I drew a square of side 5 together with its circumcircle. At the middle of the arc between a side and the circle, I drew a smaller square with two vertices on the side, and the other two on the circle.
What was the area of the smaller square?
Half the diagonal of the large square, (5/2)*sqrt(2), is the radius of the circumcircle. Let x be the size of the side of the smaller square.
From the center of the circle to the point where the smaller square touches the circle is of course (5/2)*sqrt(2), as is the sum 5/2 + x + a smaller sliver extending to the circle. Let's say that that distance measure lies along a horizontal line from the center of the circle, through the center of the small square and beyond. Call the angle between the ray to the small-circle vertex intersection with the circle and the horizontal line theta.
sin theta = (x/2) / ((5/2)*sqrt(2))
cos theta = ((5/2)+x) / ((5/2)*sqrt(2))
The sum of the squares of these should equal 1.
If I did my algebra right the simplified quadratic equation is
5*x^2 + 20*x - 25 = 0
and of course we want the positive solution, (-20 + sqrt(400 + 500)) / 10, or
(sqrt(900) - 20) / 10 = 1
The area of a 1x1 square is still 1.
BTW: the first time through I didn't do my algebra right (when I squared x/2 I got x^2 / 2), making the coefficient of the square term of the quadratic 6 instead of 5. But I checked my numeric answer with a construction in Geometers' Sketchpad and noticed a large enough discrepancy to go back and check my math.
Edited on December 6, 2014, 7:42 am
Posted by Charlie
on 2014-12-05 15:20:18