Difference between revisions of "2004 AMC 10B Problems/Problem 22"
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As this is a right triangle, the center of the circumcircle is in the middle of the hypotenuse, at <math>(6,2.5)</math>. | As this is a right triangle, the center of the circumcircle is in the middle of the hypotenuse, at <math>(6,2.5)</math>. | ||
− | The radius <math>r</math> of the inscribed circle can be computed using the well-known identity <math>\frac{rP}2=S</math>, where <math>S</math> is the area of the triangle and <math>P</math> its perimeter. In our case, <math>S=5\cdot 12 | + | The radius <math>r</math> of the inscribed circle can be computed using the well-known identity <math>\frac{rP}2=S</math>, where <math>S</math> is the area of the triangle and <math>P</math> its perimeter. In our case, <math>S=\frac{5\cdot 12}{2}=30</math> and <math>P=5+12+13=30</math>. Thus, <math>r=2</math>. As the inscribed circle touches both legs, its center must be at <math>(r,r)=(2,2)</math>. |
The distance of these two points is then <math>\sqrt{ (6-2)^2 + (2.5-2)^2 } = \sqrt{16.25} = \sqrt{\frac{65}4} = \boxed{\frac{\sqrt{65}}2}</math>. | The distance of these two points is then <math>\sqrt{ (6-2)^2 + (2.5-2)^2 } = \sqrt{16.25} = \sqrt{\frac{65}4} = \boxed{\frac{\sqrt{65}}2}</math>. |
Revision as of 16:11, 7 January 2017
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
This is obviously 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 .
See also
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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