Question

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

答案：717

知识点：切线圆内接四边形正弦定理坐标几何

难度评级：3370

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 $XY = AT \sin A$ 。使用正弦定理 $\scriptsize\left(\sin A = \frac{11}{R},\ \sin B = \frac{AC}{2R},\ \sin C = \frac{AB}{2R}\right)$ ，题设条件变为 $\begin{aligned} &\frac{64\left(AB^2 + AC^2\right) + 121\,AT^2}{R^2} \\ &= 1143. \end{aligned}$

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 $\begin{aligned} AB^2 + AC^2 &= 2(x^2 + y^2) + 242 \\ &= \frac{484}{\sqrt{135}}\,y + 484, \end{aligned}$ $\begin{aligned} AT^2 &= x^2 + y^2 \\ &\quad {}+ 2\sqrt{135}\,y + 135 \\ &= \frac{512}{\sqrt{135}}\,y + 256. \end{aligned}$

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 $\begin{aligned} XY^2 &= AT^2 \sin^2 A \\ &= \frac{121\,AT^2}{R^2} \\ &= \frac{121 \cdot 183552}{30976} \\ &= \frac{183552}{256} = 717. \end{aligned}$

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 $A, X, T, Y$ lie on a circle with diameter $AT,$ whence $XY = AT \sin A.$ Using the law of sines $\scriptsize\left(\sin A = \frac{11}{R},\ \sin B = \frac{AC}{2R},\ \sin C = \frac{AB}{2R}\right),$ 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 $\begin{aligned} XY^2 &= AT^2 \sin^2 A \\ &= \frac{121\,AT^2}{R^2} \\ &= \frac{121 \cdot 183552}{30976} \\ &= \frac{183552}{256} = 717. \end{aligned}$

2020 AIME II 第 15 题

答案：717

难度评级：3370

解答：

其他年份的第 15 题