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

Revision as of 19:02, 28 April 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

@@ Line 72: / Line 72: @@
 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>.
-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 octogon. Since the area “op0;op9o;lopof the octogon is 1, <math>\therefore</math> the area of the rectangle is <math>\boxed{\textbf{(D)}\ \frac{1}{2}}</math>.
+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>.
 ~ghfhgvghj10

2003 AMC 12B (Problems • Answer Key • Resources)
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2003 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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