For a given triangle, we construct an equilateral triangle outward and inward on each of its sides. In how many ways can 4 points be selected from the 9 points found in the figure so that they are the vertices of a parallelogram?

Would our findings remain valid if, instead of equilateral triangles, we constructed isosceles right triangles where the hypotenuse is always the corresponding side of the original triangle?

[^0]:    ${ }^{1}$ We remind the contestants that the findings must be justified, as always. First-year students can answer the second question based on precise observations made from a detailed drawing this time.
I. Let's choose the labeling of the given triangle such that its vertices $A, B, C$ are traversed in a way that the interior of the triangle is to the left of the path, i.e., $A B C$ gives a positive traversal. Let the triangles inscribed inward on the sides be $A B C_{1}, B C A_{1}, C A B_{1}$, and the outward inscribed triangles be $B A C_{2}, C B A_{2}, A C B_{2}$ (Figures 1a - 1b). According to the assumption, the triangles $A C_{1} B_{2}, C_{2} B A_{1}$, $B_{1} A_{2} C$ are formed by rotating the triangle $A B C$ around the points $A, B, C$ respectively, in the positive rotational direction by $60^{\circ}$, hence the new positions of the sides $A B, B C, C A$ are equal and parallel segments:

$$
A C_{1} \# C_{2} B \# B_{1} A_{2}, \quad C_{1} B_{2} \# B A_{1} \# A_{2} C, \quad B_{2} A \# A_{1} C_{2} \# C B_{1} .
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_87f1c99cd971e0c20258g-1.jpg?height=450&width=1202&top_left_y=542&top_left_x=336)

Figures 1a and 1b

Any pair of segments from a triplet forms two opposite sides of a parallelogram. From each triplet, we can choose two segments in 3 different ways - by not selecting each one once in sequence, thus from the 3 triplets, we can obtain the vertices of a parallelogram in $3 \cdot 3=9$ different ways.

Similarly, the triangles $A C_{2} B_{1}, C_{1} B A_{2}, B_{2} A_{1} C$ are the $60^{\circ}$ rotations of the triangle $A B C$ around $A$, then $B$, and finally $C$, in the negative direction, hence the triplets

$$
A C_{2}, C_{1} B, B_{2} A_{1} ; \quad C_{2} B_{1}, B A_{2}, A_{1} C ; \quad B_{1} A, A_{2} C_{1}, C B_{2}
$$

any pair of which also forms a parallelogram, but these are identical to the previous ones, now obtained through the previously unmentioned side pairs.

If the triangle $A B C$ has a $60^{\circ}$ or $120^{\circ}$ angle, one of the segment pairs will lie on the same line, and one of the parallelograms will degenerate into a line segment.

II. By drawing isosceles right triangles over the sides, in each such triangle the legs are $\sqrt{2}$ times smaller than the hypotenuse, which is the corresponding side of the original triangle, and the angles at the hypotenuse are all $45^{\circ}$. Accordingly - using the above labeling again - each member of the two triangle triplets is formed by a rotation and dilation from the original triangle, the rotation angle is $45^{\circ}$, the direction is the same in both triangle triplets, and the dilation ratio is $1 / \sqrt{2}$, hence the members of the (1) and (2) segment triplets are again equal and parallel, and the statement remains valid (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_87f1c99cd971e0c20258g-1.jpg?height=583&width=727&top_left_y=1828&top_left_x=688)

![](https://cdn.mathpix.com/cropped/2024_05_02_87f1c99cd971e0c20258g-2.jpg?height=536&width=545&top_left_y=125&top_left_x=736)

Figure 3

Starting from a right triangle, degeneracies also occur here (Figure 3).

From the work of József Berács (Győr, Czuczor G. g. II. o. t.) with additions.

Note. In the case of an equilateral triangle, the point system of case I consists of only 6 points and contains 3 parallelograms. In contrast to the mentioned degeneracies, in some special cases, additional parallelograms may appear, for example, in case I, in the $30^{\circ}, 60^{\circ}, 90^{\circ}$ and $30^{\circ}, 30^{\circ}, 120^{\circ}$ triangles.