2013 AIME I Problems/Problem 12
Let be a triangle with and . A regular hexagon with side length 1 is drawn inside so that side lies on , side lies on , and one of the remaining vertices lies on . There are positive integers and such that the area of can be expressed in the form , where and are relatively prime, and c is not divisible by the square of any prime. Find .
First, find that . Draw . Now draw around such that is adjacent to and . The height of is , so the length of base is . Let the equation of be . Then, the equation of is . Solving the two equations gives . The area of is .
Cartesian Variation Solution
Also use a coordinate system, but upon drawing the diagram call the origin and be on the x-axis. It is easy to see that is the vertex on . After labeling coordinates (noting additionally that is an equilateral triangle), we see that the area is times times the ordinate of . Draw a perpendicular of , call it , and note that after using the trig functions for degrees.
Now, get the lines for and : and , whereupon we get the ordinate of to be , and the area is , so our answer is .
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