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 angles, 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>.
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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>.
  
Let the side length be called <math>x</math>.
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[[Image:2006_I_AIME-1.png]]
[[Image:Diagram1.png]]
 
  
Then <math>AB=BC=CD=DE=EF=AF=x</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>,
 
 
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  
 
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<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>.
  
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<math>2116=x^2</math>
 
<math>2116=x^2</math>
  
<math>x=46</math>
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<math>x=46</math></div>
  
Therefore, AB is 46.
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Therefore, <math>AB</math> is <math>046</math>.
  
 
== See also ==
 
== See also ==

Revision as of 16:52, 25 September 2007

Problem

In convex hexagon $ABCDEF$, all six sides are congruent, $\angle A$ and $\angle D$ are right angles, and $\angle B, \angle C, \angle E,$ 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 $x$, so $x=AB=BC=CD=DE=EF=AF$.

2006 I AIME-1.png

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

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

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

$2116=x^2$

$x=46$

Therefore, $AB$ is $046$.

See also

2006 AIME I (ProblemsAnswer KeyResources)
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
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