2017 AMC 10B Problem 21

The triangle $ABC$ is a right triangle with a right angle at $A.$ This makes $D$ the circumcenter of the triangle since it is the midpoint of the hypotenuse. Therefore, $$AD = BD = DC = 5.$$ Also, the area of $ABC$ is $$\dfrac{6\cdot 8}2 = 24.$$ The bases $BD$ and $DC$ are equal, and the two triangles share the same altitude from $A.$ Therefore, $\triangle ABD$ and $\triangle ACD$ each have area $12.$ Then, for each triangle, we have $A = rs$ where $A$ is the area, $r$ is the inradius, and $s$ is the semiperimeter. Equivalently, $12=\dfrac12rP,$ where $P$ is the perimeter, so $r=\dfrac{24}{P}.$ For $\triangle ABD,$ the inradius is $\dfrac{24}{5+5+6}=\dfrac32.$ For $\triangle ACD,$ it is $\dfrac{24}{5+5+8}=\dfrac43.$ Their sum is $\dfrac 32 + \dfrac 43 = \dfrac{17}6 .$ Thus, the correct answer is D .