In the plane (in a fixed position), there is a triangle $ABC$ and a triangle $X'Y'Z'$. Write a triangle $XYZ$ into $\triangle ABC$ such that the sides of $\triangle XYZ$ are parallel to the sides of $\triangle X'Y'Z'$. (The points $X, Y, Z$ are on the sides of $\triangle ABC$ and not on their extensions.) What are the conditions for solvability and how many solutions are there? How many solutions are there if the vertices of the sought triangle can also be on the extensions of the sides of the given $\triangle ABC$?
Assuming that the given two triangles have no parallel sides. Through the vertices of $\mathrm{X}^{\prime} \mathrm{Y}^{\prime} \mathrm{Z}_{\triangle}^{\prime}$, we draw lines parallel to the sides $a, b$, and $c$ of $\mathrm{A} \mathrm{B} \mathrm{C}_{\triangle}$. Through each vertex, 3 such parallel lines pass: $a_{x}$, $b_{x}$, $c_{x}$; $a_{y}$, $b_{y}$, $c_{y}$; $a_{z}$, $b_{z}$, $c_{z}$. Only those triples of $a, b, c$ whose indices are permutations of $x, y, z$ form a triangle with one point of $\mathrm{X}^{\prime} \mathrm{Y}^{\prime} \mathrm{Z}^{\prime}$ on each side (and not at the vertices): $a_{x}$, $b_{y}$, $c_{z}$; $a_{x}$, $b_{z}$, $c_{y}$; $a_{y}$, $b_{x}$, $c_{z}$; $a_{y}$, $b_{z}$, $c_{x}$; $a_{z}$, $b_{x}$, $c_{y}$; $a_{z}$, $b_{y}$, $c_{x}$. Therefore, we get a total of $3!=6$ triangles, all of which are similar and similarly oriented to the given $\mathrm{A} \mathrm{B} \mathrm{C}_{\triangle}$.

Since for each direction, among the 3 parallel lines through the vertices $\mathrm{X}^{\prime}, \mathrm{Y}^{\prime}$, and $\mathrm{Z}^{\prime}$, there are always 2 extreme lines that do not intersect the triangle $\mathrm{X}^{\prime} \mathrm{Y}^{\prime} \mathrm{Z}^{\prime}$ and always 1 middle line that necessarily intersects the triangle, among the 9 lines in question, there are always 3 and only 3 intersecting lines and exactly 6 non-intersecting lines.
![](https://cdn.mathpix.com/cropped/2024_05_02_87f1cc188a812e9482d5g-1.jpg?height=1000&width=1132&top_left_y=522&top_left_x=515)

Figure 3

The distribution of these 6 non-intersecting lines according to the vertices can only be of the following three types: 2, 2, 2 (Figure 1), 1, 2, 3 (Figure 2), and 0, 3, 3 (Figure 3), and accordingly, we find among the 6 triangles 2, 1, or 0 triangles that completely enclose the area of $\mathrm{X}^{\prime} \mathrm{Y}^{\prime} \mathrm{Z}_{\Delta}^{\prime}$, meaning that the points $\mathrm{X}^{\prime}, \mathrm{Y}^{\prime}, \mathrm{Z}^{\prime}$ are on the sides of one of these triangles and not on their extensions.

Suppose we find 2 such triangles: $\mathrm{A}_{1}^{\prime} \mathrm{B}_{1}^{\prime} \mathrm{C}_{1 \Delta}^{\prime}$ and $\mathrm{A}_{2}^{\prime} \mathrm{B}_{2}^{\prime} \mathrm{C}_{2 \Delta}^{\prime}$ (Figure 1). The direction of traversal in these triangles is clearly always opposite. If we transform these two triangles along with $\mathrm{X}^{\prime} \mathrm{Y}^{\prime} \mathrm{Z}_{\Delta}^{\prime}$ from the external similarity center $H_{1}$ and the internal similarity center $H_{2}$ to $\mathrm{A} \mathrm{B} \mathrm{C}_{\Delta}$, then in the latter, we obtain the 2 solutions to our problem: $\mathrm{X}_{1} \mathrm{Y}_{1} \mathrm{Z}_{1 \Delta} \Delta$ and $\mathrm{X}_{2} \mathrm{Y}_{2} \mathrm{Z}_{2 \Delta}$.

Thus, our problem has 2, 1, or 0 solutions, according to the 3 types of distribution of the 6 "non-intersecting" lines mentioned above. If the vertices of the sought triangle can also be on the extensions of the sides of the given $\mathrm{A} \mathrm{B} \mathrm{C}_{\triangle}$, then, as we have seen, there are 6 solutions, the number of which can decrease depending on how many of the 6 triangles around $\mathrm{X}^{\prime} \mathrm{Y}^{\prime} \mathrm{Z}^{\prime}$ degenerate into points.