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

m (Solution 1)
(Adding solution)
Line 14: Line 14:
 
Using the similar triangles in triangle <math>ADC</math> gives <math>AE = \frac{48}{5}</math> and <math>DE = \frac{36}{5}</math>. Quadrilateral <math>ABDF</math> is cyclic, implying that <math>\angle{B} + \angle{DFA}</math> = 180°. Therefore, <math>\angle{B} = \angle{EFA}</math>, and triangles <math>AEF</math> and <math>ADB </math> are similar. Solving the resulting proportion gives <math>EF = 4</math>. Therefore, <math>DF = ED - EF = \frac{16}{5}. \implies{\boxed{(B)}}</math>
 
Using the similar triangles in triangle <math>ADC</math> gives <math>AE = \frac{48}{5}</math> and <math>DE = \frac{36}{5}</math>. Quadrilateral <math>ABDF</math> is cyclic, implying that <math>\angle{B} + \angle{DFA}</math> = 180°. Therefore, <math>\angle{B} = \angle{EFA}</math>, and triangles <math>AEF</math> and <math>ADB </math> are similar. Solving the resulting proportion gives <math>EF = 4</math>. Therefore, <math>DF = ED - EF = \frac{16}{5}. \implies{\boxed{(B)}}</math>
  
 +
==Solution 3==
 +
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
 +
 +
<math>BF = \sqrt{(4/5x + 5)^2 + (3/5x)^2}</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 19:53, 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 $3/5$x and $DG$ is $4/5$x. Using Pythagorean theorem, we now get

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

See also

2013 AMC 12B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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