Difference between revisions of "2011 AMC 10B Problems/Problem 16"

Revision as of 04:59, 6 April 2019

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 probability that the dart lands within the center square?

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

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

2011 AMC 10B (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 21: / Line 21: @@
 <math> \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}</math>
 [[Category: Introductory Geometry Problems]]
-== Solution ==
-<center><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>
-</center>
-If the side lengths of the dart board and the side lengths of the center square are all <math>\sqrt{2},</math> then the side length of the legs of the triangles are <math>1</math>.
-<cmath>\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*}</cmath>
-Use [[Geometric probability]] by putting the area of the desired region over the area of the entire region.
-<cmath> \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}}</cmath>
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Difference between revisions of "2011 AMC 10B Problems/Problem 16"

Revision as of 04:59, 6 April 2019

Problem

See Also