2011 AMC 10B Problems/Problem 16
Contents
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.
Solution 2
area of a regular octagon = 2(1+sqrt{2})a^2 where a is the side hence the answer is obvious now.
See Also
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.