2011 AIME I Problems/Problem 14
Problem
Let be a regular octagon. Let , , , and be the midpoints of sides , , , and , respectively. For , ray is constructed from towards the interior of the octagon such that , , , and . Pairs of rays and , and , and , and and meet at , , , respectively. If , then can be written in the form , where and are positive integers. Find .
Solution
Let . Thus we have that .
Since is a regular octagon and , let .
Extend and until they intersect. Denote their intersection as . Through similar triangles & the triangles formed, we find that .
We also have that through ASA congruence (, , ). Therefore, we may let .
Thus, we have that and that . Therefore .
Squaring gives that and consequently that through the identities and .
Thus we have that . Therefore .
Alternate Solution: Let A_1A_2 = 2. Then B_1 and B_3 are the projections of M_1 and M_5 onto the line B_1B_3, so 2=B_1B_3=(M_1M_5)(cos x), where x = angle A_3M_3B_1. Then since M_1M_5 = 2+ sqrt 2, cos x = 2/(2+sqrt 2)= sqrt 2 -1, cos 2x = 2*(cos x)^2 -1 = 5 - 4 sqrt 2 = 5- sqrt 32, m+n=37.
See also
2011 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |