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

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Notice that by extending <math>AB</math> and <math>CD</math> to meet at a point <math>E</math>, <math>\triangle ACE</math> is isosceles. Now we can do a straightforward coordinate bash. Let <math>C=(0,0)</math>, <math>B=(12,0)</math>, <math>E=(12,-5)</math> and <math>A=(12,5)</math>, and the equation of line <math>CD</math> is <math>y=-\dfrac{5}{12}x</math>. Let F be the intersection point of <math>AD</math> and <math>BC</math>, and by using the Angle Bisector Theorem: <math>\dfrac{BF}{AB}=\dfrac{FC}{AC}</math> we have <math>FC=\dfrac{26}{3}</math>. Then the equation of the line <math>AF</math> through the points <math>(12,5)</math> and <math>\left(\frac{26}{3},0\right)</math> is <math>y=\frac32 x-13</math>. Hence the intersection point of <math>AF</math> and <math>CD</math> is the point <math>D</math> at the coordinates <math>\left(\dfrac{156}{23},-\dfrac{65}{23}\right)</math>. Using the distance formula, <math>AD=\sqrt{\left(12-\dfrac{156}{23}\right)^2+\left(5+\dfrac{65}{23}\right)^2}=\dfrac{60\sqrt{13}}{23}</math> for an answer of <math>60+13+23=\fbox{096}</math>.
 
Notice that by extending <math>AB</math> and <math>CD</math> to meet at a point <math>E</math>, <math>\triangle ACE</math> is isosceles. Now we can do a straightforward coordinate bash. Let <math>C=(0,0)</math>, <math>B=(12,0)</math>, <math>E=(12,-5)</math> and <math>A=(12,5)</math>, and the equation of line <math>CD</math> is <math>y=-\dfrac{5}{12}x</math>. Let F be the intersection point of <math>AD</math> and <math>BC</math>, and by using the Angle Bisector Theorem: <math>\dfrac{BF}{AB}=\dfrac{FC}{AC}</math> we have <math>FC=\dfrac{26}{3}</math>. Then the equation of the line <math>AF</math> through the points <math>(12,5)</math> and <math>\left(\frac{26}{3},0\right)</math> is <math>y=\frac32 x-13</math>. Hence the intersection point of <math>AF</math> and <math>CD</math> is the point <math>D</math> at the coordinates <math>\left(\dfrac{156}{23},-\dfrac{65}{23}\right)</math>. Using the distance formula, <math>AD=\sqrt{\left(12-\dfrac{156}{23}\right)^2+\left(5+\dfrac{65}{23}\right)^2}=\dfrac{60\sqrt{13}}{23}</math> for an answer of <math>60+13+23=\fbox{096}</math>.
  
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==Solution 4==
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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>.
 
== See Also ==
 
== See Also ==
  
 
{{AIME box|year=2009|n=II|num-b=9|num-a=11}}
 
{{AIME box|year=2009|n=II|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 01:45, 18 March 2019

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 $p$ + $q$ + $r$.


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 $AD=(\frac{5\sqrt{13}}{3})(1+x)=\frac{60\sqrt{13}}{23}$, so the answer is $60+13+23=\boxed{096}$.

See Also

2009 AIME II (ProblemsAnswer KeyResources)
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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