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Difference between revisions of "2011 AMC 12B Problems/Problem 12"

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After careful inspection of the regular octagon, we find that all of the sides of the regular octagon must be <math>1</math>.
 
After careful inspection of the regular octagon, we find that all of the sides of the regular octagon must be <math>1</math>.
  
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This gives the total area to be <math>4 * triangle area</math> + <math>4 * rectangle area</math> + <math>square area</math> = <math>4(1/4) + 4(sqrt(1/2)) + 1</math>, which simplifies to <math>2(1+sqrt(2))</math>.
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This gives the total area to be <math>4 \cdot \text{triangle area}</math> + <math>4 \cdot \text{rectangle area}</math> + <math>\text{square area}</math> = <math>4(\dfrac{1}{4}) + 4(\sqrt{\dfrac{1}{2}}) + 1</math>, which simplifies to <math>2(1+\sqrt{2})</math>.
  
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The percent area of the center square is the area of the square divided by the total area of the regular octagon. This gives <math>1/(2(1+sqrt(2))).</math> Multiplying by the conjugate gives (A).
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The percent area of the center square is the area of the square divided by the total area of the regular octagon. This gives <math>\dfrac{1}{2(1+\sqrt{2})}.</math> Multiplying by the conjugate gives answer choice <math>\boxed{(A).}</math>
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|ab=B|num-b=11|num-a=13}}
 
{{AMC12 box|year=2011|ab=B|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:11, 31 August 2024

Problem

A dart board is a regular octagon divided into regions as shown below. Suppose that a dart thrown at the board is equally likely to land anywhere on the board. What is the probability that the dart lands within the center square?

[asy] unitsize(10mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,1), B=(1,0), C=(1+sqrt(2),0), D=(2+sqrt(2),1), E=(2+sqrt(2),1+sqrt(2)), F=(1+sqrt(2),2+sqrt(2)), G=(1,2+sqrt(2)), H=(0,1+sqrt(2)); draw(A--B--C--D--E--F--G--H--cycle); draw(A--D); draw(B--G); draw(C--F); draw(E--H);[/asy]

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

Solution

Let's assume that the side length of the octagon is $x$. The area of the center square is just $x^2$. The triangles are all $45-45-90$ triangles, with a side length ratio of $1:1:\sqrt{2}$. The area of each of the $4$ identical triangles is $\left(\dfrac{x}{\sqrt{2}}\right)^2\times\dfrac{1}{2}=\dfrac{x^2}{4}$, so the total area of all of the triangles is also $x^2$. Now, we must find the area of all of the 4 identical rectangles. One of the side lengths is $x$ and the other side length is $\dfrac{x}{\sqrt{2}}=\dfrac{x\sqrt{2}}{2}$, so the area of all of the rectangles is $2x^2\sqrt{2}$. The ratio of the area of the square to the area of the octagon is $\dfrac{x^2}{2x^2+2x^2\sqrt{2}}$. Cancelling $x^2$ from the fraction, the ratio becomes $\dfrac{1}{2\sqrt2+2}$. Multiplying the numerator and the denominator each by $2\sqrt{2}-2$ will cancel out the radical, so the fraction is now $\dfrac{1}{2\sqrt2+2}\times\dfrac{2\sqrt{2}-2}{2\sqrt{2}-2}=\dfrac{2\sqrt{2}-2}{4}=\boxed{\mathrm{(A)}\ \dfrac{\sqrt{2}-1}{2}}$

Solution

Instead of doing mindless algebra, let's set the side lengths of the square to $1$.

After careful inspection of the regular octagon, we find that all of the sides of the regular octagon must be $1$.

This gives the total area to be $4 \cdot \text{triangle area}$ + $4 \cdot \text{rectangle area}$ + $\text{square area}$ = $4(\dfrac{1}{4}) + 4(\sqrt{\dfrac{1}{2}}) + 1$, which simplifies to $2(1+\sqrt{2})$.

The percent area of the center square is the area of the square divided by the total area of the regular octagon. This gives $\dfrac{1}{2(1+\sqrt{2})}.$ Multiplying by the conjugate gives answer choice $\boxed{(A).}$

See also

2011 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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

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