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

Solution 4

A triangle with sides $5,12,$ and $13$ must be right. Let the right angle be at $A$ so that $AB=5,AC=12,$ and $BC=13$ . The circumcenter $O$ must be the midpoint of $BC$ , so that $BO=CO=\frac{13}{2}.$ Let $I$ be the incenter and $D$ be the point where $BC$ is tangent to the incircle. Since $ID \perp DO, \triangle IDO$ is a right triangle. Therefore, to find $IO$ , it suffices to find $ID$ and $DO$ . The area of triangle $ABC$ is equal to the semiperimeter times the inradius, $r$ . This allows us to set up an equation involving $r$ : $\frac{5 \cdot 12}{2}= r \cdot \frac{5+12+13}{2}.$ Solving, we get $r=2$ . Now it only remains to find $DO$ . To start, note that $BD=s-b$ , where $s$ is the semiperimeter and $b$ is the length of $AC$ . Simplifing, we get $BD=3<\frac{13}{2}$ , so $D$ is on the same side of $O$ as $B$ , and $DO=BO-BD=\frac{13}{2}-3=\frac{7}{2}.$ Therefore, $IO=\sqrt{(\frac{7}{2})^2+2^2}=\sqrt{\frac{65}{4}}=\boxed{(D)\frac{\sqrt{65}}{2}}.$

2004 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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