Question

A rectangle has sides of length $a$ 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 $a$ 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 $a$ parallel and separated by a distance of $24,$ the hexagon has the same area as the original rectangle. Find $a^2.$

Answer

Answer: 720

Concepts:Pythagorean Theorem trapezoid area

Difficulty rating: 2110

Solution:

In the hexagon, each side of length $36$ has folded at its midpoint into two bars of length $18.$ The two sides of length $a$ are $24$ apart, so each bar spans a vertical distance of $12$ and hence a horizontal distance of $\sqrt{18^2 - 12^2} = \sqrt{180} = 6\sqrt{5}.$

The line through the two midpoint hinges splits the hexagon into two congruent trapezoids with parallel sides $a$ and $a + 12\sqrt{5}$ and height $12,$ so the hexagon has area $\begin{aligned} &2 \cdot \frac{a + (a + 12\sqrt{5})}{2} \cdot 12 \\ &= 24a + 144\sqrt{5}. \end{aligned}$

Setting this equal to the rectangle's area $36a$ gives $12a = 144\sqrt{5},$ so $a = 12\sqrt{5}$ and $a^2 = 720.$

2014 AIME II Problem 3

Answer: 720

Difficulty rating: 2110

Solution:

Problem 3 in Other Years