# Closing an irrational curve

For any rational number q, a finite number of congruent circular arcs each measuring 2πq radians can be assembled into a continuous (possibly self-intersecting) closed curve. There are many other possibilities, but for fractions of the form 1/n, one can simply use n copies to reconstitute a circle. And more generally for a/b, b copies can close a curve covering a circle a times.

Does there exist any irrational number p such that finite congruent 2πp-radian arcs can be assembled into a continuous closed curve with no corners?

• pi doesn't have any rational multiples. – matt Oct 23 '20 at 21:51
• "rational multiple of x" meaning "product of x with a rational number" rather than "rational number which is product of x with a positive integer" – Zomulgustar Oct 23 '20 at 21:55
• $2 pi$ is a rational multiple of $pi$, for example. – Florian F Oct 23 '20 at 22:10

You can.

You can do right-right-left-right-left, left-left-right-left-right. By symmetry you are bound to get back to a point on the same axis as where you started.

By adjusting the angle you can move the end point forward or backwards and make it coincide with the starting position. The path is irregular enough to probably end up with an irrational.

One angle that works is $$1.823476582$$ radians, or $$0.290215312 \times2\pi$$.
I have no proof that it is actually irrational, though, but it looks like.

• Nicely done! I was just about to say that you accomplish the same trick with a shorter sequence, but you beat me to it. ^_^ I'd say that earns enough brownie points to skip the formal proof of irrationality, but if you're so inclined, consider where the endpoints of each arc would have to be on the complex plane... – Zomulgustar Oct 23 '20 at 22:28
• If we scale/rotate/translate the first arc to have endpoints at 0 and 1, then each time the arc pivots in the same direction we rotate the resulting displacement by our not-yet-determined angle, or multiplying by a unit-norm z. Each time we pivot back in the opposite direction, we remove a factor of z. So each arc respectively translates us 1,z,1,z,1,z^-1,z^-2,z^-1,z^-2,z^-1 , and all the translations bring us back where they started, and thus must sum to zero. Combining like terms gives 2z+3+3z^-1+2z^-2=0, which can multiply by z^2 to give 2z^3+3z^2+3z+2=0. – Zomulgustar Oct 24 '20 at 0:37
• Factoring out the real root gives (z + 1) (2 z^2 + z + 2)=0, leaving two more unit-norm roots, which verifies there is an angle that actually works. If the angle defined by these were rational, z would be a primitive nth root of unity for some integer n, and thus also be a solution to z^n-1=0. However, every factorization of these with integer coefficients is monic (coefficient of 1 on each leading term), and thus cannot simultaneously have (2 z^2 + z + 2) as a factor. – Zomulgustar Oct 24 '20 at 0:59
• Nice but if you think of the endpoints of the curves, doing left-right-left-right makes four steps in the same direction. If however you replace each curved segment by a 1/2 unit forward, rotation on the spot, 1/2 unit forward, and recombine the 1/2 units before and after the endpoint you get steps 1, z, 1, z and your reasoning applies. – Florian F Oct 24 '20 at 14:28
• Quite right, looks like I was following the midpoints rather than the endpoints and just got lucky. Thanks for straightening that out! – Zomulgustar Oct 24 '20 at 22:15