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 $e$ ,

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

One comment

July 8, 2019 at 5:25 pm

Artem says:

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.

Stumbling Robot

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

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