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 12/2=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>.
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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 17:11, 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?

$\mathrm{(A) \ } \frac{3\sqrt{5}}{2} \qquad \mathrm{(B) \ } \frac{7}{2} \qquad \mathrm{(C) \ } \sqrt{15} \qquad \mathrm{(D) \ } \frac{\sqrt{65}}{2} \qquad \mathrm{(E) \ } \frac{9}{2}$

Solution

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 $\sqrt{ (6-2)^2 + (2.5-2)^2 } = \sqrt{16.25} = \sqrt{\frac{65}4} = \boxed{\frac{\sqrt{65}}2}$.

See also

2004 AMC 10B (ProblemsAnswer KeyResources)
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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