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

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. • If it's a one-degree sector, then points at the "corners" are much closer to each other than either is to the vertex. – Michael Hardy Aug 2 '18 at 1:58
• Sorry, yeah I am thinking of a sector which has an angle $\leq \frac{\pi}{3}$ – T.Harish Aug 2 '18 at 2:23
• If that's what you have in mind, then the corners are not the right places to make points far away from each other. – Michael Hardy Aug 2 '18 at 2:25
• If you divide the unit disk into 6 equal sectors by three diameters, wouldn't the two points at the boundary be of maximum possible distance in that sector? – T.Harish Aug 2 '18 at 2:30
• It should be $\pi/3,$ radians, not $\pi/6.$ One-sixth of a circle is $\pi/3$ radians because the full circle is $2\pi$ radians. The reason the boundary is at $\pi/3,$ i.e. at one-sixth of a circle, is that that is the point at which the chord is equal to the radius. For smaller angles, the chord is less than the radius; for larger angles, the chord is more than the radius. – Michael Hardy Aug 2 '18 at 2:46

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