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>. | ||
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+ | ===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
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. ! Undefined control sequence.)
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 |
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