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

(Adding solution)
(→‎Solution 3)
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==Solution 3==
 
==Solution 3==
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If we draw a diagram as given, but then add <math>DG</math> as an altitude to use the Pythagorean theorem, we end up with similar triangles <math>\triangle{DFG}</math> and <math>\triangle{DCE}</math>. Thus, <math>FG</math> is <math>3/5</math>x and <math>DG</math> is <math>4/5</math>x. Using Pythagorean theorem, we now get
+
If we draw a diagram as given, but then add <math>DG</math> as an altitude to use the Pythagorean theorem, we end up with similar triangles <math>\triangle{DFG}</math> and <math>\triangle{DCE}</math>. Thus, <math>FG</math> is <math>\frac{3/5} x</math> and <math>DG</math> is <math>\frac{4/5} x</math>. Using Pythagorean theorem, we now get
  
 
<math>BF = \sqrt{(4/5x + 5)^2 + (3/5x)^2}</math>
 
<math>BF = \sqrt{(4/5x + 5)^2 + (3/5x)^2}</math>
 +
 +
and <math>AF</math> can be found out noting that <math>AE</math> is just 48/5 through area times height (12(9) = 15<math>36/5</math>, similar triangles gives AE = <math>\frac{48/5}</math>), and that <math>EF</math> is just <math>\frac{36/5} - x</math>.
 +
 +
From there, <math>AF</math> is <math>\sqrt{36/5 - x)^2 + (48/5)^2)}</math>
 +
 +
Using algebra, we get that <math>2x^2 - 32/5x = 0</math>, so
 +
 +
x = <math>(16/5)</math>.
 +
 +
Thus, the answer is <math>(B) 21.</math>
 +
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2013|ab=B|num-b=18|num-a=20}}
 
{{AMC12 box|year=2013|ab=B|num-b=18|num-a=20}}

Revision as of 18:59, 3 February 2017

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, since $\angle{AFB}=\angle{AED}=90$, triangles $ABF$ and $AFE$ 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, $DF = ED - EF = \frac{16}{5}. \implies{\boxed{(B)}}$

Solution 3

If we draw a diagram as given, but then add $DG$ as an altitude to use the Pythagorean theorem, we end up with similar triangles $\triangle{DFG}$ and $\triangle{DCE}$. Thus, $FG$ is $\frac{3/5} x$ and $DG$ is $\frac{4/5} x$. Using Pythagorean theorem, we now get

$BF = \sqrt{(4/5x + 5)^2 + (3/5x)^2}$

and $AF$ can be found out noting that $AE$ is just 48/5 through area times height (12(9) = 15$36/5$, similar triangles gives AE = $\frac{48/5}$ (Error compiling LaTeX. Unknown error_msg)), and that $EF$ is just $\frac{36/5} - x$.

From there, $AF$ is $\sqrt{36/5 - x)^2 + (48/5)^2)}$

Using algebra, we get that $2x^2 - 32/5x = 0$, so

x = $(16/5)$.

Thus, the answer is $(B) 21.$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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