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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?

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The answer is

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°.

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    $\begingroup$ I think it's a perfectly fine proof, what exactly do you not like? $\endgroup$ Jun 13 '21 at 19:46
  • $\begingroup$ 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?) $\endgroup$ Jun 13 '21 at 20:06
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    $\begingroup$ @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. $\endgroup$
    – loopy walt
    Jun 13 '21 at 20:14
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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 $\frac{90}{91}$.

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