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

Revision as of 18:13, 25 September 2007

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

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 $046$ .

@@ Line 23: / Line 23: @@
 == See also ==
-{{AIME box|year=2006|n=I|before=First Question|num-a=2}}
+{{AIME box|year=2006|n=II|before=First Question|num-a=2}}
 [[Category:Intermediate Geometry Problems]]

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

Revision as of 18:13, 25 September 2007

Problem

Solution

See also