AMC 8 2017 Problem 22

AMC8 2017 年第 22 题

三角形★★★☆☆

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?

在直角三角形 ABC 中，AC=12，BC=5，且角 C 是个直角。一个半圆如下图所示内切于三角形中，那么这个半圆的半径是多少？

A. $\displaystyle \frac{7}{6}$
B. $\displaystyle \frac{13}{5}$
C. $\displaystyle \frac{59}{18}$
D. $\displaystyle \frac{10}{3}$
E. $\displaystyle \frac{60}{13}$

正确答案 · Correct Answer

解析 · Solution

Solution 1

Let 𝑂 O be the center of the inscribed semicircle on 𝐴 𝐶 ‾ AC, and let 𝐷 D be the point at which 𝐴 𝐵 ‾ AB is tangent to the semicircle. Because 𝑂 𝐷 ‾ OD is a radius of the semicircle it is perpendicular to 𝐴 𝐵 ‾ AB, making 𝑂 𝐷 ‾ OD an altitude of △ 𝐴 𝑂 𝐵 △AOB. By the Pythagorean Theorem, 𝐴 𝐵 = 13 AB=13. In the diagram, 𝑂 𝐵 ‾ OB partitions △ 𝐴 𝐵 𝐶 △ABC so that Area ⁡ ( △ 𝐴 𝐵 𝐶 ) = Area ⁡ ( △ 𝐵 𝑂 𝐶 ) + Area ⁡ ( △ 𝐴 𝑂 𝐵 ) Area(△ABC)=Area(△BOC)+Area(△AOB) Since we know △ 𝐴 𝐵 𝐶 △ABC has area 30 30, we have 30 = Area ⁡ ( △ 𝐵 𝑂 𝐶 ) + Area ⁡ ( △ 𝐴 𝑂 𝐵 ) = 1 2 ( 𝐵 𝐶 ) 𝑟 + 1 2 ( 𝐴 𝐵 ) 𝑟 = 5 2 𝑟 + 13 2 𝑟 = 9 𝑟. 30 =Area(△BOC)+Area(△AOB) = 2 1 (BC)r+ 2 1 (AB)r= 2 5 r+ 2 13 r=9r. Therefore 𝑟 = 30 9 = 10 3 r= 9 30 = 3 10 . OR OR Because 𝑂 𝐷 ‾ OD is a radius of the semicircle, it is perpendicular to 𝐴 𝐵 ‾ AB, making △ 𝐴 𝐷 𝑂 △ADO similar to △ 𝐴 𝐶 𝐵 △ACB. Because 𝐵 𝐶 ‾ BC and 𝐵 𝐷 ‾ BD are both tangent to the semicircle, they are congruent. So 𝐵 𝐷 = 5 BD=5 and 𝐴 𝐷 = 8 AD=8. It follows that 𝑟 8 = 5 12 8 r = 12 5 and so 𝑟 = 40 12 = 10 3 r= 12 40 = 3 10 . Answer: 𝐷 D .

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