7. a) Let $T_{1}$ and $T_{2}$ be the contours of two equilateral triangles parallel to each other on a plane, with a common center and sides of lengths 1 and 2, respectively. What is the distance $\rho\left(T_{1}, T_{2}\right)$ from $T_{1}$ to $T_{2}$? What is the distance $\rho\left(T_{2}, T_{1}\right)$ from $T_{2}$ to $T_{1}$?

b) Let $K_{1}$ and $K_{2}$ be the contours of two cubes parallel to each other with a common center and edges of lengths 1 and 2, respectively. What is the distance $\rho\left(K_{1}, K_{2}\right)$ from $K_{1}$ to $K_{2}$? What is the distance $\rho\left(K_{2}, K_{1}\right)$ from $K_{2}$ to $K_{1}$?

From the results of problem 7, it follows that the distance defined above from one figure to another is not symmetric: the distance $\rho\left(\Phi_{1}, \Phi_{2}\right)$ from figure $\Phi_{1}$ to figure $\Phi_{2}$ may not equal the distance $\rho\left(\Phi_{2}, \Phi_{1}\right)$ from figure $\Phi_{2}$ to figure $\Phi_{1}$. However, this "distance" satisfies another important property of ordinary distances.

i) infimum - in Latin, lower.

${ }^{2}$ ) supremum - in Latin, higher.
7. a) If $T_{1}$ is the smaller of the two triangles (Fig. 64), then the distance from point $A_{1}$ of triangle $T_{1}$ (understood as a line, not as part of the plane ${ }^{1}$)) to the nearest point of triangle $T_{2}$ will be the same for all

![](https://cdn.mathpix.com/cropped/2024_05_21_d4ffc236376258137647g-127.jpg?height=309&width=317&top_left_y=650&top_left_x=176)

Fig. 64. points $A_{1}$ - it will be equal to $\frac{\sqrt{3}}{6}$; therefore, $\rho\left(T_{1}, T_{2}\right)=\frac{\sqrt{3}}{6}$. The distance from point $A_{2}$ of triangle $T_{2}$ to the nearest point of triangle $T_{1}$ will be the greatest when point $A_{2}$ coincides with the vertex of the triangle, and in this case, it is equal to $\frac{\sqrt{3}}{3}$; therefore, $\rho\left(T_{2}, T_{1}\right)=\frac{\sqrt{3}}{3}$

b) If $K_{1}$ is the smaller of the two cubes, then the distance from point $A_{1}$ of cube $K_{1}$ (understood as a surface, not as part of the space! ²)) to the nearest point of cube $K_{2}$ will be the same for all points $A_{1}$ - it will be equal to $\frac{1}{2}$; therefore, $\rho\left(K_{1}, K_{2}\right)=\frac{1}{2}$. The distance from point $A_{2}$ of cube $K_{2}$ to the nearest point of cube $K_{1}$ will be the greatest when point $A_{2}$ coincides with the vertex of the cube, and in this last case, it is equal to $\quad \frac{\sqrt{3}}{2}$; therefore, $\rho\left(K_{2}, K_{1}\right)=\frac{\sqrt{3}}{2}$.