This question is motivated by an answer I provided to a question here on the arc length of a cycloid.

I noticed that the ratio of the circumference of the generating circle (which is also the horizontal distanced traveled through one cycle) to the arc length of the cycloid is

$$ \frac{2\pi r}{8r} = \frac{\pi}{4} . $$

If we take that generating circle and circumscribe a square around it, the ratio of the areas of the circle to the square is the same:

$$ \frac{\pi r^2}{(2r)^2} = \frac{\pi}{4} . $$

Given a circle, is there some sort of deeper relationship between the cycloid it generates and its circumscribed square? Or is this just a coincidence? I don't have much experience with roulettes, so I'm at a bit of a loss here - would appreciate any insight anybody has on whether there's some deeper geometric relationship here.

  • $\begingroup$ My vote is coincidence. The mathematical world is full of numerical coincidences. $\endgroup$ Commented Feb 26, 2021 at 19:02
  • 2
    $\begingroup$ @TedShifrin that was my inital thought too, but was tinkering with cyclogons and thought there might have been something there. Also, this isn't the appropriate venue for this, but I just have to say that my first undergraduate math course used your linear algebra textbook, it completely changed how I thought about both algebra and geometry, and I just wanted to thank you for that! $\endgroup$
    – Amaan M
    Commented Feb 27, 2021 at 1:38
  • $\begingroup$ @OscarLanzi I hope you find something, I saw your initial response but wasn't able to fully think through it, looking forward to what you come up with! $\endgroup$
    – Amaan M
    Commented Feb 27, 2021 at 1:39
  • $\begingroup$ Done, have a look. And there is a bonus! $\endgroup$ Commented Feb 28, 2021 at 16:45

1 Answer 1


Yes there is a relationship, based on a couple hidden features of cycloidal loci. And there is also a similar relationship between the four-cusped hypocycloid and a regular hexagon inscribed in the same circle, derivable from thecsame concepts.

Making a clean sweep

You already know that when a circle is rolled along a fixed straight line, a fixed point on the circle generates a cycloid loop whose arc length is four times the diameter of the circle. Since the center of the circle also slides through a length equal to (of course) $\pi$ times the diameter, the arc length of the cycloid loop may also be rendered as $4/\pi$ time the length traversed by the center.

Now suppose the straight line is replaced by a circular arc to make an epicycloidal loop (if the fixed arc is externally tangent to the roller) or a hypocycloidal loop (if internally tangent). The calculus is more complicated now, but we find that the arc length of the epicycloidal/hypocycloidal loop is still $4/\pi$ times the arc length traversed by the center of the roller.

To explain this geometrically, consider the picture below in which a cycloid is generated from point $A$ through $A'$ to $A''$. The roller sweeps through an area, which is equivalent to a rectangle whose dimensions are the length of the center's path and the diameter of the roller. Replacing the fixed line with a circular arc leaves this relationship unchanged; half the area is stretched but the other half is squeezed to balance it. (We assume that in the hypocloidal case the fixed arc has large enough radius so that its center is not in the interior of the roller. This assumption will be obeyed in what is to follow.)

enter image description here

We imagine this area as being filled with crisscrossing cycloidal loops whose points on the roller are distributed over the circumference of the same. Thus the arc length of a loop should be proportional to the swept area divided by the circumference. This quotient contains only the length of the centerline path.

Through this proportionality we may relate the length of the cycloid loop to various epicycloid or hypocycloid loops. For instance, suppose that the rolling circle from the cycloid generation is now held fixed and we make a new roller, internally tangent with half the radius. The fixed point now moves from $B$ through $B'$ to $B''$ where it re-contacts the larger circle.

enter image description here

In this case the center of the roller has moved only half its circumference and the roller itself is only half the circumference of that used for the cycloid. Therefore, the length of this path should be one-fourth the length of the cycloid loop. If the $B$ to $B''$ path corresponds to one side of a circumscribed square, then the cycloid loop will match the whole square. But, can we prove the former claim? And is there something to the hypocycloid loop in my second picture looking like a straight line?


In my younger days I played with a drawing toy in which you place a pen inside a wheel (the pen being a fixed point in the wheel) and roll it around a wheel, ring or rack that is fixed to the paper, thus producing various patterns. If you choose a wheel with $48$ teeth and roll it in a ring interior having $96$ teeth (a $1:2$ ratio), you get just an oval, which becomes longer and narrower if you place the pen near the outer edge of the wheel. Evidently if the pen could be placed at the very edge of the wheel (physically impossible because the gear-teeth are there), this oval would be elongated and flattened to a full diameter of the ring. Which would match a side of a circumscribed square ... .

To prove this mathematically, consider the figure below for the case where the roller is moved up to 90° from the initial contact.

enter image description here

Because of the $1:2$ circumference ratio the roller has the center $O$ of the fixed circle on its boundary (not in the interior, see above), and also the arc on the roller from $C$ to $B'$ has twice as much angular measure as the arc from $B$ to $C$. Now, central $\angle BOC$ has the same measure as arc $BC$, and inscribed $\angle B'OC$ has half the angular measure as arc $B'C$ -- so the two angles are congruent, and this indeed forces $B'$ to lie on the radius $\overline{OB}$.

For a greater rolling angle, first move the roller $180°$ to recontact the locus at $B''$, then roll back through the supplementary acute angle and apply the above reasoning replacing $B$ with $B''$.

Fitting a square peg around a round hole...

So our hypocycloidal loop, with a $1:2$ ratio of diameters or circum-ferences, indeed matches a diameter of the circle and thus one side of a circumscribed square, and measures one-fourth the length of the corresponding cycloid. Therefore the cycloid loop measures the full perimeter of the square.

...and a hexagonal peg inside

At the beginning of this geo-metrical trip, I mentioned that there is a similar relationship between a four-cusped hypocycloid and a regular hexagon inscribed in the same circle. I present that here in the form of a "homework" problem.

  1. A four-cusped hypocycloid is generated as the fixed-point locus obtained when a roller of diameter $r/2$ is rolled inside a fixed circle of diameter $2r$. Using the fact that the hypocloid can be rendered with the parametric equations $x=r\cos^3\theta,y=r\sin^3\theta$, apply standard calculus techniques to find that the length of the complete hypocycloid matches the perimeter of an inscribed regular hexagon.

  2. Compare the arc length traversed by the center of the roller over all four hypocycloidal loops to the corresponding arc length associated with diameter $\overline {BB''}$ above. Conclude that the length of the hypocycloid matches three diameters of the circle, which can be matched by obvious geometric means with the inscribed regular hexagon.

  • $\begingroup$ This is fantastic, thanks for the explanation! Looking forward to working through the homework problems! $\endgroup$
    – Amaan M
    Commented Mar 6, 2021 at 18:45

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