2004 AMC 10B Problem 22

1. Spot the right triangle. $5^2 + 12^2 = 25 + 144 = 169 = 13^2$, so the triangle has a right angle between the sides $5$ and $12$.
2. Put it on a grid. A right angle fits neatly in a corner of the axes. Put the right angle at $(0,0)$, and the other corners at $(12, 0)$ and $(0, 5)$. The long side ($13$) joins $(12,0)$ and $(0,5)$.
3. Center of the outer (circumscribed) circle. In a right triangle, the longest side is a diameter of the outer circle, so the center is the middle of that side:
$$\left(\frac{12 + 0}{2}, \frac{0 + 5}{2}\right) = (6,\ 2.5).$$
Check: this point is $6.5$ from each corner, which is half of $13$.
4. Radius of the inner (inscribed) circle. Call it $r$. Join the inner center to the three corners. That cuts the triangle into three smaller triangles, each with a side of the big triangle as its base and height $r$. So
$$\text{area} = \tfrac{1}{2} r \cdot 5 + \tfrac{1}{2} r \cdot 12 + \tfrac{1}{2} r \cdot 13 = \tfrac{1}{2} r \cdot 30 = 15r.$$
The area is also $\frac{1}{2} \cdot 5 \cdot 12 = 30$, so $15r = 30$ and $r = 2$.
5. Center of the inner circle. It is $r = 2$ away from each side. Two of the sides lie on the axes, so the center is $2$ right of the $y$-axis and $2$ above the $x$-axis: $(2, 2)$.
6. Distance between the centers. From $(2,2)$ to $(6, 2.5)$ is $4$ across and $0.5$ up:
$$\sqrt{4^2 + 0.5^2} = \sqrt{16 + 0.25} = \sqrt{\tfrac{65}{4}} = \frac{\sqrt{65}}{2}.$$

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

A right triangle makes both centers easy to find. The outer center is always the middle of the longest side, and the inner center sits at $(r, r)$ when the two short sides lie on the axes. After that, the question is just the distance formula. Spotting a Pythagorean triple like $5, 12, 13$ is the signal to set things up this way.

There is a formula (Euler's) for the distance $d$ between the two centers: $d^2 = R(R - 2r)$, where $R$ is the outer radius and $r$ the inner radius. Here $R = 6.5$ (half of $13$) and $r = 2$, so $d^2 = 6.5 \times 2.5 = 16.25 = \frac{65}{4}$ and $d = \frac{\sqrt{65}}{2}$. This is quick if you know the formula, but the grid method needs nothing beyond school geometry.