Difference between revisions of "2009 AIME II Problems/Problem 10"
Aimesolver (talk | contribs) (→Solution) |
|||
| Line 9: | Line 9: | ||
{{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}} | ||
Revision as of 22:36, 4 July 2013
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 
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 
= 
. Since + = 
= , 
= 
, and = 
. 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
