2014 AIME I Problems/Problem 15
Problem 15
In 
, 
, 
, and 
. Circle 
intersects 
at 
and 
, 
at and 
, and 
at 
and 
. Given that 
and 
, length 
, where 
and 
are relatively prime positive integers, and 
is a positive integer not divisible by the square of any prime. Find 
.
Solution
First we note that triangle DEF is an isosceles right triangle with hypotenuse DE the same as the diameter of "omega". We also note that triangle "DGE similar to ABC" since angle EGD is a right angle and the ratios of the sides are 3:4:5. From congruent arc intersections, we know that angle GED congruent to angle GBC, and that from similar triangles angle GED also congruent to angle GCB. Thus, triangle BGC is an isosceles triangle with BG = GC, so G is the midpoint of AC and AG = GC = 5/2. Similarly, we can find from angle chasing that BF is the angle bisector of B, so from the angle bisector theorem we have AF/AB = CF/CB, so AF = 15/7 and CF = 20/7. Lastly, we apply power of a point from points A and C with respect to "omega" and have AE*AB=AF*AG and 
, so we can compute that EB = 17/14 and DB = 31/14. From Pythagorean Theorem, we result in 
, so 
See also
| 2014 AIME I (Problems • Answer Key • Resources) | ||
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