Difference between revisions of "2006 AIME I Problems/Problem 1"
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== Solution == | == Solution == | ||
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− | + | 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> | |
− | + | Therefore, AB is 46. | |
== See also == | == See also == |
Revision as of 13:54, 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 .
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
.
Therefore, AB is 46.
See also
2006 AIME I (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 |