Claim ($p = 1$ implies concurrence) — Suppose the six points are chosen so that triangles $AB_2C_1$, $BC_2A_1$, $CA_2B_1$ all have perimeter $1$. Then lines $\overline{B_1C_2}$, $\overline{C_1A_2}$, and $\overline{A_1B_2}$ are concurrent.

Proof. The perimeter conditions mean that $\overline{B_2C_1}$, $\overline{C_2A_1}$, and $\overline{A_2B_1}$ are tangent to the incircle of $\triangle ABC$.

¶ Barycentric solution (by Carl, Krit, Milan) We show that, if the common perimeter is $1$, then the lines concur. To do this, we use barycentric coordinates. Let $A = (1:0:0)$, $B = (0:1:0)$, and $C = (0:0:1)$. Let $A_1 = (0:1-a_1:a_1)$, $A_2 = (0:a_2:1-a_2)$, $B_1 = (b_1:0:1-b_1)$, $B_2 = (1-b_2:0:b_2)$, $C_1 = (1-c_1:c_1:0)$, and $C_2 = (c_2:1-c_2:0)$. The line $B_1C_2$ is defined by the equation
$$\det \begin{bmatrix} x & y & z \\ b_1 & 0 & 1-b_1 \\ c_2 & 1-c_2 & 0 \end{bmatrix} = 0;$$
i.e.
$$x(-(1-b_1)(1-c_2)) + y((1-b_1)c_2) + z(b_1(1-c_2)) = 0.$$

Computing the equations for the other lines cyclically, we get that the lines $B_1C_2$, $C_1A_2$, and $A_1B_2$ concur if and only if
$$\det \begin{bmatrix} -(1-b_1)(1-c_2) & (1-b_1)c_2 & b_1(1-c_2) \\ c_1(1-a_2) & -(1-c_1)(1-a_2) & (1-c_1)a_2 \\ (1-a_1)b_2 & a_1(1-b_2) & -(1-a_1)(1-b_2) \end{bmatrix} = 0.$$

Let this matrix be $M$. We also define the similar matrix
$$N = \begin{bmatrix} -(1-b_2)(1-c_1) & (1-b_2)c_1 & b_2(1-c_1) \\ c_2(1-a_1) & -(1-c_2)(1-a_1) & (1-c_2)a_1 \\ (1-a_2)b_1 & a_2(1-b_1) & -(1-a_2)(1-b_1) \end{bmatrix}.$$

Claim — We have $\det M = \det N$.

Proof. To show $\det M = \det N$, it suffices to demonstrate that the determinant above is invariant under swapping subscripts of “1” and “2,” an operation we call $\Psi$.
We use the definition of the determinant as a sum over permutations. The even permutations give us the following three terms:
$$-(1-b_1)(1-c_2)(1-c_1)(1-a_2)(1-a_1)(1-b_2) = -\prod_{i=1}^{2}((1-a_i)(1-b_i)(1-c_i)) \\ (1-a_1)b_2(1-b_1)c_2(1-c_1)a_2 = ((1-a_1)(1-b_1)(1-c_1))(a_2b_2c_2) \\ c_1(1-a_2)a_1(1-b_2)b_1(1-c_2) = ((1-a_2)(1-b_2)(1-c_2))(a_1b_1c_1).$$

The first term is invariant under $\Psi$, while the second and third terms are swapped under $\Psi$. For the odd permutations, we have a contribution to the determinant of
$$\sum_{\text{cyc}} (1-b_1)(1-c_2)(1-c_1)a_2a_1(1-b_2);$$
each summand is invariant under $\Psi$. This finishes the proof of our claim. $\square$

Now, it suffices to show that, if $AB_2C_1$, $BC_2A_1$, and $CA_2B_1$ each have perimeter $1$, then
$$\det \begin{bmatrix} -(1-b_2)(1-c_1) & (1-b_2)c_1 & b_2(1-c_1) \\ c_2(1-a_1) & -(1-c_2)(1-a_1) & (1-c_2)a_1 \\ (1-a_2)b_1 & a_2(1-b_1) & -(1-a_2)(1-b_1) \end{bmatrix} = 0.$$

Indeed, we have $AB_2 = b_2$ and $AC_1 = c_1$, so by the law of cosines,
$$1 - b_2 - c_1 = 1 - AB_2 - AC_1 = B_2C_1 = \sqrt{b_2^2 + c_1^2 - b_2c_1}.$$
This gives
$$(1 - b_2 - c_1)^2 = b_2^2 + c_1^2 - b_2c_1 \implies 1 - 2b_2 - 2c_1 + 3b_2c_1 = 0.$$
Similarly,
$$1 - 2c_2 - 2a_1 + 3c_2a_1 = 0 \text{ and } 1 - 2a_2 - 2b_1 + 3a_2b_1 = 0.$$

Now,
$$N \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} -(1-b_2)(1-c_1) + (1-b_2)c_1 + b_2(1-c_1) \\ -(1-c_2)(1-a_1) + (1-c_2)a_1 + c_2(1-a_1) \\ -(1-a_2)(1-b_1) + (1-a_2)b_1 + a_2(1-b_1) \end{bmatrix} \\ = \begin{bmatrix} -1 + 2b_2 + 2c_1 - 3b_2c_1 \\ -1 + 2c_2 + 2a_1 - 3c_2a_1 \\ -1 + 2a_2 + 2b_1 - 2a_2b_1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}.$$

So it follows $\det N = 0$, as desired.

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Synthetic approach (from author) We prove the converse of the problem first:

Claim ($p = 1$ implies concurrence) — Suppose the six points are chosen so that triangles $AB_2C_1$, $BC_2A_1$, $CA_2B_1$ all have perimeter $1$. Then lines $\overline{B_1C_2}$, $\overline{C_1A_2}$, and $\overline{A_1B_2}$ are concurrent.

Proof. The perimeter conditions mean that $\overline{B_2C_1}$, $\overline{C_2A_1}$, and $\overline{A_2B_1}$ are tangent to the incircle of $\triangle ABC$.

Hence the result follows by Brianchon's theorem. $\Box$

Now suppose $p \neq 1$. Let $\overline{B'_2C'_1}$ be the dilation of $\overline{B_2C_1}$ with ratio $\frac{1}{p}$ at center $A$, and define $C'_2$, $A'_1$, $A'_2$, $B'_1$ similarly. The following diagram showcases the situation $p < 1$.

By the reasoning in the $p=1$ case, note that $\overline{B'_1C'_2}$, $\overline{C'_1A'_2}$, and $\overline{A'_1B'_2}$ are concurrent. However, $\overline{B_1C_2}$, $\overline{C_1A_2}$, $\overline{A_1B_2}$ lie in the interior of quadrilaterals $BCB'_1C'_2$, $CAC'_1A'_2$, and $ABA'_1B'_2$, and these quadrilaterals do not share an interior point, a contradiction.
Thus $p \ge 1$. Similarly, we can show $p \le 1$, and so $p=1$ is forced (and achieved if, for example, the three triangles are equilateral with side length $1/3$).

Therefore, the only possible value of the common perimeter is $\boxed{1}$.