Difference between revisions of "2011 AMC 10B Problems/Problem 16"
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<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> | <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> | ||
| + | |||
| + | == See Also== | ||
| + | |||
| + | {{AMC10 box|year=2011|ab=B|num-b=16|num-a=18}} | ||
| + | {{MAA Notice}} | ||
Revision as of 14:52, 26 November 2019
Contents
See Also
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]](http://latex.artofproblemsolving.com/6/8/1/68141b2184d1682ae8f8d03c273456191e2cb1bd.png)
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]](http://latex.artofproblemsolving.com/d/c/4/dc4877d213827aa8426c61a0f2dec96c9f470de2.png)
If the side lengths of the dart board and the side lengths of the center square are all 
then the side length of the legs of the triangles are 
.
Use Geometric probability by putting the area of the desired region over the area of the entire region.
![\[\frac{2}{4+4\sqrt{2}} = \frac{1}{2+2\sqrt{2}} \times \frac{2-2\sqrt{2}}{2-2\sqrt{2}} = \frac{2-2\sqrt{2}}{-4} = \boxed{\textbf{(A)} \frac{\sqrt{2}-1}{2}}\]](http://latex.artofproblemsolving.com/2/b/b/2bbad9c1ddee3983e69e37fd039410e18bd0b83a.png)
See Also
| 2011 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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. 
