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Difference between revisions of "2013 AMC 12B Problems/Problem 19"

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In triangle <math>ABC</math>, <math>AB=13</math>, <math>BC=14</math>, and <math>CA=15</math>. Distinct points <math>D</math>, <math>E</math>, and <math>F</math> lie on segments <math>\overline{BC}</math>, <math>\overline{CA}</math>, and <math>\overline{DE}</math>, respectively, such that <math>\overline{AD}\perp\overline{BC}</math>, <math>\overline{DE}\perp\overline{AC}</math>, and <math>\overline{AF}\perp\overline{BF}</math>. The length of segment <math>\overline{DF}</math> can be written as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?
 
In triangle <math>ABC</math>, <math>AB=13</math>, <math>BC=14</math>, and <math>CA=15</math>. Distinct points <math>D</math>, <math>E</math>, and <math>F</math> lie on segments <math>\overline{BC}</math>, <math>\overline{CA}</math>, and <math>\overline{DE}</math>, respectively, such that <math>\overline{AD}\perp\overline{BC}</math>, <math>\overline{DE}\perp\overline{AC}</math>, and <math>\overline{AF}\perp\overline{BF}</math>. The length of segment <math>\overline{DF}</math> can be written as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?
  
−
<math> \textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}}\ 27\qquad\textbf{(E)}\ 30</math>
+
<math> \textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 30</math>
 +
[[Category: Introductory Geometry Problems]]
  
−
==Solution==
+
==Diagram==
−
The 13-14-15 triangle is very commonly seen in competition problems, since the altitude from the point opposite the side of length 14 (<math>A</math>, in this case) divides the triangle into 9-12-15 and 5-12-13 right triangles. This means that <math>DA = 12</math>, <math>DB = 5</math>, and <math>DC = 9</math>.
+
<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);
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label("$9$",D--C,NE);
 +
label(Label("$13$",Rotate(B-A)), B--A);
 +
</asy>
  
−
We now proceed by coordinate geometry. Place the origin of the system at <math>D</math>, let the positive x-axis be <math>\overrightarrow{DC}</math>, and the positive y-axis be <math>\overrightarrow{DA}</math>. Then consider <math>\overline{DE}</math>. It is perpendicular to <math>\overline{AC}</math>, and <math>\overline{AC}</math> has slope <math>\frac{-12}{9} = \frac{-4}{3}</math>. Thus <math>\overleftrightarrow{DE}</math> is governed by the equation <math>y = \frac{3}{4}x</math> (recall that perpendicular lines' slopes are negative reciprocals of each other). This means that <math>F</math> must lie at a point given by <math>(4x, 3x)</math>.
+
==Solution 1==
 +
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.  
  
 +
So, the ratio of the longer leg to the hypotenuse of all three similar triangles is <math>\tfrac{12}{15} = \tfrac{4}{5}</math>, and the ratio of the shorter leg to the hypotenuse is <math>\tfrac{9}{15} = \tfrac{3}{5}</math>. It follows that <math>AF=(\tfrac{4}{5}\cdot 13), BF=(\tfrac{3}{5}\cdot 13)</math>.
  
−
Now consider the vectors <math>\overrightarrow{FB}</math> and <math>\overrightarrow{FA}</math>. Since <math>B</math> lies at <math>(-5, 0)</math> and <math>A</math> at <math>(0, 12)</math>, the vectors must be <math>\langle -4x - 5, -3x \rangle</math> and <math>\langle -4x, 12 - 3x \rangle</math>, respectively. If <math>\overline{AF}\perp\overline{BF}</math>, then <math>\overrightarrow{FB}</math> and <math>\overrightarrow{FA}</math> must be orthogonal, and their dot product must be zero. Therefore:
+
Let <math>x=DF</math>. By [[Ptolemy's Theorem|Ptolemy's Theorem]], we have <cmath>13x+\left(5\cdot 13\cdot \frac{4}{5}\right)= 12\cdot 13\cdot \frac{3}{5} \qquad \Leftrightarrow \qquad 13x+52=93.6.</cmath> 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>.
  
−
<cmath> \langle -4x - 5, -3x \rangle \cdot \langle -4x, 12 - 3x \rangle = 0 </cmath>
+
~Edits by BakedPotato66
−
<cmath> 16x^2 + 20x - 36x + 9x^2 = 0 </cmath>
 
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<cmath> x(25x - 16) = 0 </cmath>
 
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<cmath> x = 0 \vee x = \frac{16}{25} </cmath>
 
  
 +
==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{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>.
  
−
The first solution corresponds to <math>(0, 0)</math>, or point <math>D</math>. The other must be point <math>F</math> (since it is given that <math>D</math> and <math>F</math> are distinct). The value of <math>DF</math> is equal to the distance from <math>(0, 0)</math> to <math>(4x, 3x)</math>, and this is clearly <math>5x</math>. Therefore <math>DF = 5 * \frac{16}{25} = \frac{16}{5}</math>, and it is evident that <math>m = 16</math> and <math>n = 5</math>, thus our answer is <math>m + n = 16 + 5 = 21</math>. Answer choice <math>\boxed{\textbf{(B)} \ 21}</math> is correct.
+
<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>
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:South South]
 +
 
 +
==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>.
 +
 
 +
==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>.
 +
 
 +
 
 +
==Solution 5 (Coord Bash)==
 +
If we draw the diagram like above, but make <math>BC</math> the base, we can set <math>B(0,0), C(14,0)</math> and <math>A(5,12)</math>, as <math>AD</math> can quickly be verified to be <math>12</math> by Pythagorean triples or similar triangles. Construct <math>X</math> on <math>BC</math> such that <math>EX \perp BC</math>. This implies <math>\triangle ADC \sim \triangle EXC</math> as <math>AD \parallel EX</math>, and <math> \angle ACD = \angle ECX </math>. Also construct <math>FY</math> such that <math>FY \perp BC</math>.
 +
 
 +
 
 +
Line <math>\overline {AC}</math> has a slope of <math>-\frac{4}{3}</math> by slope formula. Since <math>D = (5,0)</math> and <math>DE \perp AC,</math>, the equation of <math> DE = \frac{3}{4}x - \frac{15}{4}</math>.
 +
 
 +
Furthermore, <math>F</math> can now be expressed as <math>(x,\frac{3}{4}x - \frac{15}{4}).</math> Since we know <math>AF \perp BF,</math> we can solve for <math>x</math> with the perpendicular slope formula like so: <cmath>\frac{\frac{3}{4}x - \frac{63}{4}}{x - 5} \cdot \frac{\frac{3}{4}x - \frac {15}{4}}{x} = -1</cmath> <cmath>\frac{x-21}{x-5} \cdot \frac{x-5}{x} = -\frac{16}{9}</cmath> <cmath>\frac{x-21}{x} = -\frac{16}{9}</cmath> <cmath> 9x = 189 -16x </cmath> <cmath>x = \frac{189}{25}</cmath>
 +
 
 +
Plugging <math>\frac{189}{25}</math> into <math>F</math>, we get <math>F(\frac{189}{25},\frac{48}{25}).</math> Since <math>FY \perp DY</math>, we get that <math>\triangle FYD</math> has side lengths of <math>FY = \frac{48}{25}</math>  
 +
 
 +
and <math>DY = BY - BD = BY - 5 = \frac{64}{25}</math>.
 +
 
 +
Clearly, <math>\triangle FYD</math> is a <math>3 - 4 - 5</math> pythagorean triple, so <math>DF = \frac{80}{25} = \frac{16}{5}</math>.
 +
 
 +
<math>16 + 5 = m + n = \boxed{\textbf{(B) }21}</math>.
 +
 
 +
~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
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2013|ab=B|num-b=18|num-a=20}}
 
{{AMC12 box|year=2013|ab=B|num-b=18|num-a=20}}
 +
 
{{AMC10 box|year=2013|ab=B|num-b=22|num-a=24}}
 
{{AMC10 box|year=2013|ab=B|num-b=22|num-a=24}}
−
 
+
{{MAA Notice}}
−
[[Category:Introductory Geometry Problems]]
 

Latest revision as of 21:24, 10 October 2024

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


Solution 5 (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

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=229560"