2013 AMC 12B Problems/Problem 19: Difference between revisions

Revision as of 22:04, 12 November 2022

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

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

Video Solution (Similar Triangles)

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

@@ Line 58: / Line 58: @@
 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>.
+==Video Solution (Similar Triangles)==
+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

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