2011 AMC 10B Problems/Problem 16

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Problem

A dart board is a regular octagon divided into regions as shown. 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

[asy] unitsize(10mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=1;  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)); pair I=(1,1), J=(1+sqrt(2),1), K=(1+sqrt(2),1+sqrt(2)), L=(1,1+sqrt(2));  draw(A--B--C--D--E--F--G--H--cycle); draw(A--D); draw(B--G); draw(C--F); draw(E--H);  pair[] ps={A,B,C,D,E,F,G,H,I,J,K,L}; dot(ps); label("$A$",A,W); label("$B$",B,S); label("$C$",C,S); label("$D$",D,E); label("$E$",E,E); label("$F$",F,N); label("$G$",G,N); label("$H$",H,W); label("$I$",I,NE); label("$J$",J,NW); label("$K$",K,SW); label("$L$",L,SE); label("$\sqrt{2}$",midpoint(B--C),S); label("$1$",midpoint(A--I),N); [/asy]

If the side lengths of the dart board and the side lengths of the center square are all $\sqrt{2},$ then the side length of the legs of the triangles are $1$.

\begin{align*} \text{area of center square} &: \sqrt{2} \times \sqrt{2} = 2\\ \text{total area} &: (\sqrt{2})^2 + 4(1 \times \sqrt{2}) + 4(1 \times 1 \times \frac{1}{2}) = 2 + 4\sqrt{2} + 2 = 4 + 4\sqrt{2} \end{align*}

Use Geometric probability by putting the area of the desired region over the area of the entire region.

\[\frac{2}{4+4\sqrt{2}} = \frac{1}{2+2\sqrt{2}} \times \frac{2-2\sqrt{2}}{2-2\sqrt{2}} = \frac{2-2\sqrt{2}}{-4} = \boxed{\textbf{(A)} \frac{\sqrt{2}-1}{2}}\]


Explanation: The area of the octagon consists of the area of the triangles, the rectangles, and the square in the middle. Assume the octagon has side length $1$. The triangles are right isosceles triangles with the hypotenuse 1, so their side length is $\frac{\sqrt{2}}{2}$ and the area of one triangle is $\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2}}{2}\cdot\frac{1}{2}=\frac{1}{4}$. The area of all 4 triangles is then just $1$. The rectangles share one side with the octagon, and another with the triangle. The octagon has side length 1, and the triangle has side length $\frac{\sqrt{2}}{2}$ as found earlier. So the area of one rectangle is $\frac{\sqrt{2}}{2}$. The area of all 4 is $4\cdot\frac{\sqrt{2}}{2}=2\sqrt{2}$. Finally, the area of the square in the middle is $1\cdot1=1$. The total area is $1+1+2\sqrt{2}=2+2\sqrt{2}$. We want the area of the square over the area of the octagon, which is $\frac{1}{2+2\sqrt{2}}$. Rationalize by multiplying both numerator and denominator by $2-2\sqrt{2}$: $\frac{1}{2+2\sqrt{2}}\cdot\frac{2-2\sqrt{2}}{2-2\sqrt{2}}=\frac{2-2\sqrt{2}}{\left(2+2\sqrt{2}\right)\left(2-2\sqrt{2}\right)}$. By the difference of squares, the denominator reduces to $-4$ and the fraction is $\frac{2-2\sqrt{2}}{-4}=\frac{\sqrt{2}-1}{2}$ which is $\boxed{\textbf{(A) } \frac{\sqrt{2}-1}{2}}$.

~Explanation by JH. L

Solution 2

Area of a regular octagon is $2(1+\sqrt{2})a^2$ where $a$ is the side. Hence the answer is obvious now.

See Also

2011 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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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