2011 AIME I Problems/Problem 14
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 .
Diagram
All distances are to scale.
Solution 1
We use coordinates. Let the octagon have side length and center . Then all of its vertices have the form or .
By symmetry, is a square. Thus lines and are parallel, and its side length is the distance between these two lines. However, this is given to be the side length of the octagon, or .
Suppose the common slope of the lines is and let . Then, we want to find
It can easily be seen that the equations of the lines are By the distance between parallel lines formula, a corollary of the point to line distance formula, the distance between these two lines is Since we want this to equal , we have Since we have . Thus The answer is .
Solution 2
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 3
Let . Then and are the projections of and onto the line , so , where . Then since , , and .
Solution 4
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 5
Assume that Denote the center , and the midpoint of and as . Then we have thatThus, by the cosine double-angle theorem,so .
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 | ||
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