Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.6 AIME, harder Find the answer

3. On the sides $\overline{B C}$ and $\overline{C D}$ of the rectangle $A B C D$ , points $G$ and $E$ are given such that $|D E|=|B G|$ and $|D E|=\frac{1}{2}|E C|$ . Point $F$ is inside the rectangle $A B C D$ such that $C E F G$ is also a rectangle.

If the area of the hexagon $A B G F E D$ (the shaded region in the figure) is equal to the area of the rectangle $C E F G$ (the unshaded part in the figure), how many times longer is $\overline{B C}$ than $\overline{D E}$ ?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

First method:

Let point $H$ be such that $E D H F$ and $A B G H$ are rectangles. Denote their areas by $P_{1}$ and $P_{2}$ , respectively, and the area of rectangle $C E F G$ by $P_{3}$ .

We have

$P_{1}=x \cdot y, \quad P_{2}=3 x \cdot x, \quad P_{3}=2 x \cdot y$

From the condition, we know that $P_{1}+P_{2}=P_{3}$ , which leads to the equation: $x \cdot y+3 x \cdot x=2 x \cdot y$ .

On the left side, apply the distributive property of multiplication over addition, and then divide both sides of the equation by $x$ (which is permissible since the length $x$ is not 0):

$x \cdot(y+3 x)=2 x \cdot y$

$y+3 x=2 y, \text{i.e.} \quad$ we get $y=3 x$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.