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Difference between revisions of "2006 AIME II Problems/Problem 1"
(→‎Solution)
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Therefore, <math>AB</math> is <math>\boxed{046}</math>.
 
Therefore, <math>AB</math> is <math>\boxed{046}</math>.
 +
 +
== Solution ==
 +
 +
Because <math>\angle
 +
B</math>, <math>\angle C</math>, <math>\angle E</math>, and <math>\angle F</math> are congruent, the degree-measure of each of them is <math>\displaystyle
 +
{{720-2\cdot90}\over4}= 135</math>. Lines <math>BF</math> and <math>CE</math> divide the hexagonal region into two right triangles and a rectangle. Let <math>AB=x</math>. Then <math>BF=x\sqrt2</math>. Thus \begin{align*}
 +
2116(\sqrt2+1)&=[ABCDEF]\\
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&=2\cdot {1\over2}x^2+x\cdot x\sqrt2=x^2(1+\sqrt2),
 +
\end{align*}so <math>x^2=2116</math>, and <math>x=\boxed{46}</math>.
 +
 +
[asy]
 +
pair A,B,C,D,I,F;
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A=(0,0);
 +
B=(7,0);
 +
F=(0,7);
 +
I=(6,13);
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D=(13,13);
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C=(13,6);
 +
dot(A);
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dot(B);
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dot(C);
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dot(D);
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dot(I);
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dot(F);
 +
draw(A--B--C--D--I--F--cycle,linewidth(0.7));
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label("{\tiny <math>A</math>}",A,S);
 +
label("{\tiny <math>B</math>}",B,S);
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label("{\tiny <math>C</math>}",C,E);
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label("{\tiny <math>D</math>}",D,N);
 +
label("{\tiny <math>E</math>}",I,N);
 +
label("{\tiny <math>F</math>}",F,W);
 +
[/asy]
  
 
== See also ==
 
== See also ==

Revision as of 01:09, 3 May 2016

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 $\boxed{046}$.

Solution

Because $\angle B$, $\angle C$, $\angle E$, and $\angle F$ are congruent, the degree-measure of each of them is $\displaystyle {{720-2\cdot90}\over4}= 135$. 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,I,F; A=(0,0); B=(7,0); F=(0,7); I=(6,13); D=(13,13); C=(13,6); dot(A); dot(B); dot(C); dot(D); dot(I); dot(F); draw(A--B--C--D--I--F--cycle,linewidth(0.7)); label("{\tiny $A$}",A,S); label("{\tiny $B$}",B,S); label("{\tiny $C$}",C,E); label("{\tiny $D$}",D,N); label("{\tiny $E$}",I,N); label("{\tiny $F$}",F,W); [/asy]

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

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