AMC10 2019 A

AMC10 2019 A · Q21

AMC10 2019 A · Q21. It mainly tests Triangles (properties), 3D geometry (volume).

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?

一个以 $O$ 为球心的球半径为 6。一个边长为 15、15 和 24 的三角形位于空间中，使得它的每条边都与该球相切。求 $O$ 与该三角形确定的平面之间的距离。

(A) $2\sqrt{3}$ $2\sqrt{3}$

(B) 4 4

(D) $2\sqrt{5}$ $2\sqrt{5}$

(E) 5 5

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Let $\triangle ABC$ be the given triangle, with $AB=24$ and $AC=BC=15$, and let $D$ be the midpoint of $AB$. The length of the altitude to $AB$ is $CD=\sqrt{15^2-12^2}=9$. The area of $\triangle ABC$ is $\frac12\cdot24\cdot9=108$. The plane of the triangle intersects the sphere in a circle, which is the inscribed circle for $\triangle ABC$. Let $r$ and $I$ be the radius and the center of the inscribed circle, respectively. The semiperimeter of the triangle is $\frac12(AB+BC+AC)=27$, so $r=\frac{108}{27}=4$. In right triangle $DIO$ the hypotenuse $\overline{OD}$ has length $6$ (the radius of the sphere) and $DI=r=4$, so $OI=\sqrt{36-16}=2\sqrt5$, the requested distance between the center of the sphere and the plane determined by $\triangle ABC$.

答案（D）：设$\triangle ABC$为所给三角形，且$AB=24$、$AC=BC=15$，令$D$为$AB$的中点。到底边$AB$的高为$CD=\sqrt{15^2-12^2}=9$。$\triangle ABC$的面积为$\frac12\cdot24\cdot9=108$。三角形所在平面与球相交成一圆，该圆即为$\triangle ABC$的内切圆。设$r$与$I$分别为内切圆半径与圆心。三角形的半周长为$\frac12(AB+BC+AC)=27$，因此$r=\frac{108}{27}=4$。在直角三角形$DIO$中，斜边$\overline{OD}$长为$6$（球的半径），且$DI=r=4$，所以$OI=\sqrt{36-16}=2\sqrt5$，这就是球心到$\triangle ABC$所确定平面的所求距离。

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