2017 AMC 10B Problems/Problem 21

Problem

In $\triangle ABC$ , $AB=6$ , $AC=8$ , $BC=10$ , and $D$ is the midpoint of $\overline{BC}$ . What is the sum of the radii of the circles inscribed in $\triangle ADB$ and $\triangle ADC$ ?

$\textbf{(A)}\ \sqrt{5}\qquad\textbf{(B)}\ \frac{11}{4}\qquad\textbf{(C)}\ 2\sqrt{2}\qquad\textbf{(D)}\ \frac{17}{6}\qquad\textbf{(E)}\ 3$

Solution

We note that by the converse of the Pythagorean Theorem, $\triangle ABC$ is a right triangle with a right angle at $A$ . Therefore, $AD = BD = CD = 5$ , and $[ADB] = [ADC] = 12$ . Since $A = rs$ , the inradius of $\triangle ADB$ is $\frac{12}{(5+5+6)/2} = \frac 32$ , and the inradius of $\triangle ADC$ is $\frac{12}{(5+5+8)/2} = \frac 43$ . Adding the two together, we have $\boxed{\textbf{(D) } \frac{17}6}$ .

edit by asdf334: what do you mean by A = rs?

edit by Lej: Area of Incircle = (inradius)(Semiperimeter)

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2017 AMC 10B Problems/Problem 21

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