Let $ABC$ be an equilateral triangle with side length $1$. Points $A_1$ and $A_2$ are chosen on side $BC$, points $B_1$ and $B_2$ are chosen on side $CA$, and points $C_1$ and $C_2$ are chosen on side $AB$ such that $BA_1<BA_2$, $CB_1<CB_2$, and $AC_1<AC_2$. 
Suppose that the three line segments $B_1C_2$, $C_1A_2$, $A_1B_2$ are concurrent, and the perimeters of triangles $AB_2C_1$, $BC_2A_1$, and $CA_2B_1$ are all equal. Find all possible values of this common perimeter.

[i]Ankan Bhattacharya[/i]
1. **Setup and Initial Conditions:**
   - Let $ABC$ be an equilateral triangle with side length $1$.
   - Points $A_1$ and $A_2$ are chosen on side $BC$ such that $BA_1 < BA_2$.
   - Points $B_1$ and $B_2$ are chosen on side $CA$ such that $CB_1 < CB_2$.
   - Points $C_1$ and $C_2$ are chosen on side $AB$ such that $AC_1 < AC_2$.
   - The line segments $B_1C_2$, $C_1A_2$, and $A_1B_2$ are concurrent.
   - The perimeters of triangles $AB_2C_1$, $BC_2A_1$, and $CA_2B_1$ are all equal.

2. **Trisection and Concurrency:**
   - To show that the common perimeter can be $1$, we can trisect each side of the equilateral triangle.
   - Let $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, and $C_2$ trisect the sides $BC$, $CA$, and $AB$ respectively.
   - This means $BA_1 = A_1A_2 = A_2C = \frac{1}{3}$, and similarly for the other points.

3. **Calculation of Perimeters:**
   - The perimeter of $\triangle AB_2C_1$ is:
     \[
     AB_2 + B_2C_1 + C_1A = \frac{2}{3} + \frac{1}{3} + \frac{2}{3} = 1
     \]
   - Similarly, the perimeters of $\triangle BC_2A_1$ and $\triangle CA_2B_1$ are also $1$.

4. **Concurrency at the Center:**
   - The line segments $B_1C_2$, $C_1A_2$, and $A_1B_2$ concur at the centroid of $\triangle ABC$.
   - This is because the points trisecting the sides of an equilateral triangle and the lines connecting these points to the opposite vertices will always meet at the centroid.

5. **Proof of Uniqueness:**
   - Assume $p \neq 1$ and consider the excircles $\omega_A$, $\omega_B$, and $\omega_C$ of triangles $AB_2C_1$, $BC_2A_1$, and $CA_2B_1$ respectively.
   - By tangent segments being equal, if $p$ is the common perimeter:
     \[
     AT_A + AS_A = AB_2 + B_2T_A + AC_1 + C_1S_A = AB_2 + AC_1 + B_2U_A + C_1U_A = AB_2 + AC_1 + B_2C_1 = p
     \]
   - By symmetry, we obtain $AT_A + AS_A = BT_B + BS_B = CT_C + CS_C = p$, so $\omega_A$, $\omega_B$, and $\omega_C$ are congruent.

6. **Contradiction for $p \neq 1$:**
   - Define $B_2'$, $C_1'$, $C_2'$, $A_1'$, $A_2'$, and $B_1'$ such that $B_2'C_1'$ is tangent to the incircle $\omega$ of $\triangle ABC$ and $B_2'C_1' \parallel B_2C_1$.
   - Consider the dilation of factor $\frac{1}{p}$ centered at $A$. This sends $\omega_A \mapsto \omega$, $B_2 \mapsto B_2'$, and $C_1 \mapsto C_2'$.
   - Perform the same dilation at $B$ and $C$.
   - This implies $C_1A_2 \parallel C_1'A_2'$ and similar parallel lines.
   - Let $P$ be the concurrency point of $B_1'C_2'$, $C_1'A_2'$, and $A_1'B_2'$ by Brianchon's theorem.
   - If $p < 1$, then $X$ lies on the strict interior of $\triangle PB_2'C_1'$, $Y$ in $\triangle PC_2'A_1'$, and $Z$ in $\triangle PA_2'B_1'$, leading to a contradiction.
   - If $p > 1$, the same argument applies, leading to a contradiction.

7. **Conclusion:**
   - Therefore, the only possible value for the common perimeter is $1$.

The final answer is $\boxed{1}$.