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Difference between revisions of "2006 AIME A Problems/Problem 1"

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<math>x=46</math>
 
<math>x=46</math>
  
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Enter 046 in the answer circle.
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The answer is therefore 046.
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--[[User:Someperson01|Someperson01]] 21:27, 13 July 2006 (EDT)
 
  
 
== See also ==
 
== See also ==
 
*[[2006 AIME II Problems]]
 
*[[2006 AIME II Problems]]

Revision as of 16:35, 15 July 2006

Problem

In convex hexagon $ABCDEF$, all six sides are congruent, $\angle A$ and $\angle D$ are right angles, and $\angle B, \angle C, \angle D, \angle E,$ 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 $x$. Diagram1.png

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

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

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

$2116=x^2$

$x=46$

The answer is therefore 046.

See also

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