Difference between revisions of "2011 AIME I Problems/Problem 14"
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== Solution == | == Solution == | ||
+ | ===Solution 1=== | ||
Let <math>\theta=\angle M_1 M_3 B_1</math>. Thus we have that <math>\cos 2 \angle A_3 M_3 B_1=\cos(2\theta + \frac{\pi}{2})=-\sin2\theta</math>. | Let <math>\theta=\angle M_1 M_3 B_1</math>. Thus we have that <math>\cos 2 \angle A_3 M_3 B_1=\cos(2\theta + \frac{\pi}{2})=-\sin2\theta</math>. | ||
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Thus we have that <math>\cos 2 \angle A_3 M_3 B_1=5-4\sqrt2=5-\sqrt{32}</math>. Therefore <math>m+n=5+32=\boxed{037}</math>. | Thus we have that <math>\cos 2 \angle A_3 M_3 B_1=5-4\sqrt2=5-\sqrt{32}</math>. Therefore <math>m+n=5+32=\boxed{037}</math>. | ||
− | + | ===Solution 2=== | |
− | 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= | + | Let <math>A_1A_2 = 2</math>. Then <math>B_1</math> and <math>B_3</math> are the projections of <math>M_1</math> and <math>M_5</math> onto the line <math>B_1B_3</math>, so <math>2=B_1B_3=M_1M_5\cos x</math>, where <math>x = \angle A_3M_3B_1</math>. Then since <math>M_1M_5 = 2+\sqrt{2}, \cos x = \dfrac{2}{2+\sqrt{2}}= \sqrt{2} -1</math>,<math>\cos 2x = 2\cos^2 x -1 = 5 - 4\sqrt{2} = 5-\sqrt{32}</math>, and <math>m+n=\boxed{037}</math>. |
== See also == | == See also == | ||
{{AIME box|year=2011|n=I|num-b=13|num-a=15}} | {{AIME box|year=2011|n=I|num-b=13|num-a=15}} |
Revision as of 16:54, 3 July 2011
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
Solution 1
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
.
Solution 2
Let . Then
and
are the projections of
and
onto the line
, so
, where
. Then since
,
, and
.
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 |