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"

(→‎Solution 2)
m (→‎Solution 2)
Line 47: Line 47:
  
 
==Solution 2==
 
==Solution 2==
−
From solution 1, we know that <math>AD = 12</math> and <math>DC = 9</math>. Since <math>\triangle ADC \sim \triangle DEC</math>, we can figure out that <math>DE = \frac{36}{5}</math>. We also know what <math>AC</math> is so we can figure what <math>AE</math> is: <math>AE = 15 - \frac{16}{5} = \frac{48}{5}</math> . Quadrilateral <math>ABDF</math> is cyclic, implying that <math>\angle{B} + \angle{DFA}</math> = 180°. Therefore, <math>\angle{B} = 180 - \angle{DFA} = \angle{EFA}</math>, and triangles <math>\triangle AEF \sim \triangle ADB</math>. Solving the resulting proportion gives <math>EF = 4</math>. Therefore, <math>DF = ED - EF = \frac{16}{5}</math>. <math>m + n = 16 + 5 = 21</math> and our answer is <math>\boxed{\textbf{(B) } 21}</math>.  
+
From solution 1, we know that <math>AD = 12</math> and <math>DC = 9</math>. Since <math>\triangle ADC \sim \triangle DEC</math>, we can figure out that <math>DE = \frac{36}{5}</math>. We also know what <math>AC</math> is so we can figure what <math>AE</math> is: <math>AE = 15 - \frac{27}{5} = \frac{48}{5}</math> . Quadrilateral <math>ABDF</math> is cyclic, implying that <math>\angle{B} + \angle{DFA}</math> = 180°. Therefore, <math>\angle{B} = 180 - \angle{DFA} = \angle{EFA}</math>, and triangles <math>\triangle AEF \sim \triangle ADB</math>. Solving the resulting proportion gives <math>EF = 4</math>. Therefore, <math>DF = ED - EF = \frac{16}{5}</math>. <math>m + n = 16 + 5 = 21</math> and our answer is <math>\boxed{\textbf{(B) } 21}</math>.  
  
 
<math>\triangle ADC \sim \triangle DEC</math> because of <math>AA \sim</math> . <math>\angle{ADC} = \angle{DEC} = 90°</math>. Lets say <math>\angle{ADE} = x</math>. So  <math>\angle{EDC} = 90 - x</math> and <math>\angle{DEC} = 180 - 90 - (90 - x) = x</math> so <math>\angle{ADE} = \angle{DEC}</math>
 
<math>\triangle ADC \sim \triangle DEC</math> because of <math>AA \sim</math> . <math>\angle{ADC} = \angle{DEC} = 90°</math>. Lets say <math>\angle{ADE} = x</math>. So  <math>\angle{EDC} = 90 - x</math> and <math>\angle{DEC} = 180 - 90 - (90 - x) = x</math> so <math>\angle{ADE} = \angle{DEC}</math>

Revision as of 12:04, 16 June 2023

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 $m+n$?

$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 30$

Diagram

[asy] size(200); defaultpen(linewidth(0.4)+fontsize(10)); pen s = linewidth(0.8)+fontsize(8); pair A,B,C,D,E0,F,G; A = origin; C = (15,0); B = IP(CR(A,13),CR(C,14)); D = foot(A,C,B); E0 = foot(D,A,C); F = OP(CR((A+B)/2,length(B-A)/2), D--E0); draw(A--C--B--A, black+0.8); draw(B--F--A--D--E0); dot("$A$",A,W); dot("$B$",B,N); dot("$C$",C,E); dot("$D$",D,NE); dot("$E$",E0,S); dot("$F$",F,E); draw(rightanglemark(B,D,A,15)); draw(rightanglemark(B,F,A,15)); draw(rightanglemark(D,E0,A,15)); label("$5$",D--B,NE); label("$9$",D--C,NE); label(Label("$13$",Rotate(B-A)), B--A); [/asy]

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 $\tfrac{12}{15} = \tfrac{4}{5}$, and the ratio of the shorter leg to the hypotenuse is $\tfrac{9}{15} = \tfrac{3}{5}$. It follows that $AF=(\tfrac{4}{5}\cdot 13), BF=(\tfrac{3}{5}\cdot 13)$.

Let $x=DF$. By Ptolemy's Theorem, we have \[13x+\left(5\cdot 13\cdot \frac{4}{5}\right)= 12\cdot 13\cdot \frac{3}{5} \qquad \Leftrightarrow \qquad 13x+52=93.6.\] 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

From solution 1, we know that $AD = 12$ and $DC = 9$. Since $\triangle ADC \sim \triangle DEC$, we can figure out that $DE = \frac{36}{5}$. We also know what $AC$ is so we can figure what $AE$ is: $AE = 15 - \frac{27}{5} = \frac{48}{5}$ . Quadrilateral $ABDF$ is cyclic, implying that $\angle{B} + \angle{DFA}$ = 180°. Therefore, $\angle{B} = 180 - \angle{DFA} = \angle{EFA}$, and triangles $\triangle AEF \sim \triangle ADB$. Solving the resulting proportion gives $EF = 4$. Therefore, $DF = ED - EF = \frac{16}{5}$. $m + n = 16 + 5 = 21$ and our answer is $\boxed{\textbf{(B) } 21}$.

$\triangle ADC \sim \triangle DEC$ because of $AA \sim$ . $\angle{ADC} = \angle{DEC} = 90°$. Lets say $\angle{ADE} = x$. So $\angle{EDC} = 90 - x$ and $\angle{DEC} = 180 - 90 - (90 - x) = x$ so $\angle{ADE} = \angle{DEC}$

~South

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, $m+n = 16+5 = 21$ $\boxed{\textbf{(B)}}$.

Video Solution

https://youtu.be/XZBKnobK-JU?t=3064

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=194528"