Question

空间中有一个以 $O$ 为球心、半径为 $6$ 的球，以及一个边长为 $15, 15$ 和 $24$ 的三角形。三角形的每一条边都与球相切。求 $O$ 到该三角形所在平面的距离。

A sphere with center $O$ has radius $6.$ A triangle with sides of length $15, 15,$ and $24$ is situated in space so that each of its sides is tangent to the sphere. What is the distance between $O$ and the plane determined by the triangle?

$2\sqrt{3}$

$4$

$3\sqrt{2}$

$2\sqrt{5}$

$5$

答案：D

知识点：球内切圆、内心与内切圆半径勾股定理筝形

难度评级：1970

解答：

取垂直于三角形平面的截面，如下图。 $24$ $\frac12\cdot24\cdot9=108$ $27$ $\sqrt{15^2-12^2}=9,$ $r=\frac{108}{27}=4.$

由勾股定理， $d=OP$ ，并且 $P$ 四边形 $4$ 是风筝形，所以从而 $O$ 。利用相似三角形得设 $6$ 为球心到该平面的距离。 $d=2\sqrt5$ 也是从 $O$ 到 $P$ 的距离。 $d^2+4^2=6^2.$

所以正确答案是 D。

The altitude to the side of length $24$ is $\sqrt{15^2-12^2}=9,$ so the triangle has area $\frac12\cdot24\cdot9=108$ and semiperimeter $27$ . Its inradius is therefore $r=\frac{108}{27}=4.$

Let $P$ be the perpendicular projection of $O$ onto the triangle's plane, and let $d=OP$ . Because all three side-lines are tangent to the sphere, $P$ is the incenter and its perpendicular distance to each side is $4$ . The distance from $O$ to each side-line is the sphere's radius, $6$ , so the Pythagorean theorem gives $d^2+4^2=6^2.$ Hence $d=2\sqrt5.$

Thus, D is the correct answer.

2019 AMC 10A 第 21 题

答案：D

难度评级：1970

解答：

其他年份的第 21 题