How many (convex) pentagons are there, each of whose angles is $108^{\circ}$, and among whose sides there are 1, 2, and 3 units long?
All known sides are either consecutive to each other or two of them are adjacent, and the third one is between the two other (unknown) sides.

In both cases, the construction is uniquely determined by which of the three consecutive sides we choose as the middle one, or as the one separated from the others. Thus, we need to examine at most $2 \cdot 3=6$ constructions, but it is possible that some choices do not result in a pentagon.

I. Let the middle side be $A B=1$ unit. At its endpoints, we measure a $108^{\circ}$ angle, and on the new sides, we measure $B C=3, E A=2$ units. By measuring $108^{\circ}$ angles again at the endpoints, the intersection points of these sides give the 5th vertex (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_78eeb15fe3c9f93eb761g-1.jpg?height=585&width=445&top_left_y=737&top_left_x=276)

Figure 1
![](https://cdn.mathpix.com/cropped/2024_05_02_78eeb15fe3c9f93eb761g-1.jpg?height=858&width=1114&top_left_y=522&top_left_x=710)

$0^{\circ}$

The construction proceeds similarly if the middle segment is 2 or 3 units (Figures 2 and 3). Thus, we obtain 3 different pentagons.

II. Now let's examine the case where the 2-unit side stands alone at $E D$:

$A B=1, B C=3$ (Figure 1) $E^{\prime} D^{\prime}=2$, clearly $E^{\prime} D^{\prime} \| E D$. If we measure the distance $E D_{1}=2$ units on $E D$ and draw a line parallel to $E A$ through the endpoint, it will cut the ray $C D$ at the appropriate point $D^{\prime}$.

This provides a general procedure for constructing the remaining cases when the 1-unit or 3-unit side stands alone.

Figure 2 shows the construction when the 1-unit side stands alone.
![](https://cdn.mathpix.com/cropped/2024_05_02_78eeb15fe3c9f93eb761g-1.jpg?height=472&width=1148&top_left_y=1835&top_left_x=477)

![](https://cdn.mathpix.com/cropped/2024_05_02_78eeb15fe3c9f93eb761g-2.jpg?height=699&width=632&top_left_y=125&top_left_x=733)

From Figure 3, we can read that the 3-unit side cannot stand alone, because if $E D_{1}=3$, the line parallel to $E A$ through $D_{1}$ will not intersect the ray $C D$. This can be proven as follows: draw a line parallel to $E D$ through $C$, forming an isosceles trapezoid $E D C C_{1}$ with $C_{1} C \| E D$. In the triangle $A C_{1} C$, the angle at $C_{1}$ is obtuse, so the opposite side is the largest, i.e., $A C > C_{1} C$. In the triangle $A B C$, $A C > C_{1} C > E D$. Therefore, a pentagon can only be formed if $E D < 3$.

Thus, the number of distinct pentagons is: 5.

Note. Some contestants did not draw the pentagons but tried to count the different pentagons in some way among all possible cases.

Some non-competitive solutions are not independent work, and the class, and possibly the school, name are missing. Since we only accept solutions from elementary school students and first-grade students marked with the letter $E$, it is particularly important that contestants also write their class next to their name and school.