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(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.

PrincessEev
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simulacra
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    There's a subtle difference between "fits inside" and "inscribes." It can indeed be shown that a square which fits inside a circle of radius $r$ has an area $A \leq 2r^2$, with $A = 2r^2$ occurring when the square is inscribed. That said, there is rarely ever a "unique" proof of some fact; take the Pythagorean theorem as an example: books full just of hundreds of proofs of it have been published. – PrincessEev Jan 07 '19 at 19:41
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    As for one proof I'd make. I'd make the assumption the inscribed square has diagonal $2r$, show said square is indeed inscribed, then show a square can have no other diagonal (w.r.t. to its length), and then calculate the area of said square. (This is the barest of bones of the proof, going into depth on said proof probably isn't going to be very useful.) – PrincessEev Jan 07 '19 at 19:44
  • As Eevee Trainer stated, there are many proofs of what you are asking. I have just given a different one here, plus I can also think of $1$ or $2$ more with some effort. However, like Eevee also stated, I'm not sure how useful that will be for you. – John Omielan Jan 07 '19 at 19:49

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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.

Shubham Johri
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    Please ignore my previous comment as I just saw the title say "biggest square that can be inscribed within a circle". However, the text in the question differs slightly from this by saying "fit inside" which implied to me that it's the set of all squares contained in a circle. – John Omielan Jan 07 '19 at 19:29
  • Yes, there is a dichotomy in the title and first line. Although it can be resolved by noting that the area $\frac12d_1d_2$ is maximized when the vertices lie on the circle, for the largest line segment that fits in a circle is its diameter – Shubham Johri Jan 07 '19 at 19:40
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Square inscribed in a circleIn 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 $r^2 + r^2 = 2r^2$.

John Omielan
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