Question

在锐角三角形 $ABC$ 中，点 $P$ 与 $Q$ 分别是从 $C$ 到 $\overline{AB}$ 、从 $B$ 到 $\overline{AC}$ 的垂足。直线 $PQ$ 与 $\triangle ABC$ 的外接圆交于两个不同的点 $X$ 和 $Y$ 。已知 $XP = 10$ 、 $PQ = 25$ 、 $QY = 15$ 。 $AB \cdot AC$ 的值可写为 $m\sqrt{n}$ ，其中 $m$ 与 $n$ 是正整数，且 $n$ 不被任何质数的平方整除。求 $m + n$ 。

In acute triangle $ABC,$ points $P$ and $Q$ are the feet of the perpendiculars from $C$ to $\overline{AB}$ and from $B$ to $\overline{AC},$ respectively. Line $PQ$ intersects the circumcircle of $\triangle ABC$ in two distinct points, $X$ and $Y.$ Suppose $XP = 10,$ $PQ = 25,$ and $QY = 15.$ The value of $AB \cdot AC$ can be written in the form $m\sqrt{n},$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m + n.$

答案

答案：574

知识点：圆幂相似余弦定理

难度评级：3370

解答：

记 $b = AC$ 、 $c = AB$ 、 $a = BC$ 、 $k = \cos A$ 。直角三角形 $APC$ 与 $AQB$ 给出 $AP = bk$ 、 $AQ = ck$ ，所以三角形 $APQ$ 与三角形 $ACB$ 相似，比例为 $k$ ，从而 $PQ = ak = 25$ 。直线上的点顺序为 $X, P, Q, Y$ ，所以 $P$ 的幂给出 $XP \cdot PY = 10 \cdot 40 = 400$ $= AP \cdot PB$ ，而 $Q$ 的幂给出 $YQ \cdot QX = 15 \cdot 35 = 525$ $= AQ \cdot QC$ 。令 $u = bk$ 、 $v = ck$ ，这些式子为也就是 $w - u^2 = 400$ 、 $w - v^2 = 525$ ，其中 $w = \frac{uv}{k} = bck$ 。 $\begin{aligned} &u(c - u) = 400, \\ &\qquad v(b - v) = 525, \end{aligned}$

由余弦定理， $a^2 = b^2 + c^2 - 2bck$ ，所以 $a^2k^2 = u^2 + v^2 - 2uvk = 625$ 。代入 $u^2 = w - 400$ 、 $v^2 = w - 525$ ，以及 $uv = wk$ ，得 $2w - 925 - 2wk^2 = 625$ ，所以 $wk^2 = w - 775$ 。于是化简得 $150w = 210000$ ，所以 $w = 1400$ ，并且 $k^2 = \frac{1400 - 775}{1400} = \frac{25}{56}$ 。 $\begin{aligned} (uv)^2 &= w^2k^2 \\ &= w(w - 775) \\ &= (w - 400)(w - 525), \end{aligned}$

因此 $k = \frac{5}{2\sqrt{14}}$ ，且所以 $m + n = 560 + 14 = 574$ 。 $\begin{aligned} bc = \frac{w}{k} &= 1400 \cdot \frac{2\sqrt{14}}{5} \\ &= 560\sqrt{14}, \end{aligned}$

Write $b = AC,$ $c = AB,$ $a = BC,$ and $k = \cos A.$ Right triangles $APC$ and $AQB$ give $AP = bk$ and $AQ = ck,$ so triangle $APQ$ is similar to triangle $ACB$ with ratio $k,$ whence $PQ = ak = 25.$ The points on the line occur in the order $X, P, Q, Y,$ so the power of $P$ gives $XP \cdot PY = 10 \cdot 40 = 400$ $= AP \cdot PB,$ and the power of $Q$ gives $YQ \cdot QX = 15 \cdot 35 = 525$ $= AQ \cdot QC.$ With $u = bk$ and $v = ck$ these read $\begin{aligned} &u(c - u) = 400, \\ &\qquad v(b - v) = 525, \end{aligned}$ that is, $w - u^2 = 400$ and $w - v^2 = 525,$ where $w = \frac{uv}{k} = bck.$

By the law of cosines, $a^2 = b^2 + c^2 - 2bck,$ so $a^2k^2 = u^2 + v^2 - 2uvk = 625.$ Substituting $u^2 = w - 400$ and $v^2 = w - 525,$ and $uv = wk,$ gives $2w - 925 - 2wk^2 = 625,$ so $wk^2 = w - 775.$ Then $\begin{aligned} (uv)^2 &= w^2k^2 \\ &= w(w - 775) \\ &= (w - 400)(w - 525), \end{aligned}$ which simplifies to $150w = 210000,$ so $w = 1400$ and $k^2 = \frac{1400 - 775}{1400} = \frac{25}{56}.$

Thus $k = \frac{5}{2\sqrt{14}}$ and $\begin{aligned} bc = \frac{w}{k} &= 1400 \cdot \frac{2\sqrt{14}}{5} \\ &= 560\sqrt{14}, \end{aligned}$ so $m + n = 560 + 14 = 574.$

2019 AIME II 第 15 题

答案：574

难度评级：3370

解答：

其他年份的第 15 题