Difference between revisions of "2006 AIME A Problems/Problem 1"
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== Solution == | == 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 == | == See also == | ||
*[[2006 AIME II Problems]] | *[[2006 AIME II Problems]] |
Revision as of 21:27, 13 July 2006
Problem
In convex hexagon , all six sides are congruent,
and
are right angles, and
and
are congruent. The area of the hexagonal region is
Find
.
Solution
Let the side length be called .
Then .
The diagonal .
Then the areas of the triangles AFB and CDE in total are ,
and the area of the rectangle BCEF equals
Then we have to solve the equation
.
Enter 046 in the answer circle.
--Someperson01 21:27, 13 July 2006 (EDT)