## Task 4 - 260814

Let $A B C D E F$ be a straight triangular prism. All three side faces $A B E D, B C F E, C A D F$ as well as the base and top faces $A B C$ and $D E F$ are mutually perimeter-equal. Given the length $h$ of the segment $A D$.

Determine the lengths of the segments $B C, C A$, and $A B$ in dependence on $h$!
The side faces of the prism are rectangles, so they have equal opposite sides. Therefore, we have

$$
\overline{A D}=\overline{B E}=\overline{C F}=h
$$

and for the sought lengths

$$
a=\overline{B C}=\overline{E F}, \quad b=\overline{C A}=\overline{F D}, \quad c=\overline{A B}=\overline{D E}
$$

From the equality of the perimeters of the side faces, we get

$$
2 a+2 h=2 b+2 h=2 c+2 h
$$

and from the equality of the perimeters of the side faces with the base or top face, we get

$$
2 a+2 h=a+b+c
$$

From (1), it follows that $a=b=c$, and therefore from (2), $2 a+2 h=3 h$, so $a=2 h$.

The sought lengths are thus $\overline{B C}=\overline{C A}=\overline{A B}=2 h$.

Solutions of the 1st Round 1986 adopted from [5]

### 5.28.2 2nd Round 1986, Class 8