1. Subscribertalzamir
    Art, not a Toil
    60.13N / 25.01E
    Joined
    19 Sep '11
    Moves
    45043
    11 May '13 13:08
    A rectangle 2" long by 4" wide has an area of 8 square inches and a perimeter length of 12". Find three other polygons with the same area and perimeter length.
  2. Subscribertalzamir
    Art, not a Toil
    60.13N / 25.01E
    Joined
    19 Sep '11
    Moves
    45043
    12 May '13 13:53
    While at it.. let's make it a thousand more polygons with the same area and perimeter.
  3. Subscribertalzamir
    Art, not a Toil
    60.13N / 25.01E
    Joined
    19 Sep '11
    Moves
    45043
    15 May '13 09:541 edit
    One method.. start with a square, 3" x 3". It has the right perimeter, but the area is one square inch too big. Remove from the bottom right corner a rectangle, 1" wide by x" high where 0 < x < 1. Repeat for the bottom right corner, 1" wide by (1-x)" inches high. The total area of the two missing rectangles is one square inch, which adjusts the square into an irregular octagon that looks a bit like the letter T, with the desired area and perimeter. With different values of x you can get any number of solutions. Say,
    x = 0.001 x n, where n = 1 .. 1000, gives the desired 1,000 variants.