Question

三个半径分别为 $11$ 、 $13$ 、 $19$ 的球两两外切。一个平面与这三个球相交，得到三个全等圆，它们的圆心分别为 $A$ 、 $B$ 、 $C$ ，并且三个球的球心都在该平面的同一侧。已知 $AB^2 = 560$ 。求 $AC^2$ 。

Three spheres with radii $11,$ $13,$ and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A,$ $B,$ and $C,$ respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560.$ Find $AC^2.$

答案

答案：756

知识点：球立体几何勾股定理

难度评级：2560

解答：

设三个球心到平面的高度分别为 $h_1, h_2, h_3$ 。每个截圆的圆心是对应球心到平面的垂足，公共截圆半径 $\rho$ 满足 $\rho^2 = 11^2 - h_1^2$ $= 13^2 - h_2^2$ $= 19^2 - h_3^2$ 。

前两个球相切，所以球心相距 $11 + 13 = 24$ 。投影到平面上，有 $AB^2 = 24^2 - (h_2 - h_1)^2$ 。因此 $(h_2 - h_1)^2 = 576 - 560 = 16$ 。截圆全等给出 $h_2^2 - h_1^2 = 169 - 121 = 48$ ，所以 $h_2 - h_1 = 4$ 且 $h_2 + h_1 = 12$ （另一个符号会给出负的和），得到 $h_1 = 4$ 、 $h_2 = 8$ ，并且 $\rho^2 = 121 - 16 = 105$ 。于是 $h_3^2 = 361 - 105 = 256$ ，所以 $h_3 = 16$ 。

第一和第三个球心相距 $11 + 19 = 30$ ，所以 $\begin{aligned} AC^2 &= 30^2 - (h_3 - h_1)^2 \\ &= 900 - 144 = 756. \end{aligned}$

Let the sphere centers be at heights $h_1, h_2, h_3$ above the plane. Each circle's center is the foot of the perpendicular from the sphere's center, and the common circle radius $\rho$ satisfies $\rho^2 = 11^2 - h_1^2$ $= 13^2 - h_2^2$ $= 19^2 - h_3^2.$

The first two spheres are tangent, so their centers are $11 + 13 = 24$ apart, and projecting onto the plane, $AB^2 = 24^2 - (h_2 - h_1)^2.$ Thus $(h_2 - h_1)^2 = 576 - 560 = 16.$ Congruence gives $h_2^2 - h_1^2 = 169 - 121 = 48,$ so $h_2 - h_1 = 4$ and $h_2 + h_1 = 12$ (the other sign gives a negative sum), yielding $h_1 = 4,$ $h_2 = 8,$ and $\rho^2 = 121 - 16 = 105.$ Then $h_3^2 = 361 - 105 = 256,$ so $h_3 = 16.$

The first and third centers are $11 + 19 = 30$ apart, so $\begin{aligned} AC^2 &= 30^2 - (h_3 - h_1)^2 \\ &= 900 - 144 = 756. \end{aligned}$

2022 AIME I 第 10 题

答案：756

难度评级：2560

解答：

其他年份的第 10 题