2013 AMC 12B Problems/Problem 19

The following problem is from both the 2013 AMC 12B #19 and 2013 AMC 10B #23, so both problems redirect to this page.

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 1

Since $\angle{AFB}=\angle{ADB}=90^{\circ}$ , quadrilateral $ABDF$ is cyclic. It follows that $\angle{ADE}=\angle{ABF}$ . In addition, since $\angle{AFB}=\angle{AED}=90$ , triangles $ABF$ and $AFE$ 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 $DF=\frac{16}{5}\implies{16+5=21}\implies{\boxed{(B)}}$

Solution 2

Using the similar triangles in triangle $ADC$ gives $AE = \frac{48}{5}$ and $DE = \frac{36}{5}$ . Quadrilateral $ABDF$ is cyclic, implying that $\angle{B} + \angle{DFA}$ = 180°. Therefore, $\angle{B} = \angle{EFA}$ , and triangles $AEF$ and $ADB$ are similar. Solving the resulting proportion gives $EF = 4$ . Therefore, $DF = ED - EF = \frac{16}{5}. \implies{\boxed{(B)}}$

Solution 3

If we draw a diagram as given, but then add $DG$ as an altitude to use the Pythagorean theorem, we end up with similar triangles $\triangle{DFG}$ and $\triangle{DCE}$ . Thus, $FG$ is $3/5$ x and $DG$ is $4/5$ x. Using Pythagorean theorem, we now get

$BF = \sqrt{(4/5x + 5)^2 + (3/5x)^2}$

2013 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

2013 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2013 AMC 12B Problems/Problem 19

Contents

Problem

Solution 1

Solution 2

Solution 3

See also