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2016 AMC 10A Problem 15

2016 AMC 10A Problem 15AMC 10 Step by Step

Name: Date: September 13, 2026

Circles · difficulty 2 of 5 · about 2 min · position P11-15

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?

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

Answer

Key insight

The dough's radius is 3 (center cookie radius plus one outer cookie's diameter), so the scrap area is 9pi - 7pi = 2pi and its radius is sqrt(2).

Solution

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

Why this works

"Same thickness" turns a volume statement into an area statement, and the only real work is reading the dough's radius off the tangency structure. Tangent circles make distances additive along a line through their centers: the center cookie's radius plus a full outer cookie's diameter spans exactly one dough radius.

The trap

Taking the dough radius as 2 (the distance to the outer cookies' centers) rather than 3.

Common mistakes

Taking the dough radius as 2 (the distance to the outer cookies' centers) rather than 3.
Equating perimeters or scaling the radius by area ratio incorrectly, e.g. answering $2$ because the scrap area is "twice a cookie."

Techniques

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