A proof of that would be great !
Why is this interesting ?
Some motivation :
Consider a convex hexagon with integer sides.
Now lets assume this hexagon has the property that there is a point T in it,
such that the distance from the point T to all the vertices is also integer.
This means that the hexagon is made up of 6 integer sided triangles.
integer sided triangles have angles of the type arcsin( w_1 sqrt(v) ) = arccos( w_2 ) where w_1 , w_2 are positive rationals and v is a positive integer.
This can be proven for instance with heron's formula (this is where the square root comes from ) together with the sine rule or cosine rule , and considering the area and the height of the triangle etc
This in turn implies that the angles made around the point T are also of this type.
Now the sum of angles around T must make a full rotation , so they must sum to 2 pi.
This means that we should get for the corners around T :
equation T :
arcsin( r_1 sqrt(v_1) ) + arcsin( r_2 sqrt(v_2) ) +arcsin( r_3 sqrt(v_3) ) +arcsin( r_4 sqrt(v_4) ) +arcsin( r_5 sqrt(v_5) ) + arcsin( r_6 sqrt(v_6) ) = 2 pi
where again r_n and v_n are of the type above ; r_n positive rationals and v_n squarefree positive integers.
If we require that v_n are primes we get that :
Now it seems that no matter how we constructed our convex integer hexagon , in equation T there are at most two distinct sqrt v_n ; sqrt A and sqrt B with A,B being distinct primes.
And it seems the analogue is true for pentagons or any other n-gon.
So we wonder about the nature of
sum arcsin( r(j) sqrt( c(j) ) )
or
sum arccos( r_2(j) )
being equal to 2 pi.
and we conjecture that there be at most 2 distinct values for c(j) if we want it to equal 2 pi.