Difference between revisions of "2016 AMC 10A Problems/Problem 15"

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==Solution==
 
==Solution==
 
The big cookie has radius <math>3</math>, since the center of the center cookie is the same as that of the large cookie.  The difference in areas of the big cookie and the seven small ones is <math>2 \pi</math>.  The scrap cookie has this area, so its radius must be <math>\boxed{\textbf{(A) }\sqrt 2}</math>.
 
The big cookie has radius <math>3</math>, since the center of the center cookie is the same as that of the large cookie.  The difference in areas of the big cookie and the seven small ones is <math>2 \pi</math>.  The scrap cookie has this area, so its radius must be <math>\boxed{\textbf{(A) }\sqrt 2}</math>.
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==See Also==
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{{AMC10 box|year=2016|ab=A|num-b=14|num-a=16}}
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{{MAA Notice}}

Revision as of 22:49, 3 February 2016

Problem

Seven cookies of radius $1$ inch are cut from a circle of cookie dough, as shown. Neighboring cookies are tangent, and all except the center cookie are tangent to the edge of the dough. The leftover scrap is reshaped to form another cookie of the same thickness. What is the radius in inches of the scrap cookie?

[asy] draw(circle((0,0),3)); draw(circle((0,0),1)); draw(circle((1,sqrt(3)),1)); draw(circle((-1,sqrt(3)),1));  draw(circle((-1,-sqrt(3)),1)); draw(circle((1,-sqrt(3)),1));  draw(circle((2,0),1)); draw(circle((-2,0),1)); [/asy]

$\textbf{(A) } \sqrt{2} \qquad \textbf{(B) } 1.5 \qquad \textbf{(C) } \sqrt{\pi} \qquad \textbf{(D) } \sqrt{2\pi} \qquad \textbf{(E) } \pi$

Solution

The big cookie has radius $3$, since the center of the center cookie is the same as that of the large cookie. The difference in areas of the big cookie and the seven small ones is $2 \pi$. The scrap cookie has this area, so its radius must be $\boxed{\textbf{(A) }\sqrt 2}$.

See Also

2016 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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

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