## Wednesday, August 26, 2009

### Fox 132

1. First we watch the biggest square in the middle.

The three triangles around this square are similar.
Call the side of the square DEFG, with DG on AB, E on AC and F on BC.
Then: EC = 2/sqrt(5)*x and AE = sqrt(5)*x
AE+EC = AC = 2 so x = 2/7*sqrt(5) and area DEFG = x^2 = 20/49.

Now we watch any right-angled triangle similar to ABC,with a square in the right angle.
When the square had sides x, then it's easy to see that the two remaining triangles are similar with the original triangle, one with factor 2/3 and the other with factor 1/3.
And it follows that the side x = hypothenusa/(3/2*sqrt(5)) = 2/(3sqrt(5)) * hypothenusa.

Now we distinguish between the left and right side of the figure.

Right side:
In every step the triangle gets smaller with factor 1/3, so every square gets smaller with factor 1/3. The first square has side x0 = 2/(3sqrt(5))*BF = 2sqrt(5)/21 and area (x0)^2=20/441.
The areas of the squares on the right side give a geometric sequence, with area An = 20/441*(1/9)^n.
The total area of the squares on the right side is equal to Sum(n=0 to infinity)An = 20/441*9/8 = 5/98

Left side:
The first square has side 2/(3sqrt(5))*AE = 4sqrt(5)/21 and area 80/441.
The area give a geometric sequence: Bn = 80/441*(4/9)^n so the total area of the squares on the lift side is equal to 80/441*9/5 = 16/49

The total sum of the areas of all squares = 20/49 + 16/49 + 5/98 = 77/98 = 11/14