2013 AMC10B - Problem 23

Geometry Level 4

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 $\overline{AD} \bot\overline{BC}$ , $\overline{DE} \bot\overline{AC}$ , and $\overline{AF} \bot \overline{BF}$ . The length of segment $\overline{DF}$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m + n$ ?

27 21 24 30 18

2 solutions

Hana Wehbi
Aug 11, 2017

$\angle{AFB}=\angle{ADB}=90^{\circ}$ , quadrilateral $ABDF$ is cyclic. It follows that $\angle{ADE}=\angle{ABF}$ .

$\angle{AFB}=\angle{AED}=90\implies$ triangles $ABF \text{ and} ADE$ are similar.

It follows that $AF=(13)(\frac{4}{5}), BF=(13)(\frac{3}{5})$ .

By Ptolemy, we have $13DF+(5)(13)(\frac{4}{5})=(12)(13)(\frac{3}{5})$ . Divide the whole equation by $13$ to obtain

$DF=\frac{16}{5}\implies{16+5= \boxed{21}}.$

Ptolemy Theorem gives a relationship between the side lengths and the diagonals of a cyclic quadrilateral; it is the equality case of Ptolemy's Inequality. Ptolemy's Theorem frequently shows up as an intermediate step in problems involving inscribed figures. The statement is as follows:

Given a cyclic quadrilateral $ABCD$ with side lengths ${a},{b},{c},{d}$ and diagonals ${e},{f}$ :

$ac+bd=ef.$

Niranjan Khanderia
Nov 24, 2017

2013 AMC10B - Problem 23

2 solutions

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