The size of the biggest square that can be inscribed within a circle

Question

(a) Is the size the biggest square that can be fit inside a circle always $2r^2$?

(b) How do you do when you show that is indeed the case?

I want to compare my boyfriend's proofs with that of other people, since he thinks his proof is unique.

Answer 1

All squares inscribed in a given circle have equal size. There is no 'biggest' one. You can easily prove that their area is $2r^2$ by noting that the area of a rhombus is given by $\frac12d_1d_2$, where $d_1,d_2$ are the lengths of its diagonals. In your case, $d_1=d_2=2r$, as the diagonals are diameters of the circle.

Answer 2

In the diagram above, $ABCD$ are the points of the inscribed square on the circle, $O$ is the center and let $r$ be the radius. Note that $AC$ is the diameter as $\angle ABC$ is a right angle (since if the angle subtending a chord is a right angle, the chord must be the diameter), so $AC$ must be of length $2r$. Also, $OB$ is perpendicular to $AC$ (since the diagonals of a square are perpendicular to each other) and of length $r$. Thus, it's the height of $\triangle ABC$ using $AC$ as the base, so the triangle area is $\frac{2r \times r}{2} = r^2$. As the area of $\triangle ADC$ is the same, the total area of $ABCD$ is $r^2 + r^2 = 2r^2$.