AMC8 2016

AMC8 2016 · Q25

AMC8 2016 · Q25. It mainly tests Pythagorean theorem, Area & perimeter.

A semicircle is inscribed in an isosceles triangle with base 16 and height 15 so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?

一个半圆内接于一个底边长16、高15的等腰三角形中，使得半圆的直径包含在三角形的底边上，如图所示。求半圆的半径。

(A) $4\sqrt{3}$ $4\sqrt{3}$

(B) $120/17$ $120/17$

(D) $(17\sqrt{2})/2$ $(17\sqrt{2})/2$

(E) $(17\sqrt{3})/2$ $(17\sqrt{3})/2$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Let O be the midpoint of base AB of triangle ABC and the center of the semicircle. Triangle OBC is a right triangle with $OB=8$ and $OC=15$, and so, by the Pythagorean Theorem, $BC=17$. Let E be the point where the semicircle intersects $BC$, so radius $OE$ is perpendicular to $BC$. Then triangles $OEB$ and $COB$ are similar, and therefore $\frac{OE}{CO}=\frac{OB}{CB}$. Hence, $\frac{OE}{15}=\frac{8}{17}$ and so $OE=\frac{120}{17}$.

设 O 为底边 AB 的中点，也是半圆的圆心。三角形 OBC 为直角三角形，且 $OB=8$, $OC=15$，由勾股定理得 $BC=17$。设 E 为半圆与 $BC$ 的交点，则半径 $OE$ 垂直于 $BC$。于是三角形 $OEB$ 与 $COB$ 相似，因此 $\frac{OE}{CO}=\frac{OB}{CB}$。故 $\frac{OE}{15}=\frac{8}{17}$，从而 $OE=\frac{120}{17}$。

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