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
Solution 1
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.
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 | ||
