# Minimizing the edge length of a polygon preserving its angles

I am trying to minimize the edge lengths of a polygon while keeping the angles the same. I can achieve this geometrically (iteratively), however, I am looking for some related papers that can solve the problem algebraically. • @douglipinski gives an answer below, but that clearly can't be what you were interested in. What is it that you would like to keep constant while doing your minimization procedure? (Because if you don't want to keep anything constant, you'd just shrink the polygon to a point and that's what gives you the minimal edge length with preserved angles.) – Wolfgang Bangerth Feb 10 '15 at 4:47
• Thanks for the answers. What I am looking for is to find the geometric degrees of freedom of the polygon to shrink. What I keep constant is the length of the edge with minimum length and scale the other edges correspondingly. Therefore, the constants are one edge (or a specified length for the shortest edge) and the angles. – masouda Feb 10 '15 at 8:30
• Could you please add some more details directly to your question (including the extra info in your comment). Are you allowing zero length sides? If not what is the constraint on side length? If so, my intuition is that any polygon can be shrunk to either a triangle or a single line (if two sides happen to be parallel) while preserving angles. In that case are you looking for the resulting triangle with the minimum perimeter? Can the polygons be self-intersecting? – Doug Lipinski Feb 10 '15 at 14:16
• I think he wants to ensure that all shrunken edges are never shorter than the previously shortest edge. – Wolfgang Bangerth Feb 11 '15 at 2:22
• Thanks for all your comments and clarifications. Let me give a very simple example: consider you have a rectangle; you can minimize the rectangle edge to a square with edge length equal to the minimum length of the shortest edge of the rectangle. This ensures that I still have the angles of the polygon constant and shrunk all the edges to be at least equal to the shortest edge. Now, the polygon can be convex or concave. – masouda Feb 11 '15 at 11:11

You can do this easily if you have the $(x,y)$ coordinates of the vertices of the polygon with perimeter $L$ and area $A$. Simply multiply all the vertices by the same scaling factor, say $\alpha$, to get a new set of vertices and a polygon with perimeter length $\alpha L$ and area $\alpha^2 A$. You can avoid also translating the polygon with this rescaling by placing it with its center of area at the origin.