Fermat point for quadrilaterals

[littlemathquark]

TL;DR

Fermat point for quadrilaterals

Where is the Fermat point for quadrilaterals?
I think this point on the intersection of diagonals for convex quadrilaterals and it can be proof. But if quadrilateral would be concav, where is Fermat point and proof?

 

[littlemathquark]

I think it must be on the collapsed vertice.

 

[fresh_42]

TL;DR Summary: Fermat point for quadrilaterals

Where is the Fermat point for quadrilaterals?
I think this point on the intersection of diagonals for convex quadrilaterals and it can be proof. But if quadrilateral would be concav, where is Fermat point and proof?

I doubt that it is unique. You can make three circles intersect in a single point,

but can you do that with four circles?

 

[littlemathquark]

Do you doubt for convex or concav quadrilateral? It's unique for convex stiuation surely because I know how can I proof. I think it's unique also for concav. Ok 3 circle to find Fermat-Toricelli point for triangle but
Why do you think of four circle for quadrilateral?

 

[fresh_42]

Do you doubt for convex or concav quadrilateral? It's unique for convex stiuation surely because I know how can I proof. I think it's unique also for concav. Ok 3 circle to find Fermat-Toricelli point for triangle but
Why do you think of four circle for quadrilateral?

You asked about an arbitrary quadrilateral. It is ultimately an optimization problem and whether the set of feasible points is convex or not is crucial for optimization problems. Even with the restriction to convex sets, it is not obvious and you have to prove it. I suspect a finite set of solutions, but not necessarily of cardinality one. But that's a shot in the dark and I don't have a counterexample. I only meant that the Fermat point is already an unproven statement.

 

[littlemathquark]

I mention about plane quadrilateral, otherwise may not exist that point, you are right.

 

[fresh_42]

I mention about plane quadrilateral, otherwise may not exist that point, you are right.

It isn't obvious even in a plane. You have four arbitrary circles for which the sum of their diameters has to be minimal in a unique intersection point of all four. But as I already said, it is difficult to discuss geometry with only words.

 

[littlemathquark]

It isn't obvious even in a plane. You have four arbitrary circles for which the sum of their diameters has to be minimal in a unique intersection point of all four. But as I already said, it is difficult to discuss geometry with only words.

I want to show that point of intersection of diagonals can take Fermat point for plane convex qudrilaterals:

Let point O be intersection of diagonals. Let ABCD be a convex plane qudrilateral and take a M point and connect that point to the vertices of qudrilateral. İt's true the inequalities $AC<MA+MC$ in AMC triangle and $BD<MB+MC$ in BMD triangle. If we add that inequalities

$AC+BD=AO+OC+BO+OD<AM+CM+BM+DM$ so if M=O then sum of AC+BD must be minimum.

I am not able to proof for concave quadrilateral.