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

(Solution)
Line 4: Line 4:
 
== Solution ==
 
== 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.
+
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 = <math>\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}}

Revision as of 17:46, 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 (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, 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 = $\boxed{096}$.

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