Question

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 $\triangle ADC?$

$\sqrt{5}$

$\dfrac{11}{4}$

$2\sqrt{2}$

$\dfrac{17}{6}$

$3$

Answer: D

Concepts:incircle, incenter, and inradius right triangle triangle area

Difficulty rating: 1720

Solution:

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 .

2017 AMC 10B Problem 21

Answer: D

Difficulty rating: 1720

Solution:

Problem 21 in Other Years