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

Revision as of 20:27, 13 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 $x$ .

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

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

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

$2116=x^2$

$x=46$

Enter 046 in the answer circle.

--Someperson01 21:27, 13 July 2006 (EDT)

@@ Line 4: / Line 4: @@
 == Solution ==
+Let the side length be called <math>x</math>.
+[[Image:Diagram1.png]]
+Then <math>AB=BC=CD=DE=EF=AF=x</math>.
+The diagonal <math>BF=\sqrt{AB^2+AF^2}=\sqrt{x^2+x^2}=x\sqrt{2}</math>.
+Then the areas of the triangles AFB and CDE in total are <math>\frac{x^2}{2}\cdot 2</math>,
+and the area of the rectangle BCEF equals <math>x\cdot x\sqrt{2}=x^2\sqrt{2}</math>
+Then we have to solve the equation
+<math>2116(\sqrt{2}+1)=x^2\sqrt{2}+x^2</math>.
+<math>2116(\sqrt{2}+1)=x^2(\sqrt{2}+1)</math>
+<math>2116=x^2</math>
+<math>x=46</math>
+Enter 046 in the answer circle.
+--[[User:Someperson01|Someperson01]] 21:27, 13 July 2006 (EDT)
 == See also ==
 *[[2006 AIME II Problems]]

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

Revision as of 20:27, 13 July 2006

Problem

Solution

See also