Difference between revisions of "2006 AIME I Problems/Problem 1"
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== Problem == | == Problem == | ||
− | In convex hexagon <math>ABCDEF</math>, all six sides are congruent, <math>\angle A</math> and <math>\angle D</math> are right | + | In [[convex polygon|convex]] [[hexagon]] <math>ABCDEF</math>, all six sides are congruent, <math>\angle A</math> and <math>\angle D</math> are [[right angle]]s, and <math>\angle B, \angle C, \angle E,</math> and <math>\angle F</math> are [[congruent]]. The area of the hexagonal region is <math>2116(\sqrt{2}+1).</math> Find <math>AB</math>. |
== Solution == | == Solution == | ||
+ | Let the side length be called <math>x</math>, so <math>x=AB=BC=CD=DE=EF=AF</math>. | ||
− | + | [[Image:2006_I_AIME-1.png]] | |
− | [[Image: | ||
− | + | 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>, | ||
and the area of the rectangle BCEF equals <math>x\cdot x\sqrt{2}=x^2\sqrt{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 | Then we have to solve the equation | ||
− | + | <div style="text-align:center;"> | |
<math>2116(\sqrt{2}+1)=x^2\sqrt{2}+x^2</math>. | <math>2116(\sqrt{2}+1)=x^2\sqrt{2}+x^2</math>. | ||
Line 22: | Line 18: | ||
<math>2116=x^2</math> | <math>2116=x^2</math> | ||
− | <math>x=46</math> | + | <math>x=46</math></div> |
− | Therefore, AB is | + | Therefore, <math>AB</math> is <math>046</math>. |
== See also == | == See also == |
Revision as of 16:52, 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
.
![$x=46$](http://latex.artofproblemsolving.com/1/7/7/177d9e27be3fbdeb68593a8cef48eed1e0bde215.png)
Therefore, is
.
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 |