Difference between revisions of "2009 AIME II Problems/Problem 10"
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== 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 = <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>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>. |
== 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 19:44, 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 
be the intersection of 
and 
. By the Angle Bisector Theorem, 
/
= 
/
, so = 
and = 
, and + 
= = 
, so 
= 
, and = 
. Let 
be the altitude from 
to . It can be seen that triangle 
is similar to triangle 
, and triangle 
is similar to triangle 
. If 
= 
, then 
= 
, 
= 
, and 
= (*sqrt())*
. Since + = 
= , = 
, and = (
*sqrt ())/
. The answer is + + = 
.
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 | ||
