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2006 AIME II Problems/Problem 1

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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, so x=AB=BC=CD=DE=EF=AF.

Image:2006_II_AIME-1.png

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 046.

See also

2006 AIME II (ProblemsResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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