2009 AMC 10A Problem 19

P16-20~3 minPrint

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

Circle $A$ has radius $100$ . Circle $B$ has an integer radius $r<100$ and remains internally tangent to circle $A$ as it rolls once around the circumference of circle $A$ . The two circles have the same points of tangency at the beginning and end of circle $B$ 's trip. How many possible values can $r$ have?

Show solution without answering

Key insight

Rolling without slipping, circle B has turned 100/r times when it gets back to the start; the same point is touching again only if r divides 100.

Solution

1. When a circle rolls without slipping, the length of the path it rolls along equals the length of its own edge that has touched the path.
2. Circle $B$ rolls all the way around circle $A$ , a path of length $2\pi \times 100$ . Circle $B$ 's edge has length $2\pi r$ . So circle $B$ goes around
$\frac{2\pi \times 100}{2\pi r} = \frac{100}{r}\ \text{times}.$
3. Try examples. If $r = 50$ , circle $B$ turns exactly $2$ times, so the starting point of $B$ is touching again at the end. If $r = 30$ , it turns $3\frac13$ times, so a different point of $B$ is touching at the end.
4. So the condition is that $\frac{100}{r}$ is a whole number, which means $r$ divides $100$ .
5. The divisors of $100$ are $1, 2, 4, 5, 10, 20, 25, 50, 100$ . That is $9$ numbers.
6. The problem says $r < 100$ , so remove $100$ . That leaves $8$ values.

The answer is $\boxed{\textbf{(B)}\ 8}$ .

Why this works

Rolling matches lengths: the distance rolled equals the part of the circle's edge used up. So "the same point touches again" means the big circle's edge is a whole number of small circle edges long, which is a divisibility condition. Then count divisors, and read the inequality $r < 100$ carefully.

Concise solution

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

As circle $B$ rolls without slipping, the arc of $A$ it has covered always equals the arc of $B$ that has been in contact. One trip around $A$ uses up arc length $2\pi \cdot 100$ on $A$ , so the same total arc of $B$ has touched $A$ .

The same point of $B$ is back at the contact point exactly when this arc is a whole number of $B$ 's circumferences:
$\frac{2\pi \cdot 100}{2\pi r} = \frac{100}{r} \text{ is an integer},$
i.e. $r$ divides $100$ .

$100 = 2^2 \cdot 5^2$ has $(2+1)(2+1) = 9$ positive divisors: $1, 2, 4, 5, 10, 20, 25, 50, 100$ . The condition $r < 100$ removes $100$ , leaving $8$ values.

The answer is $\boxed{\textbf{(B)}\ 8}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 1: Integers: The Basics (§1.4, §1.5)
Introduction to Geometry, Chapter 11: Circles
Introduction to Number Theory, Chapter 5: Divisor Problems (§5.2, §5.3)
Introduction to Geometry, Chapter 12: Circles and Angles