Question

两个外切圆 $\omega_1$ 和 $\omega_2$ 的圆心分别为 $O_1$ 和 $O_2$ ，第三个圆 $\Omega$ 经过 $O_1$ 和 $O_2$ ，并与 $\omega_1$ 交于 $B$ 和 $C$ 与 $\omega_2$ 交于 $A$ 和 $D$ ，如图所示。已知 $AB = 2$ 、 $O_1O_2 = 15$ 、 $CD = 16$ ，且 $ABO_1CDO_2$ 是一个凸六边形。求这个六边形的面积。

Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2,$ respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D,$ as shown. Suppose that $AB = 2,$ $O_1O_2 = 15,$ $CD = 16,$ and $ABO_1CDO_2$ is a convex hexagon. Find the area of this hexagon.

答案

答案：140

知识点：弦三角恒等式面积分割

难度评级：3700

解答：

六边形的六个顶点都在 $\Omega$ ：上： $O_1$ 和 $O_2$ 由题设在其上， $A, B, C, D$ 是与 $\Omega$ 的交点。设 $\Omega$ 的半径为 $R$ ，并设边 $AB$ 、 $BO_1$ 、 $O_1C$ 、 $CD$ 、 $DO_2$ 、 $O_2A$ 所截的弧分别为 $2\alpha, 2\beta, 2\beta, 2\gamma, 2\delta, 2\delta$ （两个 $\beta$ 来自弦 $BO_1 = O_1C = r_1$ ，即 $\omega_1$ 的半径；同理 $O_2A = O_2D = r_2$ ），所以 $\alpha + 2\beta + \gamma + 2\delta = \pi$ 。每条弦都等于 $2R\sin(\text{half its arc})$ ： $2R\sin\alpha = 2$ 、 $2R\sin\gamma = 16$ ，而弦 $\overline{O_1O_2}$ 对应的弧为 $2\beta + 2\gamma + 2\delta$ ，给出 $2R\sin(\alpha + \sigma) = 15$ ，其中 $\sigma = \beta + \delta$ 外切给出另一个等于 $15$ ：的方程： $r_1 + r_2 = 2R(\sin\beta + \sin\delta) = 15$ 。

因为 $\gamma = \pi - \alpha - 2\sigma$ ，所以 $\sin\gamma = \sin(\alpha + 2\sigma)$ 。和差化积给出因此 $\cos\sigma = \frac{3}{5}$ ，类似地， $14 = 4R\cos(\alpha + \sigma)\sin\sigma$ 给出 $R\cos(\alpha + \sigma) = \frac{35}{8}$ 。结合 $2R\sin(\alpha + \sigma) = 15$ ，得到 $4R^2 = 225 + \frac{1225}{16}$ ，所以 $R^2 = \frac{4825}{64}$ 。另外 $15 = 2R(\sin\beta + \sin\delta)$ $= 4R\sin\frac{\sigma}{2}\cos\frac{\beta - \delta}{2}$ ，且 $\sin\frac{\sigma}{2} = \frac{1}{\sqrt{5}}$ ，所以 $\cos\frac{\beta - \delta}{2} = \frac{15\sqrt{5}}{4R}$ ，从而 $\cos(\beta - \delta) = \frac{1125}{8R^2} - 1$ 。 $\begin{aligned} 18 &= 2R\left[\sin(\alpha + 2\sigma) + \sin\alpha\right] \\ &= 4R\sin(\alpha + \sigma)\cos\sigma \\ &= 30\cos\sigma, \end{aligned}$

连接 $\Omega$ 的圆心和六个顶点，可将六边形分成六个三角形，所以面积为 $\frac{1}{2}R^2$ ${}\cdot \small \left[\sin 2\alpha + 2\sin 2\beta + 2\sin 2\delta + \sin 2\gamma\right]$ 。现在 $R^2(\sin 2\beta + \sin 2\delta)$ $= 2R^2 \sin\sigma\cos(\beta - \delta)$ $= \frac{8}{5}\left(\frac{1125}{8} - R^2\right)$ $= \frac{835}{8}$ ，而 $\sin 2\alpha + \sin 2\gamma$ $= -2\sin 2\sigma \cos\bigl(2(\alpha + \sigma)\bigr)$ $= -2 \cdot \frac{24}{25} \cdot \left(-\frac{95}{193}\right)$ ，因为 $\cos\bigl(2(\alpha+\sigma)\bigr) = 1 - \frac{2 \cdot 225}{4R^2}$ $= -\frac{95}{193}$ ，所以 $\frac{1}{2}R^2(\sin 2\alpha + \sin 2\gamma)$ $= \frac{1}{2} \cdot \frac{4825}{64} \cdot \frac{912}{965}$ $= \frac{285}{8}$ 面积为 $\frac{835}{8} + \frac{285}{8} = 140$ 。

All six hexagon vertices lie on $\Omega:$ $O_1$ and $O_2$ by hypothesis, and $A, B, C, D$ as intersection points with $\Omega.$ Let $R$ be the radius of $\Omega,$ and let the arcs cut off by the sides $AB,$ $BO_1,$ $O_1C,$ $CD,$ $DO_2,$ $O_2A$ be $2\alpha, 2\beta, 2\beta, 2\gamma, 2\delta, 2\delta$ (the two $\beta$ 's because chords $BO_1 = O_1C = r_1,$ the radius of $\omega_1,$ and likewise $O_2A = O_2D = r_2$ ), so $\alpha + 2\beta + \gamma + 2\delta = \pi.$ Each chord equals $2R\sin(\text{half its arc}):$ $2R\sin\alpha = 2,$ $2R\sin\gamma = 16,$ and the chord $\overline{O_1O_2}$ subtends $2\beta + 2\gamma + 2\delta,$ giving $2R\sin(\alpha + \sigma) = 15$ where $\sigma = \beta + \delta.$ External tangency gives a second equation worth $15:$ $r_1 + r_2 = 2R(\sin\beta + \sin\delta) = 15.$

Since $\gamma = \pi - \alpha - 2\sigma,$ we have $\sin\gamma = \sin(\alpha + 2\sigma).$ Sum-to-product then gives $\begin{aligned} 18 &= 2R\left[\sin(\alpha + 2\sigma) + \sin\alpha\right] \\ &= 4R\sin(\alpha + \sigma)\cos\sigma \\ &= 30\cos\sigma, \end{aligned}$ so $\cos\sigma = \frac{3}{5},$ and similarly $14 = 4R\cos(\alpha + \sigma)\sin\sigma$ gives $R\cos(\alpha + \sigma) = \frac{35}{8}.$ Combining with $2R\sin(\alpha + \sigma) = 15$ yields $4R^2 = 225 + \frac{1225}{16},$ so $R^2 = \frac{4825}{64}.$ Also $15 = 2R(\sin\beta + \sin\delta)$ $= 4R\sin\frac{\sigma}{2}\cos\frac{\beta - \delta}{2}$ with $\sin\frac{\sigma}{2} = \frac{1}{\sqrt{5}},$ so $\cos\frac{\beta - \delta}{2} = \frac{15\sqrt{5}}{4R}$ and $\cos(\beta - \delta) = \frac{1125}{8R^2} - 1.$

Joining the center of $\Omega$ to the six vertices splits the hexagon into six triangles, so its area is $\frac{1}{2}R^2$ ${}\cdot \small \left[\sin 2\alpha + 2\sin 2\beta + 2\sin 2\delta + \sin 2\gamma\right].$ Now $R^2(\sin 2\beta + \sin 2\delta)$ $= 2R^2 \sin\sigma\cos(\beta - \delta)$ $= \frac{8}{5}\left(\frac{1125}{8} - R^2\right)$ $= \frac{835}{8},$ while $\sin 2\alpha + \sin 2\gamma$ $= -2\sin 2\sigma \cos\bigl(2(\alpha + \sigma)\bigr)$ $= -2 \cdot \frac{24}{25} \cdot \left(-\frac{95}{193}\right)$ since $\cos\bigl(2(\alpha+\sigma)\bigr) = 1 - \frac{2 \cdot 225}{4R^2}$ $= -\frac{95}{193},$ so $\frac{1}{2}R^2(\sin 2\alpha + \sin 2\gamma)$ $= \frac{1}{2} \cdot \frac{4825}{64} \cdot \frac{912}{965}$ $= \frac{285}{8}.$ The area is $\frac{835}{8} + \frac{285}{8} = 140.$

2022 AIME II 第 15 题

答案：140

难度评级：3700

解答：

其他年份的第 15 题