Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2013 AMC 12B Problems/Problem 19"
Line 16: Line 16:
 
==Solution 3==
 
==Solution 3==
 
If we draw a diagram as given, but then add point <math>G</math> on <math>\overline{BC}</math> such that <math>\overline{FG}\perp\overline{BC}</math> in order 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>, where <math>x</math> is the length of <math>\overline{DF}</math>. Using the 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 base 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>.
 
If we draw a diagram as given, but then add point <math>G</math> on <math>\overline{BC}</math> such that <math>\overline{FG}\perp\overline{BC}</math> in order 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>, where <math>x</math> is the length of <math>\overline{DF}</math>. Using the 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 base 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>.
 +
 +
==Solution 4 (Power of a Point)==
 +
First, we find <math>BD = 5</math>, <math>DC = 9</math>, and <math>AD = 12</math> via the Pythagorean Theorem or by using similar triangles. Next, because <math>DE</math> is an altitude of triangle <math>ADC</math>, <math>DE = \frac{AD\cdot DC}{AC} = \frac{36}{5}</math>. Using that, we can use the Pythagorean Theorem and similar triangles to find <math>EC = \frac{27}{5}</math> and <math>AE = \frac{48}{5}</math>.
 +
 +
Points <math>A</math>, <math>B</math>, <math>D</math>, and <math>F</math> all lie on a circle whose diameter is <math>AB</math>. Let the point where the circle intersects <math>AC</math> be <math>G</math>. Using power of a point, we can write the following equation to solve for <math>AG</math>: <cmath>DC\cdot BC = CG\cdot AC</cmath> <cmath>9\cdot 14 = CG\cdot 15</cmath> <cmath>CG = 126/15</cmath> Using that, we can find <math>AG = \frac{99}{15}</math>, and using <math>AG</math>, we can find that <math>GE = 3</math>.
 +
 +
We can use power of a point again to solve for <math>DF</math>: <cmath>FE\cdot DE = GE\cdot AE</cmath> <cmath>(\frac{36}{5} – DF)\cdot \frac{36}{5} = 3 \cdot \frac{48}{5}</cmath> <cmath>\frac{36}{5} – DF = 4</cmath> <cmath>DF = \frac{16}{5} = \frac{m}{n}</cmath> Thus, <math>m+n = 16+5 = 21</math> <math>\boxed{\textbf{(B)}}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 08:50, 28 January 2021

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, triangles $ABF$ and $ADE$ are similar, and triangles $ADE$ and $ADC$ are similar. We can easily find $AD=12$, $BD = 5$, and $DC=9$ using pythagorean triples. So, the ratio of the hypotenuse to the longer leg of all three similar triangles is $\frac{15}{12} = \frac{4}{5}$, and the ratio of the hypotenuse to the shorter leg is $\frac{15}{9} = \frac{3}{5}$. It follows that $AF=(13)(\frac{4}{5}), BF=(13)(\frac{3}{5})$. By Ptolemy's Theorem, we have $13x+(5)(13)(\frac{4}{5})=(12)(13)(\frac{3}{5})\implies 13x+52=93.6$ where $x=DF$. Dividing by $13$ we get $x+4=7.2\implies x=\frac{16}{5}$ 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 point $G$ on $\overline{BC}$ such that $\overline{FG}\perp\overline{BC}$ in order to use the Pythagorean theorem, we end up with similar triangles $\triangle{DFG}$ and $\triangle{DCE}$. Thus, $FG=\tfrac35x$ and $DG=\tfrac45x$, where $x$ is the length of $\overline{DF}$. Using the 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 base 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 $\boxed{\textbf{(B)}}$.

Solution 4 (Power of a Point)

First, we find $BD = 5$, $DC = 9$, and $AD = 12$ via the Pythagorean Theorem or by using similar triangles. Next, because $DE$ is an altitude of triangle $ADC$, $DE = \frac{AD\cdot DC}{AC} = \frac{36}{5}$. Using that, we can use the Pythagorean Theorem and similar triangles to find $EC = \frac{27}{5}$ and $AE = \frac{48}{5}$.

Points $A$, $B$, $D$, and $F$ all lie on a circle whose diameter is $AB$. Let the point where the circle intersects $AC$ be $G$. Using power of a point, we can write the following equation to solve for $AG$: \[DC\cdot BC = CG\cdot AC\] \[9\cdot 14 = CG\cdot 15\] \[CG = 126/15\] Using that, we can find $AG = \frac{99}{15}$, and using $AG$, we can find that $GE = 3$.

We can use power of a point again to solve for $DF$: \[FE\cdot DE = GE\cdot AE\] \[(\frac{36}{5} – DF)\cdot \frac{36}{5} = 3 \cdot \frac{48}{5}\] \[\frac{36}{5} – DF = 4\] \[DF = \frac{16}{5} = \frac{m}{n}\] Thus, $m+n = 16+5 = 21$ $\boxed{\textbf{(B)}}$.

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_19&oldid=143594"