My name is Craig Ellis and I have a question on the 10-12 grade level. I am a student teacher intern studying to be a high school teacher so I really should be able to figure out this question myself, but for the life of me I have been unable to do so. The problem was posed by my methods teacher.
We have a circle of radius 3. inside the circle and tangent to the circle of radius 3 at one point is a circleof radius 1. The question is if we could roll the smaller circle around the inside of the larger circle how many revolutions would it take to get around to where we started.
The professor said that the answer was not three although it seems that it should be to me. I am obviously doing something wrong. I even built a model and it came out as three. Next we are to roll the circle around the outside of our circle of radius three and see how many revolutions that takes. Anyway I hope that you can make better sense of this than I can and thanks in advance for any light you are able to shed on problem.
Your answers are both correct -- it's three revolutions or two
revolutions depending on what you want to call a revolution. Before
explaining this mysterious situation, I should point out that the same sort
of question arises concerning our moon as it moves about the earth once
each month: the same face of the moon always points toward the earth, so to
us on earth it looks as if the moon does not rotate -- it just slides
around the earth on its 29.5-day journey. For an astronaut who is far
enough out in space to see both the earth and the moon, he would see the
moon rotate once about its axis as it makes one revolution in its orbit
about the earth. (To keep the two motions straight astronomers use the
special terminology that that a body ROTATES about its axis and REVOLVES in
an orbit about a body fixed in space. For the rest of us these words are
interchangable, hence the ambiguity.)
If you are still stuck, you could try a simpler example where the ratio of the radii is, for example, 2:1. Still stuck? Try the ratio being essentially 1:1, with the inside circle just pressing lightly on the outside circle as the 'point of pressure' moves around the outside circle. The situation with the ratio 1:1 and R rolling about the outside of F is probably easier to simulate with, say, a couple of quarters.
The story is clearly explained by Jack Eidswick in his lovely little article, "Two Trisectrices for the Price of One Rolling Coin," in THE COLLEGE MATHEMATICS JOURNAL, 24:5 (November, 1993) pages 422-430. For the general answer we adopt the convention that r > 0 if R is OUTSIDE F, and r < 0 INSIDE F: the number of revolutions (of the rolling circle about its center) is then f/r - 1 (where we interpret a "negative revolution" to be clockwise).