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Difference between revisions of "2013 AMC 10B Problems/Problem 23"

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Since <math>\angle{AFB}=\angle{ADB}=90</math>, quadrilateral <math>ABDF</math> is cyclic. It follows that <math>\angle{ADE}=\angle{ABF}</math>. In addition, since <math>\angle{AFB}=\angle{AED}=90</math>, triangles <math>ABF</math> and <math>ADE</math> are similar. It follows that <math>AF=(13)(\frac{4}{5}), BF=(13)(\frac{3}{5})</math>. By Ptolemy, we have <math>13DF+(5)(13)(\frac{4}{5})=(12)(13)(\frac{3}{5})</math>. Cancelling <math>13</math>, the rest is easy. We obtain <math>DF=\frac{16}{5}\implies{16+5=21}\implies{\boxed{\textbf{(B)}  21}</math>

Revision as of 16:07, 22 February 2013

Problem

In triangle $ABC$, $AB = 13$, $BC = 14$, and $CA = 15$. Distinct points $D$, $E$, and $F$ lie on segments $\overline{BC}$, $\overline{CA}$, and $\overline{DE}$, respectively, such that $\overline{AD} \perp \overline{BC}$, $\overline{DE} \perp \overline{AC}$, and $\overline{AF} \perp \overline{BF}$. The length of segment $\overline{DF}$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m + n$?


$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 30$

Solution

Since $\angle{AFB}=\angle{ADB}=90$, quadrilateral $ABDF$ is cyclic. It follows that $\angle{ADE}=\angle{ABF}$. In addition, since $\angle{AFB}=\angle{AED}=90$, triangles $ABF$ and $ADE$ are similar. It follows that $AF=(13)(\frac{4}{5}), BF=(13)(\frac{3}{5})$. By Ptolemy, we have $13DF+(5)(13)(\frac{4}{5})=(12)(13)(\frac{3}{5})$. Cancelling $13$, the rest is easy. We obtain $DF=\frac{16}{5}\implies{16+5=21}\implies{\boxed{\textbf{(B)} 21}$ (Error compiling LaTeX. Unknown error_msg)

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