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| − | == Problem ==
| + | #REDIRECT [[2006 AIME I Problems/Problem 1]] |
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| − | In convex hexagon <math>ABCDEF</math>, all six sides are congruent, <math>\angle A</math> and <math>\angle D</math> are right angles, and <math>\angle B, \angle C, \angle D, \angle E,</math> and <math>\angle F</math> are congruent. The area of the hexagonal region is <math>2116(\sqrt{2}+1).</math> Find <math>AB</math>.
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| − | == Solution ==
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| − | Let the side length be called <math>x</math>.
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| − | [[Image:Diagram1.png]]
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| − | Then <math>AB=BC=CD=DE=EF=AF=x</math>.
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| − | The diagonal <math>BF=\sqrt{AB^2+AF^2}=\sqrt{x^2+x^2}=x\sqrt{2}</math>.
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| − | Then the areas of the triangles AFB and CDE in total are <math>\frac{x^2}{2}\cdot 2</math>,
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| − | and the area of the rectangle BCEF equals <math>x\cdot x\sqrt{2}=x^2\sqrt{2}</math>
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| − | Then we have to solve the equation
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| − | <math>2116(\sqrt{2}+1)=x^2\sqrt{2}+x^2</math>.
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| − | <math>2116(\sqrt{2}+1)=x^2(\sqrt{2}+1)</math>
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| − | <math>2116=x^2</math>
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| − | <math>x=46</math>
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| − | Enter 046 in the answer circle.
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| − | --[[User:Someperson01|Someperson01]] 21:27, 13 July 2006 (EDT)
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| − | == See also ==
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| − | *[[2006 AIME II Problems]]
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