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That one got me good, so my future at College Board is as bright as yours. However, I don't think the argument by demonstration in that video is particularly convincing.

Instead, I think's its easier to note that that the _center_ of a circle of radius r travels 2 * pi * r distance over one rotation. In the problem, the center of the smaller circle has to travel further than the circumference of the bigger circle - it traces a circle whose radius is the sum of the two radii.

So, if 3 * r_small = r_big, the center of the small circle has to travel 2 * pi * (r_s + r_b) = 4 * 2 * pi * r_s, then divide by 2 * pi * r_s per rotation to get 4 rotations.

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You could also argue that it is a matter of perspective. From the perspective of either circle, A will only revolve 3 times.

Only by introducing a larger frame of reference, a grid or in the video a table, you gain an outside perspective. From this outside perspective you redefine a revolution according to some new orientation and end up with n+1 revolutions.

Or maybe the argument is backwards and I just try to justify answering 3.

The demonstration would be clearer if the "point of contact" at the start of the rotation would be marked on the smaller circle and the larger circle would be divided into 3 segments of different color. That would make it obvious that 1 rotation of the small circle doesn't trace a whole circumference of the small circle on the large one.

Without the math:

Radius is always proportional to circumference, so a circle twice the size is twice as big around.

Take the case of two identical circles. To move a point on the first circle from 12 o'clock back to 12 o'clock, it only goes halfway around the other circle, which you can prove to yourself by imagining you've wrapped a string around the circle and marked it at 12 and 6 o'clock. If you unwrap half of the string and wrap it around the other circle, then the end of the rope is at 6 o'clock. To roll the string back up by moving the circle, the top of the circle will be pointing upward again when it reaches the bottom. 1 full revolution. Now wrap the other half of the string around the other side, 6 has to go back to the bottom again to roll the string back up. 2 revolutions.

It’s interesting to read others reasoning on this. I find it hard to follow and much harder to generalize without some notation (and a diagram which this comment box struggles to reproduce)

I like to decompose it.

Hold center of small circle still, rotate big circle once counter clock wise, small one rotates 3x clockwise.

Glue them together, rotate big circle once clockwise, small circle also rotates once clockwise.

Sum them together, 0 rotations for big circle, 4 for small. I'm not at all sure how to rigorously generalize.

I don't think that argument holds water. Imagine rolling it on the inside of the circle – the center of the circle traces a path with only twice its radius, yet it still rolls 4 times around.

I don’t think that’s right. It if the radius of the inner circle is one third the outer, it only rotates twice rolling around the inside, which makes sense as it’s center traces a circle that’s the the difference of the radii.

Imagine the limiting case, as the inner circle approaches the size of the outer circle - the inner circle completes much less than one rotation per lap around the inside edge of the outer circle, and ‘seizes’ (if we’re imagining these as gears), completing zero rotations per lap when the circles are the same size. However, rolling around the outside, a circle of the same size completes two rotations.

In general the problem is like the old Spirograph toy (which I had to break out to convince myself)

I think you're right!

I made a diagram that helped me think through it visually:

https://i.imgur.com/dospt2w.png

If circle A is rolling around the edge of circle B from within, it is actually revolving around a new, smaller circle C which has the Radius Circle B - Circle A.

Is this correct? wouldn't the radius of the circle it's revolving around be significantly smaller if you had it follow the edge of the larger circle but from the inside? I don't know much about math or physics, so I could be wrong, but I think it would be significantly less, closer to two, right?

The reason that this problem is tricky, and has a counterintuitive solution is that Circle A is rolling around Circle B, and so the 'radius; of the circular path it is following is the radius of Circle B + the radius of Circle A.

I'm no expert, but some quick math:

Circle B has a radius of 9, circumference then is 56.55 Circle A (1.3) has a radius of 3, circumference is 18.84

56.55 / 18.84 = 3. This suggests that you could "unwind" circle A (say it was made of pipecleaner), and you would need 3 circle As to fully ensconce Circle B

BUT that wasn't the question. The question states that Circle A is rolling AROUND circle B.

So the circumference of the circle we are now trying to 'ensconce' is Circle A Radius + Circle B Radius = 12, so the new circumference is 75.39, and divided by Circle A Circumference, we end up with 4, which makes sense, and matches the demonstration from the video.

HOWEVER, if we are 'rolling' around the inside of circle B, then I think you're right, the radius of the circular path Circle A will take is 9 -3 = 6, and the circumference of said circle is 38, so we it only will take two rolls. I think this is correct, i do not think it will be 4 rolls.

Think of it this way: when Circle A is rolling around the inside of Circle B, the way you're picturing it is circle A following the outside of Circle B, which is why I think intuitively it feels like the answer will still be 4 revolutions. BUT a better way to think of it is that Circle A is actually revolving around a new Circle, Circle C, which has a circumference of 38. Circle A is not revolving around Circle B, it's revolving around this new, smaller circle within circle b. Does that make sense?

I made a quick diagram to illustrate my point. I did it in a wire-framing tool that has snap to grid, so it's definitely not perfect:

https://i.imgur.com/dospt2w.png

You can play around with what I made here:

https://whimsical.com/r-DoXHwe2kUdgvSiAu2yuGpE