$ABC$ is a triangle. Find a point $X$ on $BC$ such that :
area $ABX$ / area $ACX$ = perimeter  $ABX$ / perimeter $ACX$.
1. **Define the variables and apply Stewart's Theorem:**
   Let \(AB = c\), \(BC = a\), \(CA = b\), \(BX = x\), and \(AX = d\). By Stewart's theorem, we have:
   \[
   d^2 = \frac{c^2(a-x) + b^2x}{a} + (x-a)x
   \]
   This theorem relates the lengths of the sides of a triangle to the length of a cevian (a line segment from a vertex to the opposite side).

2. **Set up the given equality:**
   The given equality is:
   \[
   \frac{\text{area } ABX}{\text{area } ACX} = \frac{\text{perimeter } ABX}{\text{perimeter } ACX}
   \]
   Using the areas and perimeters, we can write:
   \[
   \frac{x}{a-x} = \frac{c + x + d}{b + a - x + d}
   \]

3. **Simplify the given equality:**
   Cross-multiplying the equality, we get:
   \[
   x(b + a - x + d) = (a - x)(c + x + d)
   \]
   Expanding both sides:
   \[
   bx + ax - x^2 + xd = ac + cx + ad - acx - x^2 - xd
   \]
   Simplifying, we get:
   \[
   bx + ax = ac + cx + ad - acx
   \]
   Rearranging terms:
   \[
   d(a - 2x) = bx + cx - ac
   \]

4. **Substitute \(d\) from Stewart's theorem:**
   From Stewart's theorem, we have:
   \[
   d^2 = \frac{c^2(a-x) + b^2x}{a} + (x-a)x
   \]
   Substituting \(d\) into the equation:
   \[
   \left(\frac{c^2(a-x) + b^2x}{a} + (x-a)x\right) \cdot (2x - a)^2 = (ac - cx - bx)^2
   \]

5. **Simplify the equation:**
   Simplifying the left-hand side:
   \[
   \left(\frac{c^2(a-x) + b^2x}{a} + (x-a)x\right) \cdot (2x - a)^2 = (ac - cx - bx)^2
   \]
   Expanding and simplifying:
   \[
   x(x-a)(4ax^2 + 4(b^2 - c^2 - a^2)x + (-ab^2 + 3ac^2 + a^3 - 2abc)) = 0
   \]
   This simplifies to:
   \[
   4ax^2 + 4(b^2 - c^2 - a^2)x + (-ab^2 + 3ac^2 + a^3 - 2abc) = 0
   \]

6. **Solve the quadratic equation:**
   The discriminant of the quadratic equation is:
   \[
   \Delta = 32ac(b+c)(a - \cos\beta(b-c)) > 0
   \]
   This indicates that there are real solutions for \(x\).

The final answer is \( \boxed{ x } \) such that \(4ax^2 + 4(b^2 - c^2 - a^2)x + (-ab^2 + 3ac^2 + a^3 - 2abc) = 0\).