AMC8 2017

AMC8 2017 · Q22

AMC8 2017 · Q22. It mainly tests Triangles (properties), Pythagorean theorem.

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，∠C 为直角。如图所示，一个半圆内接于三角形中。求该半圆的半径。

(A) $\dfrac{7}{6}$ $\dfrac{7}{6}$

(B) $\dfrac{13}{5}$ $\dfrac{13}{5}$

(D) $\dfrac{10}{3}$ $\dfrac{10}{3}$

(E) $\dfrac{60}{13}$ $\dfrac{60}{13}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Let O be the center of the inscribed semicircle on AC, and let D be the point at which AB is tangent to the semicircle. Because OD is a radius of the semicircle it is perpendicular to AB, making OD an altitude of $\triangle AOB$. By the Pythagorean Theorem, $AB = 13$. In the diagram, OB partitions $\triangle ABC$ so that Area($\triangle ABC$) = Area($\triangle BOC$) + Area($\triangle AOB$) Since we know $\triangle ABC$ has area 30, we have \[ 30 = \text{Area}(\triangle BOC) + \text{Area}(\triangle AOB) = \tfrac{1}{2}(BC)r + \tfrac{1}{2}(AB)r = \tfrac{5}{2}r + \tfrac{13}{2}r = 9r. \] Therefore $r = \tfrac{30}{9} = \tfrac{10}{3}$.

答案 (D)：设 O 为在 AC 上所内切半圆的圆心，设 D 为 AB 与该半圆的切点。由于 OD 是半圆的半径，故其垂直于 AB，从而 OD 是 $\triangle AOB$ 的高。由勾股定理，$AB = 13$。在图中，OB 将 $\triangle ABC$ 分割，使得 Area($\triangle ABC$) = Area($\triangle BOC$) + Area($\triangle AOB$) 由于已知 $\triangle ABC$ 的面积为 30，于是 \[ 30 = \text{Area}(\triangle BOC) + \text{Area}(\triangle AOB) = \tfrac{1}{2}(BC)r + \tfrac{1}{2}(AB)r = \tfrac{5}{2}r + \tfrac{13}{2}r = 9r. \] 因此 $r = \tfrac{30}{9} = \tfrac{10}{3}$。

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