2016 AMC 10A Problem 15

P11-15~2 minPrint

Circles Set up the equation/formula and compute; no special trick needed

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?

Show solution without answering

Key insight

The dough's radius is 3: the center cookie's radius plus an outer cookie's width. The scrap is 9 pi - 7 pi = 2 pi.

Solution

The scrap is the dough minus the seven cookies, so first find the size of the dough.

Walk from the center of the dough straight out to its edge, through one outer cookie. You cross the center cookie's radius ( $1$ inch), then the whole width of an outer cookie ( $2$ inches), and then you are at the edge of the dough.
So the dough has radius $1 + 2 = 3$ inches and area $\pi \cdot 3^2 = 9\pi$ .
Each cookie has area $\pi \cdot 1^2 = \pi$ , so the seven cookies use $7\pi$ .
The scrap has area $9\pi - 7\pi = 2\pi$ .
The scrap cookie has the same thickness, so its top has the same area, $2\pi$ . If its radius is $r$ , then $\pi r^2 = 2\pi$ , so $r^2 = 2$ and $r = \sqrt{2}$ .

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

Why this works

"Same thickness" means volume is area times a fixed height, so you can compare areas only. When circles touch, their lengths add up along the straight line through their centers, which is how the dough's radius comes out as radius plus diameter.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Find the radius of the dough first. Going from the center of the dough straight outward: the center cookie has radius $1$ , and an outer cookie tangent to it has diameter $2$ , ending at the edge of the dough. So the dough has radius $1 + 2 = 3$ and area $9\pi$ .

The seven cookies have total area $7\cdot\pi\cdot 1^2 = 7\pi$ , leaving scrap of area $9\pi - 7\pi = 2\pi$ .

Since the thickness is unchanged, the scrap cookie has the same area $2\pi$ , so $\pi r^2 = 2\pi$ and $r = \sqrt 2$ .

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

Review in the AoPS books

Introduction to Geometry, Chapter 11: Circles
Introduction to Geometry, Chapter 12: Circles and Angles
Introduction to Geometry, Chapter 13: Power of a Point