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Difference between revisions of "2013 AMC 12B Problems/Problem 19"

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==Solution==
 
==Solution==
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The 13-14-15 triangle is very commonly seen in competition problems, since the altitude from the point opposite the side of length 14 (<math>A</math>, in this case) divides the triangle into 9-12-15 and 5-12-13 right triangles. This means that <math>DA = 12</math>, <math>DB = 5</math>, and <math>DC = 9</math>.
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Since <math>\angle{AFB}=\angle{ADB}=90</math>, quadrilateral <math>ABDF</math> is cyclic. It follows that <math>\angle{ADE}=\angle{ABF}</math>. In addition, since <math>\angle{AFB}=\angle{AED}=90</math>, triangles <math>ABF</math> and <math>ADE</math> are similar. It follows that <math>AF=(13)(\frac{4}{5}), BF=(13)(\frac{3}{5})</math>. By Ptolemy, we have <math>13DF+(5)(13)(\frac{4}{5})=(12)(13)(\frac{3}{5})</math>. Cancelling <math>13</math>, the rest is easy. We obtain <math>DF=\frac{16}{5}\implies{16+5=21}\implies{\boxed{(B)}}</math>
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We now proceed by coordinate geometry. Place the origin of the system at <math>D</math>, let the positive x-axis be <math>\overrightarrow{DC}</math>, and the positive y-axis be <math>\overrightarrow{DA}</math>. Then consider <math>\overline{DE}</math>. It is perpendicular to <math>\overline{AC}</math>, and <math>\overline{AC}</math> has slope <math>\frac{-12}{9} = \frac{-4}{3}</math>. Thus <math>\overleftrightarrow{DE}</math> is governed by the equation <math>y = \frac{3}{4}x</math> (recall that perpendicular lines' slopes are negative reciprocals of each other). This means that <math>F</math> must lie at a point given by <math>(4x, 3x)</math>.
 
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Now consider the vectors <math>\overrightarrow{FB}</math> and <math>\overrightarrow{FA}</math>. Since <math>B</math> lies at <math>(-5, 0)</math> and <math>A</math> at <math>(0, 12)</math>, the vectors must be <math>\langle -4x - 5, -3x \rangle</math> and <math>\langle -4x, 12 - 3x \rangle</math>, respectively. If <math>\overline{AF}\perp\overline{BF}</math>, then <math>\overrightarrow{FB}</math> and <math>\overrightarrow{FA}</math> must be orthogonal, and their dot product must be zero. Therefore:
 
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<cmath> \langle -4x - 5, -3x \rangle \cdot \langle -4x, 12 - 3x \rangle = 0 </cmath>
 
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<cmath> 16x^2 + 20x - 36x + 9x^2 = 0 </cmath>
 
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<cmath> x(25x - 16) = 0 </cmath>
 
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<cmath> x = 0 \vee x = \frac{16}{25} </cmath>
 
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The first solution corresponds to <math>(0, 0)</math>, or point <math>D</math>. The other must be point <math>F</math> (since it is given that <math>D</math> and <math>F</math> are distinct). The value of <math>DF</math> is equal to the distance from <math>(0, 0)</math> to <math>(4x, 3x)</math>, and this is clearly <math>5x</math>. Therefore <math>DF = 5 * \frac{16}{25} = \frac{16}{5}</math>, and it is evident that <math>m = 16</math> and <math>n = 5</math>, thus our answer is <math>m + n = 16 + 5 = 21</math>. Answer choice <math>\boxed{\textbf{(B)} \ 21}</math> is correct.
 
  
 
== See also ==
 
== See also ==

Revision as of 14:57, 26 August 2013

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 $m+n$?

$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}}\ 27\qquad\textbf{(E)}\ 30$ (Error compiling LaTeX. Unknown error_msg)

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 $DF=\frac{16}{5}\implies{16+5=21}\implies{\boxed{(B)}}$

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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