 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.