2013 AMC 10B Problems/Problem 23

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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2013 AMC 10B Problems/Problem 23

Problem

Solution