2017 AMC 10 B

We note that by the converse of the Pythagorean Theorem, is a right triangle with a right angle at . Therefore,, and . Since , the inradius of is , and the inradius of is . Adding the two together, we have .

We have a right triangle by dividing each side lengths by to create a well known triangle. We also can know that the median of a right triangle must be equal to half the hypotenuse. Using this property, we have ,, and . Now, we can use the Heron's formula to get the area of as . Afterward, we can apply this formula again on to get the area as . Notice we want the inradius. We can use another property, which is . This states that . (This can be proved by connecting the center of the inscribed circles to the vertices and we can notice the inradius is just the heights of each of the three triangles divided) Finally, we can derive the radii of each inscribed circle. Plugging the semiperimeter and area into the formula, we have and for and , respectively. Simplifying, we have the radii lengths as and . We want the sum, so we have , or .