Difference between revisions of "2011 AIME I Problems/Problem 14"
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− | 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 \left(2\theta + \frac{\pi}{2} \right)=-\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>. | 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>. |
Revision as of 00:54, 8 January 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 .
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.