# How many turns does a circle of circumference 1m make as it rolls around an equilateral triangle of perimeter 3m?

A circle of circumference $$1$$m rolls around an equilateral triangle of side $$1$$m (perimeter $$3$$m). How many turns does the circle make as it rolls around the triangle once without slipping?

The correct answer is 4 but I have no idea why it is 4 instead of 3. I have searched and found some information on the web about “the perimeter of the triangle rolls”. I have no idea what that means. Anyone can elaborate further?

Approximate image:

Thank you!

• Tip: Think about what happens at the edges. Commented Apr 20, 2020 at 12:40
• The coin is lying down in the same plane as the triangle. What happens at the corners ? Commented Apr 20, 2020 at 13:07
• Thanks!Corners or vertices, I didn't mean edges.@Empy2 Commented Apr 20, 2020 at 13:10
• @J Muzhen I voted to reopen this question. At every corner the circle makes a third of a turn. I made a picture using tikz to explain in detail, which I will load up when and if the question is reopened Commented Apr 21, 2020 at 9:50
• There is a case for the correct answer to be $3\frac23$ $-$ the question doesn't say that the circle has to end up in the same place as it started, just that it has to roll all the way around the triangle. It might have started at a corner, and stopped as soon as it touched that corner again. Commented Apr 21, 2020 at 12:30

To count the turns, mark a fixed radius on the circle (in red below), then count how many times it turns around the center during the movement. While rolling along a side, the fixed radius turns at a constant (angular) rate, as expected. But watch what happens when you turn the corner. Right before the turn, the radius is perpendicular to the bottom side of the triangle. Right after the turn, the radius becomes perpendicular to the right side of the triangle. Therefore, the turn introduced an instantaneous additional rotation, and it's simple geometry to see that it's $$120^\circ$$. There is $$3$$ corners, so it adds up to one additional complete revolution by the time the circle rolls back to the starting position. Add that to the $$3$$ turns due to rolling, and it's the correct answer of $$4$$ turns total.
Once you visualize this right, it becomes obvious that there is nothing special about the triangle. If the circle rolled around a rectangle, instead, there would be $$4$$ turns of $$90^\circ$$ each, which would once again add up to one complete additional revolution. It doesn't even have to be a polygon, the same would happen around another circle, or around any simple closed curve, for that matter.
At each corner the circle makes a third of a turn. From the position where it touches the corner and is tangent to the side of the triangle to the position "on top of the triangle" ( i.e. where the tangent is perpendicular to the height of the triangle) it makes a sixth of a turn, since from the picture You can see that it turns by an angle of $$\alpha=60^{\circ}$$. This happens again on the other side of the corner. So if the circle ends up in the same position as where it started it makes $$3$$ turns,$$1$$ per side, plus $$3\cdot\frac{1}{3}$$ for the corners, which makes $$4$$. However as @TonyK pointed out it might end at the other side of the corner as where it started and in this case it would have made just $$3+\frac{2}{3}$$ turns. The picture shows what happens when the circle passes from the position tangent to the side of the triangle to the top:
• In this case the answer is clearly :"$4$ turns!" Commented Apr 22, 2020 at 10:37
• I corrected "3 per side" to "3 turns,1 per side", what I meant with "from the picture You can see" is, the interior angel of the triangle is $60^{\circ}$ since it is equilateral, so the side angles are $60^{\circ}$, too, and thus the angle between the radii, since it is one of the side angles rotated by $90^{\circ}$, thus the circle makes a sixth of a turn between these positions as $\frac{60}{360}=\frac{1}{6}$. Commented Apr 22, 2020 at 11:28