Is the ratio of the perimeter of any shape with circular curves to its diameter result in an irrational number?

I suppose it would depend one what would be defined as "circular curves", as well as where you would measure the diameter on a shape that is not infinitely radially symmetrical like a circle. So I'd like to keep the question open to different possibilities compared with others.

What brings me to ask to the question in the first place is imagining the irrational or transcendental property emerging from curves.

One scenario to consider would be a square with rounded corners. If it was a ratio of the shape's perimeter to the the diameter measured from one corner to a corner diagonally opposite of it, would it result in an irrational number?

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How about the same scenario, but it's the ratio of the shape's perimeter to the diameter measured from the center of one flat side to the opposite flat side?

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The reason I have not figured this out for myself is that I don't know how to get the perimeter from a square with curved corners, or where to measure from in the case where the diameter is measured from the corners. (I don't know how to get the exact points on the curve.)

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    $\begingroup$ What do you mean by circular curves? Any curve? Any continuous border that is not linear? And how do you define a diameter if the shape is not a circle? Perhaps it would be a more satisfying question and jumping off point to ask this question of "shapes of constant width". $\endgroup$ Commented Mar 19, 2022 at 17:13
  • $\begingroup$ I brought up both of those questions. It feels like I have to keep it vague since more rules would have to be introduced compared to getting the result of pi. $\endgroup$ Commented Mar 19, 2022 at 17:15
  • $\begingroup$ I'm not sure what other shapes have constant width besides circles and Reuleaux shapes, but a Reuleaux triangle may be one to consider. $\endgroup$ Commented Mar 19, 2022 at 17:19
  • $\begingroup$ Actually I see that in the case of the Reuleaux triangle, the ratio to get pi for a circle works similarly here, which results in pi again. But part of my equation to measure the perimeter is pi * width, so that's not surprising. $\endgroup$ Commented Mar 19, 2022 at 17:27
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    $\begingroup$ Take your square with circular corners example. Let $r$ be the ratio you're interested in. If you round the corners a bit more, and let $r'$ be the new ratio, then $r \neq r'.$ Regardless of whether $r$ and/or $r'$ are rational or irrational, there will be a rational number $q$ between $r$ and $r',$ and thus at some point during the process of rounding the corners the ratio will equal $q$ (by an intuitive continuity assumption), and so the ratio can be rational. (@Hagen von Eitzen's answer, written while I was writing my comment, seems to be essentially the same idea.) $\endgroup$ Commented Mar 19, 2022 at 17:43

1 Answer 1


If within the class of curves you want to consider, there exists a 1-parametric family of curves (i.e., you can sometimes continuously deform such a curve without making it “invalid”) and not all curves in thus family have the same perimeter/diameter ratio, then the answer is definitely: No, some such curves have a rational ratio.

Note that this is specifically true for your example class of curves: squares with rounded corners. Indeed, such a shape with straight edge parts of length $a$ and rounded corners if radius $r$ has perimeter $4a+2\pi r$ and diameter $a\sqrt2+2r$ so that the ratio is $$\frac{4a+2\pi r}{a\sqrt2+2r}$$ Or with $t:=\frac ar$, $$\frac{4t+2\pi}{t\sqrt2+2}.$$ For $t=0$, this is $\pi$, and as $t\to\infty$, the ratio approaches $2\sqrt2$. Any value in between, including infinitely many rational values, is obtained for a suitable $t$. One such rational value is $3$. Can you find the $ t$ for that?

  • $\begingroup$ For the top equation I used a value of 2.5 for a, which gives me about 2.9581445341. For the bottom one, I used 1 as a value of t, and I get almost 3, but not quite. (3.0118752443) $\endgroup$ Commented Mar 19, 2022 at 18:07
  • $\begingroup$ I understand your question, but solving this algebra is a little beyond me embarrassingly enough, but I'm reviewing how to do it now. I think of these things, but have no background in math. $\endgroup$ Commented Mar 19, 2022 at 18:28

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