For the following use a compass and straightedge procedure.

Given a unit square ABCD construct 4 circles centered at each corner and having diagonal radius. Form the smallest of the squares defined by the points where these circles intersect.

Use this process on the new square to create a third square.

1. Find and provide the area [in surd format] of each square so formed.

2. Following this process find a general expression for the area of the n^th square.

The distance of one corner of the second square is sqrt(2) distant from one of the two far corners of the original square. This can be considered the hypotenuse of a right triangle that includes half the original square's side, farthest from the vertex of the second square, as a leg. The remaining leg has length sqrt(2 - 1/4). Twice this value is 1 unit more than the diagonal length of the second square, as there is overlap of 1 with the congruent figure on the other side of the diagram.

Thus the diagonal of the second square is 2*sqrt(7/4) - 1 = sqrt(7) - 1.  To get the side length of this square, we need to divide by sqrt(2).

So the side length is (sqrt(7) - 1) / sqrt(2), and therefore the area is (sqrt(7) - 1)^2 / 2.

The third square then of course has the same ratio to the second, and so is (sqrt(7) - 1)^4 / 4.

In general, then, the nth square has area (sqrt(7) - 1)^(2*(n-1)) / 2^(n-1), where the necessity for the n-1 rather than n results from the naming of the "second" square being the first iteration of the transformation.

Corrected per brianjn's comment (switched exponent and base in denominator of general formula).

Edited on January 31, 2011, 10:18 am

Posted by Charlie on 2011-01-30 20:03:38