2013 AMC 12B Problems/Problem 19

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

Let $\angle{ADF} = \theta,$ and $DF = x.$ Noting that $AD$ is the altitude from $A$ to $BC,$ $AD=12,$ $BD = 5,$ and $DC = 9.$ Notice that since $DE$ is perpendicular to $AC$ , $\triangle{ADC}$ is similar to $\triangle{AEC}.$ Therefore since $\triangle{ADC}$ is a $3$ - $4$ - $5$ triangle, $\cos{\theta}=\frac{3}{5}.$ Applying Law of Cosines on $\triangle{ADF}$ we have, $AF^2 = 144+x^2 - 2\cdot 12 \cdot x \cdot \cos{\theta} = 144+x^2 - \frac{72x}{5}.$ Next notice that $\cos{\angle{BDF}} = \cos{(90 + \theta)} = -\sin{\theta} = \frac{-4}{5}.$ Now we apply Law of Cosines on $\triangle{BDF},$ $BF^2 = 25 + x^2 - 2 \cdot 5 \cdot x \cdot \cos{\angle{BDF}} = 25 + x^2 + 8x.$ Finally using Pythagorean Theorem on $\triangle{ABF}$ we have, $AB^2 = AF^2 + BF^2.$ $13^2 = 144+x^2 - \frac{72x}{5} + 25 + x^2 + 8x.$ $0 = 2x^2 - \frac{32x}{5}.$ Solving for the nonnegative value for $x$ we find $x = DF = \frac{16}{5} \implies \boxed{21}.$

~mathkiddus

Solution 4

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

Solution 6 (Coord Bash)

If we draw the diagram like above, but make $BC$ the base, we can set $B(0,0), C(14,0)$ and $A(5,12)$ , as $AD$ can quickly be verified to be $12$ by Pythagorean triples or similar triangles. Construct $X$ on $BC$ such that $EX \perp BC$ . This implies $\triangle ADC \sim \triangle EXC$ as $AD \parallel EX$ , and $\angle ACD = \angle ECX$ . Also construct $FY$ such that $FY \perp BC$ .

Line $\overline {AC}$ has a slope of $-\frac{4}{3}$ by slope formula. Since $D = (5,0)$ and $DE \perp AC,$ , the equation of $DE = \frac{3}{4}x - \frac{15}{4}$ .

Furthermore, $F$ can now be expressed as $(x,\frac{3}{4}x - \frac{15}{4}).$ Since we know $AF \perp BF,$ we can solve for $x$ with the perpendicular slope formula like so: $\frac{\frac{3}{4}x - \frac{63}{4}}{x - 5} \cdot \frac{\frac{3}{4}x - \frac {15}{4}}{x} = -1$ $\frac{x-21}{x-5} \cdot \frac{x-5}{x} = -\frac{16}{9}$ $\frac{x-21}{x} = -\frac{16}{9}$ $9x = 189 -16x$ $x = \frac{189}{25}$

Plugging $\frac{189}{25}$ into $F$ , we get $F(\frac{189}{25},\frac{48}{25}).$ Since $FY \perp DY$ , we get that $\triangle FYD$ has side lengths of $FY = \frac{48}{25}$

and $DY = BY - BD = BY - 5 = \frac{64}{25}$ .

Clearly, $\triangle FYD$ is a $3 - 4 - 5$ pythagorean triple, so $DF = \frac{80}{25} = \frac{16}{5}$ .

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

~JT0543164

Video Solution by Pi Academy

https://youtu.be/2wrOFzxYCcM?si=J-u-_5Yb8o3hXrAp

~ Pi Academy

Video Solution 2

https://www.youtube.com/watch?v=X0YJfFiBy0g

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

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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