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Difference between revisions of "2007 AIME II Problems/Problem 15"

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== Solution ==
 
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[[Image:2007 AIME II-15.png]]
 
=== Solution 1 ===
 
=== Solution 1 ===
 
First, apply [[Heron's formula]] to find that the area is <math>\sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84</math>. Also the semiperimeter is <math>21</math>. So the [[inradius]] is <math>\frac{A}{s} = \frac{84}{21} = 4</math>.  
 
First, apply [[Heron's formula]] to find that the area is <math>\sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84</math>. Also the semiperimeter is <math>21</math>. So the [[inradius]] is <math>\frac{A}{s} = \frac{84}{21} = 4</math>.  

Revision as of 09:33, 5 April 2007

Problem

Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$, $\omega_{B}$ to $BC$ and $BA$, $\omega_{C}$ to $CA$ and $CB$, and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$. If the sides of triangle $ABC$ are $13,$ $14,$ and $15,$ the radius of $\omega$ can be represented in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution

2007 AIME II-15.png

Solution 1

First, apply Heron's formula to find that the area is $\sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84$. Also the semiperimeter is $21$. So the inradius is $\frac{A}{s} = \frac{84}{21} = 4$.

Now consider the incenter I. Let the radius of one of the small circles be $r$. Let the centers of the three little circles tangent to the sides of $\triangle ABC$ be $X$, $Y$, and $Z$. Let the centre of the circle tangent to those three circles be P. A homothety centered at $I$ takes $XYZ$ to $ABC$ with factor $\frac{4 - r}{4}$. The same homothety takes $P$ to the circumcentre of $\triangle ABC$, so $\frac{PX}R = \frac{2r}R = \frac{4 - r}4$, where $R$ is the circumradius of $\triangle ABC$. The circumradius of $\triangle ABC$ can be easily computed by $R = \frac a{2\sin A}$, so doing that reveals $R = \frac{65}8$. Then $\frac{2r}{\frac{65}{8}} = \frac{(4-r)}4 \Longrightarrow \frac{16r}{65} = \frac{1 - r}4 \Longrightarrow \frac{129r}{260} = 1 \Longrightarrow r = \frac{260}{129}$, so the answer is $389$.

Solution 2

2007 AIME II-15b.gif

Consider a 13-14-15 triangle. $A=84.$ [By Heron's Formula or by 5-12-13 and 9-12-15 right triangles.]

The inradius is $r=\frac{A}{s}=\frac{84}{21}=4$, where $s$ is the semiperimeter. Scale the triangle with the inradius by a linear scale factor, $u.$

The circumradius is $R=\frac{abc}{4rs}=\frac{13\cdot 14\cdot 15}{4\cdot 4\cdot 21}=\frac{65}{8},$ where $a,$ $b,$ and $c$ are the side-lengths. Scale the triangle with the circumradius by a linear scale factor, $v$.

Cut and combine the triangles, as shown. Then solve for 4u:

$\frac{65}{8}v=8u$
$v=\frac{64}{65}u$
$\displaystyle u+v=1$
$u+\frac{64}{65}u=1$
$\frac{129}{65}u=1$
$4u=\frac{260}{129}$

The solution is $260+129=389$.

See also

2007 AIME II (Problems • Answer Key • Resources)
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