2
$\begingroup$

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. enter image description here

$\endgroup$
  • $\begingroup$ If it's a one-degree sector, then points at the "corners" are much closer to each other than either is to the vertex. $\endgroup$ – Michael Hardy Aug 2 '18 at 1:58
  • $\begingroup$ Sorry, yeah I am thinking of a sector which has an angle $\leq \frac{\pi}{3}$ $\endgroup$ – T.Harish Aug 2 '18 at 2:23
  • $\begingroup$ If that's what you have in mind, then the corners are not the right places to make points far away from each other. $\endgroup$ – Michael Hardy Aug 2 '18 at 2:25
  • $\begingroup$ 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? $\endgroup$ – T.Harish Aug 2 '18 at 2:30
  • 1
    $\begingroup$ 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. $\endgroup$ – Michael Hardy Aug 2 '18 at 2:46
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.