Problem: 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 $B_1C_2$, $C_1A_2$, and $A_1B_2$
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.

Solution: The only possible value of the common perimeter, denoted $p$, is $1$.

\paragraph{Synthetic approach (from author).}
We prove the converse of the problem first:
\begin{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 $\ol{B_1C_2}$, $\ol{C_1A_2}$, and $\ol{A_1B_2}$ are concurrent.
\end{claim*}
\begin{proof}
  The perimeter conditions mean that
  $\ol{B_2C_1}$, $\ol{C_2A_1}$, and $\ol{A_2B_1}$
  are tangent to the incircle of $\triangle ABC$.

  \begin{center}
    \begin{asy}
      unitsize(48);
      pair pole(pair p, pair q) {
        return 2*p*q/(p+q);
      }
      pair A = 2 * dir( 90);
      pair B = 2 * dir(210);
      pair C = 2 * dir(330);
      pair D = dir(100);
      pair E = dir(190);
      pair F = dir(320);

      pair A1 = pole(dir(270), E);
      pair A2 = pole(dir(270), F);
      pair B1 = pole(dir( 30), F);
      pair B2 = pole(dir( 30), D);
      pair C1 = pole(dir(150), D);
      pair C2 = pole(dir(150), E);

      draw(A--B--C--cycle);
      filldraw(A--B2--C1--cycle, opacity(0.1)+lightgreen, black);
      filldraw(B--C2--A1--cycle, opacity(0.1)+lightgreen, black);
      filldraw(C--A2--B1--cycle, opacity(0.1)+lightgreen, black);
      fill(A1--A2--B1--B2--C1--C2--cycle, opacity(0.1)+yellow);
      draw(A1--B2^^B1--C2^^C1--A2, dashed);
      draw(unitcircle, blue);

      dot("$A$", A, dir(A));
      dot("$B$", B, dir(B));
      dot("$C$", C, dir(C));
      dot(midpoint(A--B));
      dot(midpoint(B--C));
      dot(midpoint(C--A));

      dot("$A_1$", A1, dir(270));
      dot("$A_2$", A2, dir(270));
      dot("$B_1$", B1, dir( 30));
      dot("$B_2$", B2, dir( 30));
      dot("$C_1$", C1, dir(150));
      dot("$C_2$", C2, dir(150));
    \end{asy}
  \end{center}

  Hence the result follows by \emph{Brianchon's theorem}.
\end{proof}

Now suppose $p \neq 1$.
Let $\ol{B_2'C_1'}$ be the dilation of $\ol{B_2C_1}$
with ratio $\tfrac1p$ at center $A$,
and define $C_2'$, $A_1'$, $A_2'$, $B_1'$ similarly.
The following diagram showcases the situation $p < 1$.
\begin{center}
  \begin{asy}
    unitsize(64);
    pair pole(pair p, pair q) {
      return 2*p*q/(p+q);
    }

    real fac = 0.7;

    pair A = 2 * dir( 90);
    pair B = 2 * dir(210);
    pair C = 2 * dir(330);

    pair D = dir(100);
    pair E = dir(190);
    pair F = dir(320);

    pair A1p = pole(dir(270), E);
    pair A2p = pole(dir(270), F);
    pair B1p = pole(dir( 30), F);
    pair B2p = pole(dir( 30), D);
    pair C1p = pole(dir(150), D);
    pair C2p = pole(dir(150), E);

    pair A1 = B + fac * (A1p - B);
    pair A2 = C + fac * (A2p - C);
    pair B1 = C + fac * (B1p - C);
    pair B2 = A + fac * (B2p - A);
    pair C1 = A + fac * (C1p - A);
    pair C2 = B + fac * (C2p - B);

    draw(unitcircle, gray(0.7));
    draw(A--B--C--cycle, gray(0.6));
    draw(B2p--C1p^^C2p--A1p^^A2p--B1p, blue+dashed);
    draw(B2--C1^^C2--A1^^A2--B1, red+1.4);

    draw(A1p--B2p^^B1p--C2p^^C1p--A2p, blue+dotted);
    draw(A1--B2^^B1--C2^^C1--A2, orange+0.8);

    dot("$A$", A, dir(A));
    dot("$B$", B, dir(B));
    dot("$C$", C, dir(C));

    dot("$A_1'$", A1p, dir(270));
    dot("$A_2'$", A2p, dir(270));
    dot("$B_1'$", B1p, dir( 30));
    dot("$B_2'$", B2p, dir( 30));
    dot("$C_1'$", C1p, dir(150));
    dot("$C_2'$", C2p, dir(150));

    dot("$A_1$", A1, dir(270));
    dot("$A_2$", A2, dir(270));
    dot("$B_1$", B1, dir( 30));
    dot("$B_2$", B2, dir( 30));
    dot("$C_1$", C1, dir(150));
    dot("$C_2$", C2, dir(150));
  \end{asy}
\end{center}
By the reasoning in the $p = 1$ case,
note that $\ol{B_1'C_2'}$, $\ol{C_1'A_2'}$, and $\ol{A_1'B_2'}$ are concurrent.
However, $\ol{B_1C_2}$, $\ol{C_1A_2}$, $\ol{A_1B_2}$
lie in the interior of quadrilaterals
$BCB_1'C_2'$, $CAC_1'A_2'$, and $ABA_1'B_2'$,
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 $1/3$).

\paragraph{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.
\[ x\big(-(1-b_1)(1-c_2)\big)
  + y\big((1-b_1)c_2\big)
  + z\big(b_1(1-c_2)\big) = 0. \]
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
\[ N = \begin{bmatrix}
    -(1-b_2)(1-c_1) & (1-b_2)c_1 & b_2(1-c_1)\\
    c_2(1-a_1) & -(1-c_2)(1-a_1) & (1-c_2)a_1\\
    (1-a_2)b_1 & a_2(1-b_1) & -(1-a_2)(1-b_1)
  \end{bmatrix}. \]
Geometrically, $\det N=0$ if and only if
$B_2'C_1'$, $C_2'A_1'$, and $A_2'B_1'$ concur,
where for a point $P$ on a side of triangle $ABC$,
$P'$ denotes its reflection over that side's midpoint.
\begin{claim*}
  We have $\det M = \det N$.
\end{claim*}
\begin{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:
  \begin{align*}
    -(1-b_1)(1-c_2)(1-c_1)(1-a_2)(1-a_1)(1-b_2)
      &= -\prod_{i=1}^2\big((1-a_i)(1-b_i)(1-c_i)\big)\\
    (1-a_1)b_2(1-b_1)c_2(1-c_1)a_2
      &= \big((1-a_1)(1-b_1)(1-c_1)\big)\big(a_2b_2c_2\big)\\
    c_1(1-a_2)a_1(1-b_2)b_1(1-c_2)
      &= \big((1-a_2)(1-b_2)(1-c_2)\big)\big(a_1b_1c_1\big).
  \end{align*}
  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.
\end{proof}

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, $1-2c_2-2a_1+3c_2a_1=0$ and $1-2a_2-2b_1+3a_2b_1=0$.

Now,
\begin{align*}
  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}.
\end{align*}
So it follows $\det N = 0$, as desired.