Do there exist six points $X_1,X_2,Y_1, Y_2,Z_1,Z_2$ in the plane such that all of the triangles $X_iY_jZ_k$ are similar for $1\leq i, j, k \leq 2$?
Proposed by Morteza Saghafian
To determine if there exist six points \(X_1, X_2, Y_1, Y_2, Z_1, Z_2\) in the plane such that all of the triangles \(X_iY_jZ_k\) are similar for \(1 \leq i, j, k \leq 2\), we need to analyze the conditions under which this can happen.

1. **Initial Triangle and Parallelogram Construction**:
   - Start with a generic triangle \(X_1Y_2Z_2\).
   - Construct point \(X_2\) such that \(X_1Y_2X_2Z_2\) forms a parallelogram. This ensures \(\triangle X_1Y_2Z_2 \equiv \triangle X_2Y_2Z_2\).

2. **Spiral Similarity and Point Construction**:
   - Construct point \(Z_1\) such that \(\triangle X_2Y_2Z_1 \sim \triangle Z_2Y_2X_1\). This is achieved by considering the spiral similarity centered at \(Y_2\) that maps \(Z_2\) to \(X_2\) and defining \(Z_1\) as the image of \(X_1\).
   - By definition, \(\triangle X_2Y_2Z_1\) is similar to \(\triangle X_1Y_2Z_2\).

3. **Proving Similarity of \(\triangle X_1Y_2Z_1\)**:
   - We need to show \(\triangle X_1Y_2Z_1 \sim \triangle Y_2Z_2X_1\).
   - Notice that \(\angle X_1Y_2Z_1 = \angle X_1Y_2X_2 - \angle Z_1Y_2X_2\).
   - Using angle properties and similarity, we get:
     \[
     \angle X_1Y_2Z_1 = \angle X_1Y_2Z_2 + \angle Y_2Z_2X_1 - \angle Z_1Y_2X_2 = \angle Y_2Z_2X_1
     \]
   - We also need to show \(\frac{X_1Y_2}{Y_2Z_1} = \frac{Y_2Z_2}{Z_2X_1}\).
   - Let \(x, y, z\) be the lengths of the sides opposite to \(X_1, Y_2, Z_2\) in \(\triangle X_1Y_2Z_2\). Then \(Y_2X_2 = X_1Z_2 = y\).
   - The ratio of similarity \(\triangle X_2Y_2Z_1 \sim \triangle Z_2Y_2X_1\) is \(\frac{y}{x}\). Therefore, \(Y_2Z_1 = \frac{y}{x} \cdot z\).
   - Thus, \(\frac{X_1Y_2}{Y_2Z_1} = \frac{z}{\frac{yz}{x}} = \frac{x}{y} = \frac{Y_2Z_2}{Z_2X_1}\).

4. **Constructing the Sixth Point \(Y_1\)**:
   - Construct \(Y_1\) such that \(Y_2X_2Y_1Z_1\) forms a parallelogram. This implies \(Z_1Y_1Z_2X_1\) is also a parallelogram.
   - We have \(\triangle X_1Y_2Z_1 \equiv \triangle Z_2X_2Y_1\) and \(\triangle X_2Y_2Z_1 \equiv \triangle X_2Y_1Z_1\).

5. **Final Similarity Check**:
   - The remaining triangles \(X_1Y_1Z_2\) and \(X_1Y_1Z_1\) are congruent.
   - For \(X_1Y_1Z_2\) to be similar to \(X_1Y_2Z_2\), we need \(\angle X_1Z_2Y_1 = \angle Y_2X_1Z_2\), which holds if \(\angle X = \angle Z + 2 \angle Y\).
   - We also need \(\frac{X_1Z_2}{Z_2Y_1} = \frac{Y_2X_1}{X_1Z_2}\), leading to \(\frac{y}{Z_2Y_1} = \frac{z}{y} \Rightarrow Z_2Y_1 = \frac{y^2}{z}\).
   - Given \(Z_2Y_1 = X_1Z_1\) and \(\triangle X_1Y_2Z_1 \sim \triangle Y_2Z_2X_1\), we have \(X_1Z_1 = \frac{z^2}{x}\).
   - Thus, \(\frac{y^2}{z} = \frac{z^2}{x}\), and setting \(x=1\), we get \(y^2 = z^3\), implying \(y = t^3\) and \(z = t^2\) for some \(t > 0\).

The final answer is \( \boxed{ y = t^3 } \) and \(z = t^2\) for some \(t > 0\).