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?
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,
Solution 3
Construct such that , , and . Since this is a pythagorean triple, . By a property of circumcircles and right triangles, the circumcenter, , lies on the midpoint of , so . Turning to the incircle, we find that the inradius is , using the formula , where is the area of the triangle, is the inradius, and is the semiperimeter. We then denote the incenter , along with the points of tangency , , and . Because by a property of tangency, , and so is a square. Then, since , . As , , and because by HL, . Therefore, . Because , pythagorean theorem gives
Solution 4
A triangle with sides and must be right. Let the right angle be at so that and . The circumcenter must be the midpoint of , so that Let be the incenter and be the point where is tangent to the incircle. Since is a right triangle. Therefore, to find , it suffices to find and . The area of triangle is equal to the semiperimeter times the inradius, . This allows us to set up an equation involving : Solving, we get . Now it only remains to find . To start, note that , where is the semiperimeter and is the length of . Simplifing, we get , so is on the same side of as , and Therefore,
See also
2004 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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