Question

一个边长为 $5, 12$ 和 $13$ 的三角形有内切圆和外接圆。这两个圆的圆心之间的距离是多少？

A triangle with sides of $5, 12,$ and $13$ has both an inscribed and a circumscribed circle. What is the distance between the centers of those circles?

$\dfrac{3\sqrt{5}}{2}$

$\dfrac{7}{2}$

$\sqrt{15}$

$\dfrac{\sqrt{65}}{2}$

$\dfrac{9}{2}$

答案：D

知识点：内切圆、内心与内切圆半径外接圆、外心与外接圆半径直角三角形坐标几何

难度评级：1770

解答：

因为 $5^2 + 12^2 = 13^2$ ，这是直角三角形。将顶点放在 $(0, 0)$ 、 $(5, 0)$ 、 $(0, 12)$ ，则外心为斜边中点 $\left(\tfrac52, 6\right)$ 。

内切圆半径满足 $(12 - r) + (5 - r) = 13$ ，所以 $r = 2$ ，内心为 $(2, 2)$ 。

两圆心之间的距离为 $\begin{gathered} \sqrt{\left(\tfrac52 - 2\right)^2 + (6 - 2)^2} \\ = \sqrt{\tfrac14 + 16} \\ = \dfrac{\sqrt{65}}{2}. \end{gathered}$

所以正确答案是 D。

Since $5^2 + 12^2 = 13^2,$ the triangle is right. Place it at $(0, 0),$ $(5, 0),$ $(0, 12).$ The circumcenter is the midpoint of the hypotenuse, $\left(\tfrac52, 6\right).$

The inradius satisfies $(12 - r) + (5 - r) = 13,$ so $r = 2$ and the incenter is $(2, 2).$

The distance is $\begin{gathered} \sqrt{\left(\tfrac52 - 2\right)^2 + (6 - 2)^2} \\ = \sqrt{\tfrac14 + 16} \\ = \dfrac{\sqrt{65}}{2}. \end{gathered}$

Thus, the correct answer is D.

2004 AMC 10B 第 22 题

答案：D

难度评级：1770

解答：

其他年份的第 22 题