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2006 AIME II Problems/Problem 1: Difference between revisions

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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>
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>
{{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
{{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
<math>\begin{align*}
<cmath>\begin{align*}
2116(\sqrt2+1)&=[ABCDEF]\\
2116(\sqrt2+1)&=[ABCDEF]\\
&=2\cdot {1\over2}x^2+x\cdot x\sqrt2=x^2(1+\sqrt2),
&=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>.
\end{align*}</cmath>so <math>x^2=2116</math>, and <math>x=\boxed{46}</math>.


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


Revision as of 15:08, 30 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 $x=AB=BC=CD=DE=EF=AF$.

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 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,E); label("{\tiny $D$}",D,N); label("{\tiny $E$}",E,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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