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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, triangles <math>ABF</math> and <math>ADE</math> are similar, and triangles <math>ADE</math> and <math>ADC</math> are similar. We can easily find <math>AD=12</math>, <math>BD = 5</math>, and <math>DC=9</math> using pythagorean triples. So, the ratio of the hypotenuse to the longer leg of all three similar triangles is <math>\frac{15}{12} = \frac{4}{5}</math>, and the ratio of the hypotenuse to the shorter leg is <math>\frac{15}{9} = \frac{3}{5}</math>. 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>13x+(5)(13)(\frac{4}{5})=(12)(13)(\frac{3}{5})\implies 13x+52=93.6</math> where <math>x=DF</math>. Dividing by <math>13</math> we get <math>x+4=7.2\implies x=\frac{16}{5}</math> so our answer is <math>16+5=\boxed{21\,\textbf{(B)}}</math>.
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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>, so <math>\triangle ABF \sim \triangle ADE</math> are similar. In addition, <math>\triangle ADE \sim \triangle ACD</math>. We can easily find <math>AD=12</math>, <math>BD = 5</math>, and <math>DC=9</math> using Pythagorean triples.  
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So, the ratio of the longer leg to the hypotenuse of all three similar triangles is <math>\frac{12}{15} = \frac{4}{5}</math>, and the ratio of the shorter leg to the hypotenuse is <math>\frac{9}{15} = \frac{3}{5}</math>. It follows that <math>AF=(13)(\tfrac{4}{5}), BF=(13)(\tfrac{3}{5})</math>.
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Let <math>x=DF</math>. By [[Ptolemy's Theorem|Ptolemy's Theorem]], we have <math>13x+(5)(13)(\frac{4}{5})=(12)(13)(\frac{3}{5})\implies 13x+52=93.6</math>. Dividing by <math>13</math> we get <math>x+4=7.2\implies x=\frac{16}{5}</math> so our answer is <math>\boxed{\textbf{(B) }21}</math>.
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~Edits by BakedPotato66
  
 
==Solution 2==
 
==Solution 2==

Revision as of 10:54, 24 July 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 $\angle{ADE}=\angle{ABF}$, so $\triangle ABF \sim \triangle ADE$ are similar. In addition, $\triangle ADE \sim \triangle ACD$. We can easily find $AD=12$, $BD = 5$, and $DC=9$ using Pythagorean triples.

So, the ratio of the longer leg to the hypotenuse of all three similar triangles is $\frac{12}{15} = \frac{4}{5}$, and the ratio of the shorter leg to the hypotenuse is $\frac{9}{15} = \frac{3}{5}$. It follows that $AF=(13)(\tfrac{4}{5}), BF=(13)(\tfrac{3}{5})$.

Let $x=DF$. By Ptolemy's Theorem, we have $13x+(5)(13)(\frac{4}{5})=(12)(13)(\frac{3}{5})\implies 13x+52=93.6$. Dividing by $13$ we get $x+4=7.2\implies x=\frac{16}{5}$ so our answer is $\boxed{\textbf{(B) }21}$.

~Edits by BakedPotato66

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

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