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Difference between revisions of "2009 AIME II Problems/Problem 10"
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== Solution ==
 
== Solution ==
  
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Let <math>O</math> be the intersection of <math>BC</math> and <math>AD</math>. By the [[Angle Bisector Theorem]], <math>5</math>/<math>BO</math> = <math>13</math>/<math>CO</math>, so <math>BO</math> = <math>5x</math> and <math>CO</math> = <math>13x</math>, and <math>BO</math> + <math>OC</math> = <math>BC</math> = <math>12</math>, so <math>x</math> = <math>2/3</math>, and <math>OC</math> = <math>26/3</math>. Let <math>P</math> be the altitude from <math>D</math> to <math>OC</math>. It can be seen that triangle <math>DOP</math> is similar to triangle <math>AOB</math>, and triangle <math>DPC</math> is similar to triangle <math>ABC</math>. If <math>DP</math> = <math>15y</math>, then <math>CP</math> = <math>36y</math>, <math>OP</math> = <math>10y</math>, and <math>OD</math> = (<math>5</math>*sqrt(<math>13</math>))*<math>y</math>. Since <math>OP</math> + <math>CP</math> = <math>46y</math> = <math>26/3</math>, <math>y</math> = <math>13/69</math>, and <math>AD</math> = (<math>60</math>*sqrt (<math>13</math>))/<math>23</math>. The answer is <math>60</math> + <math>13</math> + <math>23</math> = <math>\boxed{096}</math>.
+
Let <math>O</math> be the intersection of <math>BC</math> and <math>AD</math>. By the [[Angle Bisector Theorem]], <math>\frac {5}{BO}</math> = <math>\frac {13}{CO}, so </math>BO<math> = </math>5x<math> and </math>CO<math> = </math>13x<math>, and </math>BO<math> + </math>OC<math> = </math>BC<math> = </math>12<math>, so </math>x<math> = </math>\frac {2}{3}<math>, and </math>OC<math> = </math>\frac {26}{3}<math>. Let </math>P<math> be the altitude from </math>D<math> to </math>OC<math>. It can be seen that triangle </math>DOP<math> is similar to triangle </math>AOB<math>, and triangle </math>DPC<math> is similar to triangle </math>ABC<math>. If </math>DP<math> = </math>15y<math>, then </math>CP<math> = </math>36y<math>, </math>OP<math> = </math>10y<math>, and </math>OD<math> = </math>5y\sqrt {13}<math>. Since </math>OP<math> + </math>CP<math> = </math>46y<math> = </math>\frac {26}{3}<math>, </math>y<math> = </math>\frac {13}{69}<math>, and </math>AD<math> = </math>\frac {60\sqrt{13}}{23}<math>. The answer is </math>60<math> + </math>13<math> + </math>23<math> = </math>\boxed{096}$.
  
 
== 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}}

Revision as of 20:52, 17 April 2009

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

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 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$AD$=$\frac {60\sqrt{13}}{23}$. The answer is$60$+$13$+$23$=$\boxed{096}$.

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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