Difference between revisions of "2006 AIME II Problems/Problem 1"
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Let the side length be called <math>x</math>, so <math>x=AB=BC=CD=DE=EF=AF</math>. | Let the side length be called <math>x</math>, so <math>x=AB=BC=CD=DE=EF=AF</math>. | ||
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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>, | 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>, | ||
Revision as of 19:37, 25 September 2007
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 
, so 
.
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
.
Therefore, is 
.
See also
| 2006 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
