Question

半径分别为 $961$ 和 $625$ 的圆 $\omega_1$ 与 $\omega_2$ 相交于不同的两点 $A$ 与 $B$ 。第三个圆 $\omega$ 与 $\omega_1$ 和 $\omega_2$ 都外切。设直线 $AB$ 与 $\omega$ 相交于两点 $P$ 和 $Q$ ，且小弧 $\widehat{PQ}$ 的度数为 $120^\circ$ 。求 $\omega_1$ 与 $\omega_2$ 的圆心之间的距离。

Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625,$ respectively, intersect at distinct points $A$ and $B.$ A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2.$ Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^\circ.$ Find the distance between the centers of $\omega_1$ and $\omega_2.$

答案

答案：672

知识点：根轴圆幂相切圆弦

难度评级：3270

解答：

设 $O$ 与 $r$ 为 $\omega$ 的圆心和半径， $O_1, O_2$ 为另外两个圆心。外切给出 $OO_1 = r + 961$ ，所以 $O$ 关于 $\omega_1$ 的幂为 $OO_1^2 - 961^2 = r^2 + 2 \cdot 961r$ ；同理，关于 $\omega_2$ 的幂为 $r^2 + 2 \cdot 625r$ 。两者之差为 $2r(961 - 625) = 672r$ 。

对任意点 $X$ ，差 $\mathrm{pow}_{\omega_1}(X) - \mathrm{pow}_{\omega_2}(X)$ $= (XO_1^2 - XO_2^2)$ $- (961^2 - 625^2)$ 是关于 $X$ 的线性函数，且在根轴直线 $AB$ 上为零；它沿垂直于 $AB$ 的方向变化率为 $2 \cdot O_1O_2$ 。因此这个差等于 $2 \cdot O_1O_2 \cdot \operatorname{dist}(O, AB)$ 。另一方面，圆 $\omega$ 的弦 $PQ$ 对应 $120^\circ$ 的圆心角，所以 $\operatorname{dist}(O, AB) = r\cos 60^\circ = \frac{r}{2}$ 。

因此 $672r = 2 \cdot O_1O_2 \cdot \frac{r}{2} = O_1O_2 \cdot r$ ，两圆心距离为 $672$ 。

Let $O$ and $r$ be the center and radius of $\omega,$ and $O_1, O_2$ the other centers. External tangency gives $OO_1 = r + 961,$ so the power of $O$ with respect to $\omega_1$ is $OO_1^2 - 961^2 = r^2 + 2 \cdot 961r;$ similarly its power with respect to $\omega_2$ is $r^2 + 2 \cdot 625r.$ The difference is $2r(961 - 625) = 672r.$

For any point $X,$ the difference $\mathrm{pow}_{\omega_1}(X) - \mathrm{pow}_{\omega_2}(X)$ $= (XO_1^2 - XO_2^2)$ $- (961^2 - 625^2)$ is a linear function of $X$ that vanishes on the radical axis, which is line $AB;$ its rate of change perpendicular to $AB$ is $2 \cdot O_1O_2.$ So the difference equals $2 \cdot O_1O_2 \cdot \operatorname{dist}(O, AB).$ Meanwhile the chord $PQ$ of $\omega$ subtends a $120^\circ$ central angle, so $\operatorname{dist}(O, AB) = r\cos 60^\circ = \frac{r}{2}.$

Therefore $672r = 2 \cdot O_1O_2 \cdot \frac{r}{2} = O_1O_2 \cdot r,$ and the distance between the centers is $672.$

2021 AIME I 第 13 题

答案：672

难度评级：3270

解答：

其他年份的第 13 题