2012 AMC 12A Problems/Problem 18

Problem

Triangle $ABC$ has $AB=27$ , $AC=26$ , and $BC=25$ . Let $I$ denote the intersection of the internal angle bisectors of $\triangle ABC$ . What is $BI$ ?

$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 5+\sqrt{26}+3\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{26}\qquad\textbf{(D)}\ \frac{2}{3}\sqrt{546}\qquad\textbf{(E)}\ 9\sqrt{3}$

Solution

Inscribe circle $C$ of radius $r$ inside triangle $ABC$ so that it meets $AB$ at $Q$ , $BC$ at $R$ , and $AC$ at $S$ . Note that angle bisectors of triangle $ABC$ are concurrent at the center $O$ of circle $C$ . Let $x=QB$ , $y=RC$ and $z=AS$ . Note that $BR=x$ , $SC=y$ and $AQ=z$ . Hence $x+z=27$ , $x+y=25$ , and $z+y=26$ . Subtracting the last 2 equations we have $x-z=-1$ and adding this to the first equation we have $x=13$ .

By Herons formula for the area of a triangle we have that the area of triangle $ABC$ is $\sqrt{39(14)(13)(12)}$ . On the other hand the area is given by $(1/2)25r+(1/2)26r+(1/2)27r$ . Then $39r=\sqrt{39(14)(13)(12)}$ so that $r^2=56$ .

Since the radius of circle $O$ is perpendicular to $BC$ at $R$ , we have by the pythagorean theorem $BO^2=BI^2=r^2+x^2=56+169=225$ so that $BI=15$ .

2012 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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2012 AMC 12A Problems/Problem 18

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