AMC10 2004 B

AMC10 2004 B · Q22

AMC10 2004 B · Q22. It mainly tests Triangles (properties), Circle theorems.

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?

一个边长为 5、12 和 13 的三角形既有内切圆也有外接圆。那两个圆心的距离是多少？

(A) \frac{3\sqrt{5}}{2} \frac{3\sqrt{5}}{2}

(B) \frac{7}{2} \frac{7}{2}

(D) \frac{\sqrt{65}}{2} \frac{\sqrt{65}}{2}

(E) \frac{9}{2} \frac{9}{2}

Answer

Correct choice: (D)

正确答案：（D）

Solution

(D) The triangle is a right triangle that can be placed in a coordinate system with vertices at $(0,0)$, $(5,0)$, and $(0,12)$. The center of the circumscribed circle is the midpoint of the hypotenuse, which is $(5/2,6)$. To determine the radius $r$ of the inscribed circle notice that the hypotenuse of the triangle is $$(12-r)+(5-r)=13 \quad \text{so} \quad r=2.$$ So the center of the inscribed circle is $(2,2)$, and the distance between the two centers is $$\sqrt{\left(\frac{5}{2}-2\right)^2+(6-2)^2}=\frac{\sqrt{65}}{2}.$$

（D）该三角形是一个直角三角形，可放置在坐标系中，顶点为 $(0,0)$、$(5,0)$ 和 $(0,12)$。外接圆的圆心是斜边的中点，即 $(5/2,6)$。为求内切圆的半径 $r$，注意该三角形的斜边满足 $$(12-r)+(5-r)=13 \quad \text{所以} \quad r=2.$$ 因此内切圆的圆心为 $(2,2)$，两圆心之间的距离为 $$\sqrt{\left(\frac{5}{2}-2\right)^2+(6-2)^2}=\frac{\sqrt{65}}{2}.$$

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