1993 AIME Problems/Problem 14
A rectangle that is inscribed in a larger rectangle (with one vertex on each side) is called unstuck if it is possible to rotate (however slightly) the smaller rectangle about its center within the confines of the larger. Of all the rectangles that can be inscribed unstuck in a 6 by 8 rectangle, the smallest perimeter has the form , for a positive integer . Find .
Solution: Put the rectangle on the coordinate plane so its vertices are at , for all four combinations of positive and negative. Then by symmetry, the other rectangle is also centered at the origin, .
Note that such a rectangle is unstuck if its four vertices are in or on the edge of all four quadrants, and it is not the same rectangle as the big one. Let the four vertices of this rectangle be , , and for nonnegative . Then this is a rectangle, so , or , so .
Reflect across the side of the rectangle containing to . Then is constant, and the perimeter of the rectangle is equal to . The midpoint of is , and since and , always lies below .
If is positive, it can be decreased to . This causes to decrease as well, to , where and is still positive. If and are held in place as everything else moves, then moves units up and units left to , which must lie within . Then we must have , and the perimeter of the rectangle is decreased. Therefore, the minimum perimeter must occur with , so .
By the distance formula, this minimum perimeter is
Note that the diagonal of the rectangle with minimum perimeter must have the diagonal along the middle segment of length 8 of the rectangle (any other inscribed rectangle can be rotated a bit, then made smaller; this one can't because then the rectangle cannot be inscribed since its longest diagonal is less than 8 in length). Then since a rectangle must have right angles, we draw a circle of radius 4 around the center of the rectangle. Picking the two midpoints on the sides of length 6 and opposite intersection points on the segments of length 8, we form a rectangle. Let and be the sides of the rectangle. Then since both are twice the area of the same right triangle, and . So , so .
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