The only possible value of the common perimeter, denoted $p$, is 1 .
『 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 $A B_{2} C_{1}, B C_{2} A_{1}, C A_{2} B_{1}$ all have perimeter 1 . Then lines $\overline{B_{1} C_{2}}, \overline{C_{1} A_{2}}$, and $\overline{A_{1} B_{2}}$ are concurrent.

Proof. The perimeter conditions mean that $\overline{B_{2} C_{1}}, \overline{C_{2} A_{1}}$, and $\overline{A_{2} B_{1}}$ are tangent to the incircle of $\triangle A B C$.
Hence the result follows by Brianchon's theorem.
Now suppose $p \neq 1$. Let $\overline{B_{2}^{\prime} C_{1}^{\prime}}$ be the dilation of $\overline{B_{2} C_{1}}$ with ratio $\frac{1}{p}$ at center $A$, and define $C_{2}^{\prime}, A_{1}^{\prime}, A_{2}^{\prime}, B_{1}^{\prime}$ similarly. The following diagram showcases the situation $p<1$.
By the reasoning in the $p=1$ case, note that $\overline{B_{1}^{\prime} C_{2}^{\prime}}, \overline{C_{1}^{\prime} A_{2}^{\prime}}$, and $\overline{A_{1}^{\prime} B_{2}^{\prime}}$ are concurrent. However, $\overline{B_{1} C_{2}}, \overline{C_{1} A_{2}}, \overline{A_{1} B_{2}}$ lie in the interior of quadrilaterals $B C B_{1}^{\prime} C_{2}^{\prime}, C A C_{1}^{\prime} A_{2}^{\prime}$, and $A B A_{1}^{\prime} B_{2}^{\prime}$, and these quadrilaterals do not share an interior point, a contradiction.

Thus $p \geq 1$. Similarly, we can show $p \leq 1$, and so $p=1$ is forced (and achieved if, for example, the three triangles are equilateral with side length $1 / 3$ ).

『 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 $A_{1}=\left(0: 1-a_{1}: a_{1}\right), A_{2}=\left(0: a_{2}: 1-a_{2}\right)$, $B_{1}=\left(b_{1}: 0: 1-b_{1}\right), B_{2}=\left(1-b_{2}: 0: b_{2}\right), C_{1}=\left(1-c_{1}: c_{1}: 0\right)$, and $C_{2}=\left(c_{2}: 1-c_{2}: 0\right)$. The line $B_{1} C_{2}$ is defined by the equation
$$
\begin{array}{c}
\operatorname{det}\left[\begin{array}{ccc}
x & y & z \\
b_{1} & 0 & 1-b_{1} \\
c_{2} & 1-c_{2} & 0
\end{array}\right]=0 ; \\
x\left(-\left(1-b_{1}\right)\left(1-c_{2}\right)\right)+y\left(\left(1-b_{1}\right) c_{2}\right)+z\left(b_{1}\left(1-c_{2}\right)\right)=0 .
\end{array}
$$

Computing the equations for the other lines cyclically, we get that the lines $B_{1} C_{2}, C_{1} A_{2}$, and $A_{1} B_{2}$ concur if and only if
$$
\operatorname{det}\left[\begin{array}{ccc}
-\left(1-b_{1}\right)\left(1-c_{2}\right) & \left(1-b_{1}\right) c_{2} & b_{1}\left(1-c_{2}\right) \\
c_{1}\left(1-a_{2}\right) & -\left(1-c_{1}\right)\left(1-a_{2}\right) & \left(1-c_{1}\right) a_{2} \\
\left(1-a_{1}\right) b_{2} & a_{1}\left(1-b_{2}\right) & -\left(1-a_{1}\right)\left(1-b_{2}\right)
\end{array}\right]=0 .
$$

Let this matrix be $M$. We also define the similar matrix
$$
N=\left[\begin{array}{ccc}
-\left(1-b_{2}\right)\left(1-c_{1}\right) & \left(1-b_{2}\right) c_{1} & b_{2}\left(1-c_{1}\right) \\
c_{2}\left(1-a_{1}\right) & -\left(1-c_{2}\right)\left(1-a_{1}\right) & \left(1-c_{2}\right) a_{1} \\
\left(1-a_{2}\right) b_{1} & a_{2}\left(1-b_{1}\right) & -\left(1-a_{2}\right)\left(1-b_{1}\right)
\end{array}\right] .
$$

Geometrically, det $N=0$ if and only if $B_{2}^{\prime} C_{1}^{\prime}, C_{2}^{\prime} A_{1}^{\prime}$, and $A_{2}^{\prime} B_{1}^{\prime}$ concur, where for a point $P$ on a side of triangle $A B C, P^{\prime}$ denotes its reflection over that side's midpoint.

Claim - We have $\operatorname{det} M=\operatorname{det} N$.
Proof. To show $\operatorname{det} M=\operatorname{det} N$, 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:
$$
\begin{aligned}
-\left(1-b_{1}\right)\left(1-c_{2}\right)\left(1-c_{1}\right)\left(1-a_{2}\right)\left(1-a_{1}\right)\left(1-b_{2}\right) & =-\prod_{i=1}^{2}\left(\left(1-a_{i}\right)\left(1-b_{i}\right)\left(1-c_{i}\right)\right) \\
\left(1-a_{1}\right) b_{2}\left(1-b_{1}\right) c_{2}\left(1-c_{1}\right) a_{2} & =\left(\left(1-a_{1}\right)\left(1-b_{1}\right)\left(1-c_{1}\right)\right)\left(a_{2} b_{2} c_{2}\right) \\
c_{1}\left(1-a_{2}\right) a_{1}\left(1-b_{2}\right) b_{1}\left(1-c_{2}\right) & =\left(\left(1-a_{2}\right)\left(1-b_{2}\right)\left(1-c_{2}\right)\right)\left(a_{1} b_{1} c_{1}\right) .
\end{aligned}
$$

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
$$
\sum_{\text {cyc }}\left(1-b_{1}\right)\left(1-c_{2}\right)\left(1-c_{1}\right) a_{2} a_{1}\left(1-b_{2}\right) ;
$$
each summand is invariant under $\Psi$. This finishes the proof of our claim.
Now, it suffices to show that, if $A B_{2} C_{1}, B C_{2} A_{1}$, and $C A_{2} B_{1}$ each have perimeter 1, then
$$
\operatorname{det}\left[\begin{array}{ccc}
-\left(1-b_{2}\right)\left(1-c_{1}\right) & \left(1-b_{2}\right) c_{1} & b_{2}\left(1-c_{1}\right) \\
c_{2}\left(1-a_{1}\right) & -\left(1-c_{2}\right)\left(1-a_{1}\right) & \left(1-c_{2}\right) a_{1} \\
\left(1-a_{2}\right) b_{1} & a_{2}\left(1-b_{1}\right) & -\left(1-a_{2}\right)\left(1-b_{1}\right) .
\end{array}\right]=0 .
$$

Indeed, we have $A B_{2}=b_{2}$ and $A C_{1}=c_{1}$, so by the law of cosines,
$$
1-b_{2}-c_{1}=1-A B_{2}-A C_{1}=B_{2} C_{1}=\sqrt{b_{2}^{2}+c_{1}^{2}-b_{2} c_{1}} \text {. }
$$

This gives
$$
\left(1-b_{2}-c_{1}\right)^{2}=b_{2}^{2}+c_{1}^{2}-b_{2} c_{1} \Longrightarrow 1-2 b_{2}-2 c_{1}+3 b_{2} c_{1}=0 .
$$

Similarly, $1-2 c_{2}-2 a_{1}+3 c_{2} a_{1}=0$ and $1-2 a_{2}-2 b_{1}+3 a_{2} b_{1}=0$.
Now,
$$
\begin{aligned}
N\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right] & =\left[\begin{array}{l}
-\left(1-b_{2}\right)\left(1-c_{1}\right)+\left(1-b_{2}\right) c_{1}+b_{2}\left(1-c_{1}\right) \\
-\left(1-c_{2}\right)\left(1-a_{1}\right)+\left(1-c_{2}\right) a_{1}+c_{2}\left(1-a_{1}\right) \\
-\left(1-a_{2}\right)\left(1-b_{1}\right)+\left(1-a_{2}\right) b_{1}+a_{2}\left(1-b_{1}\right)
\end{array}\right] \\
& =\left[\begin{array}{l}
-1+2 b_{2}+2 c_{1}-3 b_{2} c_{1} \\
-1+2 c_{2}+2 a_{1}-3 c_{2} a_{1} \\
-1+2 a_{2}+2 b_{1}-2 a_{2} b_{1}
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right] .
\end{aligned}
$$
So it follows $\operatorname{det} N=0$, as desired.
Let $A B C$ be an equilateral triangle with side length 1 . Points $A_{1}$ and $A_{2}$ are chosen on side $B C$, points $B_{1}$ and $B_{2}$ are chosen on side $C A$, and points $C_{1}$ and $C_{2}$ are chosen on side $A B$ such that $B A_{1}<B A_{2}, C B_{1}<C B_{2}$, and $A C_{1}<A C_{2}$.

Suppose that the three line segments $B_{1} C_{2}, C_{1} A_{2}$, and $A_{1} B_{2}$ are concurrent, and the perimeters of triangles $A B_{2} C_{1}, B C_{2} A_{1}$, and $C A_{2} B_{1}$ are all equal. Find all possible values of this common perimeter.