1997 AIME Problems/Problem 15
The sides of rectangle have lengths and . An equilateral triangle is drawn so that no point of the triangle lies outside . The maximum possible area of such a triangle can be written in the form , where , , and are positive integers, and is not divisible by the square of any prime number. Find .
Solution 1 (Coordinate Bash)
Consider points on the complex plane . Since the rectangle is quite close to a square, we figure that the area of the equilateral triangle is maximized when a vertex of the triangle coincides with that of the rectangle. Set one vertex of the triangle at , and the other two points and on and , respectively. Let and . Since it's equilateral, then , so , and expanding we get .
We can then set the real and imaginary parts equal, and solve for . Hence a side of the equilateral triangle can be found by . Using the area formula , the area of the equilateral triangle is . Thus .
This is a trigonometric re-statement of the above. Let ; by alternate interior angles, . Let and the side of the equilateral triangle be , so by the Pythagorean Theorem. Now . This reduces to .
Thus, the area of the triangle is , which yields the same answer as above.
Since and , it follows that . Rotate triangle degrees clockwise. Note that the image of is . Let the image of be . Since angles are preserved under rotation, . It follows that . Since , it follows that quadrilateral is cyclic with circumdiameter and thus circumradius . Let be its circumcenter. By Inscribed Angles, . By the definition of circle, . It follows that triangle is equilateral. Therefore, . Applying the Law of Cosines to triangle , . Squaring and multiplying by yields
-Solution by thecmd999
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