2004 AMC 10B Problem 22

Since $5^2 + 12^2 = 13^2$, the triangle is right-angled. Put the right angle at the origin with the legs on the axes: vertices $(0,0)$, $(12,0)$, $(0,5)$.

Circumcenter: for a right triangle the hypotenuse is a diameter of the circumcircle, so the circumcenter is the midpoint of the hypotenuse, $\left(6, \tfrac{5}{2}\right)$.

Incenter: the inradius is $r = \dfrac{\text{area}}{\text{semiperimeter}} = \dfrac{30}{15} = 2$ (equivalently $r = \frac{5 + 12 - 13}{2}$ for a right triangle). The incenter is $r$ away from both legs, which lie on the axes, so it is at $(2, 2)$.

Distance between the centers:
$$\sqrt{(6 - 2)^2 + \left(\tfrac{5}{2} - 2\right)^2} = \sqrt{16 + \tfrac{1}{4}} = \sqrt{\tfrac{65}{4}} = \frac{\sqrt{65}}{2}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{\sqrt{65}}{2}}$.

A right triangle makes both centers easy to locate: the circumcenter by Thales' theorem, the incenter because the two legs are perpendicular tangent lines, so the center is $(r, r)$. Coordinates then reduce "distance between two special points" to arithmetic. Recognizing a Pythagorean triple is the trigger for this whole setup.

Euler's formula for the distance $OI$ between circumcenter and incenter: $OI^2 = R(R - 2r)$. Here $R = \frac{13}{2}$ (half the hypotenuse) and $r = 2$, so $OI^2 = \frac{13}{2}\left(\frac{13}{2} - 4\right) = \frac{13}{2} \cdot \frac{5}{2} = \frac{65}{4}$.

• Placing the circumcenter somewhere other than the midpoint of the hypotenuse, or computing r as area over perimeter (giving 1) instead of area over semiperimeter.
• Putting the incenter at the centroid $(4, \tfrac{5}{3})$, or the circumcenter at the centroid, and getting a distance that matches none of the choices.