Difference between revisions of "2011 AIME I Problems/Problem 14"
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− | Let <math> | + | == Problem == |
+ | Let <math>A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8</math> be a regular octagon. Let <math>M_1</math>, <math>M_3</math>, <math>M_5</math>, and <math>M_7</math> be the midpoints of sides <math>\overline{A_1 A_2}</math>, <math>\overline{A_3 A_4}</math>, <math>\overline{A_5 A_6}</math>, and <math>\overline{A_7 A_8}</math>, respectively. For <math>i = 1, 3, 5, 7</math>, ray <math>R_i</math> is constructed from <math>M_i</math> towards the interior of the octagon such that <math>R_1 \perp R_3</math>, <math>R_3 \perp R_5</math>, <math>R_5 \perp R_7</math>, and <math>R_7 \perp R_1</math>. Pairs of rays <math>R_1</math> and <math>R_3</math>, <math>R_3</math> and <math>R_5</math>, <math>R_5</math> and <math>R_7</math>, and <math>R_7</math> and <math>R_1</math> meet at <math>B_1</math>, <math>B_3</math>, <math>B_5</math>, <math>B_7</math> respectively. If <math>B_1 B_3 = A_1 A_2</math>, then <math>\cos 2 \angle A_3 M_3 B_1</math> can be written in the form <math>m - \sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers. Find <math>m + n</math>. | ||
+ | |||
+ | == Solution == | ||
+ | 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>. | ||
+ | |||
+ | Since <math>A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8</math> is a regular octagon and <math>B_1 B_3 = A_1 A_2</math>, let <math>k=A_1 A_2 = A_2 A_3 = B_1 B_3</math>. | ||
+ | |||
+ | |||
+ | Extend <math>\overline{A_1 A_2}</math> and <math>\overline{A_5 A_6}</math> until they intersect. Denote their intersection as <math>I_1</math>. Through similar triangles & the <math>45-45-90</math> triangles formed, we find that <math>M_1 M_3=\frac{k}{2}(2+\sqrt2)</math>. | ||
+ | |||
+ | We also have that<math>\triangle M_7 B_7 M_1 =\triangle M_1 B_1 M_3</math> through ASA congruence (<math>\angle B_7 M_7 M_1 =\angle B_1 M_1 M_3</math>, <math>M_7 M_1 = M_1 M_3</math>, <math>\angle B_7 M_1 M_7 =\angle B_1 M_3 M_1</math>). Therefore, we may let <math>n=M_1 B_7 = M_3 B_1</math>. | ||
+ | |||
+ | Thus, we have that <math>\sin\theta=\frac{n+k}{\frac{k}{2}(2+\sqrt2)}</math> and that <math>\cos\theta=\frac{n}{\frac{k}{2}(2+\sqrt2)}</math>. Therefore <math>\sin\theta-\cos\theta=\frac{k}{\frac{k}{2}(2+\sqrt2)}=\frac{2}{2+\sqrt2}=2-\sqrt2</math>. | ||
+ | |||
+ | Squaring gives that <math>\sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta = 6-4\sqrt2</math> and consequently that <math>-2\sin\theta\cos\theta = 5-4\sqrt2 = -\sin2\theta</math> through the identities <math>\sin^2\theta + \cos^2\theta = 1</math> and <math>\sin2\theta = 2\sin\theta\cos\theta</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>. |
Revision as of 14:40, 20 March 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
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
.