3. Given a rectangle $A B C D$ with perimeter $o$. In its plane, find the set of all points whose sum of distances from the lines $A B, B C, C D, D A$ is equal to $\frac{2}{3} o$.
SOLUTION. The required value of the sum of four distances can be written in the form

$$
\frac{2}{3} o=\frac{1}{6} o+\frac{1}{2} o=\frac{1}{6} o+|A B|+|B C| \text {. }
$$

For any point in the strip determined by the lines $A B$ and $C D$, it holds that the sum of its distances from these two parallel lines is equal to their distance, i.e., $|B C|$. For any point outside this strip, the sum of the two considered distances is equal to the sum of the value $|B C|$ and twice the distance from the closer of the two parallel lines. Similar two statements hold for the sum of distances of any point from the parallel lines $B C$ and $A D$ in relation to their distance $|A B|$. With regard to the expression (1), we can make the first two conclusions.

(1) In the strip between the lines $A B$ and $C D$, the sought points are precisely those whose sum of distances from the lines $B C$ and $A D$ is equal to $\frac{1}{6} o+|A B|$. These are the points that lie outside the strip determined by the lines $B C$ and $A D$ and have a distance from the closer of them equal to $\frac{1}{6} o: 2=\frac{1}{12} o$. The set of sought points in the strip between $A B$ and $C D$ thus consists of two segments $B_{1} C_{1}$ and $A_{1} D_{1}$ as shown in Fig. 1. Their endpoints $A_{1}, B_{1}$ lie on the line $A B$ outside the segment $A B$ such that $\left|A A_{1}\right|=\left|B B_{1}\right|=\frac{1}{12} o$; the endpoints $C_{1}, D_{1}$ lie on the line $C D$ outside the segment $C D$ such that $\left|C C_{1}\right|=\left|D D_{1}\right|=\frac{1}{12} o$.

![](https://cdn.mathpix.com/cropped/2024_04_17_beef4017b9d326766e3bg-5.jpg?height=668&width=1034&top_left_y=568&top_left_x=517)

Fig. 1

(2) In the strip between the lines $B C$ and $A D$, the sought points are precisely those whose sum of distances from the lines $A B$ and $C D$ is equal to $\frac{1}{6} o+|B C|$. These are the points that lie outside the strip determined by the lines $A B$ and $C D$ and have a distance from the closer of them equal to $\frac{1}{12} o$. The set of sought points in the strip between $B C$ and $A D$ thus consists of two segments $A_{2} B_{2}$ and $C_{2} D_{2}$, with the endpoints $B_{2}, C_{2}$ lying on the line $B C$ outside the segment $B C$ such that $\left|B B_{2}\right|=\left|C C_{2}\right|=\frac{1}{12} o$ and the endpoints $A_{2}, D_{2}$ lying on the line $A D$ outside the segment $A D$ such that $\left|A A_{2}\right|=\left|D D_{2}\right|=\frac{1}{12} o$.

It remains to find the sought points outside the union of the two considered strips, i.e., points lying in one of the four right angles $A_{1} A A_{2}, B_{1} B B_{2}, C_{1} C C_{2}$, or $D_{1} D D_{2}$. From the above considerations, it follows that in each of these angles, we are looking for precisely those points whose sum of distances from both sides of the angle is equal to the value $\frac{1}{12} o$. Considering the symmetry, we will only show that such points in the angle $A_{1} A A_{2}$ fill the segment $A_{1} A_{2}$; in the other three angles, they will be the segments $B_{1} B_{2}, C_{1} C_{2}, D_{1} D_{2}$ (Fig. 1).

First, note that the points $A_{1}, A_{2}$ are the only points on the sides of the angle $A_{1} A A_{2}$ that have the required property. For any interior point $X$ of the angle $A_{1} A A_{2}$, let $d_{1}, d_{2}$ be the distances of point $X$ from the sides $A A_{1}$ and $A A_{2}$, respectively. We are looking for precisely those points $X$ for which $d_{1}+d_{2}=\frac{1}{12} o$ (Fig. 2). We will solve this "equation" by considering the area $S$ of the figure $A A_{1} X A_{2}$, which is either a triangle or a convex or concave quadrilateral.

![](https://cdn.mathpix.com/cropped/2024_04_17_beef4017b9d326766e3bg-5.jpg?height=408&width=417&top_left_y=1875&top_left_x=1411)

Fig. 2 The area $S$ is always equal to the sum of the areas of two triangles $A A_{1} X$ and $A A_{2} X$:

$$
S=S_{A A_{1} X}+S_{A A_{2} X}=\frac{1}{2}\left|A A_{1}\right| d_{1}+\frac{1}{2}\left|A A_{2}\right| d_{2}=\frac{1}{2} \cdot \frac{1}{12} o \cdot\left(d_{1}+d_{2}\right) .
$$

The equation $d_{1}+d_{2}=\frac{1}{12} o$ is satisfied precisely when the area $S$ has the same value as the area $S_{0}$ of the right triangle $A A_{1} A_{2}$, whose both legs have the same length $\frac{1}{12} o$.

The sought points $X$ are thus precisely those for which the figure $A A_{1} X A_{2}$ is a triangle; if $A A_{1} X A_{2}$ is a convex or concave quadrilateral, it is clear that $S>S_{0}$ or $S<S_{0}$. The sought points $X$ of the angle $A A_{1} A_{2}$ therefore indeed form the segment $A_{1} A_{2}$.

Answer: The sought set is the union of eight segments that form the boundary of the octagon $A_{1} A_{2} B_{2} B_{1} C_{1} C_{2} D_{2} D_{1}$.

Note. From Fig. 2, it is also clear that the equation $d_{1}+d_{2}=c$, where $c=\left|A A_{1}\right|=\left|A A_{2}\right|$, will be satisfied precisely when $\left|X_{1} A_{1}\right|=d_{1}$ and $\left|X_{2} A_{2}\right|=d_{2}$, i.e., precisely when both triangles $X X_{1} A_{1}$ and $X X_{2} A_{2}$ are isosceles. This clearly happens precisely when the angle $A_{1} X A_{2}$ is right, because $\left|\nless A A_{1} A_{2}\right|=45^{\circ}$.

GUIDING AND ADDITIONAL PROBLEMS: