Total area for a natural nested set of convex polygons.Suppose we have a convex polygon P 0 with n given vertices, and we want to "nest" polygons P j for j &gt; 0 by taking the midpoints between edges of P j − 1 as the vertices. For a regular polygon the total area of all the polygons will form a geometric series that is pretty easy to solve for (at least in terms of trig functions) in terms of the area A of the original polygon P 0 . However, what if P 0 is not regular? If we know the area of P 0 , is there a formula for the sum of areas of all the nested polygons, in terms of n and the original area? Or do we need more information, like the vertices of P 0 ? If we need to know the vertices, are there formulas in terms of the vertices at least for some small values of n, like n = 3 , 4?

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Total area for a natural nested set of convex polygons.Suppose we have a convex polygon       P    0   with n given vertices, and we want to &quot;nest&quot; polygons       P    j   for   j  &amp;gt;  0 by taking the midpoints between edges of       P          j      −      1       as the vertices. For a regular polygon the total area of all the polygons will form a geometric series that is pretty easy to solve for (at least in terms of trig functions) in terms of the area A of the original polygon       P    0  . However, what if       P    0   is not regular? If we know the area of       P    0  , is there a formula for the sum of areas of all the nested polygons, in terms of n and the original area? Or do we need more information, like the vertices of       P    0  ? If we need to know the vertices, are there formulas in terms of the vertices at least for some small values of n, like   n  =  3  ,  4?

brulotfao · Accepted Answer

Step 1Let the vertices of the polygon be             p        1    ,            p        2    ,  …  ,            p        n   in counterclockwise order, and for convenience define             p              n      +      1        =            p        1    ,            p              n      +      2        =            p        2    , and so on. Then the (signed) area of the polygon is given by   A  =      1    2        ∑          i      =      1        n    (            p        i    ∧            p              i      +      1        )  , where       p    ∧      q    =      |                                        p            x                                                q            x                                                            p            y                                                q            y                                |    =      p    x        q    y    −      p    y        q    x  . Note that       p    ∧      p    =  0.The nested polygon has vertices       1    2    (            p        1    +            p        2    )  ,      1    2    (            p        2    +            p        3    )  ,  …  ,      1    2    (            p        n    +            p              n      +      1        ), and so its area is                              A          ′                                    =                  1          2                          ∑                      i            =            1                    n                          (                                                                      p                                i                            +                                                p                                                  i                  +                  1                                                      2                    ∧                                                                      p                                                  i                  +                  1                                            +                                                p                                                  i                  +                  2                                                      2                    )                                                  =                                                                              1                  8                                                  ∑                                      i                    =                    1                                    n                                (                                                      p                                    i                                ∧                                                      p                                                        i                    +                    1                                                  )                            ⏟                                            A                          /                        4                          +                  1          8                          ∑                      i            =            1                    n                (                              p                    i                ∧                              p                                i            +            2                          )        +                  1          8                          ∑                      i            =            1                    n                                                                    (                                                      p                                                        i                    +                    1                                                  ∧                                                      p                                                        i                    +                    1                                                  )                            ⏟                                0                +                                                                              1                  8                                                  ∑                                      i                    =                    1                                    n                                (                                                      p                                                        i                    +                    1                                                  ∧                                                      p                                                        i                    +                    2                                                  )                            ⏟                                            A                          /                        4                                                            =                  1          2                A        +                  1          4                          (                      1            2                                ∑                          i              =              1                        n                                              p                        i                    ∧                                    p                                      i              +              2                                )                .            Step 2The area of the nested polygon is       1    2   times the area of the original polygon, and plus       1    4   times the total &quot;area&quot; of the polygon(s) obtained by connecting each vertex to the next-to-next one.Geometrically, connecting vertices by twos gives two cases. When n is even, you get two polygons             p        1              p        3    ⋯            p              n      −      1       and             p        2              p        4    ⋯            p        n  . When n is odd, you get a single &quot;doubly-wrapped&quot; polygon             p        1              p        3    ⋯            p        n              p        2              p        4    ⋯            p              n      −      1      , such as a pentagram. In this case, the formula double-counts the area of the central region, which is why I&#039;ve put &quot;area&quot; in quotes above.For small n, life is pretty simple, though:When   n  =  3, the polygon             p        1              p        3              p        2   is just the same triangle with its vertices connected in backwards order, so it has the opposite (clockwise) orientation and negative signed area. Then you get       A    ′    =      1    2    A  +      1    4    (  −  A  )  =      1    4    A.When   n  =  4, the &quot;polygons&quot;             p        1              p        3   and             p        2              p        4   are the diagonals of the quadrilateral and have zero area. So       A    ′    =      1    2    A.In both these cases, the areas of the sequence of nested polygons form a geometric series no matter whether the original polygon was regular or not, and so you can easily compute the total area of the entire series.When   n  ≥  5, it turns out that the area of the alternating polygon(s) also depends on the arrangement of the vertices, not just the area of the original polygon. So one cannot determine A′ from A alone.

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