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Suppose that a goat lives inside a disc and is tethered by a rope fixed at a point $P$ on the circumference of the disc. The rope is just long enough to let the goat graze half the area of the disc. This implies that the grazing boundary is an arc of a circle which subtends an angle $\alpha$ at $P$, where $\alpha$ is the unique solution in $(0,\pi)$ of $$\sin\alpha-\alpha\cos\alpha=\frac\pi2\ .$$

My question: is $\alpha$ transcendental? (Obvious guess: yes.) Can anyone prove it?

I have tried assuming $\alpha$ is algebraic, so $i\alpha$ is algebraic, so by Lindemann's Theorem $e^{i\alpha}$ is transcendental, looking for a contradiction; but have not got anywhere.

If anyone wants to know where the equation came from, one of many sources is https://www.youtube.com/watch?v=ZdQFN2XKeKI.

David
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It doesn't seem likely that a proof for the transcendence of $\alpha$ is currently possible using existing results.

$\alpha$ is the least positive solution to $\sin(\alpha)=\frac{\frac{\pi}{2}+\alpha\sqrt{\alpha^2+1-\frac{\pi^2}{4}}}{\alpha^2+1}$, so if $\alpha$ is algebraic, $\sin(\alpha)$ must be an algebraic function of $\pi$, or equivalently, $\alpha$ is an algebraic function of $\pi$ and $\sin(\alpha)$

The transcendence of $\alpha$ would follow from the algebraic independence of $\sin(\beta)$ and $\pi$ for any algebraic $\beta$, or $e^\beta$ and $\pi$. Unfortunately, even much simpler examples of the transcendence of constants $\alpha_1$ that are algebraic functions of $\pi$ and some $e^\beta$, like $\alpha_1=\pi e$ or $\alpha_1=\pi+e$, are currently out of reach.

David Chew
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  • On the other hand, it is very easy to prove that if $\alpha=\cos\alpha$ then $\alpha$ is transcendental. (Easy subject to assuming Lindemann's Theorem, anyway.) – David Aug 27 '24 at 03:22
  • Here is another interesting example. The constant $\alpha=\cos(\beta)+\sqrt{(\pi+\beta)^2-\sin(\beta)^2}$ where $\beta$ is the least positive solution to $\tan(\beta)=\beta+\pi$ can be proven transcendental via Lindemann's Theorem, while $\beta$ cannot (Context here). To summarize, this can be done by showing $\alpha=-\sec(\beta_1)$ where $\beta_1$ is a solution to $\beta_1=\arctan(\beta_1)$, so $\beta_1$ is transcendental and $\alpha$ is an algebraic function of a transcendental. Here, more complexity made transcendence easier to prove. – David Chew Aug 27 '24 at 18:49