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

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Therefore, <math>AB</math> is <math>\boxed{46}</math>.
 
Therefore, <math>AB</math> is <math>\boxed{46}</math>.
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~removal of extraneous zeros by K124659
  
 
== Solution 2 ==
 
== Solution 2 ==
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label("{\tiny $A$}",A,S);
 
label("{\tiny $A$}",A,S);
 
label("{\tiny $B$}",B,S);
 
label("{\tiny $B$}",B,S);
−
label("{\tiny $C$}",C,E);
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label("{\tiny $C$}",C,dir(0));
 
label("{\tiny $D$}",D,N);
 
label("{\tiny $D$}",D,N);
 
label("{\tiny $E$}",E,N);
 
label("{\tiny $E$}",E,N);
 
label("{\tiny $F$}",F,W);
 
label("{\tiny $F$}",F,W);
 
</asy>
 
</asy>
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 +
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~minor asymptote edit by Yiyj1
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~removal of extraneous zeros by K124659
  
 
== See also ==
 
== See also ==

Latest revision as of 11:07, 31 August 2024

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 II 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 $\boxed{46}$. ~removal of extraneous zeros by K124659

Solution 2

Because $\angle B$, $\angle C$, $\angle E$, and $\angle F$ are congruent, the degree-measure of each of them is ${{720-2\cdot90}\over4}= 135$. Lines $BF$ and $CE$ divide the hexagonal region into two right triangles and a rectangle. Let $AB=x$. Then $BF=x\sqrt2$. Thus \begin{align*} 2116(\sqrt2+1)&=[ABCDEF]\\ &=2\cdot {1\over2}x^2+x\cdot x\sqrt2=x^2(1+\sqrt2), \end{align*}so $x^2=2116$, and $x=\boxed{46}$.

[asy] pair A,B,C,D,E,F; A=(0,0); B=(7,0); C=(13,6); E=(6,13); D=(13,13); F=(0,7); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); draw(A--B--C--D--E--F--cycle,linewidth(0.7)); label("{\tiny $A$}",A,S); label("{\tiny $B$}",B,S); label("{\tiny $C$}",C,dir(0)); label("{\tiny $D$}",D,N); label("{\tiny $E$}",E,N); label("{\tiny $F$}",F,W); [/asy]


~minor asymptote edit by Yiyj1 ~removal of extraneous zeros by K124659

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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