Difference between revisions of "2011 AIME I Problems/Problem 14"
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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>. | 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 | + | 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>. | 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>. |
Revision as of 17:14, 4 July 2020
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 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 thatThus, 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.