Difference between revisions of "2011 AIME I Problems/Problem 14"
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Since we are considering a regular hexagon, <math>M_3</math> is directly opposite to <math>M_7</math> and <math>\angle A_3M_3B_1=90 ^\circ +\phi</math>. All that's left is to calculate <math>\cos 2\angle A_3M_3B_1=\cos^2(90^\circ+\phi)-\sin^2(90^\circ+\phi)=\sin^2\phi-\cos^2\phi</math>. By drawing a right triangle or using the Pythagorean identity, <math>\cos^2\phi=\frac{2+2\sqrt2}{3+2\sqrt2}</math> and <math>\cos 2\angle A_3M_3B_1=\frac{-1-2\sqrt2}{3+2\sqrt2}=5-4\sqrt2=5-\sqrt{32}</math>, so <math>m+n=\boxed{037}</math>. | Since we are considering a regular hexagon, <math>M_3</math> is directly opposite to <math>M_7</math> and <math>\angle A_3M_3B_1=90 ^\circ +\phi</math>. All that's left is to calculate <math>\cos 2\angle A_3M_3B_1=\cos^2(90^\circ+\phi)-\sin^2(90^\circ+\phi)=\sin^2\phi-\cos^2\phi</math>. By drawing a right triangle or using the Pythagorean identity, <math>\cos^2\phi=\frac{2+2\sqrt2}{3+2\sqrt2}</math> and <math>\cos 2\angle A_3M_3B_1=\frac{-1-2\sqrt2}{3+2\sqrt2}=5-4\sqrt2=5-\sqrt{32}</math>, so <math>m+n=\boxed{037}</math>. | ||
+ | |||
+ | |||
+ | ===Solution 4=== | ||
+ | Assume that <math>A_1A_2=1.</math> | ||
+ | Denote the center <math>O</math>, and the midpoint of <math>B_1</math> and <math>B_3</math> as <math>B_2</math>. Then we have that<cmath>\cos\angle A_3M_3B_1=\cos(\angle A_3M_3O+\angleOM_3B_1)=-\sin(\angle OM_3B_1)=-\frac{OB_2}{OM_3}=-\frac{\sqrt2/4}{1/2+\sqrt2/2}=-\frac{1}{\sqrt2+1}=1-\sqrt2.</cmath>Thus, by the cosine double-angle theorem,<cmath>\cos2\angle A_3M_3B_1=2(1-\sqrt2)^2-1=5-\sqrt{32},</cmath>so <math>m+n=\boxed{037}</math>. | ||
==Diagram== | ==Diagram== |
Revision as of 19:30, 10 March 2019
Contents
[hide]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
.
Solution 3
Notice that and
are parallel (
is a square by symmetry and since the rays are perpendicular) and
the distance between the parallel rays. If the regular hexagon as a side length of
, then
has a length of
. Let
be on
such that
is perpendicular to
, and
. The distance between
and
is
, so
.
Since we are considering a regular hexagon, is directly opposite to
and
. All that's left is to calculate
. By drawing a right triangle or using the Pythagorean identity,
and
, so
.
Solution 4
Assume that
Denote the center , and the midpoint of
and
as
. Then we have that
\[\cos\angle A_3M_3B_1=\cos(\angle A_3M_3O+\angleOM_3B_1)=-\sin(\angle OM_3B_1)=-\frac{OB_2}{OM_3}=-\frac{\sqrt2/4}{1/2+\sqrt2/2}=-\frac{1}{\sqrt2+1}=1-\sqrt2.\] (Error compiling LaTeX. Unknown error_msg)
Thus, by the cosine double-angle theorem,so
.
Diagram
All distances are to scale.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.