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2004 AMC 10B Problems/Problem 22

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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) \ ...

Solution

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=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}.

See also

2004 AMC 10B (ProblemsResources)
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
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