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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.
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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>.
  
  

Revision as of 20:58, 23 February 2013

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

The 13-14-15 triangle is very commonly seen in competition problems, since the altitude from the point opposite the side of length 14 ($A$, in this case) divides the triangle into 9-12-15 and 5-12-13 right triangles. This means that $DA = 12$, $DB = 5$, and $DC = 9$.


We now proceed by coordinate geometry. Place the origin of the system at $D$, let the positive x-axis be $\overrightarrow{DC}$, and the positive y-axis be $\overrightarrow{DA}$. Then consider $\overline{DE}$. It is perpendicular to $\overline{AC}$, and $\overline{AC}$ has slope $\frac{-12}{9} = \frac{-4}{3}$. Thus $\overleftrightarrow{DE}$ is governed by the equation $y = \frac{3}{4}x$ (recall that perpendicular lines' slopes are negative reciprocals of each other). This means that $F$ must lie at a point given by $(4x, 3x)$.


Now consider the vectors $\overrightarrow{FB}$ and $\overrightarrow{FA}$. Since $B$ lies at $(-5, 0)$ and $A$ at $(0, 12)$, the vectors must be $\langle -4x - 5, -3x \rangle$ and $\langle -4x, 12 - 3x \rangle$, respectively. If $\overline{AF}\perp\overline{BF}$, then $\overrightarrow{FB}$ and $\overrightarrow{FA}$ must be orthogonal, and their dot product must be zero. Therefore:

\[\langle -4x - 5, -3x \rangle \cdot \langle -4x, 12 - 3x \rangle = 0\] \[16x^2 + 20x - 36x + 9x^2 = 0\] \[x(25x - 16) = 0\] \[x = 0 \vee x = \frac{16}{25}\]


The first solution corresponds to $(0, 0)$, or point $D$. The other must be point $F$ (since it is given $D$ and $F$ are distinct). The value of $DF$ is equal to the distance from $(0, 0)$ to $(4x, 3x)$, and this is clearly $5x$. Therefore $DF = 5 * \frac{16}{25} = \frac{16}{5}$, and it is evident that $m = 16$ and $n = 5$, thus our answer is $m + n = 16 + 5 = 21$. Answer choice $\textbf{(B)}$ is correct.

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
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