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Difference between revisions of "2009 AIME II Problems/Problem 10"

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==Problem==
 
==Problem==
 
Four lighthouses are located at points <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>. The lighthouse at <math>A</math> is <math>5</math> kilometers from the lighthouse at <math>B</math>, the lighthouse at <math>B</math> is <math>12</math> kilometers from the lighthouse at <math>C</math>, and the lighthouse at <math>A</math> is <math>13</math> kilometers from the lighthouse at <math>C</math>. To an observer at <math>A</math>, the angle determined by the lights at <math>B</math> and <math>D</math> and the angle determined by the lights at <math>C</math> and <math>D</math> are equal. To an observer at <math>C</math>, the angle determined by the lights at <math>A</math> and <math>B</math> and the angle determined by the lights at <math>D</math> and <math>B</math> are equal. The number of kilometers from <math>A</math> to <math>D</math> is given by <math>\frac {p\sqrt{q}}{r}</math>, where <math>p</math>, <math>q</math>, and <math>r</math> are relatively prime positive integers, and <math>r</math> is not divisible by the square of any prime. Find <math>p</math> + <math>q</math> + <math>r</math>.  
 
Four lighthouses are located at points <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>. The lighthouse at <math>A</math> is <math>5</math> kilometers from the lighthouse at <math>B</math>, the lighthouse at <math>B</math> is <math>12</math> kilometers from the lighthouse at <math>C</math>, and the lighthouse at <math>A</math> is <math>13</math> kilometers from the lighthouse at <math>C</math>. To an observer at <math>A</math>, the angle determined by the lights at <math>B</math> and <math>D</math> and the angle determined by the lights at <math>C</math> and <math>D</math> are equal. To an observer at <math>C</math>, the angle determined by the lights at <math>A</math> and <math>B</math> and the angle determined by the lights at <math>D</math> and <math>B</math> are equal. The number of kilometers from <math>A</math> to <math>D</math> is given by <math>\frac {p\sqrt{q}}{r}</math>, where <math>p</math>, <math>q</math>, and <math>r</math> are relatively prime positive integers, and <math>r</math> is not divisible by the square of any prime. Find <math>p</math> + <math>q</math> + <math>r</math>.  
  
 +
==Diagram==
 +
<asy>
 +
size(120); pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(10); pair B=(0,0), A=(5,0), C=(0,13), E=(-5,0), O = incenter(E,C,A), D=IP(A -- A+3*(O-A),E--C); D(A--B--C--cycle); D(A--D--C); D(D--E--B, linetype("4 4")); MP("5", (A+B)/2, f); MP("13", (A+C)/2, NE,f); MP("A",D(A),f); MP("B",D(B),f); MP("C",D(C),N,f); MP("A'",D(E),f); MP("D",D(D),NW,f); D(rightanglemark(C,B,A,20)); D(anglemark(D,A,E,35));D(anglemark(C,A,D,30));
 +
</asy>
 +
-asjpz
 +
 +
==Video Solution by Punxsutawney Phil==
 +
https://www.youtube.com/watch?v=f3zEesJh4Ws
  
 
== Solution 1==
 
== Solution 1==
Line 32: Line 39:
 
==Solution 4==
 
==Solution 4==
  
−
After drawing a good diagram, we reflect <math>ABC</math> over the line <math>BC</math>, forming a new point that we'll call <math>A'</math>. Also, let the intersection of <math>AD</math> and <math>BC</math> be point <math>E</math>. Point <math>D</math> lies on line <math>A'C</math>. Since line <math>AD</math> bisects <math>\angle{CAB}</math>, we can use the Angle Bisector Theorem. <math>AA'=10</math> and <math>AC=13</math>, so <math>\frac{CD}{A'D}=\frac{13}{10}</math>. Letting the segments be <math>13x</math> and <math>10x</math> respectively, we now have <math>13x+10x=13</math>. Therefore, <math>x=\frac{13}{23}</math>. By the Pythagorean Theorem, <math>AE=\frac{5\sqrt{13}}{3}</math>. Using the Angle Bisector Theorem on <math>\angle{ACD}</math>, we have that <math>ED=\frac{5x\sqrt{13}}{3}</math>. Substituting in <math>x=\frac{13}{23}</math>, we have that <math>AD=(\frac{5\sqrt{13}}{3})(1+x)=\frac{60\sqrt{13}}{23}</math>, so the answer is <math>60+13+23=\boxed{096}</math>.
+
After drawing a good diagram, we reflect <math>ABC</math> over the line <math>BC</math>, forming a new point that we'll call <math>A'</math>. Also, let the intersection of <math>AD</math> and <math>BC</math> be point <math>E</math>. Point <math>D</math> lies on line <math>A'C</math>. Since line <math>AD</math> bisects <math>\angle{CAB}</math>, we can use the Angle Bisector Theorem. <math>AA'=10</math> and <math>AC=13</math>, so <math>\frac{CD}{A'D}=\frac{13}{10}</math>. Letting the segments be <math>13x</math> and <math>10x</math> respectively, we now have <math>13x+10x=13</math>. Therefore, <math>x=\frac{13}{23}</math>. By the Pythagorean Theorem, <math>AE=\frac{5\sqrt{13}}{3}</math>. Using the Angle Bisector Theorem on <math>\angle{ACD}</math>, we have that <math>ED=\frac{5x\sqrt{13}}{3}</math>. Substituting in <math>x=\frac{13}{23}</math>, we have that <math>AD=\left(\frac{5\sqrt{13}}{3}\right)(1+x)=\frac{60\sqrt{13}}{23}</math>, so the answer is <math>60+13+23=\boxed{096}</math>.
 +
 
 +
-RootThreeOverTwo
 +
 
 +
==Solution 5 (Angle Bisector Theorem + Law of Cosines)==
 +
Let <math> CD</math> and <math> AB</math> meet at <math> A'</math>; then <math> \overline{AD}</math> is an angle bisector of isosceles <math> \triangle AA'C</math>. Then by the Angle Bisector Theorem, <math> A'D=\frac {10}{10+13} \cdot 13=\frac {130}{23}</math>, and <math> \cos \angle DA'A=\frac {5}{13}</math>. By the Law of Cosines on <math> \triangle AA'D</math>, we have <cmath>AD^2=10^2+\left(\frac {130}{23}\right)^2-2 \cdot 10 \cdot \frac {130}{23} \cdot \frac {5}{13}=\frac {2^4 \cdot 3^2 \cdot 5^2 \cdot 13}{23^2} \Longrightarrow AD=\frac {60\sqrt {13}}{23}</cmath> and the answer is <math> 60+13+23=\boxed{096}</math>.
 +
 
 +
 
 +
<asy>
 +
size(120); pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(10); pair B=(0,0), A=(5,0), C=(0,13), E=(-5,0), O = incenter(E,C,A), D=IP(A -- A+3*(O-A),E--C); D(A--B--C--cycle); D(A--D--C); D(D--E--B, linetype("4 4")); MP("5", (A+B)/2, f); MP("13", (A+C)/2, NE,f); MP("A",D(A),f); MP("B",D(B),f); MP("C",D(C),N,f); MP("A'",D(E),f); MP("D",D(D),NW,f); D(rightanglemark(C,B,A,20)); D(anglemark(D,A,E,35));D(anglemark(C,A,D,30));
 +
</asy>
 +
 
 +
-azjps
 +
 
 +
==Solution 6 (Law of Sines)==
 +
<asy>pathpen = linewidth(0.7);
 +
pen f = fontsize(10);
 +
size(144);
 +
pair B=(0,0);
 +
pair A=(5,0);
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pair C=(0,12);
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pair E=(-5,0);
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pair abA = A+unit(C-A)+unit(B-A);
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pair D  = extension(A,abA,C,E);
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D(A--C);
 +
D(C--B);
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D(A--B);
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D(C--D);
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D(A--D);
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D(D--E,dashed);
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D(B--E,dashed);
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D(rightanglemark(C,B,A,20),red);
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MP("A",D(A),plain.SE,f);
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MP("B",D(B),plain.S,f);
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MP("C",D(C),plain.N,f);
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MP("D",D(D),plain.NW,f);
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MP("E",D(E),plain.SW,f);
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MP("5",(A+B)/2,plain.S,f);
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MP("12",(B+C)/2,plain.E,f);
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MP("13",(A+C)/2,plain.NE,f);
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MP("\alpha",A,4*(unit(D-A)+unit(C-A))/2,f);
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MP("\alpha",A,4*(unit(B-A)+unit(D-A))/2,f);
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MP("2\alpha",E,4*(unit(C-E)+unit(B-E))/2,f);
 +
MP("\beta",C,6*(unit(A-C)+unit(B-C))/2,f);
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MP("\beta",C,6*(unit(E-C)+unit(B-C))/2,f);</asy>
 +
 
 +
Using the law of sines on <math> \bigtriangleup ADE</math>,
 +
 
 +
<cmath>\begin{align*}
 +
AD  &= AE \cdot \dfrac{\sin(2\alpha)}{\sin(180^\circ-3\alpha)}\\
 +
    &= AE \cdot \dfrac{\sin(2\alpha)}{\sin(\alpha+2\alpha)}\\
 +
    &= 10 \cdot \dfrac{\sin(2\alpha)}{\sin(\alpha)\cos(2\alpha)+\cos(\alpha)\sin(2\alpha)}\\
 +
    &= 10 \cdot \dfrac{\sin(2\alpha)}{\sqrt{\dfrac{1-\cos(2\alpha)}{2}}\cdot\cos(2\alpha)+\sqrt{\dfrac{1+\cos(2\alpha)}{2}}\cdot\sin(2\alpha)}\\
 +
    &= 10 \cdot \dfrac{\dfrac{12}{13}}{\sqrt{\dfrac{8}{26}}\cdot\dfrac{5}{13}+\sqrt{\dfrac{18}{26}}\cdot\dfrac{12}{13}}\\
 +
    &= 10 \cdot \dfrac{12\sqrt{13}}{10+36} = \dfrac{60\sqrt{13}}{23}
 +
\end{align*}</cmath>
 +
 
 +
<math> \therefore</math> the answer is <math> 60+13+23=\boxed{096}</math>.
  
−
(Solution by RootThreeOverTwo)
+
-m1sterzer0
  
 
== See Also ==
 
== See Also ==

Latest revision as of 12:56, 18 December 2022

Problem

Four lighthouses are located at points $A$, $B$, $C$, and $D$. The lighthouse at $A$ is $5$ kilometers from the lighthouse at $B$, the lighthouse at $B$ is $12$ kilometers from the lighthouse at $C$, and the lighthouse at $A$ is $13$ kilometers from the lighthouse at $C$. To an observer at $A$, the angle determined by the lights at $B$ and $D$ and the angle determined by the lights at $C$ and $D$ are equal. To an observer at $C$, the angle determined by the lights at $A$ and $B$ and the angle determined by the lights at $D$ and $B$ are equal. The number of kilometers from $A$ to $D$ is given by $\frac {p\sqrt{q}}{r}$, where $p$, $q$, and $r$ are relatively prime positive integers, and $r$ is not divisible by the square of any prime. Find $p$ + $q$ + $r$.

Diagram

[asy] size(120); pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(10); pair B=(0,0), A=(5,0), C=(0,13), E=(-5,0), O = incenter(E,C,A), D=IP(A -- A+3*(O-A),E--C); D(A--B--C--cycle); D(A--D--C); D(D--E--B, linetype("4 4")); MP("5", (A+B)/2, f); MP("13", (A+C)/2, NE,f); MP("A",D(A),f); MP("B",D(B),f); MP("C",D(C),N,f); MP("A'",D(E),f); MP("D",D(D),NW,f); D(rightanglemark(C,B,A,20)); D(anglemark(D,A,E,35));D(anglemark(C,A,D,30)); [/asy] -asjpz

Video Solution by Punxsutawney Phil

https://www.youtube.com/watch?v=f3zEesJh4Ws

Solution 1

Let $O$ be the intersection of $BC$ and $AD$. By the Angle Bisector Theorem, $\frac {5}{BO}$ = $\frac {13}{CO}$, so $BO$ = $5x$ and $CO$ = $13x$, and $BO$ + $OC$ = $BC$ = $12$, so $x$ = $\frac {2}{3}$, and $OC$ = $\frac {26}{3}$. Let $P$ be the foot of the altitude from $D$ to $OC$. It can be seen that triangle $DOP$ is similar to triangle $AOB$, and triangle $DPC$ is similar to triangle $ABC$. If $DP$ = $15y$, then $CP$ = $36y$, $OP$ = $10y$, and $OD$ = $5y\sqrt {13}$. Since $OP$ + $CP$ = $46y$ = $\frac {26}{3}$, $y$ = $\frac {13}{69}$, and $AD$ = $\frac {60\sqrt{13}}{23}$ (by the pythagorean theorem on triangle $ABO$ we sum $AO$ and $OD$). The answer is $60$ + $13$ + $23$ = $\boxed{096}$.

Solution 2

Extend $AB$ and $CD$ to intersect at $P$. Note that since $\angle ACB=\angle PCB$ and $\angle ABC=\angle PBC=90^{\circ}$ by ASA congruency we have $\triangle ABC\cong \triangle PBC$. Therefore $AC=PC=13$.

By the angle bisector theorem, $PD=\dfrac{130}{23}$ and $CD=\dfrac{169}{23}$. Now we apply Stewart's theorem to find $AD$:

\begin{align*}13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=13\cdot 13\cdot \dfrac{130}{23}+10\cdot 10\cdot \dfrac{169}{23}\\ 13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=\dfrac{169\cdot 130+169\cdot 100}{23}\\ 13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=1690\\ AD^2&=130-\dfrac{130\cdot 169}{23^2}\\ AD^2&=\dfrac{130\cdot 23^2-130\cdot 169}{23^2}\\ AD^2&=\dfrac{130(23^2-169)}{23^2}\\ AD^2&=\dfrac{130(360)}{23^2}\\ AD&=\dfrac{60\sqrt{13}}{23}\\ \end{align*}

and our final answer is $60+13+23=\boxed{096}$.

Solution 3

Notice that by extending $AB$ and $CD$ to meet at a point $E$, $\triangle ACE$ is isosceles. Now we can do a straightforward coordinate bash. Let $C=(0,0)$, $B=(12,0)$, $E=(12,-5)$ and $A=(12,5)$, and the equation of line $CD$ is $y=-\dfrac{5}{12}x$. Let F be the intersection point of $AD$ and $BC$, and by using the Angle Bisector Theorem: $\dfrac{BF}{AB}=\dfrac{FC}{AC}$ we have $FC=\dfrac{26}{3}$. Then the equation of the line $AF$ through the points $(12,5)$ and $\left(\frac{26}{3},0\right)$ is $y=\frac32 x-13$. Hence the intersection point of $AF$ and $CD$ is the point $D$ at the coordinates $\left(\dfrac{156}{23},-\dfrac{65}{23}\right)$. Using the distance formula, $AD=\sqrt{\left(12-\dfrac{156}{23}\right)^2+\left(5+\dfrac{65}{23}\right)^2}=\dfrac{60\sqrt{13}}{23}$ for an answer of $60+13+23=\fbox{096}$.

Solution 4

After drawing a good diagram, we reflect $ABC$ over the line $BC$, forming a new point that we'll call $A'$. Also, let the intersection of $AD$ and $BC$ be point $E$. Point $D$ lies on line $A'C$. Since line $AD$ bisects $\angle{CAB}$, we can use the Angle Bisector Theorem. $AA'=10$ and $AC=13$, so $\frac{CD}{A'D}=\frac{13}{10}$. Letting the segments be $13x$ and $10x$ respectively, we now have $13x+10x=13$. Therefore, $x=\frac{13}{23}$. By the Pythagorean Theorem, $AE=\frac{5\sqrt{13}}{3}$. Using the Angle Bisector Theorem on $\angle{ACD}$, we have that $ED=\frac{5x\sqrt{13}}{3}$. Substituting in $x=\frac{13}{23}$, we have that $AD=\left(\frac{5\sqrt{13}}{3}\right)(1+x)=\frac{60\sqrt{13}}{23}$, so the answer is $60+13+23=\boxed{096}$.

-RootThreeOverTwo

Solution 5 (Angle Bisector Theorem + Law of Cosines)

Let $CD$ and $AB$ meet at $A'$; then $\overline{AD}$ is an angle bisector of isosceles $\triangle AA'C$. Then by the Angle Bisector Theorem, $A'D=\frac {10}{10+13} \cdot 13=\frac {130}{23}$, and $\cos \angle DA'A=\frac {5}{13}$. By the Law of Cosines on $\triangle AA'D$, we have \[AD^2=10^2+\left(\frac {130}{23}\right)^2-2 \cdot 10 \cdot \frac {130}{23} \cdot \frac {5}{13}=\frac {2^4 \cdot 3^2 \cdot 5^2 \cdot 13}{23^2} \Longrightarrow AD=\frac {60\sqrt {13}}{23}\] and the answer is $60+13+23=\boxed{096}$.


[asy] size(120); pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(10); pair B=(0,0), A=(5,0), C=(0,13), E=(-5,0), O = incenter(E,C,A), D=IP(A -- A+3*(O-A),E--C); D(A--B--C--cycle); D(A--D--C); D(D--E--B, linetype("4 4")); MP("5", (A+B)/2, f); MP("13", (A+C)/2, NE,f); MP("A",D(A),f); MP("B",D(B),f); MP("C",D(C),N,f); MP("A'",D(E),f); MP("D",D(D),NW,f); D(rightanglemark(C,B,A,20)); D(anglemark(D,A,E,35));D(anglemark(C,A,D,30)); [/asy]

-azjps

Solution 6 (Law of Sines)

[asy]pathpen = linewidth(0.7); pen f = fontsize(10); size(144); pair B=(0,0); pair A=(5,0); pair C=(0,12); pair E=(-5,0); pair abA = A+unit(C-A)+unit(B-A); pair D = extension(A,abA,C,E); D(A--C); D(C--B); D(A--B); D(C--D); D(A--D); D(D--E,dashed); D(B--E,dashed); D(rightanglemark(C,B,A,20),red); MP("A",D(A),plain.SE,f); MP("B",D(B),plain.S,f); MP("C",D(C),plain.N,f); MP("D",D(D),plain.NW,f); MP("E",D(E),plain.SW,f); MP("5",(A+B)/2,plain.S,f); MP("12",(B+C)/2,plain.E,f); MP("13",(A+C)/2,plain.NE,f); MP("\alpha",A,4*(unit(D-A)+unit(C-A))/2,f); MP("\alpha",A,4*(unit(B-A)+unit(D-A))/2,f); MP("2\alpha",E,4*(unit(C-E)+unit(B-E))/2,f); MP("\beta",C,6*(unit(A-C)+unit(B-C))/2,f); MP("\beta",C,6*(unit(E-C)+unit(B-C))/2,f);[/asy]

Using the law of sines on $\bigtriangleup ADE$,

\begin{align*} AD &= AE \cdot \dfrac{\sin(2\alpha)}{\sin(180^\circ-3\alpha)}\\ &= AE \cdot \dfrac{\sin(2\alpha)}{\sin(\alpha+2\alpha)}\\ &= 10 \cdot \dfrac{\sin(2\alpha)}{\sin(\alpha)\cos(2\alpha)+\cos(\alpha)\sin(2\alpha)}\\ &= 10 \cdot \dfrac{\sin(2\alpha)}{\sqrt{\dfrac{1-\cos(2\alpha)}{2}}\cdot\cos(2\alpha)+\sqrt{\dfrac{1+\cos(2\alpha)}{2}}\cdot\sin(2\alpha)}\\ &= 10 \cdot \dfrac{\dfrac{12}{13}}{\sqrt{\dfrac{8}{26}}\cdot\dfrac{5}{13}+\sqrt{\dfrac{18}{26}}\cdot\dfrac{12}{13}}\\ &= 10 \cdot \dfrac{12\sqrt{13}}{10+36} = \dfrac{60\sqrt{13}}{23} \end{align*}

$\therefore$ the answer is $60+13+23=\boxed{096}$.

-m1sterzer0

See Also

2009 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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