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?
Let the equation of the circle be x^2+y^2=25/2 (eq1).
Then the coordinates of the point where the small square touches the circle in the 1st quadrant are: (x, y)
2x and y-2.5 are the sides of the small square.
2x= y-2.5 (eq2) squared and subtracted from 4* (eq1) leads to
y^2-y-8.75=0 , (eq3)
(y-3,5)*(y+2,5)=0
The positive answer is y=3.5, so 2x=1 and y-2,5=1
and the area of the smaller square is 1*1=1.
Edited on December 7, 2014, 10:27 pm