Let $ABC$ a triangle be given  with $BC = a$, $CA = b$, $AB = c$ ($a \neq b \neq c \neq a$). In  plane ($ABC$) take the points $A'$, $B'$, $C'$ such that:

[b]I.[/b] The pairs of points $A$ and $A'$, $B$ and $B'$, $C$ and $C'$ either all lie in one side either all lie in different sides under the lines $BC$, $CA$, $AB$ respectively;

[b]II.[/b] Triangles $A'BC$, $B'CA$, $C'AB$ are similar isosceles triangles.

Find the value of angle $A'BC$ as function of $a, b, c$ such that lengths $AA', BB', CC'$ are not sides of an triangle. (The word "triangle" must be understood in its ordinary meaning: its vertices are not collinear.)
1. **Restate the Problem in Terms of Barycentric Coordinates:**
   We are given a triangle \(ABC\) with sides \(BC = a\), \(CA = b\), and \(AB = c\). We need to find the angle \(\phi\) such that the segments \(AA^\phi\), \(BB^\phi\), and \(CC^\phi\) are not the sides of a non-degenerate triangle. The points \(A^\phi\), \(B^\phi\), and \(C^\phi\) are the apices of isosceles triangles \(BA^\phi C\), \(CB^\phi A\), and \(AC^\phi B\) respectively, with base angles \(\phi\).

2. **Lemma 1:**
   We first prove a lemma that will be useful in solving the problem.
   
   **Lemma 1:** Let \(ABC\) be a triangle, and let \(X\), \(Y\), \(Z\) be points in its plane such that the triangles \(BXC\), \(CYA\), and \(AZB\) are all directly similar. Then, the segments \(AX\), \(BY\), and \(CZ\) always form the sides of a (possibly degenerate) triangle. This triangle degenerates if and only if the lines \(AX\), \(BY\), and \(CZ\) are all parallel.

   **Proof of Lemma 1:**
   We use complex numbers to prove this lemma. Let \(a\), \(b\), \(c\), \(x\), \(y\), and \(z\) be the affixes of the points \(A\), \(B\), \(C\), \(X\), \(Y\), and \(Z\) respectively. Since the triangles \(BXC\), \(CYA\), and \(AZB\) are all directly similar, we have:
   \[
   \frac{x-c}{b-c} = \frac{y-a}{c-a} = \frac{z-b}{a-b} = k
   \]
   for some complex number \(k\). This implies:
   \[
   x = kb + (1-k)c, \quad y = kc + (1-k)a, \quad z = ka + (1-k)b
   \]
   Now, consider the sum:
   \[
   (a-x) + (b-y) + (c-z) = \left(a - (kb + (1-k)c)\right) + \left(b - (kc + (1-k)a)\right) + \left(c - (ka + (1-k)b)\right) = 0
   \]
   Hence, by the triangle inequality:
   \[
   |a-x| + |b-y| \geq |(a-x) + (b-y)| = |z-c|
   \]
   Similarly, we get:
   \[
   |b-y| + |c-z| \geq |a-x| \quad \text{and} \quad |c-z| + |a-x| \geq |b-y|
   \]
   Thus, the segments \(AX\), \(BY\), and \(CZ\) form the sides of a (possibly degenerate) triangle. This triangle degenerates if and only if at least one of the triangle inequalities becomes an equality, which happens if and only if the lines \(AX\), \(BY\), and \(CZ\) are all parallel. This completes the proof of Lemma 1.

3. **Application of Lemma 1:**
   The triangles \(BA^\phi C\), \(CB^\phi A\), and \(AC^\phi B\) are isosceles triangles with the same base angles \(\phi\). Therefore, they are all directly similar, and Lemma 1 applies. Consequently, the segments \(AA^\phi\), \(BB^\phi\), and \(CC^\phi\) always form the sides of a (possibly degenerate) triangle. This triangle degenerates if and only if the lines \(AA^\phi\), \(BB^\phi\), and \(CC^\phi\) are all parallel.

4. **Finding the Angle \(\phi\):**
   According to the reference [1], the lines \(AA^\phi\), \(BB^\phi\), and \(CC^\phi\) concur at the Kiepert perspector \(K(\phi)\) of triangle \(ABC\). These lines are parallel if and only if \(K(\phi)\) lies on the line at infinity. The homogeneous barycentric coordinates of \(K(\phi)\) are:
   \[
   K(\phi) = \left(\frac{1}{S_A + S_\phi} : \frac{1}{S_B + S_\phi} : \frac{1}{S_C + S_\phi}\right)
   \]
   This point lies on the line at infinity if and only if:
   \[
   \frac{1}{S_A + S_\phi} + \frac{1}{S_B + S_\phi} + \frac{1}{S_C + S_\phi} = 0
   \]
   This simplifies to:
   \[
   (S_B S_C + S_C S_A + S_A S_B) + 2(S_A + S_B + S_C)S_\phi + 3S_\phi^2 = 0
   \]
   Using the formulas \(S_B S_C + S_C S_A + S_A S_B = S^2\) and \(S_A + S_B + S_C = S_\omega\), we get:
   \[
   S^2 + 2S_\omega S_\phi + 3S_\phi^2 = 0
   \]
   Solving this quadratic equation for \(S_\phi\), we obtain:
   \[
   S_\phi = \frac{-S_\omega \pm \sqrt{S_\omega^2 - 3S^2}}{3}
   \]
   Since \(S_\theta = S \cdot \cot \theta\) for any angle \(\theta\), this yields:
   \[
   \cot \phi = \frac{-\cot \omega \pm \sqrt{\cot^2 \omega - 3}}{3}
   \]
   Using \(\cot \omega = \frac{a^2 + b^2 + c^2}{4\Delta}\), where \(\Delta\) is the area of triangle \(ABC\), and applying Heron's formula for \(\Delta\), we get:
   \[
   \cot \phi = \frac{-\left(a^2 + b^2 + c^2\right) \pm 2\sqrt{a^4 + b^4 + c^4 - b^2 c^2 - c^2 a^2 - a^2 b^2}}{12\Delta}
   \]

The final answer is \( \boxed{ \cot \phi = \frac{-\left(a^2 + b^2 + c^2\right) \pm 2\sqrt{a^4 + b^4 + c^4 - b^2 c^2 - c^2 a^2 - a^2 b^2}}{12\Delta} } \)