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

(→‎Solution)
(→‎See Also)
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== See Also==
 
== See Also==
  
−
{{AMC10 box|year=2011|ab=B|num-b=15|num-a=17}}
+
== Solution ==
−
{{MAA Notice}}
+
 
 +
<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>

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]

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

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]

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.

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

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