = 1000$ ，且 $a \lt 500$

Write $r = \frac{a}{b}$ in lowest terms; then $a + b = 1000$ with $a \lt 500$

大提示：

$\gcd(a, 1000 - a) = \gcd(a, 1000)$ ，所以只需数出满足 $a \lt 500$ 且与 $1000$ 互质的数

$\gcd(a, 1000 - a) = \gcd(a, 1000),$ so count $a \lt 500$ that are coprime to $1000$

解答：

将 $r = \frac{a}{b}$ 写成最简分数，且 $a + b = 1000$ ；由于 $0 \lt r \lt 1$ ，需要 $1 \le a \le 499$ 。又因为 $\gcd(a, b) = \gcd(a, 1000 - a)$ $= \gcd(a, 1000)$ ，这个分数为最简分数当且仅当 $a$ 与 $1000$ 互质。

欧拉函数给出 $\varphi(1000) = 1000 \cdot \frac{1}{2} \cdot \frac{4}{5} = 400$ ，即在 $[1, 999]$ 中，与 $1000$ 互质的整数共有这么多个。它们按 $a \leftrightarrow 1000 - a$ 成对出现（注意 $a = 500$ 与 $1000$ 不互质），所以其中恰有 $200$ 个小于 $500$ 。答案为 $200$ 。

Write $r = \frac{a}{b}$ in lowest terms with $a + b = 1000;$ since $0 \lt r \lt 1,$ we need $1 \le a \le 499.$ Because $\gcd(a, b) = \gcd(a, 1000 - a)$ $= \gcd(a, 1000),$ the fraction is in lowest terms exactly when $a$ is coprime to $1000.$

There are $\varphi(1000) = 1000 \cdot \frac{1}{2} \cdot \frac{4}{5} = 400$ integers in $[1, 999]$ coprime to $1000,$ and they pair up as $a \leftrightarrow 1000 - a$ (note $a = 500$ is not coprime to $1000$ ), so exactly $200$ of them are less than $500.$ The answer is $200.$

4.

Jon 和 Steve 在一条与两条并排东西向铁轨平行的小路上骑自行车。Jon 以每小时 $20$ 英里的速度向东骑，Steve 以每小时 $20$ 英里的速度向西骑。两列长度相等、方向相反且速度恒定但不同的火车分别经过这两位骑车人。每列火车经过 Jon 都正好需要 $1$ 分钟。向西行驶的火车经过 Steve 所需时间是向东行驶的火车经过 Steve 所需时间的 $10$ 倍。每列火车的长度为 $\frac{m}{n}$ 英里，其中 $m$ 和 $n$ 是互质的正整数。求 $m + n$ 。

Jon and Steve ride their bicycles on a path that parallels two side-by-side train tracks running in the east/west direction. Jon rides east at $20$ miles per hour, and Steve rides west at $20$ miles per hour. Two trains of equal length, traveling in opposite directions at constant but different speeds, each pass the two riders. Each train takes exactly $1$ minute to go past Jon. The westbound train takes $10$ times as long as the eastbound train to go past Steve. The length of each train is $\frac{m}{n}$ miles, where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

知识点：相对速度方程组

难度评级：2300

小提示：

经过 Jon 需要一分钟会确定两列火车的速度：有 $v_1 - 20 = v_2 + 20 = 60L$ ，其中 $L$ 表示火车长度的英里数

Passing Jon in one minute forces both train speeds: $v_1 - 20 = v_2 + 20 = 60L$ with $L$ in miles

大提示：

经过 Steve 时，西行火车所用时间是东行火车的 $10$ 倍： $v_1 + 20 = 10(v_2 - 20)$

Passing Steve, the westbound time is $10$ times the eastbound: $v_1 + 20 = 10(v_2 - 20)$

解答：

设东行和西行火车的速度分别为每小时 $v_1$ 和 $v_2$ 英里，共同长度为 $L$ 英里。火车经过骑车人的时间等于 $L$ 除以相对速度。经过向东以 $20$ 英里每小时骑行的 Jon 用 $\frac{1}{60}$ 小时，因此 $\frac{L}{v_1 - 20} = \frac{L}{v_2 + 20} = \frac{1}{60}\text{，}$ 所以 $v_1 = 60L + 20$ ，且 $v_2 = 60L - 20$ 。

相对于向西以 $20$ 英里每小时骑行的 Steve，两列火车的相对速度分别为 $v_1 + 20$ 和 $v_2 - 20$ ，并且西行火车所用时间是东行火车的 $10$ 倍： $\frac{L}{v_2 - 20} = \frac{10L}{v_1 + 20}$ ，所以 $v_1 + 20 = 10(v_2 - 20)$ 。代入得 $60L + 40 = 600L - 400$ ，因此 $540L = 440$ ， $L = \frac{22}{27}$ 。

因为 $\gcd(22, 27) = 1$ ，答案为 $22 + 27 = 49$ 。

Let the eastbound and westbound trains have speeds $v_1$ and $v_2$ miles per hour and common length $L$ miles. A train passes a rider in time $L$ divided by their relative speed. Passing Jon (riding east at $20$ ) in $\frac{1}{60}$ hour gives $\frac{L}{v_1 - 20} = \frac{L}{v_2 + 20} = \frac{1}{60},$ so $v_1 = 60L + 20$ and $v_2 = 60L - 20.$

Relative to Steve (riding west at $20$ ), the speeds are $v_1 + 20$ and $v_2 - 20,$ and the westbound train takes $10$ times as long: $\frac{L}{v_2 - 20} = \frac{10L}{v_1 + 20},$ so $v_1 + 20 = 10(v_2 - 20).$ Substituting, $60L + 40 = 600L - 400,$ so $540L = 440$ and $L = \frac{22}{27}.$

Since $\gcd(22, 27) = 1,$ the answer is $22 + 27 = 49.$

5.

令集合 $S = \{P_1, P_2, \ldots, P_{12}\}$ 由一个正 $12$ 边形的十二个顶点组成。若 $Q$ 是 $S$ 的一个子集，并且存在一个圆，使得 $Q$ 中所有点都在圆内，而 $S$ 中不属于 $Q$ 的所有点都在圆外，则称这样的子集为可聚子集。有多少个可聚子集？（注意空集也是可聚子集。）

Let the set $S = \{P_1, P_2, \ldots, P_{12}\}$ consist of the twelve vertices of a regular $12$ -gon. A subset $Q$ of $S$ is called communal if there is a circle such that all points of $Q$ are inside the circle, and all points of $S$ not in $Q$ are outside of the circle. How many communal subsets are there? (Note that the empty set is a communal subset.)

答案

知识点：子集正多边形基本计数

难度评级：2390

小提示：

一个子集能被某个圆切分出来，当且仅当它的顶点在这个 $12$ 边形上连续排列

A subset can be cut off by a circle exactly when its vertices are consecutive on the $12$ -gon

大提示：

对每个大小从 $1$ 到 $11$ 的子集，都有 $12$ 段连续顶点；不要忘记空集和整个 $S$

For each size from $1$ to $11$ there are $12$ runs of consecutive vertices; don’t forget the empty set and all of $S$

解答：

子集 $Q$ 是可聚子集的充要条件是它的顶点在 $12$ 边形周围连续。确实，一个分离圆与这个 $12$ 边形的外接圆最多相交于两点，所以在它内部的顶点形成一段连续弧。反过来，任意一段连续顶点都可以用一条直线与剩余顶点分开，再在这条直线适当一侧取一个足够大的圆，就能恰好包含这段顶点。

对于每个大小 $k$ （其中 $1 \le k \le 11$ ），都有 $12$ 段由 $k$ 个连续顶点组成的子集（每个顶点都可作为起点），共得到 $12 \cdot 11 = 132$ 个子集；此外空集和整个 $S$ 也都是可聚子集。总数为 $132 + 2 = 134$ 。

A subset $Q$ is communal exactly when its vertices are consecutive around the $12$ -gon. Indeed, a separating circle meets the circumcircle of the $12$ -gon in at most two points, so the vertices inside it form a contiguous arc. Conversely, any run of consecutive vertices can be separated from the remaining vertices by a line, and a sufficiently large circle on the proper side of that line contains exactly that run.

For each size $k$ with $1 \le k \le 11$ there are $12$ runs of $k$ consecutive vertices (one starting at each vertex), giving $12 \cdot 11 = 132$ subsets, and the empty set and all of $S$ are also communal. The total is $132 + 2 = 134.$

6.

图像 $y = 3(x-h)^2 + j$ 和 $y = 2(x-h)^2 + k$ 的 $y$ 轴截距分别为 $2013$ 和 $2014$ ，并且每个图像都有两个正整数 $x$ 轴截距。求 $h$ 。

The graphs $y = 3(x-h)^2 + j$ and $y = 2(x-h)^2 + k$ have $y$ -intercepts of $2013$ and $2014,$ respectively, and each graph has two positive integer $x$ -intercepts. Find $h.$

答案

知识点：二次方程韦达定理质因数分解

难度评级：2450

小提示：

令 $x = 0$ ，得到 $3h^2 + j = 2013$ 和 $2h^2 + k = 2014$ ；于是 $x$ 轴截距分别是 $3x^2 - 6hx + 2013$ 与 $2x^2 - 4hx + 2014$ 的根

Set $x = 0$ to get $3h^2 + j = 2013$ and $2h^2 + k = 2014;$ the $x$ -intercepts are then roots of $3x^2 - 6hx + 2013$ and $2x^2 - 4hx + 2014$

大提示：

这些正整数根的乘积分别为 $671 = 11 \cdot 61$ 和 $1007 = 19 \cdot 53$ ，并且两对根必须有相同的和 $2h$

The positive integer roots multiply to $671 = 11 \cdot 61$ and $1007 = 19 \cdot 53,$ and both pairs must have the same sum $2h$

解答：

令 $x = 0$ ，得到 $3h^2 + j = 2013$ 和 $2h^2 + k = 2014$ 。展开后，第一个图像为 $y = 3x^2 - 6hx + 2013$ ，它的根是正整数，和为 $2h$ ，积为 $\frac{2013}{3} = 671 = 11 \cdot 61$ 。类似地，第二个图像为 $y = 2x^2 - 4hx + 2014$ ，其整数根的和为 $2h$ ，积为 $\frac{2014}{2} = 1007 = 19 \cdot 53$ 。

第一对根为 $\{11, 61\}$ 或 $\{1, 671\}$ ，所以 $2h = 72$ 或 $672$ ；第二对根为 $\{19, 53\}$ 或 $\{1, 1007\}$ ，所以 $2h = 72$ 或 $1008$ 。唯一共同值是 $2h = 72$ ，因此 $h = 36$ ，这确实给出 $x$ 轴截距 $11$ 、 $61$ 和 $19$ 、 $53$ 。

Setting $x = 0$ gives $3h^2 + j = 2013$ and $2h^2 + k = 2014.$ Expanding, the first graph is $y = 3x^2 - 6hx + 2013,$ whose roots are positive integers with sum $2h$ and product $\frac{2013}{3} = 671 = 11 \cdot 61.$ Similarly the second is $y = 2x^2 - 4hx + 2014,$ with integer roots of sum $2h$ and product $\frac{2014}{2} = 1007 = 19 \cdot 53.$

The first pair of roots is $\{11, 61\}$ or $\{1, 671\},$ so $2h = 72$ or $672;$ the second pair is $\{19, 53\}$ or $\{1, 1007\},$ so $2h = 72$ or $1008.$ The only common value is $2h = 72,$ so $h = 36,$ which indeed gives $x$ -intercepts $11,$ $61$ and $19,$ $53.$

7.

设 $w$ 和 $z$ 是复数，满足 $|w| = 1$ 且 $|z| = 10$ 。令 $\theta = \arg\left(\tfrac{w-z}{z}\right)$ 。 $\tan^2 \theta$ 的最大可能值可写成 $\frac{p}{q}$ ，其中 $p$ 和 $q$ 是互质的正整数。求 $p + q$ 。（注意， $\arg(w)$ 对于 $w \ne 0$ 表示复平面中从 $0$ 指向 $w$ 的射线与正实轴所成角的大小。）

Let $w$ and $z$ be complex numbers such that $|w| = 1$ and $|z| = 10.$ Let $\theta = \arg\left(\tfrac{w-z}{z}\right).$ The maximum possible value of $\tan^2 \theta$ can be written as $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p + q.$ (Note that $\arg(w),$ for $w \ne 0,$ denotes the measure of the angle that the ray from $0$ to $w$ makes with the positive real axis in the complex plane.)

答案

知识点：复数圆切线最优化

难度评级：2560

小提示：

$\frac{w-z}{z} = \frac{w}{z} - 1$ 位于半径为 $\frac{1}{10}$ 、圆心为 $-1$ 的圆上

$\frac{w-z}{z} = \frac{w}{z} - 1$ lies on the circle of radius $\frac{1}{10}$ centered at $-1$

大提示：

当从原点出发的射线与该圆相切时，辐角达到极值，所以它与实轴所成角的正弦为 $\frac{1}{10}$

The argument is extremal when the ray from the origin is tangent to that circle, so the sine of the angle with the real axis is $\frac{1}{10}$

解答：

因为 $\frac{w-z}{z} = \frac{w}{z} - 1$ ，而 $\frac{w}{z}$ 可以是任意模长为 $\frac{1}{10}$ 的复数，所以点 $\zeta = \frac{w-z}{z}$ 的轨迹是半径为 $\frac{1}{10}$ 、圆心为 $-1$ 的圆。

由于 $\tan^2\theta$ 在 $\theta$ 改变 $180^\circ$ 时不变，我们要求的是从原点指向该圆的射线与实轴所成的最大角 $\alpha$ 。极端位置的射线与圆相切，此时 $\sin \alpha = \frac{\frac{1}{10}}{1} = \frac{1}{10}$ 。

因此 $\begin{aligned} \tan^2\theta &= \frac{\sin^2\alpha}{1 - \sin^2\alpha} = \frac{\frac{1}{100}}{\frac{99}{100}} \\ &= \frac{1}{99} \end{aligned}\text{，}$ 所以 $p + q = 1 + 99 = 100$ 。

Since $\frac{w-z}{z} = \frac{w}{z} - 1,$ and $\frac{w}{z}$ can be any complex number of modulus $\frac{1}{10},$ the point $\zeta = \frac{w-z}{z}$ ranges over the circle of radius $\frac{1}{10}$ centered at $-1.$

Because $\tan^2\theta$ is unchanged when $\theta$ shifts by $180^\circ,$ we want the largest angle $\alpha$ that a ray from the origin to this circle makes with the real axis. The extreme rays are tangent to the circle, where $\sin \alpha = \frac{\frac{1}{10}}{1} = \frac{1}{10}.$

Then $\begin{aligned} \tan^2\theta &= \frac{\sin^2\alpha}{1 - \sin^2\alpha} = \frac{\frac{1}{100}}{\frac{99}{100}} \\ &= \frac{1}{99}, \end{aligned}$ so $p + q = 1 + 99 = 100.$

8.

正整数 $N$ 和 $N^2$ 都以同一串四位数字 $abcd$ 结尾（用 $10$ 进制表示），其中数字 $a$ 不为零。求三位数 $abc$ 。

The positive integers $N$ and $N^2$ both end in the same sequence of four digits $abcd$ when written in base $10,$ where digit $a$ is not zero. Find the three-digit number $abc.$

答案

知识点：模运算中国剩余定理数字

难度评级：2710

小提示：

条件说明 $N^2 \equiv N \pmod{10^4}$ ，也就是 $N(N-1) \equiv 0 \pmod{2^4 \cdot 5^4}$

The condition says $N^2 \equiv N \pmod{10^4},$ i.e. $N(N-1) \equiv 0 \pmod{2^4 \cdot 5^4}$

大提示：

$N$ 与 $N-1$ 互质，所以 $16$ 和 $625$ 各自必须整除其中一个；四种情况中只有一种的首位数字 $a$ 非零

$N$ and $N-1$ are coprime, so $16$ and $625$ each divide one of them; only one of the four cases has nonzero leading digit $a$

解答：

条件为 $N^2 \equiv N \pmod{10^4}$ ，即 $N(N-1) \equiv 0 \pmod{2^4 \cdot 5^4}$ 。由于相邻整数互质， $16$ 必须整除 $N$ 或 $N - 1$ 中的一个， $625$ 也必须整除其中一个。这在模 $10000$ 意义下给出四种情况： $N \equiv 0$ 、 $N \equiv 1$ 、 $N \equiv 625$ （它是 $0$ 模 $625$ ，是 $1$ 模 $16$ ），以及 $N \equiv 9376$ （它是 $0$ 模 $16$ ，是 $1$ 模 $625$ ）。

后四位 $abcd$ 必须满足 $a \ne 0$ ，这排除了 $0000$ 、 $0001$ 和 $0625$ 。因此 $abcd = 9376$ ，例如 $9376^2 = 87909376$ ，所以 $abc = 937$ 。

The condition is $N^2 \equiv N \pmod{10^4},$ that is, $N(N-1) \equiv 0 \pmod{2^4 \cdot 5^4}.$ Since consecutive integers are coprime, $16$ divides one of $N,$ $N - 1$ and $625$ divides one of them. This gives four cases modulo $10000:$ $N \equiv 0,$ $N \equiv 1,$ $N \equiv 625$ (which is $0$ mod $625$ and $1$ mod $16$ ), and $N \equiv 9376$ (which is $0$ mod $16$ and $1$ mod $625$ ).

The last four digits $abcd$ must have $a \ne 0,$ which rules out $0000,$ $0001,$ and $0625.$ So $abcd = 9376$ — for instance $9376^2 = 87909376$ — and $abc = 937.$

9.

设 $x_1 \lt x_2 \lt x_3$ 是方程 $\sqrt{2014}\,x^3 - 4029x^2 + 2 = 0$ 的三个实根。求 $x_2(x_1 + x_3)$ 。

Let $x_1 \lt x_2 \lt x_3$ be the three real roots of the equation $\sqrt{2014}\,x^3 - 4029x^2 + 2 = 0.$ Find $x_2(x_1 + x_3).$

答案

知识点：多项式因式分解韦达定理

难度评级：2560

小提示：

因为 $4029 = 2 \cdot 2014 + 1$ ，这个三次多项式可分解为 $\left(\sqrt{2014}\,x - 1\right)$ 乘以一个二次多项式

Since $4029 = 2 \cdot 2014 + 1,$ the cubic factors as $\left(\sqrt{2014}\,x - 1\right)$ times a quadratic

大提示：

二次式 $x^2 - 2\sqrt{2014}\,x - 2$ 有一个负根和一个很大的正根，所以中间的根是 $\frac{1}{\sqrt{2014}}$

The quadratic $x^2 - 2\sqrt{2014}\,x - 2$ has one negative root and one large positive root, so the middle root is $\frac{1}{\sqrt{2014}}$

解答：

记 $a = \sqrt{2014}$ ，则方程为 $ax^3 - (2a^2 + 1)x^2 + 2 = 0$ 。它可分解为 $\left(ax - 1\right)\left(x^2 - 2ax - 2\right) = 0\text{，}$ 展开即可验证。因此一个根是 $\frac{1}{a}$ ，另外两个根为 $a \pm \sqrt{a^2 + 2}$ ，它们的积为 $-2$ ，和为 $2a$ 。

因为 $a - \sqrt{a^2+2} \lt 0$ $\lt \frac{1}{a} \lt a + \sqrt{a^2+2}$ ，所以中间根为 $x_2 = \frac{1}{a}$ ，且 $x_1 + x_3 = 2a$ 。因此 $x_2(x_1 + x_3) = \frac{1}{a} \cdot 2a = 2$ 。

Write $a = \sqrt{2014},$ so the equation is $ax^3 - (2a^2 + 1)x^2 + 2 = 0.$ It factors as $\left(ax - 1\right)\left(x^2 - 2ax - 2\right) = 0,$ as expanding confirms. So one root is $\frac{1}{a},$ and the other two are $a \pm \sqrt{a^2 + 2},$ with product $-2$ and sum $2a.$

Since $a - \sqrt{a^2+2} \lt 0$ $\lt \frac{1}{a} \lt a + \sqrt{a^2+2},$ the middle root is $x_2 = \frac{1}{a},$ and $x_1 + x_3 = 2a.$ Therefore $x_2(x_1 + x_3) = \frac{1}{a} \cdot 2a = 2.$

10.

一个半径为 $1$ 的圆盘与一个半径为 $5$ 的圆盘外切。设 $A$ 为两圆盘的切点， $C$ 为小圆盘的圆心， $E$ 为大圆盘的圆心。保持大圆盘不动，让小圆盘沿大圆盘外侧滚动，直到小圆盘转过 $360^\circ$ 。也就是说，若小圆盘圆心移动到点 $D$ ，而小圆盘上原本位于 $A$ 的点现在移动到点 $B$ ，则 $\overline{AC}$ 平行于 $\overline{BD}$ 。于是 $\sin^2(\angle BEA) = \frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质的正整数。求 $m + n$ 。

A disk with radius $1$ is externally tangent to a disk with radius $5.$ Let $A$ be the point where the disks are tangent, $C$ be the center of the smaller disk, and $E$ be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until the smaller disk has turned through an angle of $360^\circ.$ That is, if the center of the smaller disk has moved to the point $D,$ and the point on the smaller disk that began at $A$ has now moved to point $B,$ then $\overline{AC}$ is parallel to $\overline{BD}.$ Then $\sin^2(\angle BEA) = \frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

知识点：圆变换坐标几何三角学

难度评级：2920

小提示：

在半径为 $5$ 的圆外滚动时，圆盘在固定坐标系中每转过 $6^\circ$ ，圆心就扫过 $1^\circ$ （绕 $E$ ），所以圆心总共扫过 $60^\circ$

Rolling outside a radius- $5$ circle, the disk turns $6^\circ$ in the ground frame for each $1^\circ$ its center sweeps about $E,$ so the center sweeps $60^\circ$

大提示：

转过完整一圈后，圆盘恢复原来的朝向，所以 $B = D + (A - C)$ ；把 $E$ 放在原点并计算

After one full turn the disk regains its original orientation, so $B = D + (A - C);$ place $E$ at the origin and compute

解答：

将 $E$ 放在原点，并取 $C = (6, 0)$ ，于是 $A = (5, 0)$ 。半径为 $1$ 的圆沿半径为 $5$ 的固定圆外侧无滑动滚动时，若其圆心扫过角 $\varphi$ （绕 $E$ ），滚动接触会使圆盘相对于圆心连线转过 $5\varphi$ ，而这条圆心连线本身的转动又增加 $\varphi$ ，所以圆盘在固定坐标系中共转过 $6\varphi$ 。转过 $360^\circ$ 因此意味着 $\varphi = 60^\circ$ ，所以 $D = 6(\cos 60^\circ, \sin 60^\circ)$ $= (3, 3\sqrt{3})$ 。

完整转过 $360^\circ$ 后，圆盘回到原来的朝向，因此从圆心到标记点的向量不变： $B = D + (A - C)$ $= (3 - 1, 3\sqrt{3})$ $= (2, 3\sqrt{3})$ 。（特别地， $\overline{BD}$ 平行于 $\overline{AC}$ ，正如题目所述。）

射线 $EA$ 是正 $x$ 轴，所以 $\begin{aligned} \sin^2(\angle BEA) &= \frac{(3\sqrt{3})^2}{2^2 + (3\sqrt{3})^2} \\ &= \frac{27}{31} \end{aligned}\text{，}$ 因而 $m + n = 27 + 31 = 58$ 。

Place $E$ at the origin with $C = (6, 0),$ so $A = (5, 0).$ When a circle of radius $1$ rolls without slipping outside a fixed circle of radius $5$ and its center sweeps an angle $\varphi$ about $E,$ the rolling contact turns the disk through $5\varphi$ relative to the line of centers, and the revolution of that line adds $\varphi$ more, so the disk turns $6\varphi$ in the ground frame. Turning through $360^\circ$ therefore means $\varphi = 60^\circ,$ so $D = 6(\cos 60^\circ, \sin 60^\circ)$ $= (3, 3\sqrt{3}).$

Having turned through a full $360^\circ,$ the disk is back in its original orientation, so the vector from its center to the marked point is unchanged: $B = D + (A - C)$ $= (3 - 1, 3\sqrt{3})$ $= (2, 3\sqrt{3}).$ (In particular $\overline{BD}$ is parallel to $\overline{AC},$ as the problem states.)

The ray $EA$ is the positive $x$ -axis, so $\begin{aligned} \sin^2(\angle BEA) &= \frac{(3\sqrt{3})^2}{2^2 + (3\sqrt{3})^2} \\ &= \frac{27}{31}, \end{aligned}$ and $m + n = 27 + 31 = 58.$

11.

一个棋子从点 $(0, 0)$ （位于 $xy$ 坐标网格中）出发，然后连续移动六次。每次移动都沿某条坐标轴平行方向移动 $1$ 个单位。每次移动都从四个可能方向中随机选择，并且彼此独立。棋子最终落在图像 $|y| = |x|$ 上的概率为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质的正整数。求 $m + n$ 。

A token starts at the point $(0, 0)$ of an $xy$ -coordinate grid and then makes a sequence of six moves. Each move is $1$ unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability that the token ends at a point on the graph of $|y| = |x|$ is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

知识点：随机游走独立事件容斥原理

难度评级：2990

小提示：

追踪 $u = x + y$ 和 $v = x - y$ ：每一步都会让 $u$ 和 $v$ 各自独立且等概率地改变 $\pm 1$

Track $u = x + y$ and $v = x - y:$ each move changes each of $u$ and $v$ by $\pm 1,$ independently and uniformly

大提示：

棋子最终在 $|y| = |x|$ 上当且仅当 $u = 0$ 或 $v = 0$ ，每个事件的概率都是 $\frac{\binom{6}{3}}{2^6}$ ；相加后减去重叠部分

The token ends on $|y| = |x|$ when $u = 0$ or $v = 0,$ each with probability $\frac{\binom{6}{3}}{2^6};$ add these and subtract the overlap

解答：

使用对角坐标 $u = x + y$ 和 $v = x - y$ 。四种移动都会使 $u$ 改变 $\pm 1$ ，也使 $v$ 改变 $\pm 1$ ，并且四种移动正好等可能地实现四种符号组合，所以 $u$ 和 $v$ 是相互独立的六步 $\pm 1$ 随机游走。棋子最终在 $|y| = |x|$ 上，恰好等价于 $y = \pm x$ ，也就是 $u = 0$ 或 $v = 0$ 。

$u = 0$ 和 $v = 0$ 各自都要求三次 $+1$ 和三次 $-1$ ，概率为 $\frac{\binom{6}{3}}{2^6} = \frac{20}{64} = \frac{5}{16}$ 。由独立性和容斥原理，所求概率为 $\begin{aligned} &\frac{5}{16} + \frac{5}{16} - \left(\frac{5}{16}\right)^2 \\ &= \frac{160 - 25}{256} = \frac{135}{256} \end{aligned}\text{。}$

因此 $m + n = 135 + 256 = 391$ 。

Work in the diagonal coordinates $u = x + y$ and $v = x - y.$ Each of the four moves changes $u$ by $\pm 1$ and $v$ by $\pm 1,$ and the four moves realize all four sign combinations equally often — so $u$ and $v$ perform independent six-step $\pm 1$ walks. The token ends on $|y| = |x|$ exactly when $y = \pm x,$ that is, when $u = 0$ or $v = 0.$

Each of $u = 0$ and $v = 0$ requires three $+1$ s and three $-1$ s, with probability $\frac{\binom{6}{3}}{2^6} = \frac{20}{64} = \frac{5}{16}.$ By independence and inclusion-exclusion, the probability is $\begin{aligned} &\frac{5}{16} + \frac{5}{16} - \left(\frac{5}{16}\right)^2 \\ &= \frac{160 - 25}{256} = \frac{135}{256}. \end{aligned}$

Thus $m + n = 135 + 256 = 391.$

12.

设 $A = \{1, 2, 3, 4\}$ ，并随机选择两个函数 $f$ 和 $g$ （二者不一定不同），它们都从 $A$ 映射到 $A$ 。 $f$ 的值域与 $g$ 的值域不相交的概率为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质的正整数。求 $m$ 。

Let $A = \{1, 2, 3, 4\},$ and let $f$ and $g$ be randomly chosen (not necessarily distinct) functions from $A$ to $A.$ The probability that the range of $f$ and the range of $g$ are disjoint is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m.$

答案

知识点：函数基本概率分类讨论

难度评级：2990

小提示：

若 $f$ 的值域有 $k$ 个元素，则恰有 $(4-k)^4$ 个函数 $g$ 的值域与它不相交

If the range of $f$ has $k$ elements, exactly $(4-k)^4$ functions $g$ have range disjoint from it

大提示：

按值域大小计数函数：数量依次为 $4$ 个（值域大小 $1$ ）、 $84$ 个（值域大小 $2$ ）、 $144$ 个（值域大小 $3$ ）、 $24$ 个（值域大小 $4$ ）

Count functions by range size: $4$ with size $1,$ $84$ with size $2,$ $144$ with size $3,$ $24$ with size $4$

解答：

按 $f$ 的值域分类。若它有 $k$ 个元素，则 $g$ 的值域与其不相交，当且仅当 $g$ 把 $A$ 映到剩下的 $4 - k$ 个元素中；满足条件的映射方式有 $(4-k)^4$ 种，全部共有 $4^4 = 256$ 个函数 $g$ 。

按值域大小计数 $f$ ：常值函数有 $4$ 个；另有 $\binom{4}{2}(2^4 - 2) = 84$ 个函数的值域大小为 $2$ ；另有 $\binom{4}{3} \cdot 36 = 144$ 个函数的值域大小为 $3$ （从四个元素满射到三个元素有 $36$ 个）；双射有 $4! = 24$ 个。有利的有序函数对数量为 $\begin{aligned} &4 \cdot 3^4 + 84 \cdot 2^4 \\ &\quad {}+ 144 \cdot 1^4 + 24 \cdot 0^4 \\ &= 324 + 1344 + 144 = 1812 \end{aligned}\text{。}$

概率为 $\frac{1812}{4^8} = \frac{1812}{65536} = \frac{453}{16384}$ ，由于 $16384$ 是 $2$ 的幂，而 $453 = 3 \cdot 151$ 为奇数，此分数已是最简。因此 $m = 453$ 。

Condition on the range of $f.$ If it has $k$ elements, then the range of $g$ is disjoint from it exactly when $g$ maps $A$ into the remaining $4 - k$ elements, which happens for $(4-k)^4$ of the $4^4 = 256$ functions $g.$

Count functions $f$ by range size: $4$ constant functions; $\binom{4}{2}(2^4 - 2) = 84$ with range size $2;$ $\binom{4}{3} \cdot 36 = 144$ with range size $3$ (there are $36$ surjections from four elements onto three); and $4! = 24$ bijections. The number of favorable pairs is $\begin{aligned} &4 \cdot 3^4 + 84 \cdot 2^4 \\ &\quad {}+ 144 \cdot 1^4 + 24 \cdot 0^4 \\ &= 324 + 1344 + 144 = 1812. \end{aligned}$

The probability is $\frac{1812}{4^8} = \frac{1812}{65536} = \frac{453}{16384},$ and since $16384$ is a power of $2$ while $453 = 3 \cdot 151$ is odd, this is in lowest terms. Thus $m = 453.$

13.

在正方形 $ABCD$ 中，点 $E$ 、 $F$ 、 $G$ 和 $H$ 分别在边 $\overline{AB}$ 、 $\overline{BC}$ 、 $\overline{CD}$ 和 $\overline{DA}$ 上，且 $\overline{EG} \perp \overline{FH}$ ，并且 $EG = FH = 34$ 。线段 $\overline{EG}$ 与 $\overline{FH}$ 交于点 $P$ ，四边形 $AEPH$ 、 $BFPE$ 、 $CGPF$ 和 $DHPG$ 的面积之比为 $269 : 275 : 405 : 411$ 。求正方形 $ABCD$ 的面积。

On square $ABCD,$ points $E,$ $F,$ $G,$ and $H$ lie on sides $\overline{AB},$ $\overline{BC},$ $\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG = FH = 34.$ Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P,$ and the areas of the quadrilaterals $AEPH,$ $BFPE,$ $CGPF,$ and $DHPG$ are in the ratio $269 : 275 : 405 : 411.$ Find the area of square $ABCD.$

答案

知识点：坐标几何正方形（几何）梯形方程组

难度评级：3500

小提示：

因为 $269 + 411 = 275 + 405$ ，线段 $EG$ 经过正方形中心；对梯形 $ABFH$ 作同样处理来定位 $FH$

Since $269 + 411 = 275 + 405,$ segment $EG$ passes through the center of the square; treat the trapezoid $ABFH$ the same way to locate $FH$

大提示：

设边长为 $s$ ，并设竖向变化为 $\delta$ （沿 $EG$ ）；长度条件给出 $s^2 + \delta^2 = 1156$ ，而面积比 $269 : 1360$ （对于 $AEPH$ ）会确定乘积 $s\delta$

With side $s$ and rise $\delta$ of $EG,$ the length gives $s^2 + \delta^2 = 1156,$ and the ratio $269 : 1360$ for $AEPH$ pins down the product $s\delta$

解答：

取 $B = (0,0)$ 、 $C = (s, 0)$ 、 $D = (s, s)$ 、 $A = (0, s)$ ，并设 $E = (0, e)$ 、 $F = (f, 0)$ 、 $G = (s, g)$ 、 $H = (h, s)$ 。区域 $AEPH$ 和 $DHPG$ 合起来形成梯形 $AEGD$ 面积为 $\frac{s\,((s - e) + (s - g))}{2}$ ；由于 $\frac{269 + 411}{1360} = \frac{1}{2}$ ，这正好是 $s^2$ 的一半，迫使 $e + g = s$ ，即 $EG$ 经过中心。同样， $AEPH$ 和 $BFPE$ 形成梯形 $ABFH$ 面积为 $\frac{s(f + h)}{2} = \frac{269 + 275}{1360}\,s^2 = \frac{2}{5}s^2$ ，所以 $f + h = \frac{4s}{5}$ 。方向向量 $(s, g - e)$ 与 $(h - f, s)$ 垂直，给出 $h - f = e - g$ 。令 $\delta = g - e$ ，得到 $E = \left(0, \frac{s - \delta}{2}\right)$ 、 $G = \left(s, \frac{s + \delta}{2}\right)$ 、 $F = \left(\frac{2s}{5} + \frac{\delta}{2}, 0\right)$ 、 $H = \left(\frac{2s}{5} - \frac{\delta}{2}, s\right)$ ，且 $EG = 34$ 给出 $s^2 + \delta^2 = 1156$ 。

求出直线 $EG$ 与 $FH$ 的交点，并用 $s^2 + \delta^2 = 1156$ 化简，可得 $\begin{aligned} P &= \\ &\left(\frac{s}{2} - \frac{s^3}{11560},\; \frac{s}{2} - \frac{s^2\delta}{11560}\right) \end{aligned}\text{，}$ 再对 $A$ 、 $E$ 、 $P$ 、 $H$ 使用鞋带公式，得到 $[AEPH] = \frac{s^2}{5} - \frac{s^3\delta}{231200}$ 。令它等于 $\frac{269}{1360}s^2$ ，得到 $\frac{3s^2}{1360} = \frac{s^3\delta}{231200}$ ，所以 $s\delta = 510$ 。

现在 $s^2 + \delta^2 = 1156$ ，且 $s^2\delta^2 = 260100$ ，所以 $s^2$ 和 $\delta^2$ 是方程 $t^2 - 1156t + 260100 = 0$ 的两个根，分别为 $\frac{1156 \pm 544}{2} = 850$ 和 $306$ 。由于 $|\delta| = |g - e| \lt s$ ，面积为 $s^2 = 850$ 。

Place $B = (0,0),$ $C = (s, 0),$ $D = (s, s),$ $A = (0, s),$ with $E = (0, e),$ $F = (f, 0),$ $G = (s, g),$ $H = (h, s).$ The regions $AEPH$ and $DHPG$ together form the trapezoid $AEGD,$ of area $\frac{s\,((s - e) + (s - g))}{2};$ since $\frac{269 + 411}{1360} = \frac{1}{2},$ this is half of $s^2,$ forcing $e + g = s,$ i.e. $EG$ passes through the center. Likewise $AEPH$ and $BFPE$ form the trapezoid $ABFH$ of area $\frac{s(f + h)}{2} = \frac{269 + 275}{1360}\,s^2 = \frac{2}{5}s^2,$ so $f + h = \frac{4s}{5}.$ Perpendicularity of the directions $(s, g - e)$ and $(h - f, s)$ gives $h - f = e - g.$ Writing $\delta = g - e,$ we get $E = \left(0, \frac{s - \delta}{2}\right),$ $G = \left(s, \frac{s + \delta}{2}\right),$ $F = \left(\frac{2s}{5} + \frac{\delta}{2}, 0\right),$ $H = \left(\frac{2s}{5} - \frac{\delta}{2}, s\right),$ and $EG = 34$ gives $s^2 + \delta^2 = 1156.$

Intersecting lines $EG$ and $FH$ (and simplifying with $s^2 + \delta^2 = 1156$ ) yields $\begin{aligned} P &= \\ &\left(\frac{s}{2} - \frac{s^3}{11560},\; \frac{s}{2} - \frac{s^2\delta}{11560}\right), \end{aligned}$ and the shoelace formula on $A,$ $E,$ $P,$ $H$ then gives $[AEPH] = \frac{s^2}{5} - \frac{s^3\delta}{231200}.$ Setting this equal to $\frac{269}{1360}s^2$ leaves $\frac{3s^2}{1360} = \frac{s^3\delta}{231200},$ so $s\delta = 510.$

Now $s^2 + \delta^2 = 1156$ and $s^2\delta^2 = 260100,$ so $s^2$ and $\delta^2$ are the roots of $t^2 - 1156t + 260100 = 0,$ which are $\frac{1156 \pm 544}{2} = 850$ and $306.$ Since $|\delta| = |g - e| \lt s,$ the area is $s^2 = 850.$

14.

设 $m$ 是方程 $\begin{aligned} &\frac{3}{x-3} + \frac{5}{x-5} \\ &\quad {}+ \frac{17}{x-17} + \frac{19}{x-19} \\ &= x^2 - 11x - 4 \end{aligned}$ 的最大实数解。存在正整数 $a$ 、 $b$ 和 $c$ ，使得 $m = a + \sqrt{b + \sqrt{c}}$ 。求 $a + b + c$ 。

Let $m$ be the largest real solution to the equation $\begin{aligned} &\frac{3}{x-3} + \frac{5}{x-5} \\ &\quad {}+ \frac{17}{x-17} + \frac{19}{x-19} \\ &= x^2 - 11x - 4. \end{aligned}$ There are positive integers $a,$ $b,$ and $c$ such that $m = a + \sqrt{b + \sqrt{c}}.$ Find $a + b + c.$

答案

知识点：分式方程换元法对称性（代数）

难度评级：3060

小提示：

两边都加上 $4$ ，把每个分式各配上一个一：每一项都会变成 $\frac{x}{x-k}$ 。然后代入 $t = x - 11$ 。

Add $4$ to both sides, splitting one unit to each fraction: each becomes $\frac{x}{x-k}.$ Then substitute $t = x - 11.$

大提示：

对称的两对分式合并为 $\frac{2}{t^2 - 64} + \frac{2}{t^2 - 36} = 1$ ，这是关于 $t^2$ 的二次方程

Symmetric pairs combine to $\frac{2}{t^2 - 64} + \frac{2}{t^2 - 36} = 1,$ a quadratic in $t^2$

解答：

两边都加上 $4$ ，把每个分式各配上一个一：因为 $\frac{k}{x-k} + 1 = \frac{x}{x-k}$ ，方程变为 $\begin{aligned} &\scriptsize x\left(\frac{1}{x-3} + \frac{1}{x-5} + \frac{1}{x-17} + \frac{1}{x-19}\right) \\ &= x^2 - 11x = x(x - 11) \end{aligned}\text{。}$ 除了 $x = 0$ 之外，可以除以 $x$ ，并令 $t = x - 11$ ，使这些分式成对对称： $\frac{2t}{t^2 - 64} + \frac{2t}{t^2 - 36} = t\text{。}$

除了 $t = 0$ 之外，除以 $t$ 得 $\frac{2}{t^2 - 64} + \frac{2}{t^2 - 36} = 1$ 。令 $u = t^2$ ，清除分母得到 $2(u - 36) + 2(u - 64)$ $= (u - 36)(u - 64)$ ，也就是 $u^2 - 104u + 2504 = 0$ ，所以 $u = 52 \pm \sqrt{200}$ 。

最大解为 $m = 11 + \sqrt{52 + \sqrt{200}} \approx 19.1$ ，它大于其他候选解 $0$ 、 $11$ ，以及 $11 \pm \sqrt{52 \pm \sqrt{200}}$ 。因此 $a + b + c$ $= 11 + 52 + 200 = 263$ 。

Add $4$ to both sides, giving one unit to each fraction: since $\frac{k}{x-k} + 1 = \frac{x}{x-k},$ the equation becomes $\begin{aligned} &\scriptsize x\left(\frac{1}{x-3} + \frac{1}{x-5} + \frac{1}{x-17} + \frac{1}{x-19}\right) \\ &= x^2 - 11x = x(x - 11). \end{aligned}$ Besides $x = 0,$ we can divide by $x$ and substitute $t = x - 11,$ which pairs the fractions symmetrically: $\frac{2t}{t^2 - 64} + \frac{2t}{t^2 - 36} = t.$

Besides $t = 0,$ dividing by $t$ gives $\frac{2}{t^2 - 64} + \frac{2}{t^2 - 36} = 1.$ With $u = t^2,$ clearing denominators gives $2(u - 36) + 2(u - 64)$ $= (u - 36)(u - 64),$ i.e. $u^2 - 104u + 2504 = 0,$ so $u = 52 \pm \sqrt{200}.$

The largest solution is $m = 11 + \sqrt{52 + \sqrt{200}} \approx 19.1,$ which exceeds the other candidates $0,$ $11,$ and $11 \pm \sqrt{52 \pm \sqrt{200}}.$ Therefore $a + b + c$ $= 11 + 52 + 200 = 263.$

15.

在 $\triangle ABC$ 中， $AB = 3$ 、 $BC = 4$ 、 $CA = 5$ 。圆 $\omega$ 与 $\overline{AB}$ 交于 $E$ 和 $B$ ，与 $\overline{BC}$ 交于 $B$ 和 $D$ ，与 $\overline{AC}$ 交于 $F$ 和 $G$ 。已知 $EF = DF$ ，且 $\frac{DG}{EG} = \frac{3}{4}$ 。长度 $DE = \frac{a\sqrt{b}}{c}$ ，其中 $a$ 和 $c$ 是互质的正整数， $b$ 是不被任何素数平方整除的正整数。求 $a + b + c$ 。

In $\triangle ABC,$ $AB = 3,$ $BC = 4,$ and $CA = 5.$ Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B,$ $\overline{BC}$ at $B$ and $D,$ and $\overline{AC}$ at $F$ and $G.$ Given that $EF = DF$ and $\frac{DG}{EG} = \frac{3}{4},$ length $DE = \frac{a\sqrt{b}}{c},$ where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. Find $a + b + c.$

答案

知识点：圆圆周角圆内接四边形坐标几何

难度评级：3500

小提示：

$\angle B = 90^\circ$ 内接于 $\omega$ ，所以 $DE$ 是直径，并且 $\angle EFD = \angle EGD = 90^\circ$

$\angle B = 90^\circ$ is inscribed in $\omega,$ so $DE$ is a diameter and $\angle EFD = \angle EGD = 90^\circ$

大提示：

将 $E$ 和 $D$ 到直线 $AC$ 的距离表示为 $DE$ 的倍数；这个表示式可利用圆内接四边形 $EFGD$ 的角得到，再与坐标计算比较

Express the distances from $E$ and $D$ to line $AC$ as multiples of $DE$ using the angles of cyclic quadrilateral $EFGD,$ then compare with coordinates

解答：

因为 $3^2 + 4^2 = 5^2$ ，角 $B$ 是直角；又因为 $\angle EBD = 90^\circ$ 内接于 $\omega$ ，弦 $ED$ 是直径。因此 $\angle EFD = \angle EGD = 90^\circ$ 。由 $EF = DF$ ，三角形 $EFD$ 是等腰直角三角形，所以 $EF = DF = \frac{DE}{\sqrt{2}}$ ，且 $\angle FED = 45^\circ$ ；由 $DG : EG = 3 : 4$ 和 $DG^2 + EG^2 = DE^2$ ，得到 $DG = \frac{3}{5}DE$ 且 $EG = \frac{4}{5}DE$ 。在 $\omega$ 上顺序为 $E$ 、 $F$ 、 $G$ 、 $D$ （ $F$ 与 $G$ 都在与 $B$ 相对的弧上，并且 $\angle FED = 45^\circ$ 大于 $\angle GED = \arcsin\frac{3}{5}$ ，所以 $F$ 离 $D$ 更远）。

直线 $AC$ 就是直线 $FG$ ，因此从 $E$ 和 $D$ 到它的距离可由圆内接四边形 $EFGD$ 的角得到。在 $F$ 处， $\angle EFG = 180^\circ - \angle GDE$ ，且 $\sin\angle GDE = \frac{EG}{DE} = \frac{4}{5}$ ，所以从 $E$ 到直线的距离为 $EF \sin\angle EFG = \frac{DE}{\sqrt{2}} \cdot \frac{4}{5}$ $= \frac{2\sqrt{2}}{5}DE$ 。在 $G$ 处， $\angle FGD = 180^\circ - \angle FED = 135^\circ$ ，所以从 $D$ 到直线的距离为 $DG \sin\angle FGD = \frac{3}{5}DE \cdot \frac{\sqrt{2}}{2}$ $= \frac{3\sqrt{2}}{10}DE$ 。

现在取 $B = (0,0)$ 、 $C = (4,0)$ 、 $A = (0,3)$ ，于是 $E = (0, e)$ 、 $D = (d, 0)$ ，直线 $AC$ 为 $3x + 4y = 12$ ，且 $DE^2 = d^2 + e^2$ 。令 $k = \frac{\sqrt{2}}{2}DE$ ，两个距离公式变为 $\frac{12 - 4e}{5} = \frac{2\sqrt{2}}{5}DE$ 和 $\frac{12 - 3d}{5} = \frac{3\sqrt{2}}{10}DE$ ，从而 $e = 3 - k$ ， $d = 4 - k$ 。于是 $DE^2 = 2k^2$ 变成 $2k^2 = (3-k)^2 + (4-k)^2$ $= 2k^2 - 14k + 25$ ，所以 $k = \frac{25}{14}$ ，且 $DE = \sqrt{2}\,k = \frac{25\sqrt{2}}{14}$ 。因此 $a + b + c = 25 + 2 + 14 = 41$ 。

Since $3^2 + 4^2 = 5^2,$ angle $B$ is right, and as $\angle EBD = 90^\circ$ is inscribed in $\omega,$ the chord $ED$ is a diameter. Hence $\angle EFD = \angle EGD = 90^\circ.$ From $EF = DF,$ triangle $EFD$ is an isosceles right triangle, so $EF = DF = \frac{DE}{\sqrt{2}}$ and $\angle FED = 45^\circ;$ from $DG : EG = 3 : 4$ and $DG^2 + EG^2 = DE^2$ we get $DG = \frac{3}{5}DE$ and $EG = \frac{4}{5}DE.$ On $\omega$ the order is $E,$ $F,$ $G,$ $D$ (both $F$ and $G$ lie on the arc opposite $B,$ and $\angle FED = 45^\circ$ exceeds $\angle GED = \arcsin\frac{3}{5},$ so $F$ is farther from $D$ ).

Line $AC$ is the line $FG,$ so the distances from $E$ and $D$ to it follow from the angles of cyclic quadrilateral $EFGD.$ At $F:$ $\angle EFG = 180^\circ - \angle GDE$ and $\sin\angle GDE = \frac{EG}{DE} = \frac{4}{5},$ so the distance from $E$ is $EF \sin\angle EFG = \frac{DE}{\sqrt{2}} \cdot \frac{4}{5}$ $= \frac{2\sqrt{2}}{5}DE.$ At $G:$ $\angle FGD = 180^\circ - \angle FED = 135^\circ,$ so the distance from $D$ is $DG \sin\angle FGD = \frac{3}{5}DE \cdot \frac{\sqrt{2}}{2}$ $= \frac{3\sqrt{2}}{10}DE.$

Now place $B = (0,0),$ $C = (4,0),$ $A = (0,3),$ so $E = (0, e),$ $D = (d, 0),$ line $AC:$ $3x + 4y = 12,$ and $DE^2 = d^2 + e^2.$ Setting $k = \frac{\sqrt{2}}{2}DE,$ the two distance formulas read $\frac{12 - 4e}{5} = \frac{2\sqrt{2}}{5}DE$ and $\frac{12 - 3d}{5} = \frac{3\sqrt{2}}{10}DE,$ which give $e = 3 - k$ and $d = 4 - k.$ Then $DE^2 = 2k^2$ becomes $2k^2 = (3-k)^2 + (4-k)^2$ $= 2k^2 - 14k + 25,$ so $k = \frac{25}{14}$ and $DE = \sqrt{2}\,k = \frac{25\sqrt{2}}{14}.$ Therefore $a + b + c = 25 + 2 + 14 = 41.$