2016 AMC 10A Problems/Problem 15

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 $3^2\pi-7\pi=9\pi-7\pi=2 \pi$ . The scrap cookie has this area, so its radius must be $\boxed{\textbf{(A) }\sqrt 2}$ .

Video Solution 1

https://youtu.be/dHY8gjoYFXU?t=1290

~IceMatrix

Video Solution 2 by OmegaLearn

https://youtu.be/NsQbhYfGh1Q?t=899

~ pi_is_3.14

2016 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2016 AMC 10A Problems/Problem 15

Contents

Problem

Solution

Video Solution 1

Video Solution 2 by OmegaLearn

See Also