2017 AMC 10B Problem 21

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

First recognize the

6,8,10

triangle as right

Use area equals inradius times semiperimeter for the two smaller triangles

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 .

Problem 21 in other years