1997 AIME Problems/Problem 15
Problem
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 .
Solution 2
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.
Solution 3
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
See also
1997 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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.