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 |