The only possible value of the common perimeter, denoted $p$, is $1$. 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 $\triangle ABC$. [asy][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, invisible, black); filldraw(B--C2--A1--cycle, invisible, black); filldraw(C--A2--B1--cycle, invisible, black); 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)); [/asy][/asy] Hence the result follows by Brianchon's theorem. $\blacksquare$ Now suppose $p \neq 1$. Let $\overline{B_2'C_1'}$ be the dilation of $\overline{B_2C_1}$ with ratio $\tfrac1p$ at center $A$, and define $C_2'$, $A_1'$, $A_2'$, $B_1'$ similarly. The following diagram showcases the situation $p < 1$. [asy][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)); [/asy][/asy] By the reasoning in the $p = 1$ case, note that $\overline{B_1'C_2'}$, $\overline{C_1'A_2'}$, and $\overline{A_1'B_2'}$ are concurrent. However, $\overline{B_1C_2}$, $\overline{C_1A_2}$, $\overline{A_1B_2}$ 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 $1/3$).

The answer is 1 only. Let \(\omega\) denote the incircle of \(\triangle ABC\). In fact, for any choice of triangles \(AB_2C_1\), \(BC_2A_2\), \(CA_2B_1\) with perimeter 1, we note that the tangent from \(A\) to the \(A\)-excircle of \(\triangle AB_2C_1\) has length equal to the semiperimeter \(\frac12\), implying the \(A\)-excircle of \(\triangle AB_2C_1\) coincides with \(\omega\). Then hexagon \(A_1A_2B_1B_2C_1C_2\) has an incircle \(\omega\), implying the concurrence by Brianchon. [asy][asy] size(7cm); defaultpen(fontsize(10pt)); pair D,EE,F,A,B,C,X,Y,Z,A1p,A2p,B1p,B2p,C1p,C2p,A1,A2,B1,B2,C1,C2,X; real p=.6; D=dir(270); EE=dir(30); F=dir(150); A=2/(1/EE+1/F); B=2/(1/F+1/D); C=2/(1/D+1/EE); X=dir(95); Y=dir(195); Z=dir(330); A1p=2/(1/D+1/Y); A2p=2/(1/D+1/Z); B1p=2/(1/EE+1/Z); B2p=2/(1/EE+1/X); C1p=2/(1/F+1/X); C2p=2/(1/F+1/Y); A1=B+.6*(A1p-B); A2=C+.6*(A2p-C); B1=C+.6*(B1p-C); B2=A+.6*(B2p-A); C1=A+.6*(C1p-A); C2=B+.6*(C2p-B); X=extension(A1p,B2p,C1p,A2p); draw(B1p--C2p,gray+dashed); draw(C1p--A2p,gray+dashed); draw(A1p--B2p,gray+dashed); draw(B1--C2,dashed); draw(C1--A2,dashed); draw(A1--B2,dashed); draw(A2--B1); draw(B2--C1); draw(C2--A1); draw(A2p--B1p,gray); draw(B2p--C1p,gray); draw(C2p--A1p,gray); draw(unitcircle); draw(A--B--C--cycle); dot("\(A\)",A,N); dot("\(B\)",B,SW); dot("\(C\)",C,SE); dot("\(A_1\)",A1,S); dot("\(A_1'\)",A1p,S); dot("\(A_2\)",A2,S); dot("\(A_2'\)",A2p,S); dot("\(B_1\)",B1,EE); dot("\(B_1'\)",B1p,EE); dot("\(B_2\)",B2,EE); dot("\(B_2'\)",B2p,EE); dot("\(C_1\)",C1,F); dot("\(C_1'\)",C1p,F); dot("\(C_2\)",C2,F); dot("\(C_2'\)",C2p,F); [/asy][/asy] Otherwise, let \(p\ne1\) denote the common perimeter (for contradiction) and consider \(B_2'\in\overline{AC}\) and \(C_1'\in\overline{AB}\) with \(\overline{B_2C_1}\parallel\overline{B_2'C_1'}\) and \(B_2C_1/B_2'C_1'=p\), so the perimeter of \(\triangle AB_2'C_1'\) is 1. Construct \(\triangle BC_2'A_1'\) and \(\triangle CA_2'B_1'\) analogously. By the above, \(\overline{B_1'C_2'}\), \(\overline{C_1'A_2'}\), \(\overline{A_1'B_2'}\) concur at a point \(X\). If \(p<1\), then \(\overline{B_1C_2}\) is contained in the interior or boundary of quadrilateral \(BCB_1C_2\), and symmetrically \(\overline{C_1A_2}\) and \(\overline{A_1B_2}\) are contained in \(CAC_1A_2\) and \(ABA_1B_2\), The only common point to these three quadrilaterals is \(X\), which cannot be the concurrence point, contradiction. If \(p>1\), we apply an analogous argument with \(\overline{B_1C_2}\) contained in \(\triangle AB_1'C_2'\) and so on.

@\ crazyeyemoody907 wrote: 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 $AB_2C_1$, $BC_2A_2$, and $CA_2B_1$ are all equal. Find all possible values of this common perimeter. Ankan Bhattacharya perhaps it should say "triangles $AB_2C_1, BC_2,A_1, CA_2B_1$" instead of $BC_2A_2$?

NoctNight wrote: crazyeyemoody907 wrote: 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 $AB_2C_1$, $BC_2A_2$, and $CA_2B_1$ are all equal. Find all possible values of this common perimeter. Ankan Bhattacharya perhaps it should say "triangles $AB_2C_1, BC_2,A_1, CA_2B_1$" instead of $BC_2A_2$?

Similar to many above, though approached slightly differently. Answer: $\boxed{1}$. To show $1$ is attainable, take $A_1,A_2,B_1,B_2,C_1,C_2$ so that they trisect each side - then the perimeters are $\frac 13\times 3=1$ and $B_1C_2, C_1A_2, A_1B_2$ concur at the centre of $\triangle ABC$. The main proof proceeds as follows. Key idea: Let $\omega_A,\omega_B,\omega_C$ be the excircles of triangles $AB_2C_1, BC_2A_1, CA_2B_1$. Let $AB_2, AC_1, B_2C_1$ be tangent to $\omega_A$ at $T_A, S_A,U_A$ respectively and define $T_B,S_B,U_B$ and $T_C,S_C,U_C$ analogously. 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 $\omega_A,\omega_B,\omega_C$ are congruent. Assume $p\neq 1$. Define $B_2',C_1'$ on sides $CA,AB$ such that $B_2'C_1'$ is tangent to the incircle $\omega$ of $\triangle ABC$ and $B_2'C_1'\parallel B_2C_1$. Define $C_2', A_1'$ and $A_2', B_1'$ analogously. Consider the dilation of factor $\frac 1p$ centred at $A$. This sends: $\omega_A\mapsto\omega$ $B_2\mapsto B_2'$ $C_1\mapsto C_2'$ and perform the same dilation at $B$ and $C$. Then we have $$\frac{AC_1}{AC_1'}=p=\frac{CA_2}{CA_2'}$$meaning that $C_1A_2\parallel C_1'A_2'$. We obtain similar parallel lines. Let $P$ be the concurrency point of $B_1'C_2', C_1'A_2', A_1'B_2'$ by Brianchon's theorem. Let $X=A_2C_1\cap B_2A_1, Y=B_2A_1\cap C_2B_1$ and $Z=C_2B_1\cap A_2C_1$. 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 $\triangle PA_2'B_1'$ so $X\neq Y\neq Z$, a contradiction. If $p>1$ then we apply the same argument, except $X,Y,Z$ each lie inside the other three triangles of the hexagon.

Let $p$ be the value of the common perimeter. We have the following main Claim. Claim: $p=1$ works. Proof: Indeed, let $I_A$ be the $A-$ excenter of triangle $AC_1B_2$. Then, if $P$ is the projection of $I_A$ on $AB$, then $2AP=AC_2+AB_1+C_2B_1=1,$ hence $AP=1/2$, and so $P$ is the midpoint of $AB$, implying that $I_A$ coincides with the center of triangle $ABC$. Hence, the incircle of $ABC$ is the $A-$ excircle of $AC_1B_2$. Similarly, it is the $B-$ excircle of $BC_2A_1$ and the $C-$ excircle of $CA_2B_1$. Hence, the hexagon $A_1A_2B_1B_2C_1C_2$ is circumscribed around the incircle of triangle $ABC$. Therefore, by Brianchon's theorem, $A_2C_1,A_1B_2$ and $B_1C_2$ concur $\blacksquare$ Back to the problem, assume a $p \neq 1$ worked. We have two cases. Case 1: $p>1$. Then, take points $A_1',A_2',B_1',B_2',C_1',C_2'$ in the interior of $BA_1,CA_2,CB_1,AB_2,AC_1,BC_2'$ such that $A_1'C_2' \parallel A_1C_2, B_1A_2 \parallel B_1'A_2'$ and $C_1B_2 \parallel C_1'B_2',$ and the perimeter of triangles $AC_1'B_2',BC_2'A_1',CB_1'A_2'$ is equal to $1$. By the above Claim's arguments, lines $A_1'B_2',B_1',C_2',C_1'A_2'$ concur. Let $T$ be their common point, and let $P$ be the common point of $A_1B_2,B_1C_2,C_1A_2$. Then, $T$ cannot belong in the interior of triangles $BC_1A_2,CB_2A_1,$ hence $T$ belongs in the interior of quadrilateral $AC_1PB_2$. Similarly, $T$ belongs in the interior of quadrilaterals $BC_1PA_1$ and $CB_2PA_2,$ which is a contradiction. Case 2: $p<1$. We obtain a contradiction in a similar manner. Hence, the only possible answer is $p=1$, which is achieved for all points $A_1,A_2,B_1,B_2,C_1,C_2$ such that the perimeter of the coressponding triangles is $1$ (for example, if $A_i,B_i,C_i$ trisect $BC,AC,AB$ respectively for $i \in \{1,2 \}$).

yay another application of brianchon!! The answer is $p=1$ only. Claim: $p=1$ implies concurrency. Proof. Consider the $A$-excircle of $\triangle AC_1B_2.$ From standard properties, the tangent from $A$ to this circle is half the semiperimeter of $\triangle AC_1B_2,$ which is $\tfrac{1}{2}.$ In similar fashion we find this holds for $\triangle BA_1C_2$ and $\triangle CB_1A_2$ as well. Then these excircles coincide at the incircle of $\triangle ABC.$ Apply Brianchon on $A_1A_2B_1B_2C_1C_2$ to find its principle diagonals concur. Claim: $p \neq 1$ fail. Proof. Consider $p < 1.$ Take a homothety of $C_1B_2$ with ratio $\tfrac{1}{p}$ centered at $A,$ and similarly do so with $A_1C_2$ and $B_1A_2$ with their respective vertices. Identify these new points with $C_1'B_2',$ and analogously for the other two segments. We know $C_1'A_2', B_1'C_2', A_1'B_2'$ concur from the previous claim. Observe that $B_1C_2$ is contained in $BCB_1'C_2',$ and similarly for the other two vertices. But these quadrilaterals share no points, contradiction. For $p>1,$ a similar argument works by observing $C_1A_2, B_1C_2, A_1B_2$ are contained in $BC_1'A_2',AB_1'C_2', CA_1'B_2',$ but these triangles share no points, contradiction again.

Goofy. We claim the answer is $p=1$ only. Claim: If $AB_2A_1$, $BC_2A_1$, and $CA_2B_1$ all have perimiter $1$ then $B_1C_2$, $C_1A_2$, and $A_1B_2$ concur. First, we will show hexagon $A_1A_2B_1B_2C_1C_2$ has an incircle, which is sufficient. Consider the excircle of $\Delta_A := AB_2C_1$. Since the semiperimeter of $\Delta$ is $1/2$ then the distance from $A$ to excircle touchpoints of $AB_2$ and $AC_1$ is exactly $1/2$. Note that this implies that the excircles of $\Delta_A$ and $\Delta_B$, $\Delta_C$ (defined symmetrically) are all equal. Thus, implying the result. $\square$ Now, assume $p \not = 1$ has the concurrency property. Take a homothety at each vertex taking each triangle to one with perimeter $1$, the concurrency condition now holds so can't hold for any $p \not = 1$ as desired.