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2013 AMC 12B Problems/Problem 19: Difference between revisions

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


<math> \textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}}\ 27\qquad\textbf{(E)}\ 30</math>
<math> \textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 30</math>
[[Category: Introductory Geometry Problems]]
[[Category: Introductory Geometry Problems]]


Revision as of 22:57, 10 March 2015

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 $\angle{AFB}=\angle{ADB}=90$, 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 $DF=\frac{16}{5}\implies{16+5=21}\implies{\boxed{(B)}}$

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, $DF = ED - EF = \frac{16}{5}. \implies{\boxed{(B)}}$

See also

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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