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 $\boxed{\textbf{(A) }\sqrt 2}$ .

Solution 2 (More detail)

We are given the small cookies have a radius of $1$ . This means the radius of the larger cookie is $3 \times 1=3$ . The area of the larger cookie is $\pi 3^2$ , which is $9\pi$ . The area of the $7$ smaller cookies is $7 \times \pi 1^2$ , which is $7\pi$ . The scrap dough is calculated by subtracting the area of the large cookie by the are of the $7$ small cookies, which is $9\pi-7\pi=2\pi$ . To get the radius, we first divide by $\pi$ , then take the square root of the result, which in this case is $\boxed{\textbf{(A) }\sqrt 2}$ . ~shunyipanda

Video Solution (CREATIVE THINKING)

https://youtu.be/1ocp50QWOzU

~Education, the Study of Everything

Video Solution 1

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

Video Solution

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

~ pi_is_3.14

Video Solution 3( Easy Explanation)

https://youtu.be/Xa61-YgFo4Y

2016 AMC 10A (Problems • Answer Key • Resources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

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