Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2004 AMC 10B Problems/Problem 22"

m (→‎Solution)
(→‎Solution)
Line 4: Line 4:
 
<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>.
Line 13: Line 13:
  
 
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 02:02, 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?

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

Solution 1

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

Solution 2

We directly apply Euler’s Theorem, which states that if the circumcenter is $O$ and the incenter $I$, and the inradius is $r$ and the circumradius is $R$, then \[OI^2=R(R-2r)\]

We notice that this is a right triangle, and hence has area $30$. We then find the inradius with the formula $A=rs$, where $s$ denotes semiperimeter. We easily see that $s=15$, so $r=2$.

We now find the circumradius with the formula $A=\frac{abc}{4R}$. Solving for $R$ gives $R=\frac{13}{2}$.

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, $OI=\frac{\sqrt{65}}{2}\implies \boxed{D}$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_22&oldid=93877"