|
|
| Line 1: |
Line 1: |
| − | == 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?
| |
| − |
| |
| − | <center><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>
| |
| − | </center>
| |
| − |
| |
| − | <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]]
| |
| − |
| |
| | == See Also== | | == See Also== |
| | | | |
Revision as of 04:59, 6 April 2019
See Also
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]](//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.
![\[\frac{2}{4+4\sqrt{2}} = \frac{1}{2+2\sqrt{2}} \times \frac{2-2\sqrt{2}}{2-2\sqrt{2}} = \frac{2-2\sqrt{2}}{-4} = \boxed{\textbf{(A)} \frac{\sqrt{2}-1}{2}}\]](//latex.artofproblemsolving.com/2/b/b/2bbad9c1ddee3983e69e37fd039410e18bd0b83a.png)
![[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)
then the side length of the legs of the triangles are 
.
![\[\frac{2}{4+4\sqrt{2}} = \frac{1}{2+2\sqrt{2}} \times \frac{2-2\sqrt{2}}{2-2\sqrt{2}} = \frac{2-2\sqrt{2}}{-4} = \boxed{\textbf{(A)} \frac{\sqrt{2}-1}{2}}\]](http://latex.artofproblemsolving.com/2/b/b/2bbad9c1ddee3983e69e37fd039410e18bd0b83a.png)
