AIME 2014 II · 第 3 题

AIME 2014 II — Problem 3

专题: Contest Math
难度: L4
来源: AIME

题目详情

Problem

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 $a^2$ .

$AIME diagram$

解析

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.

$AIME diagram$

By Heron's Formula, the area of each isosceles triangle is $\sqrt{(30)(12)(12)(6)}=\sqrt{180\times 12^2}=72\sqrt{5}$ . So the area of both is $144\sqrt{5}$ . From the rectangle, our original area is $36a$ . The area of the rectangle in the hexagon is $24a$ . So we have

24a+144\sqrt{5}=36a\implies 12a=144\sqrt{5}\implies a=12\sqrt{5}\implies a^2=\boxed{720}.

Solution 2

Alternatively, use basic geometry. First, scale everything down by dividing everything by 6. Let $a/6=p$ . 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 $\sqrt{3^2-2^2}=\sqrt{5}$ given 2 as the base of the constituent right triangles. The two end triangles form a large rectangle of area $\sqrt{5}$ x $4$ . Then, the area of the hexagon is $4p+4\sqrt{5}$ , and the area of the rectangle is $6p$ . Equating them, $p=2\sqrt{5}$ . Multiply by scale factor of 6 and square it to get $36(20)= 720 \implies a^2=\boxed{720}$ .

~BJHHar