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, triangles $ABF$ and $ADE$ are similar, and triangles $ADE$ and $ADC$ are similar. We can easily find $AD=12$ , $BD = 5$ , and $DC=9$ using pythagorean triples. So, the ratio of the hypotenuse to the longer leg of all three similar triangles is $\frac{15}{12} = \frac{4}{5}$ , and the ratio of the hypotenuse to the shorter leg is $\frac{15}{9} = \frac{3}{5}$ . It follows that $AF=(13)(\frac{4}{5}), BF=(13)(\frac{3}{5})$ . By Ptolemy's Theorem, we have $13x+(5)(13)(\frac{4}{5})=(12)(13)(\frac{3}{5})\implies 13x+52=93.6$ where $x=DF$ . Dividing by $13$ we get $x+4=7.2\implies x=\frac{16}{5}$ so our answer is $16+5=\boxed{21\,\textbf{(B)}}$ .

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, $DF = ED - EF = \frac{16}{5}$ and our answer is $\boxed{\textbf{(B)}}$ .

Solution 3

If we draw a diagram as given, but then add point $G$ on $\overline{BC}$ such that $\overline{FG}\perp\overline{BC}$ in order to use the Pythagorean theorem, we end up with similar triangles $\triangle{DFG}$ and $\triangle{DCE}$ . Thus, $FG=\tfrac35x$ and $DG=\tfrac45x$ , where $x$ is the length of $\overline{DF}$ . Using the Pythagorean theorem, we now get $BF = \sqrt{\left(\frac45x+ 5\right)^2 + \left(\frac35x\right)^2}$ and $AF$ can be found out noting that $AE$ is just $\tfrac{48}5$ through base times height (since $12\cdot 9 = 15 \cdot \tfrac{36}5$ , similar triangles gives $AE = \tfrac{48}5$ ), and that $EF$ is just $\tfrac{36}5 - x$ . From there, $AF = \sqrt{\left(\frac{36}5 - x\right)^2 + \left(\frac{48}5\right)^2}.$ Now, $BF^2 + AF^2 = 169$ , and squaring and adding both sides and subtracting a 169 from both sides gives $2x^2 - \tfrac{32}5x = 0$ , so $x = \tfrac{16}5$ . Thus, the answer is $\boxed{\textbf{(B)}}$ .

Solution 4 (Power of a Point)

First, we find $BD = 5$ , $DC = 9$ , and $AD = 12$ via the Pythagorean Theorem or by using similar triangles. Next, because $DE$ is an altitude of triangle $ADC$ , $DE = \frac{AD\cdot DC}{AC} = \frac{36}{5}$ . Using that, we can use the Pythagorean Theorem and similar triangles to find $EC = \frac{27}{5}$ and $AE = \frac{48}{5}$ .

Points $A$ , $B$ , $D$ , and $F$ all lie on a circle whose diameter is $AB$ . Let the point where the circle intersects $AC$ be $G$ . Using power of a point, we can write the following equation to solve for $AG$ : $DC\cdot BC = CG\cdot AC$ $9\cdot 14 = CG\cdot 15$ $CG = 126/15$ Using that, we can find $AG = \frac{99}{15}$ , and using $AG$ , we can find that $GE = 3$ .

We can use power of a point again to solve for $DF$ : $FE\cdot DE = GE\cdot AE$ $(\frac{36}{5} – DF)\cdot \frac{36}{5} = 3 \cdot \frac{48}{5}$ $\frac{36}{5} – DF = 4$ $DF = \frac{16}{5} = \frac{m}{n}$ Thus, $m+n = 16+5 = 21$ $\boxed{\textbf{(B)}}$ .

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

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