I thought of a problem, and would like to know your approach. As hinted at in the title already, what the biggest square possible to fit in the standard sphere $S^1$. I calculated it basically mechanincally. It is basically the square with length $2*\sin\pi/4=\frac{2}{\sqrt2}$. But i would like to approach it more technically. What I have done is, to think of the radius of the sphere and then using trigonometric identities to show it is $\sin\pi/4$
1 Answer
No need for trig identities. To fit a maximal square inside the unit circle, half of the diagonal must be equal to the radius. If $x$ is half of the sidelengths of the square we must thus have $2x^2 = 1$, so $x = \frac{1}{\sqrt{2}}$ and the sidelengths are $\frac{2}{\sqrt{2}}$.
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$\begingroup$ I need a more technical approach. Thanks a lot for your kind answer. $\endgroup$ Oct 13, 2022 at 14:39