2004 AMC 10B Problem 22

Question

A triangle with sides of

5,

12,

and

13

has both an inscribed and a circumscribed circle. What is the distance between the centers of those circles?

\dfrac{3\sqrt{5}}{2}

\dfrac{7}{2}

\sqrt{15}

\dfrac{\sqrt{65}}{2}

\dfrac{9}{2}

Place the right triangle with vertices

(0, 0),

(5, 0),

and

(0, 12)

The circumcenter is the midpoint of the hypotenuse; a right triangle’s inradius is

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

Since

5^2 + 12^2 = 13^2,

the triangle is right. Place it at

(0, 0),

(5, 0),

(0, 12).

The circumcenter is the midpoint of the hypotenuse,

\left(\tfrac52, 6\right).

The inradius satisfies

(12 - r) + (5 - r) = 13,

so

r = 2

and the incenter is

(2, 2).

The distance is

\begin{gathered} \sqrt{\left(\tfrac52 - 2\right)^2 + (6 - 2)^2} \\ = \sqrt{\tfrac14 + 16} \\ = \dfrac{\sqrt{65}}{2}. \end{gathered}

Thus, the correct answer is D.

Problem 22 in other years