ciao mario

?

Yes îs good !I solve it but is so long to post it

grazie so you confirm my solution?

Yes is good

where did you find this problem?

I don't know i give This with a friend

aha I think it's related with equilateral triangles it's 6 because 360/60=6

This was a question from IGMO (Instagram Maths Olympiad)

rethinkink about it I rekon the solution is actually 5, because... points are plotted so that the distances between any two are distinct... So we can't have a regular exagon, can we?

This is a nice problem! The answer is 5. A construction basically takes a regular pentagon and its center, but moves the points ever-so-slightly such that the distances are unique. We can easily verify that this works. Now we will show that 6 is impossible. Suppose that 6 segments are connected to a point $P$, one of the angles formed by two adjacent segments must be $\leq 60^\circ$ (since the minimum value must be less than the average, which is $60^\circ$). Call this angle $\theta$, and let the segments be $AP$ and $BP$. Also let $AP=a,BP=b,a<b$. Let $c$ be the length of $AB$. Then from Law of Cosines we have: $$c^2=a^2+b^2-2ab \cos \theta$$Since $\theta \leq 60^\circ$, this means that $\cos \theta \geq \frac{1}{2}$, so we have: $$c^2 \leq a^2+b^2-ab$$Since $a<b$, we have $a^2-ab<0$. This means that: $$c^2 \leq a^2+b^2-ab < b^2$$But then the segment $BP$ is not the shortest one which can be drawn from $B$, since $BA$ is shorter. Thus we have a contradiction, so 6 cannot be achieved.