2004 AMC 10B Problems/Problem 22: Difference between revisions

Revision as of 20:25, 31 July 2018

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 obviously 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 notice 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, $OI=\frac{\sqrt{65}}{2}\implies \boxed{D}$

2004 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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2004 AMC 10B Problems/Problem 22: Difference between revisions

Revision as of 20:25, 31 July 2018

Contents

Problem

Solution 1

Solution 2

See also

@@ Line 6: / Line 6: @@
 ==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 obviously a right triangle. Pick a coordinate system so that the right angle is at <math>(0,0)</math> and the other two vertices are at <math>(12,0)</math> and <math>(0,5)</math>.