Find the two points with the maximum distance in a sector of a unit disc

Question

Find two points $P,Q$ in a given sector which has an natural angle $= \frac{\pi}{3}$ of a unit disc such that they attain the maximum possible distance between them. Prove where they should be formally.

I have the intuition that they should be on the two corners of the given sector like shown in figure. I have tried circumscribing this in a regular hexagon as it seems which sector does not matter. I have tried formulating as optimization problem using a suitable coordinate system, but unable to prove formally.

Answer 1

Hint: the answer will be different depending on how the opening angle of your sector relates to $\pi/3$.

Answer 2

If either point is $\,P \equiv O\,$ then any $\,Q\,$ on the arc will be at maximum distance $\,= 1\,$. Otherwise let $\,OP = a\,$, $\,OQ = b\,$ with $\,a,b \in (0,1]\,$, and $\,\angle POQ = \varphi \le \pi/3 \,$. By the law of cosines:

$$ PQ^2 = a^2 + b^2 - 2 ab \cos \varphi \le a^2 + b^2 - 2 ab \cos \pi/3 = a^2+b^2-ab = \frac{a^3+b^3}{a+b} \le 1 $$

The last inequality follows because $\,a^3 \le a\,$ and $\,b^3 \le b\,$ for $\,0 \lt a,b \le 1\,$. Equality is attained iff $\,\varphi = \pi/3\,$ and $\,a=b=1\,$ i.e. $\,P,Q\,$ are the endpoints of the arc.