Midpoint polygons?A midpoint polygon is formed by taking the midpoints of each side of a polygon, and making a new polygon out of those points. The end result is the midpoint polygon inscribed in the polygon you started off with.Given an arbitrary polygon, is it always the midpoint polygon of some other polygon? And if so, is this other polygon unique? Is there a simple construction to find it? And also, are there any special cases of the original question which are easier to answer?

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Kaleb Burns · Accepted Answer

Step 1The midpoint transformation can be written in matrix form   Q  =  A  P, where Q are the new vertices, P are the old vertices, and the matrix A is a circulant matrix whose first row is 1/2, 1/2, 0,…,0.Step 2The relevant property is that the rank of A is   n  −  d, where d is the degree of   gcd  (      x          n      −      1        +  1  ,      x    n    −  1  ). When n is even, this gcd has at least the factor   x  +  1 and so A is singular and the midpoint transformation cannot be inverted.The gcd is actually a divisor of   x  +  1  =  x  (      x          n      −      1        +  1  )  −  (      x    n    −  1  ) and so is either 1 or   x  +  1. Therefore, A is singular iff n is even.

Macy Villanueva · Accepted Answer

Step 1Note that the midpoints of a quadrilateral form a parallelogram. So if a quadrilateral is not parallelogram it cannot be the midpoint polygon of another quadrilateral.Let the vertices of the polygon be of coordinates   (      x    1    ,      y    1    )  …  (      x    n    ,      y    n    ) Suppose that we start from (a,b) and we incrementally construct the broken line   (  a  ,  b  )  ,  (      a    1    ,      b    1    )  …  (      a    n    ,      b    n    ) having midpoints as the original polygon:      x    1    =            a      +              a        1              2  ,       y    1    =            b      +              b        1              2   so      a    1    =  −  a  +  2      x    1  ,       b    1    =  −  b  +  2      y    1        a    2    =  −      a    1    +  2      x    2  ,       b    2    =  −      b    1    +  2      y    2    …      a    n    =  −      a          n      −      1        +  2      x    n  ,       b    n    =  −      b          n      −      1        +  2      y    n  .Step 2To have a closed polygon one needs       a    n    =  a and       b    n    =  b.So for n odd you get an unique solution:  a  =      x    1    −      x    2    +      x    3    −  …  +  (  −  1      )          n      +      1            x    n    b  =      y    1    −      y    2    +      y    3    −  …  +  (  −  1      )          n      +      1            y    n  For n even, the two linear systems are not of full rank (have either an infinity of zero solutions). You may check when each of the case happens with Rouche&#039;s theoremSo when n is even, the system has solution iff:      x    1    −      x    2    +      x    3    −  …  −      x    n    =  0      y    1    −      y    2    +      y    3    −  …  −      y    n    =  0In which case it has an infinity of solutions, starting with any point of the plane.Of course depending on your unspecified requirements on the polygon (non-degenerate, non-self-intersect, convex), you might have to impose additional restrictions.

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