Consider a given square with edges of length . Prove that the inscribed square with minimum area has edges of length .

(Note: The picture is of an arbitrary inscribed square to illustrate what and are. It is definitely * not * the particular inscribed square with minimal area that we are looking for.)

*Proof.* We start with a square with edges of length . Let where and are the lengths of the two sections of created by the point at which the corner of the inscribed square meets the edge of the outer square. Let denote the length of the edge of the inscribed square. Then, implies . So,

Let this be our function . Then we take the derivative,

From this we have when and

Therefore, is decreasing when and increasing when . Hence, has a minimum at . Using our equation for , we have

Finally, solving for the length of the edge ,

I think the hardest part of this is not the derivative calculation at all – the hardest part is to formulate the problem. Additionally, it should be shown that the triangles that form the sides of the inscribed square are equal – there is no guarantee that x on one one side of the outer rectangle is equal to the x-part on the other side of the outer rectangle: the conclusion about e^2 = x^2 + y^2 would be wrong without the proof of triangle equality. And I do not think this is obvious at all – hardly remembered the triangle equality formulas from school time.