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

 
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#REDIRECT [[2006 AIME A Problems/Problem 1]]
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== 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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== Solution ==
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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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[[Image:2006_II_AIME-1.png]]
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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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Then we have to solve the equation
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<div style="text-align:center;">
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<math>2116(\sqrt{2}+1)=x^2\sqrt{2}+x^2</math>.
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<math>2116(\sqrt{2}+1)=x^2(\sqrt{2}+1)</math>
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<math>2116=x^2</math>
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<math>x=46</math></div>
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Therefore, <math>AB</math> is <math>\boxed{46}</math>.
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~removal of extraneous zeros by K124659
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== Solution 2 ==
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Because <math>\angle
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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>
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{{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
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<cmath>\begin{align*}
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2116(\sqrt2+1)&=[ABCDEF]\\
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&=2\cdot {1\over2}x^2+x\cdot x\sqrt2=x^2(1+\sqrt2),
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\end{align*}</cmath>so <math>x^2=2116</math>, and <math>x=\boxed{46}</math>.
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<asy>
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pair A,B,C,D,E,F;
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A=(0,0);
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B=(7,0);
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C=(13,6);
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E=(6,13);
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D=(13,13);
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F=(0,7);
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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(E);
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dot(F);
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draw(A--B--C--D--E--F--cycle,linewidth(0.7));
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label("{\tiny $A$}",A,S);
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label("{\tiny $B$}",B,S);
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label("{\tiny $C$}",C,dir(0));
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label("{\tiny $D$}",D,N);
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label("{\tiny $E$}",E,N);
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label("{\tiny $F$}",F,W);
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</asy>
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~minor asymptote edit by Yiyj1
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~removal of extraneous zeros by K124659
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== See also ==
 +
{{AIME box|year=2006|n=II|before=First Question|num-a=2}}
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 +
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

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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