2014 AIME II Problems/Problem 3
Contents
Problem
A rectangle has sides of length and 36. A hinge is installed at each vertex of the rectangle, and at the midpoint of each side of length 36. The sides of length can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with the sides of length parallel and separated by a distance of 24, the hexagon has the same area as the original rectangle. Find .
Solution 1
When we squish the rectangle, the hexagon is composed of a rectangle and two isosceles triangles with side lengths 18, 18, and 24 as shown below.
By Heron's Formula, the area of each isosceles triangle is . So the area of both is . From the rectangle, our original area is . The area of the rectangle in the hexagon is . So we have
Solution 2
Alternatively, use basic geometry. First, scale everything down by dividing everything by 6. Let . Then, the dimensions of the central rectangle in the hexagon is p x 4, and the original rectangle is 6 x p. By Pythagorean theorem and splitting the end triangles of the hexagon into two right triangles, the altitude of the end triangles is given 2 as the base of the constituent right triangles. The two end triangles form a large rectangle of area x . Then, the area of the hexagon is , and the area of the rectangle is . Equating them, . Multiply by scale factor of 6 and square it to get .
~BJHHar
See also
2014 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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