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

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==Solution 1==
 
==Solution 1==
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Since <math>\angle{AFB}=\angle{ADB}=90^{\circ}</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's Theorem|Ptolemy's Theorem]], 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>
+
Since <math>\angle{AFB}=\angle{ADB}=90^{\circ}</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's Theorem|Ptolemy's Theorem]], 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}</math>, so our answer is <math>16+5=\boxed{21\,\textbf{(B)}}</math>.
  
 
==Solution 2==
 
==Solution 2==
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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}</math> and our answer is <math>\boxed{\textbf{(B)}}</math>.
  
 
==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 * x</math> and <math>DG</math> is <math>4/5 *x</math>. 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=\tfrac35x</math> and <math>DG=\tfrac45x</math>. Using Pythagorean theorem, we now get <cmath>BF = \sqrt{\left(\frac45x+ 5\right)^2 + \left(\frac35x\right)^2}</cmath> and <math>AF</math> can be found out noting that <math>AE</math> is just <math>\tfrac{48}5</math> through area times height (since <math>12\cdot 9 = 15 \cdot \tfrac{36}5</math>, similar triangles gives <math>AE = \tfrac{48}5</math>), and that <math>EF</math> is just <math>\tfrac{36}5 - x</math>. From there, <cmath>AF = \sqrt{\left(\frac{36}5 - x\right)^2 + \left(\frac{48}5\right)^2}.</cmath> Now, <math>BF^2 + AF^2 = 169</math>, and squaring and adding both sides and subtracting a 169 from both sides gives <math>2x^2 - \tfrac{32}5x = 0</math>, so <math>x = \tfrac{16}5</math>. Thus, the answer is <math>\boxed{\textbf{(B)}}</math>.
−
 
 
−
<math>BF = \sqrt{(4/5 *x + 5)^2 + (3/5 * x)^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>48/5</math>), and that <math>EF</math> is just <math>36/5 - x</math>.  
 
−
 
 
−
From there, <math>AF</math> is <math>\sqrt{((36/5 - x)^2 + (48/5)^2)}</math>
 
−
 
 
−
Now, <math>BF^2 + AF^2 = 169</math>, and squaring and adding both sides and subtracting a 169 from both sides gives:
 
−
 
 
−
<math>2x^2 - 32/5x = 0</math>, so
 
−
 
 
−
x = <math>(16/5)</math>.
 
−
 
 
−
Thus, the answer is <math>{\boxed{(B)}}</math>
 
−
 
 
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==Solution 4==
 
  
 
== See also ==
 
== See also ==

Revision as of 17:04, 25 January 2018

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 $ADE$ are similar. It follows that $AF=(13)(\frac{4}{5}), BF=(13)(\frac{3}{5})$. By Ptolemy's Theorem, we have $13DF+(5)(13)(\frac{4}{5})=(12)(13)(\frac{3}{5})$. Cancelling $13$, the rest is easy. We obtain $DF=\frac{16}{5}$, so our answer is $16+5=\boxed{21\,\textbf{(B)}}$.

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, $DF = ED - EF = \frac{16}{5}$ and our answer is $\boxed{\textbf{(B)}}$.

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, $FG=\tfrac35x$ and $DG=\tfrac45x$. Using Pythagorean theorem, we now get \[BF = \sqrt{\left(\frac45x+ 5\right)^2 + \left(\frac35x\right)^2}\] and $AF$ can be found out noting that $AE$ is just $\tfrac{48}5$ through area times height (since $12\cdot 9 = 15 \cdot \tfrac{36}5$, similar triangles gives $AE = \tfrac{48}5$), and that $EF$ is just $\tfrac{36}5 - x$. From there, \[AF = \sqrt{\left(\frac{36}5 - x\right)^2 + \left(\frac{48}5\right)^2}.\] Now, $BF^2 + AF^2 = 169$, and squaring and adding both sides and subtracting a 169 from both sides gives $2x^2 - \tfrac{32}5x = 0$, so $x = \tfrac{16}5$. Thus, the answer is $\boxed{\textbf{(B)}}$.

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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