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Difference between revisions of "2003 AMC 10B Problems/Problem 23"

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{{duplicate|[[2003 AMC 12B Problems|2003 AMC 12B #15]] and [[2003 AMC 10B Problems|2003 AMC 10B #23]]}}
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==Problem==
 
==Problem==
  
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<math> \textbf{(A)}\ 1-\frac{\sqrt2}{2}\qquad\textbf{(B)}\ \frac{\sqrt2}{4}\qquad\textbf{(C)}\ \sqrt2-1\qquad\textbf{(D)}\ \frac{1}2\qquad\textbf{(E)}\ \frac{1+\sqrt2}{4} </math>
 
<math> \textbf{(A)}\ 1-\frac{\sqrt2}{2}\qquad\textbf{(B)}\ \frac{\sqrt2}{4}\qquad\textbf{(C)}\ \sqrt2-1\qquad\textbf{(D)}\ \frac{1}2\qquad\textbf{(E)}\ \frac{1+\sqrt2}{4} </math>
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==Video Solution==
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https://www.youtube.com/watch?v=LREcUjK-56U&feature=youtu.be
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==Solution 1==
 
==Solution 1==
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Area of Octagon: <math> \frac{ap}{2}=1 </math>.
 
Area of Octagon: <math> \frac{ap}{2}=1 </math>.
  
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Area of Rectangle: <math> \frac{p}{8}\times 2a=\frac{ap}{4} </math>.
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Area of Rectangle: <math> \frac{p}{8}\times 2a=\dfrac{2ap}{8}=\frac{ap}{4} </math>.
  
 
You can see from this that the octagon's area is twice as large as the rectangle's area is <math>\boxed{\textbf{(D)}\ \frac{1}{2}}</math>.
 
You can see from this that the octagon's area is twice as large as the rectangle's area is <math>\boxed{\textbf{(D)}\ \frac{1}{2}}</math>.
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==Solution 2==
 
==Solution 2==
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<asy>
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unitsize(1cm);
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defaultpen(linewidth(.8pt)+fontsize(8pt));
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pair C=dir(22.5), B=dir(67.5), A=dir(112.5), H=dir(157.5), G=dir(202.5), F=dir(247.5), E=dir(292.5), D=dir(337.5);
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draw(A--B--C--D--E--F--G--H--cycle);
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label("$A$",A,NNW);
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label("$B$",B,NNE);
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label("$C$",C,ENE);
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label("$D$",D,ESE);
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label("$E$",E,SSE);
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label("$F$",F,SSW);
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label("$G$",G,WSW);
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label("$H$",H,WNW);
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draw(A--E--F--B--C--G--H--D);
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draw(A--E--F--B--A,blue);
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draw(A--F--E--B--A,red);
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</asy>
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 +
Here is a less complicated way than that of the user below. If you draw a line segment from each vertex to the center of the octagon and draw the rectangle ABEF (in red), you can see that <math>2</math> of the triangles (in blue) share the same base and height with <math>\dfrac{1}{2}</math> the rectangle. Therefore, the rectangle's area is the same as <math>2\cdot2</math> of the <math>8</math> triangles, and is  <math>\boxed{\textbf{(D)}\ \frac{1}{2}}</math> the area of the octagon.
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==Solution 3==
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<asy>
 +
unitsize(1cm);
 +
defaultpen(linewidth(.8pt)+fontsize(8pt));
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pair C=dir(22.5), B=dir(67.5), A=dir(112.5), H=dir(157.5), G=dir(202.5), F=dir(247.5), E=dir(292.5), D=dir(337.5);
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draw(A--B--C--D--E--F--G--H--cycle);
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label("$A$",A,NNW);
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label("$B$",B,NNE);
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label("$C$",C,ENE);
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label("$D$",D,ESE);
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label("$E$",E,SSE);
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label("$F$",F,SSW);
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label("$G$",G,WSW);
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label("$H$",H,WNW);
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draw(A--D--E--H--G--B--C--F);
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</asy>
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Drawing lines <math>AD</math>, <math>BG</math>, <math>CF</math>, and <math>EH</math>, we can see that the octagon is comprised of <math>1</math> square, <math>4</math> rectangles, and <math>4</math> triangles. The triangles each are <math>45-45-90</math> triangles, and since their diagonal is length <math>x</math>, each of their sides is <math>\frac{\sqrt{2}}{2}x</math>. The area of the entire figure is, likewise, <math>x^2</math> (the square)<math>+4x^2\frac{\sqrt{2}}{2}</math> (the 4 rectangles)<math> +2\cdot(\frac{\sqrt{2}}{2}x)^2</math> (the triangles), which simplifies to <math>2x^2 + 2\sqrt{2}x^2</math>. The area of <math>ABEF</math> is just <math>x(x+\frac{2\sqrt{2}}{2}x)</math>, or <math>x^2</math> + <math>x^2\sqrt{2}</math>, which we can see is the area of <math>\frac{ABCDEFGH}{2} =  \boxed{\textbf{(D)}\ \frac{1}{2}}</math> the area of the octagon.
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==Solution 4==
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First, we are going to divide the diagram. Now we need to find the ratio of the area of the rectangle to the area of the trapezoid.
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 +
<asy> unitsize(1cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair C=dir(22.5), B=dir(67.5), A=dir(112.5), H=dir(157.5), G=dir(202.5), F=dir(247.5), E=dir(292.5), D=dir(337.5); draw(A--B--C--D--E--F--G--H--cycle); label("$A$",A,NNW); label("$B$",B,NNE); label("$C$",C,ENE); label("$D$",D,ESE); label("$E$",E,SSE); label("$F$",F,SSW); label("$G$",G,WSW); label("$H$",H,WNW);draw(A--F);draw(B--E);draw(D--G);draw(C--H);</asy>
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The area of a trapezoid is <math>\frac{b_1 + b_2}{2}h</math>
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Note that the trapezoid is made out of 2 45-45-90 triangles and a rectangle, and that <math>AH=FG=1</math>. By realizing that, the area of the trapezoid is <math>(\frac{2+\sqrt{2}}{2})(\frac{\sqrt{2}}{2}</math>). To make this product easier, note there is two trapezoids, so the new product is now this, <math>(\frac{2+\sqrt{2}}{2})(\sqrt{2}) = \sqrt{2} + 1</math>
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 +
Notice how the rectangle has side lengths <math>\sqrt{2}+1</math> and <math>1</math>, so it's area is also <math>\sqrt{2}+1</math>.
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 +
The ratio of the area of rectangle <math>ABEF</math> to the two trapezoids is <math>1:1</math>, meaning they share half the area of the octagon. Since the area of the octagon is 1, <math>\therefore</math> the area of the rectangle is <math>\boxed{\textbf{(D)}\ \frac{1}{2}}</math>.
  
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If you draw lines connecting opposite vertices and draw the rectangle ABEF, you can see that two of the triangles share the same base and height with half the rectangle. Therefore, the rectangle's area is the same as 4 of these triangles, and is  <math>\boxed{\textbf{(D)}\ \frac{1}{2}}</math> the area of the octagon
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~ghfhgvghj10
  
 
==See Also==
 
==See Also==
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{{AMC12 box|year=2003|ab=B|num-b=14|num-a=16}}
 
{{AMC10 box|year=2003|ab=B|num-b=22|num-a=24}}
 
{{AMC10 box|year=2003|ab=B|num-b=22|num-a=24}}
 +
{{MAA Notice}}

Latest revision as of 19:15, 24 August 2024

The following problem is from both the 2003 AMC 12B #15 and 2003 AMC 10B #23, so both problems redirect to this page.

Problem

A regular octagon $ABCDEFGH$ has an area of one square unit. What is the area of the rectangle $ABEF$?

[asy] unitsize(1cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair C=dir(22.5), B=dir(67.5), A=dir(112.5), H=dir(157.5), G=dir(202.5), F=dir(247.5), E=dir(292.5), D=dir(337.5); draw(A--B--C--D--E--F--G--H--cycle); label("$A$",A,NNW); label("$B$",B,NNE); label("$C$",C,ENE); label("$D$",D,ESE); label("$E$",E,SSE); label("$F$",F,SSW); label("$G$",G,WSW); label("$H$",H,WNW);[/asy]

$\textbf{(A)}\ 1-\frac{\sqrt2}{2}\qquad\textbf{(B)}\ \frac{\sqrt2}{4}\qquad\textbf{(C)}\ \sqrt2-1\qquad\textbf{(D)}\ \frac{1}2\qquad\textbf{(E)}\ \frac{1+\sqrt2}{4}$

Video Solution

https://www.youtube.com/watch?v=LREcUjK-56U&feature=youtu.be


Solution 1

Here is an easy way to look at this, where $p$ is the perimeter, and $a$ is the apothem:

Area of Octagon: $\frac{ap}{2}=1$.

Area of Rectangle: $\frac{p}{8}\times 2a=\dfrac{2ap}{8}=\frac{ap}{4}$.

You can see from this that the octagon's area is twice as large as the rectangle's area is $\boxed{\textbf{(D)}\ \frac{1}{2}}$.

Solution 2

[asy] unitsize(1cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair C=dir(22.5), B=dir(67.5), A=dir(112.5), H=dir(157.5), G=dir(202.5), F=dir(247.5), E=dir(292.5), D=dir(337.5); draw(A--B--C--D--E--F--G--H--cycle); label("$A$",A,NNW); label("$B$",B,NNE); label("$C$",C,ENE); label("$D$",D,ESE); label("$E$",E,SSE); label("$F$",F,SSW); label("$G$",G,WSW); label("$H$",H,WNW); draw(A--E--F--B--C--G--H--D); draw(A--E--F--B--A,blue); draw(A--F--E--B--A,red); [/asy]

Here is a less complicated way than that of the user below. If you draw a line segment from each vertex to the center of the octagon and draw the rectangle ABEF (in red), you can see that $2$ of the triangles (in blue) share the same base and height with $\dfrac{1}{2}$ the rectangle. Therefore, the rectangle's area is the same as $2\cdot2$ of the $8$ triangles, and is $\boxed{\textbf{(D)}\ \frac{1}{2}}$ the area of the octagon.

Solution 3

[asy] unitsize(1cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair C=dir(22.5), B=dir(67.5), A=dir(112.5), H=dir(157.5), G=dir(202.5), F=dir(247.5), E=dir(292.5), D=dir(337.5); draw(A--B--C--D--E--F--G--H--cycle); label("$A$",A,NNW); label("$B$",B,NNE); label("$C$",C,ENE); label("$D$",D,ESE); label("$E$",E,SSE); label("$F$",F,SSW); label("$G$",G,WSW); label("$H$",H,WNW); draw(A--D--E--H--G--B--C--F); [/asy] Drawing lines $AD$, $BG$, $CF$, and $EH$, we can see that the octagon is comprised of $1$ square, $4$ rectangles, and $4$ triangles. The triangles each are $45-45-90$ triangles, and since their diagonal is length $x$, each of their sides is $\frac{\sqrt{2}}{2}x$. The area of the entire figure is, likewise, $x^2$ (the square)$+4x^2\frac{\sqrt{2}}{2}$ (the 4 rectangles)$+2\cdot(\frac{\sqrt{2}}{2}x)^2$ (the triangles), which simplifies to $2x^2 + 2\sqrt{2}x^2$. The area of $ABEF$ is just $x(x+\frac{2\sqrt{2}}{2}x)$, or $x^2$ + $x^2\sqrt{2}$, which we can see is the area of $\frac{ABCDEFGH}{2} = \boxed{\textbf{(D)}\ \frac{1}{2}}$ the area of the octagon.

Solution 4

First, we are going to divide the diagram. Now we need to find the ratio of the area of the rectangle to the area of the trapezoid.

[asy] unitsize(1cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair C=dir(22.5), B=dir(67.5), A=dir(112.5), H=dir(157.5), G=dir(202.5), F=dir(247.5), E=dir(292.5), D=dir(337.5); draw(A--B--C--D--E--F--G--H--cycle); label("$A$",A,NNW); label("$B$",B,NNE); label("$C$",C,ENE); label("$D$",D,ESE); label("$E$",E,SSE); label("$F$",F,SSW); label("$G$",G,WSW); label("$H$",H,WNW);draw(A--F);draw(B--E);draw(D--G);draw(C--H);[/asy]

The area of a trapezoid is $\frac{b_1 + b_2}{2}h$ Note that the trapezoid is made out of 2 45-45-90 triangles and a rectangle, and that $AH=FG=1$. By realizing that, the area of the trapezoid is $(\frac{2+\sqrt{2}}{2})(\frac{\sqrt{2}}{2}$). To make this product easier, note there is two trapezoids, so the new product is now this, $(\frac{2+\sqrt{2}}{2})(\sqrt{2}) = \sqrt{2} + 1$

Notice how the rectangle has side lengths $\sqrt{2}+1$ and $1$, so it's area is also $\sqrt{2}+1$.

The ratio of the area of rectangle $ABEF$ to the two trapezoids is $1:1$, meaning they share half the area of the octagon. Since the area of the octagon is 1, $\therefore$ the area of the rectangle is $\boxed{\textbf{(D)}\ \frac{1}{2}}$.

~ghfhgvghj10

See Also

2003 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
2003 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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