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Find the minimal area of a square inscribed in a given square

Consider a given square with edges of length L. Prove that the inscribed square with minimum area has edges of length \frac{\sqrt{2}}{2} L.


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(Note: The picture is of an arbitrary inscribed square to illustrate what x and y 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 L. Let x + y = L where x and y are the lengths of the two sections of L created by the point at which the corner of the inscribed square meets the edge of the outer square. Let e denote the length of the edge of the inscribed square. Then, x + y = L implies y = L - x. So,

    \begin{align*}   \text{Area} = e^2 &= x^2 + y^2 & (\text{Pythagoras})\\  &= x^2 + (L-x)^2 \\  &= 2x^2 - 2Lx + L^2  \end{align*}

Let this be our function f(x). Then we take the derivative,

    \[ f'(x) = 4x - 2L. \]

From this we have f'(x) = 0 when x = \frac{L}{2} and

    \begin{align*}  f'(x) &< 0 & \text{when} && x &< \frac{L}{2}\\  f'(x) &> 0 & \text{when} && x &> \frac{L}{2}. \end{align*}

Therefore, f(x) is decreasing when x < \frac{L}{2} and increasing when x > \frac{L}{2}. Hence, f(x) has a minimum at x = \frac{L}{2}. Using our equation for y, we have

    \[ y = L - x = L - \frac{L}{2} = \frac{L}{2}. \]

Finally, solving for the length of the edge e,

    \[ e^2 = \left( \frac{L}{2} \right)^2 + \left( \frac{L}{2} \right)^2 = \frac{L^2}{2} \quad \implies \quad e = \frac{L}{\sqrt{2}} = \frac{\sqrt{2}}{2} L. \qquad \blacksquare \]

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