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Difference between revisions of "2006 AIME II Problems/Problem 1"

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== Problem ==
 
== Problem ==
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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>.
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The lengths of the sides of a [[triangle]] with positive area are <math>\log_{10} 12</math>, <math>\log_{10} 75</math>, and <math>\log_{10} n</math>, where <math>n</math> is a positive integer. Find the number of possible values for <math>n</math>.
 
  
 
== Solution ==
 
== Solution ==
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By the [[Triangle Inequality]]:
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Let the side length be called <math>x</math>, so <math>x=AB=BC=CD=DE=EF=AF</math>.
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<div style="text-align:center;">
 
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<math>\log_{10} 12 + \log_{10} n > \log_{10} 75 </math>
 
  
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<math>\log_{10} 12n > \log_{10} 75 </math>
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[[Image:2006_I_AIME-1.png]]
  
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<math> 12n > 75 </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>,
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and the area of the rectangle BCEF equals <math>x\cdot x\sqrt{2}=x^2\sqrt{2}</math>
  
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<math> n > \frac{75}{12} = \frac{25}{4} = 6.25 </math>
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Then we have to solve the equation
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</div>
 
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Also:
 
 
<div style="text-align:center;">
 
<div style="text-align:center;">
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<math>\log_{10} 12 + \log_{10} 75 > \log_{10} n </math>
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<math>2116(\sqrt{2}+1)=x^2\sqrt{2}+x^2</math>.
  
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<math>\log_{10} 12\cdot75 > \log_{10} n </math>
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<math>2116(\sqrt{2}+1)=x^2(\sqrt{2}+1)</math>
  
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<math> n < 900 </math>  
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<math>2116=x^2</math>
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</div>
 
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Combining these two inequalities:
 
  
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<cmath> 6.25 < n < 900 </cmath>
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<math>x=46</math></div>
  
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The number of possible integer values for <math>n</math> is the number of integers over the interval <math>(6.25 , 900)</math>, which is <math>893</math>.
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Therefore, <math>AB</math> is <math>046</math>.
  
 
== See also ==
 
== See also ==
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{{AIME box|year=2006|n=I|num-b=1|num-a=3}}
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{{AIME box|year=2006|n=I|before=First Question|num-a=2}}
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
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[[Category:Intermediate Algebra Problems]]
 

Revision as of 19:00, 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 (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
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