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

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Let O be the intersection of BC and AD. By the [[Angle Bisector Theorem]], 5/BO = 13/CO, so BO = 5x and CO = 13x, and BO + OC = BC = 12, so x = 2/3, and OC = 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 = (5*sqrt(13))*y. Since OP + CP = 46y = 26/3, y = 13/69, and AD = (60*sqrt (13))/23. The answer is 60 + 13 + 23 = 096.
 
Let O be the intersection of BC and AD. By the [[Angle Bisector Theorem]], 5/BO = 13/CO, so BO = 5x and CO = 13x, and BO + OC = BC = 12, so x = 2/3, and OC = 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 = (5*sqrt(13))*y. Since OP + CP = 46y = 26/3, y = 13/69, and AD = (60*sqrt (13))/23. The answer is 60 + 13 + 23 = 096.
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== See Also ==
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{{AIME box|year=2009|n=II|num-b=9|num-a=11}}

Revision as of 17:39, 17 April 2009

Four lighthouses are located at points , , , and . The lighthouse at is kilometers from the lighthouse at , the lighthouse at is kilometers from the lighthouse at , and the lighthouse at is kilometers from the lighthouse at . To an observer at , the angle determined by the lights at and and the angle determined by the lights at and are equal. To an observer at , the angle determined by the lights at and and the angle determined by the lights at and are equal. The number of kilometers from to is given by , where , , and are relatively prime positive integers, and is not divisible by the square of any prime. Find .


Solution

Let O be the intersection of BC and AD. By the Angle Bisector Theorem, 5/BO = 13/CO, so BO = 5x and CO = 13x, and BO + OC = BC = 12, so x = 2/3, and OC = 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 = (5*sqrt(13))*y. Since OP + CP = 46y = 26/3, y = 13/69, and AD = (60*sqrt (13))/23. The answer is 60 + 13 + 23 = 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