Question

三角形 $ABC$ 的边长为 $AB = 4$ 、 $BC = 5$ 、 $CA = 6$ 。点 $D$ 和 $E$ 在射线 $AB$ 上，且 $AB \lt AD \lt AE$ 。点 $F \neq C$ 是 $\triangle ACD$ 与 $\triangle EBC$ 的外接圆的一个交点，并满足 $DF = 2$ 、 $EF = 7$ 。则 $BE$ 可以表示为 $\frac{a + b\sqrt{c}}{d}$ ，其中 $a$ 、 $b$ 、 $c$ 、 $d$ 为正整数， $a$ 与 $d$ 互质，且 $c$ 不被任何质数的平方整除。求 $a + b + c + d$ 。

Triangle $ABC$ has side lengths $AB = 4,$ $BC = 5,$ and $CA = 6.$ Points $D$ and $E$ are on ray $AB$ with $AB \lt AD \lt AE.$ The point $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF = 2$ and $EF = 7.$ Then $BE$ can be expressed as $\frac{a + b\sqrt{c}}{d},$ where $a,$ $b,$ $c,$ and $d$ are positive integers such that $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a + b + c + d.$

答案

答案：32

知识点：圆内接四边形根轴圆幂余弦定理

难度评级：3370

小提示：

圆内接四边形给出 $\angle FDA = \angle FCA$ 和 $\angle FEB = \angle FCB$ ，所以 $\angle DFE = \angle ACB$ ；接着用余弦定理求 $DE$

The cyclic quadrilaterals give $\angle FDA = \angle FCA$ and $\angle FEB = \angle FCB,$ so $\angle DFE = \angle ACB;$ now the law of cosines finds $DE$

大提示：

直线 $CF$ 是根轴：若它与直线 $AB$ 交于 $G$ ，幂相等给出 $GA \cdot GD = GB \cdot GE$ 。为定位 $G$ ，使用角 $\angle GCA$ ，这个角可由三角形 $DEF$ 求出

Line $CF$ is the radical axis: where it crosses line $AB$ at $G,$ the powers give $GA \cdot GD = GB \cdot GE.$ Locate $G$ using the angle $\angle GCA$ found from triangle $DEF.$

解答：

点 $D$ 、 $E$ 位于 $B$ 以外的射线 $AB$ 上，而点 $F$ 在直线 $AB$ 的另一侧（相对于 $C$ 而言）。由于 $ACFD$ 与 $BCFE$ 为圆内接四边形，圆周角给出 $\angle FDA = \angle FCA$ 和 $\angle FEB = \angle FCB$ 。记 $\alpha = \angle FCA$ 、 $\beta = \angle FCB$ ，则三角形 $DEF$ 有 $\angle FDE = 180^\circ - \alpha$ 、 $\angle FED = \beta$ ，所以 $\angle DFE = \alpha - \beta = \angle ACB$ 。在三角形 $ABC$ 中， $\cos \angle ACB = \frac{25 + 36 - 16}{2 \cdot 5 \cdot 6} = \frac{3}{4}$ ，所以在三角形 $DFE$ 中用余弦定理得到 $\begin{aligned} DE^2 &= 2^2 + 7^2 - 2 \cdot 2 \cdot 7 \cdot \tfrac{3}{4} \\ &= 32, \\ DE &= 4\sqrt{2} \end{aligned}\text{。}$

在三角形 $DFE$ 中， $\cos \angle FDE = \frac{4 + 32 - 49}{2 \cdot 2 \cdot 4\sqrt{2}} = -\frac{13\sqrt{2}}{32}$ ，所以 $\alpha$ 为锐角，且 $\cos\alpha = \frac{13\sqrt{2}}{32}$ 、 $\sin\alpha = \sqrt{1 - \frac{338}{1024}} = \frac{7\sqrt{14}}{32}$ 。设 $G$ 为直线 $CF$ 与直线 $AB$ 的交点。在三角形 $ACG$ 中， $\angle GAC = \angle BAC$ ，且 $\cos \angle BAC = \frac{16 + 36 - 25}{2 \cdot 4 \cdot 6} = \frac{9}{16}$ 、 $\sin \angle BAC = \frac{5\sqrt{7}}{16}$ ，又 $\angle ACG = \alpha$ ，所以 $\sin(\angle BAC + \alpha)$ $= \frac{5\sqrt{7}}{16} \cdot \frac{13\sqrt{2}}{32}$ $+ \frac{9}{16} \cdot \frac{7\sqrt{14}}{32}$ $= \frac{\sqrt{14}}{4}$ ，并且 $\begin{aligned} AG &= \frac{AC \sin \alpha}{\sin(\angle BAC + \alpha)} \\ &= \frac{6 \cdot \frac{7\sqrt{14}}{32}}{\frac{\sqrt{14}}{4}} \\ &= \frac{21}{4} \end{aligned}\text{。}$

直线 $CF$ 是两圆的根轴，所以 $GA \cdot GD = GB \cdot GE$ 。设 $x = BD$ ，则 $BE = x + DE = x + 4\sqrt{2}$ ： $\begin{aligned} &\frac{21}{4}\left(x - \frac{5}{4}\right) \\ &= \frac{5}{4}\left(x - \frac{5}{4} + 4\sqrt{2}\right) \end{aligned}\text{，}$ 因为 $GD = 4 + x - \frac{21}{4}$ ，且 $GB = \frac{21}{4} - 4$ 。这给出 $16\left(x - \frac{5}{4}\right) = 20\sqrt{2}$ ，所以 $x = \frac{5 + 5\sqrt{2}}{4}$ ，并且 $BE = \frac{5 + 21\sqrt{2}}{4}$ 。因此 $a + b + c + d = 5 + 21 + 2 + 4$ $= 32$ 。

Points $D,$ $E$ lie beyond $B$ on ray $AB,$ and $F$ lies on the opposite side of line $AB$ from $C.$ Since $ACFD$ and $BCFE$ are cyclic, the inscribed angles give $\angle FDA = \angle FCA$ and $\angle FEB = \angle FCB.$ Writing $\alpha = \angle FCA$ and $\beta = \angle FCB,$ triangle $DEF$ has angles $\angle FDE = 180^\circ - \alpha$ and $\angle FED = \beta,$ so $\angle DFE = \alpha - \beta = \angle ACB.$ From triangle $ABC,$ $\cos \angle ACB = \frac{25 + 36 - 16}{2 \cdot 5 \cdot 6} = \frac{3}{4},$ so the law of cosines in triangle $DFE$ gives $\begin{aligned} DE^2 &= 2^2 + 7^2 - 2 \cdot 2 \cdot 7 \cdot \tfrac{3}{4} \\ &= 32, \\ DE &= 4\sqrt{2}. \end{aligned}$

In triangle $DFE,$ $\cos \angle FDE = \frac{4 + 32 - 49}{2 \cdot 2 \cdot 4\sqrt{2}} = -\frac{13\sqrt{2}}{32},$ so $\alpha$ is acute with $\cos\alpha = \frac{13\sqrt{2}}{32}$ and $\sin\alpha = \sqrt{1 - \frac{338}{1024}} = \frac{7\sqrt{14}}{32}.$ Let $G$ be the intersection of line $CF$ with line $AB.$ In triangle $ACG,$ $\angle GAC = \angle BAC$ has $\cos \angle BAC = \frac{16 + 36 - 25}{2 \cdot 4 \cdot 6} = \frac{9}{16},$ $\sin \angle BAC = \frac{5\sqrt{7}}{16},$ and $\angle ACG = \alpha,$ so $\sin(\angle BAC + \alpha)$ $= \frac{5\sqrt{7}}{16} \cdot \frac{13\sqrt{2}}{32}$ $+ \frac{9}{16} \cdot \frac{7\sqrt{14}}{32}$ $= \frac{\sqrt{14}}{4}$ and $\begin{aligned} AG &= \frac{AC \sin \alpha}{\sin(\angle BAC + \alpha)} \\ &= \frac{6 \cdot \frac{7\sqrt{14}}{32}}{\frac{\sqrt{14}}{4}} \\ &= \frac{21}{4}. \end{aligned}$

Line $CF$ is the radical axis of the two circles, so $GA \cdot GD = GB \cdot GE.$ With $x = BD$ and $BE = x + DE = x + 4\sqrt{2}:$ $\begin{aligned} &\frac{21}{4}\left(x - \frac{5}{4}\right) \\ &= \frac{5}{4}\left(x - \frac{5}{4} + 4\sqrt{2}\right), \end{aligned}$ since $GD = 4 + x - \frac{21}{4}$ and $GB = \frac{21}{4} - 4.$ This gives $16\left(x - \frac{5}{4}\right) = 20\sqrt{2},$ so $x = \frac{5 + 5\sqrt{2}}{4}$ and $BE = \frac{5 + 21\sqrt{2}}{4}.$ Therefore $a + b + c + d = 5 + 21 + 2 + 4$ $= 32.$

2019 AIME I 第 13 题

答案：32

难度评级：3370

小提示：

大提示：

解答：

其他年份的第 13 题