AMC 8 2016 Problem 25

AMC8 2016 年第 25 题

三角形★★★★☆

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?

如图所示，一个底边为 16，高为 15 的等腰三角形内有一个半圆与两条腰相切，并且半圆的直径与三角形的底边重合，问这个半圆的半径是多少？

A. $4 \sqrt{3}$
B. $\dfrac{120}{17}$
C.10
D. $\dfrac{17\sqrt{2}}{2}$
E. $\dfrac{17\sqrt{3}}{2}$

正确答案 · Correct Answer

解析 · Solution

解法1

常规解法 · 面积法半圆心

O

为底中点，腰

=\sqrt{8^2+15^2}=17

。半径

OE\perp

腰。看

\triangle OBC

（OB=8，高 OC=15，斜边 BC=17）：以

BC

为底，其高即

OE

，由面积

\tfrac12\cdot OB\cdot OC=\tfrac12\cdot BC\cdot OE

得

\displaystyle OE=\frac{8\cdot15}{17}=\frac{120}{17}

。

Solution 1

Elementary · area method Let

O

be the midpoint of the base and the center of the semicircle, so each leg is

\sqrt{8^2+15^2}=17

. The radius

OE

is perpendicular to

BC

. In

\triangle OBC

with OB=8, OC=15, BC=17, taking

BC

as base makes

OE

the height, so from

\tfrac12\cdot OB\cdot OC=\tfrac12\cdot BC\cdot OE

we get

\displaystyle OE=\frac{8\cdot15}{17}=\frac{120}{17}

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