Question

Let P0 be an equilateral triangle of area 10. Each side of P0 is trisected, and the corners are snipped off, creating a new polygon (in fact, a hexagon) P1. What is the area of P1? Now repeat the process to P1 – i.e. trisect each side and snip off the corners – to obtain a new polygon P2. What is the area of P2? Now repeat this process infinitely often to create an object P[infinity]. What is the area of P[infinity]?

Answer 1

Answer:

P1 = 2/3 A

P2 = 2/9 A

P2 n -->∞ = 0

Step-by-step explanation:

- Let A be the area of the original equilateral triangle. If you draw the lines connecting the 1/3 length segments, then you cut off 3 small triangles, each with area (1/3)^2= 1/9 of the original triangle so cut off 3/9= 1/3 of the original triangle, leaving (2/3)A.

- Each cut also converts each vertex into two vertices. So we now have a figure with 6 vertices of area (2/3)A. If we now cut off corners at the 1/3 length points, we are cutting off 6 triangles each with are (1/9)*(2/3)A= (2/27)A so we are cutting off total area 6(2/27)A= (4/9)A, leaving (2/3- 4/9)A= (6/9- 4/9)A= (2/9)A.

- Continue in that way to convince yourself that, at the "n"th step, you are left with :

                                            $\left(\frac{2}{3^n}\right)A.$

Taking the limit as n goes to infinity, that goes to 0.