AMC10 2017 B

AMC10 2017 B · Q21

AMC10 2017 B · Q21. It mainly tests Triangles (properties), Pythagorean theorem.

In △ABC, AB = 6, AC = 8, BC = 10, and D is the midpoint of BC. What is the sum of the radii of the circles inscribed in △ADB and △ADC ?

在△ABC中，AB=6，AC=8，BC=10，D是BC的中点。△ADB和△ADC的内切圆半径之和是多少？

(A) \sqrt{5} \sqrt{5}

(B) \dfrac{11}{4} \dfrac{11}{4}

(D) \dfrac{17}{6} \dfrac{17}{6}

(E) 3 3

Answer

Correct choice: (D)

正确答案：（D）

Solution

By the converse of the Pythagorean Theorem, ∠BAC is a right angle, so BD = CD = AD = 5, and the area of each of the small triangles is 1/2 (half the area of △ABC). The area of △ABD is equal to its semiperimeter, 1/2 · (5 + 5 + 6) = 8, multiplied by the radius of the inscribed circle, so the radius is 1/2 ÷ 8 = 3/2. Similarly, the radius of the inscribed circle of △ACD is 4/3. The requested sum is 3/2 + 4/3 = 17/6.

由勾股定理的逆命题，∠BAC是直角，因此BD=CD=AD=5，每个小三角形的面积是△ABC面积的一半。△ABD的面积等于其半周长1/2·(5+5+6)=8乘以内切圆半径，因此半径是1/2÷8=3/2。类似地，△ACD的内切圆半径是4/3。所求和是3/2+4/3=17/6。

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