1955 AMC 12 Problem 9

Attempt Problem 9 of the 1955 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1955 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

A circle is inscribed in a triangle with sides

8,

15,

and

17.

The radius of the circle is:

6

2

5

3

7

The side lengths form a Pythagorean triple

For a right triangle,

r=\frac{a+b-c}{2}

Because

8^2+15^2=17^2,

the triangle is right. Its inradius is

r=\frac{8+15-17}{2}=3.

Thus, the correct answer is D.

Problem 9 in other years