Difference between revisions of "2011 AIME I Problems/Problem 14"
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===Solution 2=== | ===Solution 2=== | ||
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+2\sqrt{2}, \cos x = \dfrac{-2}{2+2\sqrt{2}}= 1-\sqrt{2}</math>,<math>\cos 2x = 2\cos^2 x -1 = 5 - 4\sqrt{2} = 5-\sqrt{32}</math>, and <math>m+n=\boxed{037}</math>. | 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+2\sqrt{2}, \cos x = \dfrac{-2}{2+2\sqrt{2}}= 1-\sqrt{2}</math>,<math>\cos 2x = 2\cos^2 x -1 = 5 - 4\sqrt{2} = 5-\sqrt{32}</math>, and <math>m+n=\boxed{037}</math>. | ||
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+ | ===Solution 3=== | ||
+ | Notice that <math>R_3</math> and <math>R_7</math> are parallel (<math>B_1B_3B_5B_7</math> is a square by symmetry and since the rays are perpendicular) and <math>B_1B_3=B_3B_5=s=</math> the distance between the parallel rays. If the regular hexagon as a side length of <math>s</math>, then <math>M_3M_7</math> has a length of <math>s+s\sqrt{2}</math>. Let <math>X</math> be on <math>R_3</math> such that <math>M_7X</math> is perpendicular to <math>M_3X</math>, and <math>\phi=\angle M_7M_3X</math>. The distance between <math>R_3</math> and <math>R_7</math> is <math>s=M_7X</math>, so <math>\sin\phi=\frac{s}{s+s\sqrt{2}}=\frac{1}{1+\sqrt{2}}</math>. | ||
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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>. | ||
==Diagram== | ==Diagram== |
Revision as of 19:03, 31 July 2018
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 .
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.