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Mock AIME 1 2006-2007 Problems/Problem 10: Difference between revisions

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==Solution==
==Solution==
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Given <math> \frac{r}{a}+\frac{r}{b}+\frac{r}{c}=\frac{m}{n} =\frac{20}{3}</math>, so <math>m+n=23</math>.
Given <math> \frac{r}{a}+\frac{r}{b}+\frac{r}{c}=\frac{m}{n} =\frac{20}{3}</math>, so <math>m+n=23</math>.
[[File:C:\Documents and Settings\Eigenaar\Mijn documenten\Documenten van Valère\Mock06-07,10.PNG]]


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Latest revision as of 22:58, 24 April 2013

Problem

In $\triangle ABC$, $AB$, $BC$, and $CA$ have lengths $3$, $4$, and $5$, respectively. Let the incircle, circle $I$, of $\triangle ABC$ touch $AB$, $BC$, and $CA$ at $C'$, $A'$, and $B'$, respectively. Construct three circles, $A''$, $B''$, and $C''$, externally tangent to the other two and circles $A''$, $B''$, and $C''$ are internally tangent to the circle $I$ at $A'$, $B'$, and $C'$, respectively. Let circles $A''$, $B''$, $C''$, and $I$ have radii $a$, $b$, $c$, and $r$, respectively. If $\frac{r}{a}+\frac{r}{b}+\frac{r}{c}=\frac{m}{n}$ where $m$ and $n$ are positive integers, find $m+n$.

Solution

Radius $a=\frac{3}{7}$, radius $b=\frac{6}{11}$, radius $c=\frac{2}{5}$ and $r=1$, see picture.

Given $\frac{r}{a}+\frac{r}{b}+\frac{r}{c}=\frac{m}{n} =\frac{20}{3}$, so $m+n=23$.

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