2004 AMC 10B Problems/Problem 22
Contents
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?
Solution 1
This is a right triangle. Pick a coordinate system so that the right angle is at and the other two vertices are at and .
As this is a right triangle, the center of the circumcircle is in the middle of the hypotenuse, at .
The radius of the inscribed circle can be computed using the well-known identity , where is the area of the triangle and its perimeter. In our case, and . Thus, . As the inscribed circle touches both legs, its center must be at .
The distance of these two points is then .
Solution 2
We directly apply Euler’s Theorem, which states that if the circumcenter is and the incenter , and the inradius is and the circumradius is , then
We can see that this is a right triangle, and hence has area . We then find the inradius with the formula , where denotes semiperimeter. We easily see that , so .
We now find the circumradius with the formula . Solving for gives .
Substituting all of this back into our formula gives: So,
See also
2004 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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