Problem 1225

A tangent line of a convex shape is a line that has a common point with the shape, and all points of the shape lie on one side of the line. Any bounded, convex shape has a tangent line in any direction, and there are two such lines.

Let's take a point $O$ in the plane and a vector $\overrightarrow{O A}=i$. Define the following function $d(\alpha)$. Rotate $\overrightarrow{O A}$ by an angle (positive or negative, according to the sign of $\alpha$) to get $\overrightarrow{O B}$. Draw the two tangent lines of the shape that are parallel to $\overrightarrow{O B}$. The distance between these tangent lines is denoted by $d(\alpha)$.

!

By the definition of the function $d(\alpha)$, we have $d(\alpha)=d(\alpha+\pi)$. It can be shown that the function $d(\alpha)$ is continuous, and therefore the function

$$f(\alpha)=d(\alpha)-d\left(\alpha+\frac{\pi}{2}\right)$$

is also continuous.

$$\begin{aligned} f(0) & =d(0)-d\left(\frac{\pi}{2}\right), \quad f\left(\frac{\pi}{2}\right)=d\left(\frac{\pi}{2}\right)-d(\pi)= \\ & =d\left(\frac{\pi}{2}\right)-d(0) \end{aligned}$$

If $f(0)=f\left(\frac{\pi}{2}\right)=0$, that is, $d(0)=d\left(\frac{\pi}{2}\right)$, then the distances of the tangent lines corresponding to 0 and $\frac{\pi}{2}$ radians are equal, and therefore these lines form a square.

If $f(0) \neq f\left(\frac{\pi}{2}\right)$, then due to (1), the function values corresponding to 0 and $\frac{\pi}{2}$ have opposite signs, and thus there exists an angle $0<\alpha_{0}<\frac{\pi}{2}$ such that $f\left(\alpha_{0}\right)=0$. For this $\alpha_{0}$, $d\left(\alpha_{0}\right)=d\left(\alpha_{0}+\frac{\pi}{2}\right)$, so the tangent lines in the direction of $\alpha_{0}$ and the perpendicular direction form a square.

Csaba Kecskés (Budapest, Móricz Zs. Gymnasium, 4th grade)