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Two squares in a circle (Posted on 2014-12-05) Difficulty: 3 of 5

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?

See The Solution Submitted by broll    
Rating: 3.0000 (2 votes)

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Solution solution | Comment 1 of 4
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.

Typo corrected.

Edited on December 6, 2014, 7:42 am
  Posted by Charlie on 2014-12-05 15:20:18

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