# 2004 AMC 10B Problems/Problem 22

## Problem

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? $\mathrm{(A) \ } \frac{3\sqrt{5}}{2} \qquad \mathrm{(B) \ } \frac{7}{2} \qquad \mathrm{(C) \ } \sqrt{15} \qquad \mathrm{(D) \ } \frac{\sqrt{65}}{2} \qquad \mathrm{(E) \ } \frac{9}{2}$

## Solution 1 $[asy] import geometry; unitsize(0.6 cm); pair A, B, C, D, E, F, I, O; A = (5^2/13,5*12/13); B = (0,0); C = (13,0); I = incenter(A,B,C); D = (I + reflect(B,C)*(I))/2; E = (I + reflect(C,A)*(I))/2; F = (I + reflect(A,B)*(I))/2; O = (B + C)/2; draw(A--B--C--cycle); draw(incircle(A,B,C)); draw(I--D); draw(I--E); draw(I--F); draw(I--O); label("A", A, N); label("B", B, SW); label("C", C, SE); dot("D", D, S); dot("E", E, NE); dot("F", F, NW); dot("I", I, N); dot("O", O, S); [/asy]$ This is a right triangle. Pick a coordinate system so that the right angle is at $(0,0)$ and the other two vertices are at $(12,0)$ and $(0,5)$.

As this is a right triangle, the center of the circumcircle is in the middle of the hypotenuse, at $(6,2.5)$.

The radius $r$ of the inscribed circle can be computed using the well-known identity $\frac{rP}2=S$, where $S$ is the area of the triangle and $P$ its perimeter. In our case, $S=\frac{5\cdot 12}{2}=30$ and $P=5+12+13=30$. Thus, $r=2$. As the inscribed circle touches both legs, its center must be at $(r,r)=(2,2)$.

The distance of these two points is then $\sqrt{ (6-2)^2 + (2.5-2)^2 } = \sqrt{16.25} = \sqrt{\frac{65}4} = \boxed{\frac{\sqrt{65}}2}$.

## Solution 2

We directly apply Euler’s Theorem, which states that if the circumcenter is $O$ and the incenter $I$, and the inradius is $r$ and the circumradius is $R$, then $$OI^2=R(R-2r)$$

We can see that this is a right triangle, and hence has area $30$. We then find the inradius with the formula $A=rs$, where $s$ denotes semiperimeter. We easily see that $s=15$, so $r=2$.

We now find the circumradius with the formula $A=\frac{abc}{4R}$. Solving for $R$ gives $R=\frac{13}{2}$.

Substituting all of this back into our formula gives: $$OI^2= \frac{65}{4}$$ So, $OI=\frac{\sqrt{65}}{2}\implies \boxed{D}$

## Solution 3 $[asy] size(15cm); draw((0,0)--(0,5), linewidth(2)); draw((0,0)--(12,0), linewidth(2)); draw((12,0)--(0,5), linewidth(2)); draw((2,0)--(2,2), linewidth(2)); draw((2,2)--(2.770565628817799,3.8455976546592505), linewidth(2)); draw((2,2)--(6.023716614191289,2.4901180774202962), linewidth(2)); draw((2,2)--(0,5), linewidth(2)); draw((2,2)--(12,0), linewidth(2)); draw((0,2)--(2,2), linewidth(2)); label("A", (0.14164244785738467,0.25966489738837517), NE); label("B", (0.14164244785738467,5.311734560831129), NE); label("C", (12.120449134133493,0.3232129434694161), NE); label("D", (0.14164244785738467,2.324976395022205), NE); label("E", (2.111631876369636,0.3232129434694161), NE); label("F", (2.9059824523826405,4.167869731372392), NE); label("G", (6.146932802515699,2.801586740630012), NE); label("I", (2.1, 2.1), NE); [/asy]$ Construct $\triangle{ABC}$ such that $AB=5$, $AC=12$, and $BC=13$. Since this is a pythagorean triple, $\angle{A}=90$. By a property of circumcircles and right triangles, the circumcenter, $G$, lies on the midpoint of $\overline{BC}$, so $BG=\frac{13}{2}$. Turning to the incircle, we find that the inradius is $2$, using the formula $A=rs$, where $A$ is the area of the triangle, $r$ is the inradius, and $s$ is the semiperimeter. We then denote the incenter $I$, along with the points of tangency $D$, $E$, and $F$. Because $\angle{IDA}=\angle{IEA}=90$ by a property of tangency, $\angle{EID}=90$, and so $IDAE$ is a square. Then, since $IE=2$, $AD=2$. As $AB=5$, $BD=3$, and because $\triangle{BID}\cong\triangle{BIF}$ by HL, $BD=BF=3$. Therefore, $FG=\frac{7}{2}$. Because $IF=2$, pythagorean theorem gives $IG=\boxed{\frac{\sqrt{65}}{2}}$

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