Please turn on javascript in your browser to play chess.
Posers and Puzzles

Posers and Puzzles

  1. 16 Jul '13 21:13
    talzamir posted:
    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.

    I wonder if there is an answer if the area is a triangle?

    from Heron's formula
    perimeter of triangle = a+b+c (where a,b,c) are sides
    let s = p/2
    then area = sqrt(s(s-a)(s-b)(s-c))

    which looks like we might be able to find a solution to b and c for any reasonable value of A we pick....
  2. 16 Jul '13 22:46
    Nope, the maximum area of a triangle with a perimeter of 12 is obviously an equilateral triangle of side 4 which has an area of 6.9, not enough. So we end up with only imaginary solutions.

    Therefore the simplest convex solution must be a quadrilateral

    Here is one way to make infinite parallelograms with the required perimeter and area:

    Generate random side A for a rectangle of length between 2 and 4
    generate the other side, B = 6 - A
    calculate the area AB

    remembering that the area of a parallelogram is sin(x)AB where x is the smaller of the two angles at the base we need
    sin(x)AB = 8
    x = asin(8/AB)

    this gives x=90 for A=2, B=4
    x = asin(8/9) for A=3, B=3
    x = asin(8/8.75) for A=2.5, B=3.5
    etc.