2004 AMC 10B Problems/Problem 22


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 $R=\frac{13}{2}$. Additionally, we may notice that the side lengths $(5, 12, 13)$ are in a Pythagorean triple, and therefore the triangle is right for a circumradius of $\frac{13}{2}$.

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 $IG=\boxed{\frac{\sqrt{65}}{2}}$

See also

2004 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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