2017 AMC 8 第 22 题

Question

在直角三角形

ABC

中，

AC=12

、

BC=5

，且角

C

是直角。如图，一个半圆内切于该三角形。这个半圆的半径是多少？

In the right triangle

ABC,

AC=12,

BC=5,

and angle

C

is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?

\dfrac{7}{6}

\dfrac{13}{5}

\dfrac{59}{18}

\dfrac{10}{3}

\dfrac{60}{13}

答案：D

知识点：切线相似

难度评级：1640

小提示：

画出到半圆与斜边相切点的半径

Draw the radius to the point where the semicircle touches the hypotenuse.

大提示：

使用同一点引出的切线段相等，再建立相似三角形

Use tangent lengths from the same external point, then set up similar triangles.

视频讲解：

Click to load, then click again to play

文字解答：

设

O

为内切半圆圆心，

D

为半圆与

\overline{AB}

的切点。于是

BD = 5

，因为

\overline{BD}

和

\overline{BC}

都与半圆相切。由此得

AD = 8

，并记

OD = r

。因为

\overline{OD}

垂直于

\overline{AB}

，所以

\triangle ADO \sim \triangle BCA

，有

\dfrac{r}{8} = \dfrac{5}{12}

，解得

r = \dfrac{10}{3}

。

所以正确答案是 D。

Let

O

be the center of the inscribed semicircle and

D

be the tangent point of the semicircle on

\overline{AB}.

Then

BD = 5

since

\overline{BD}

and

\overline{BC}

are tangents to the semicircle. Then

AD = 8

and

OD = r.

\overline{OD}

is perpendicular to

\overline{AB}

so

\triangle ADO \sim \triangle BCA,

so

\dfrac{r}{8} = \dfrac{5}{12}.

Solving this, we get

r = \dfrac{10}{3}.

Thus, D is the correct answer.