# Difference between revisions of "2006 AIME I Problems/Problem 1"

## Problem

In convex hexagon $ABCDEF$, all six sides are congruent, $\angle A$ and $\angle D$ are right angles, and $\angle B, \angle C, \angle E,$ and $\angle F$ are congruent. The area of the hexagonal region is $2116(\sqrt{2}+1).$ Find $AB$.

## Solution

Let the side length be called $x$.

Then $AB=BC=CD=DE=EF=AF=x$.

The diagonal $BF=\sqrt{AB^2+AF^2}=\sqrt{x^2+x^2}=x\sqrt{2}$.

Then the areas of the triangles AFB and CDE in total are $\frac{x^2}{2}\cdot 2$, and the area of the rectangle BCEF equals $x\cdot x\sqrt{2}=x^2\sqrt{2}$

Then we have to solve the equation

$2116(\sqrt{2}+1)=x^2\sqrt{2}+x^2$.

$2116(\sqrt{2}+1)=x^2(\sqrt{2}+1)$

$2116=x^2$

$x=46$

Therefore, AB is 46.

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