# Vertices of a regular $13$-gon and $14$-gon on a circle with center angle $< 1°$

All vertices of a regular $$13$$-gon and all vertices of a regular $$14$$-gon lie on a circle and divide it into $$27$$ circular arcs. Is there always an arc, which corresponding center angle is less than $$1$$ degree?

yes.

The easiest way to see that is

by pigeon hole principle: There must be one side of the 13gon spanning 2 vertices of the 14gon. As 1/13-1/14=1/182 the gaps on both sides sum to less than 2°. So at least one must be less than 1°.

• I think it's a perfectly fine proof, what exactly do you not like? Jun 13, 2021 at 19:46
• Rot13 (V'z phevbhf, pna guvf or trarenyvfrq gb n erthyne k-tba naq n l-tba (k>l), ol fnlvat gung gur ynetrfg cbffvoyr inyhr bs gur fznyyrfg natyr fhograqrq ol na nep vf obhaq ol (360/k - 360/l)/2?) Jun 13, 2021 at 20:06
• @KabirKanhaArora You need to swap x and y in the formula but apart from that I don't see any reason why it wouldn't work. Note, however, that the bound will not be very sharp if x and y are not close. Jun 13, 2021 at 20:14

Let $$A$$ and $$B$$ be a vertex on the $$13$$-gon and a vertex on the $$14$$-gon, and $$O$$ be the center; then let $$\theta$$ be the angle $$AOB$$ (in radians). If we measure all angles $$AOX$$ where $$X$$ is on the $$14$$-gon, they are of the form $$\theta + k\frac{2\pi}{14}$$ for integers $$k$$, and any $$k$$ corresponds to such an angle. But angles $$AOY$$ where $$Y$$ is on the $$13$$-gon are $$k\frac{2\pi}{13}$$ (all $$k$$ work again) so angles $$XOY$$ are of the form $$\theta + 2\pi(\frac{a}{14}-\frac{b}{13}) = \theta + \frac{\pi}{91}(13a+14b)$$. But $$13a+14b$$ takes all integer values so all remainders modulo $$91$$. Then it suffices to show that for all real numbers $$x$$ there is an integer $$n$$ with $$|x\pi-\frac{n}{91}\pi|<\frac{\pi}{180}$$ or $$|x-\frac{n}{91}|<\frac{1}{180}$$ which is obvious.

Edit: since the steps are reversible, the problem is true for $$1$$ replaced by $$\frac{90}{91}+\varepsilon$$ for any $$\varepsilon>0$$, but it is false for $$\frac{90}{91}$$.