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 $m+n$ ?

$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}}\ 27\qquad\textbf{(E)}\ 30$ (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 that $D$ 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 $\boxed{\textbf{(B)} \ 21}$ is correct.

2013 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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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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2013 AMC 12B Problems/Problem 19

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