2004 AMC 10B Problems/Problem 22: Difference between revisions
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<math> \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} </math> | <math> \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} </math> | ||
==Solution== | ==Solution 1== | ||
This is obviously a right triangle. Pick a coordinate system so that the right angle is at <math>(0,0)</math> and the other two vertices are at <math>(12,0)</math> and <math>(0,5)</math>. | This is obviously a right triangle. Pick a coordinate system so that the right angle is at <math>(0,0)</math> and the other two vertices are at <math>(12,0)</math> and <math>(0,5)</math>. | ||
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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>. | ||
==Solution 2== | |||
We directly apply Euler’s Theorem, which states that if the circumcenter is <math>O</math> and the incenter <math>I</math>, and the inradius is <math>r</math> and the circumradius is <math>R</math>, then | |||
<cmath>OI^2=R(R-2r)</cmath> | |||
We notice that this is a right triangle, and hence has area <math>30</math>. We then find the inradius with the formula <math>A=rs</math>, where <math>s</math> denotes semiperimeter. We easily see that <math>s=15</math>, so <math>r=2</math>. | |||
We now find the circumradius with the formula <math>A=\frac{abc}{4R}</math>. Solving for <math>R</math> gives <math>R=\frac{13}{2}</math>. | |||
Substituting all of this back into our formula gives: | |||
\begin{align*} | |||
OI^2=R(R-2r) &= \frac{13}{2}\left(\frac{13}{2}-4\right) \\ | |||
&= \frac{13}{2}\cdot \frac{5}{2} \\ | |||
&= \frac{65}{2} | |||
\end{align*} | |||
So, <math>OI=\frac{\sqrt{65}}{2}\implies \boxed{D}</math> | |||
== See also == | == See also == | ||
Revision as of 01:02, 10 April 2018
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 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 
.
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
![\[OI^2=R(R-2r)\]](http://latex.artofproblemsolving.com/1/e/e/1ee9dc3b52bdd0ea40931dc0ca792540fcb15ebc.png)
We notice 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:
\begin{align*}
OI^2=R(R-2r) &= \frac{13}{2}\left(\frac{13}{2}-4\right) \\
&= \frac{13}{2}\cdot \frac{5}{2} \\
&= \frac{65}{2}
\end{align*}
So, 
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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