A frog trainer places one frog at each vertex of an equilateral triangle $ABC$ of unit sidelength. The trainer can make one frog jump over another along the line joining the two, so that the total length of the jump is an even multiple of the distance between the two frogs just before the jump. Let $M$ and $N$ be two points on the rays $AB$ and $AC$, respectively, emanating from $A$, such that $AM = AN = \ell$, where $\ell$ is a positive integer. After a finite number of jumps, the three frogs all lie in the triangle $AMN$ (inside or on the boundary), and no more jumps are performed.
Determine the number of final positions the three frogs may reach in the triangle $AMN$. (During the process, the frogs may leave the triangle $AMN$, only their nal positions are to be in that triangle.)
1. **Coordinate System Setup**:
   We start by placing the vertices of the equilateral triangle \(ABC\) in a coordinate system. Let \(A = (0, 0)\), \(B = (1, 0)\), and \(C = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\). This setup places the triangle in a hexagonal grid.

2. **Invariance Under Jumps**:
   The frogs can jump over each other such that the total length of the jump is an even multiple of the distance between the two frogs just before the jump. This implies that the coordinates of the frogs modulo 2 are invariant under these jumps. Therefore, the coordinates of the frogs will always be congruent to their initial positions modulo 2.

3. **Points in Triangle \(AMN\)**:
   Let \(M\) and \(N\) be points on the rays \(AB\) and \(AC\) respectively, such that \(AM = AN = \ell\), where \(\ell\) is a positive integer. We need to determine the number of final positions the three frogs may reach in the triangle \(AMN\).

4. **Counting Points Congruent to \(A\) Modulo 2**:
   The number of points in \(AMN\) that are congruent to \(A\) modulo 2 is given by:
   \[
   \frac{\left\lfloor \frac{\ell + 2}{2} \right\rfloor \left\lfloor \frac{\ell + 4}{2} \right\rfloor}{2}
   \]

5. **Counting Points Congruent to \(B\) and \(C\) Modulo 2**:
   Similarly, the number of points in \(AMN\) that are congruent to \(B\) and \(C\) modulo 2 is given by:
   \[
   \frac{\left\lfloor \frac{\ell + 1}{2} \right\rfloor \left\lfloor \frac{\ell + 3}{2} \right\rfloor}{2}
   \]
   for each of \(B\) and \(C\).

6. **Total Number of Final Positions**:
   Combining these, the total number of final positions the three frogs may reach in the triangle \(AMN\) is:
   \[
   \frac{\left\lfloor \frac{\ell + 2}{2} \right\rfloor \left\lfloor \frac{\ell + 4}{2} \right\rfloor \left(\left\lfloor \frac{\ell + 1}{2} \right\rfloor \left\lfloor \frac{\ell + 3}{2} \right\rfloor\right)^2}{8}
   \]

7. **Proof of Reachability**:
   To prove that all triangles \(DEF\) that are congruent to \(ABC \pmod 2\) can be reached, we use the fact that the area ratio \(\frac{\text{Area}(DEF)}{\text{Area}(ABC)}\) is odd. If a frog jumps over another one such that the length of the jump is exactly twice the distance between them, this jump can be reversed. By performing these kinds of jumps, we can reach a triangle where two frogs have the same y-coordinate. This process does not change the area of \(DEF\).

8. **Final Adjustments**:
   If the y-coordinates of \(D, E, F\) are \(d, e, f\) with \(\min\{|d-e|, |e-f|, |f-d|\} = |d-e| > 0\), we can perform jumps on \(D\) and \(E\) so that \(f \in [d', e']\). This reduces \(|d'-f|\) or \(|e'-f|\) to be smaller than \(|d-e|\). Continuing this process, we can reach the desired position.

9. **Special Cases**:
   If \(d = e\), let the x-coordinates of \(D, E, F\) be \(p, q, r\) with \(p < q\). If \(q - p\) is even, then \(\frac{\text{Area}(DEF)}{\text{Area}(ABC)}\) is also even, which is a contradiction. Let \(E' = (p-1, d)\). Because \(q - p + 1\) is even, \(DEF\) can be reached from \(DE'F\). Thus, we can assume \(q - p = 1\). This means we can perform reversible jumps on \(D\) and \(E\) so that \(p\) or \(q\) equals \(r\). Using the same logic, we can assume \(|f - d| = 1\).

10. **Conclusion**:
    Any unit triangle congruent to \(ABC \pmod 2\) can be reached by letting two of the frogs jump over the other one with twice the distance repeatedly.

\(\blacksquare\)

The final answer is \( \boxed{ \frac{\left\lfloor \frac{\ell + 2}{2} \right\rfloor \left\lfloor \frac{\ell + 4}{2} \right\rfloor \left(\left\lfloor \frac{\ell + 1}{2} \right\rfloor \left\lfloor \frac{\ell + 3}{2} \right\rfloor\right)^2}{8} } \)