AMC10 2007 B

AMC10 2007 B · Q11

AMC10 2007 B · Q11. It mainly tests Triangles (properties), Circle theorems.

A circle passes through the three vertices of an isosceles triangle that has two sides of length 3 and a base of length 2. What is the area of this circle?

一个圆经过一个等腰三角形的三个顶点，该等腰三角形有两个边长为3，底边长为2。这个圆的面积是多少？

(A) $2\pi$ $2\pi$

(B) $\frac{5}{2}\pi$ $\frac{5}{2}\pi$

(D) $3\pi$ $3\pi$

(E) $\frac{7}{2}\pi$ $\frac{7}{2}\pi$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Let $BD$ be an altitude of the isosceles $\triangle ABC$, and let $O$ denote the center of the circle with radius $r$ that passes through $A$, $B$, and $C$, as shown. Then $$BD=\sqrt{3^2-1^2}=2\sqrt{2}\quad \text{and}\quad OD=2\sqrt{2}-r.$$ Since $\triangle ADO$ is a right triangle, we have $$r^2=1^2+(2\sqrt{2}-r)^2=1+8-4\sqrt{2}\,r+r^2,\quad \text{and}\quad r=\frac{9}{4\sqrt{2}}=\frac{9}{8}\sqrt{2}.$$ As a consequence, the circle has area $$\left(\frac{9}{8}\sqrt{2}\right)^2\pi=\frac{81}{32}\pi.$$

答案（C）：设 $BD$ 为等腰三角形 $\triangle ABC$ 的一条高，设 $O$ 为过 $A,B,C$ 三点、半径为 $r$ 的圆的圆心，如图所示。则 $$BD=\sqrt{3^2-1^2}=2\sqrt{2}\quad \text{且}\quad OD=2\sqrt{2}-r.$$ 由于 $\triangle ADO$ 是直角三角形，有 $$r^2=1^2+(2\sqrt{2}-r)^2=1+8-4\sqrt{2}\,r+r^2,\quad \text{从而}\quad r=\frac{9}{4\sqrt{2}}=\frac{9}{8}\sqrt{2}.$$ 因此，该圆的面积为 $$\left(\frac{9}{8}\sqrt{2}\right)^2\pi=\frac{81}{32}\pi.$$

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