# Rotation angle of regular polygon that has largest taxicab maginitude between all vertices

Firstly just to apologise, I posted this on mathoverflow before realising it was focused on research level mathematics.

If I have a regular polygon that is centred at the origin. Then take the difference between all vertices.

Let $U$ = the maximum $x$ difference

Let $V$ = the maximum $y$ difference

I want to figure out the angle of rotation that creates the maximum

$$W = U + V \, .$$

So for example imagine a line (I know it's not a polygon); if the angle of rotation is $0$ then its vertices are at $( 1, 0 )$ and $( -1, 0 )$

$$U = 1 - (-1) = 2\\ V = 0 - 0 = 0$$

So $W = 2 + 0 = 2$.

The optimum is 45 degrees of rotation in which case

$$U = 0.707 - -0.707 = 1.414 \\ V = 0.707 - -0.707 = 1.414$$

So $W = 2.828$.

If my calculations are correct, an equilateral triangle's optimum is $45$ degrees as well. However its obvious to see the optimum for a square is 0 since the vertices align perfectly with $x,y$ axis (i.e., $(1,0)$, $(0,1)$, $(-1,0)$ and $(0,-1)$).

Hopefully I've explained it well enough. I can find the solution for single instance of the problem using graphs and derivatives BUT it takes a long time. I'm looking for a general purpose algorithm as this is going to be used in some software I'm writing.

Thanks in advance.

Iain

PS I forgot to mention that there'll be multiple solutions due to symmetry. So the line's optimum is $45, 135, 225, 315$. I believe the triangles is $15 + 30n$, i.e. $15, 45, 75, \dots$

PPS The vertices are generated using the following formula:

Let $n$ = number of sides

Let $c$ = 360 / $n$

For all i, 0 <= $i$ < $n$

$V_i$ = ( cos($i$.$c$), sin($i$.$c$) )

E.g. a regular triangles vertices are:

( cos(0), sin(0) ) = ( 1, 0 ) ( cos(120), sin(120) ) = ( -0.5, 0.866 ) ( cos(240), sin(240) ) = ( -0.5, 0.866 )

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@Ross thanks for the reply. I don't quite understand it though. I've appended a PPS to the original question which describes how the vertices are generated so there is no ambiguity. Firstly I agree/understand that 4n sided shapes will have an easy optimum. Using my convention 0 degrees rotation. However that leaves 4n + 1, 4n + 2 and 4n + 3 and you seem to only tackle 4n + 2. Then I don't quite understand how you only generate 4 points when n could be anything. As such I can't follow the next step which calculates the max x and y distances U and V. – Iain Fraser Jun 12 '13 at 20:27

## 1 Answer

You need to define what the angle of rotation is measured from. I would have taken zero rotation for the square to be with the sides parallel to the axes. Having a corner on the axes is clearly optimal for $4n$ sided polygons. For $4n+2$ sided polygons it will not be optimal because you gain more from rotating the flat sides than you lose by rotating the points.

To be exact, let $0$ represent having points on the $x$ axis. We start with points $(1,0), (\cos \frac {2n}{4n+2} \pi ,\sin \frac {2n}{4n+2}\pi),(-1,0), (-\cos \frac {2n}{4n+2} \pi ,-\sin \frac {2n}{4n+2}\pi)$ and if we rotate counterclockwise by $\theta$ we get $U=2 \cos \theta,V=2\sin (\frac {2n}{4n+2}\pi+\theta)$ We add these up, differentiate, set to zero, etc. $$U+V=2 \cos \theta +2\sin \left(\frac {2n}{4n+2}\pi+\theta\right)\\\frac d{d\theta}(U+V)=-2 \sin \theta+2\cos \left(\frac {2n}{4n+2}\pi+\theta\right)\\0=-2 \sin \theta +2\cos\left(\frac {2n}{4n+2}\pi \right) \cos \theta-2\sin\left( \frac {2n}{4n+2}\pi\right)\sin \theta\\\tan \theta=\frac{\cos\left(\frac {2n}{4n+2}\pi \right)} {\sin\left( \frac {2n}{4n+2}\pi\right)+2}$$

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