Nice problem!

Yay, my problem again ^_^

Really cool problem!

In general, if hexagon $ABCDEF$ has $\angle A = \angle C = \angle E$ and $\angle B = \angle D = \angle F$, then its sides can be extended to form two equilateral triangles $PQR$ and $XYZ$, as shown. [asy][asy] size(5cm); pair P = dir(90); pair Q = dir(210); pair R = dir(330); pair O = (0.1,0.2); pair z = dir(-40); pair Y = O+z*(P-O); pair X = O+z*(R-O); pair Z = O+z*(Q-O); filldraw(P--Q--R--cycle, invisible, blue); filldraw(X--Y--Z--cycle, invisible, deepcyan); pair A = extension(Q, P, Y, Z); pair B = extension(P, R, Y, Z); pair C = extension(P, R, X, Y); pair D = extension(R, Q, X, Y); pair E = extension(R, Q, X, Z); pair F = extension(Q, P, X, Z); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$R$", R, dir(R)); dot("$Y$", Y, dir(Y)); dot("$X$", X, dir(X)); dot("$Z$", Z, dir(Z)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); [/asy][/asy] The problem is solved upon proving the following claim. Claim: The angle bisectors of $\angle A$, $\angle C$, $\angle E$ are concurrent if and only if the unique spiral similarity sending $PQR$ to $YZX$ is a rotation; equivalently, the two triangles are congruent. [asy][asy]size(10cm); pair P = dir(90); pair Q = dir(210); pair R = dir(330); pair O = (0.1,0.2); pair z = dir(-40); pair Y = O+z*(P-O); pair X = O+z*(R-O); pair Z = O+z*(Q-O); filldraw(P--Q--R--cycle, invisible, blue); filldraw(X--Y--Z--cycle, invisible, deepcyan); pair A = extension(Q, P, Y, Z); pair B = extension(P, R, Y, Z); pair C = extension(P, R, X, Y); pair D = extension(R, Q, X, Y); pair E = extension(R, Q, X, Z); pair F = extension(Q, P, X, Z); draw(circumcircle(O, A, C), deepgreen); draw(circumcircle(O, C, E), deepgreen); draw(circumcircle(O, E, A), deepgreen); draw(A--O, red); draw(C--O, red); draw(E--O, red); label("$2\theta$", A, 5*dir(40), orange); label("$2\theta$", C, 5*dir(280), orange); label("$2\theta$", E, 5*dir(160), orange); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$R$", R, dir(R)); dot("$O$", O, dir(O)); dot("$Y$", Y, dir(Y)); dot("$X$", X, dir(X)); dot("$Z$", Z, dir(Z)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); /* TSQ Source: P = dir 90 Q = dir 210 R = dir 330 O = (0.1,0.2) z := dir -40 Y = O+z*(P-O) X = O+z*(R-O) Z = O+z*(Q-O) P--Q--R--cycle 0.1 lightblue / blue X--Y--Z--cycle 0.1 lightcyan / deepcyan A = extension Q P Y Z B = extension P R Y Z C = extension P R X Y D = extension R Q X Y E = extension R Q X Z F = extension Q P X Z circumcircle O A C deepgreen circumcircle O C E deepgreen circumcircle O E A deepgreen A--O red C--O red E--O red !label("$2\theta$", A, 5*dir(40), orange); !label("$2\theta$", C, 5*dir(280), orange); !label("$2\theta$", E, 5*dir(160), orange); */ [/asy][/asy] Proof. [Proof of ``if'' direction] Let $O$ be the center of the rotation. Then $O$ is equidistant form $\overline{YZ}$ and $\overline{PQ}$ by rotation; similarly for the other pairs of sides. So the bisectors meet at $O$. $\blacksquare$ Proof. [Proof of ``only if'' direction] Let $O$ be the concurrency point. Let $\angle PAB = \angle RCD = \angle QEF = 2\theta$ Since $\angle APC = \angle AYC = 60^{\circ}$ the quadrilateral $PACY$ is cyclic. But $\angle PAO + \angle PCO = (90^{\circ}+\theta) + (90^{\circ}-\theta) = 180^{\circ}$, so $PAOCY$ is cyclic. Now $\angle PAO = \angle YCO \implies OP = OY$, and $\angle POY = 2\theta$. So $O$ is the center of rotation from $\triangle PQR$ to $\triangle YXZ$ with angle $2\theta$. $\blacksquare$

Cool Geo problem with circles

You don't usually meet Fruit of Life in a geometry problem.

Wow. This problem was pretty easy. My solution is basically the same as CantonMathGuy's; except I stole Evan's diagram and I sorta proved everything here: Once again, we use Evan's diagram I have decided to steal: [asy][asy] size(5cm); pair P = dir(90); pair Q = dir(210); pair R = dir(330); pair O = (0.1,0.2); pair z = dir(-40); pair Y = O+z*(P-O); pair X = O+z*(R-O); pair Z = O+z*(Q-O); filldraw(P--Q--R--cycle, invisible, blue); filldraw(X--Y--Z--cycle, invisible, deepcyan); pair A = extension(Q, P, Y, Z); pair B = extension(P, R, Y, Z); pair C = extension(P, R, X, Y); pair D = extension(R, Q, X, Y); pair E = extension(R, Q, X, Z); pair F = extension(Q, P, X, Z); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$R$", R, dir(R)); dot("$Y$", Y, dir(Y)); dot("$X$", X, dir(X)); dot("$Z$", Z, dir(Z)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); pair J=extension(A,bisectorpoint(F,A,B),C,bisectorpoint(B,C,D)); DPA(A--J--C^^J--E,red); [/asy][/asy] Consider $P=AF\cap BC,Q=AF\cap DE,R=BC,\cap DE,X=EF\cap CD,Y=AB\cap CD,Z=AB\cap EF$. We note that \[\angle A+\angle B=\dfrac{3\angle A+3\angle B}{3}=\dfrac{\angle A+\angle B+\angle C+\angle D+\angle E+\angle F}{3}=120^{\circ}\]so $\angle APB=60^{\circ}$. By symmetry, \[\angle P=\angle Q=\angle R=\angle X=\angle Y=\angle Z=60^{\circ}\]so thus $\triangle PQR$ and $\triangle XYZ$ are equilateral. Now, we have the following claim: Claim. The angle bisectors of $A,C,E$ concur if and only if $\triangle XYZ$ and $\triangle PQR$ are congruent. Proof. We have that there exists a point $J$ in the interior that lies on the angle bisectors of $A,C,E$ if and only if they concur. Now, if we let $d(G,\ell)$ be the distance from any arbitrary point $G$ to $\ell$. Then, we get that \[d(J,DE)=d(J,EF)\]\[d(J,AF)=d(J,AB)\]\[d(J,BC)=d(J,CD)\]so thus we get that \[d(J,BC)+d(J,DE)+d(J,AF)=d(J,EF)+d(J,AB)+d(J,CD)\]However, we have the following important lemma: Lemma (Vivani's Theorem). In equilateral triangle $TUV$ and arbitrary point $W$ in the interior, we have \[d(W,UV)+d(W,UT)+d(W,TV)=d(T,UV)\]Proof. We have that \[[WUV]=\dfrac 12 (UV)d(W,UV)\]\[[WTV]=\dfrac 12 (TV)d(W,TV)\]\[[WTU]=\dfrac 12 (UT)d(W,TU)\]by the formula half times the base times the height, so thus adding, we get \[\dfrac 12(UV)d(T,UV)=[TUV]=[WUV]+[WTV]+[WTU]=\dfrac 12 (UV)(d(W,UV)+d(W,TV)+d(W,TU))\]where we used $UV=TU=TV$. Dividing by $UV$, we get the desired result. $\square$ Now, to finish off, we notice that \[d(J,BC)+d(J,DE)+d(J,AF)=d(J,PR)+d(J,QR)+d(J,PQ)=d(J,PQ)\]\[d(J,EF)+d(J,AB)+d(J,CD)=d(J,XZ)+d(J,ZY)+d(J,YX)=d(X,YZ)\]Thus, we get $d(X,YZ)=d(J,PQ)$, so thus we get that this means that $PQ=YZ$. Thus, we get $ \triangle PQR\cong\triangle XYZ$ if and only if the angle bisectors of $A,C,E$ concur. $\blacksquare$ Now, by symmetry, we conclude the angle bisectors of $A,C,E$ concur if and only if $\triangle PQR\cong\triangle XYZ$, which happens if and only if the angle bisectors of $B,D,F$ concur. This completes the proof.

Shameless mOvInG-pOiNtS Relabel the hexagon to $AFBDCE$. Note that the angle bisectors of the two consecutive angles make the angle $60^{\circ}$ so the angle bisectors of $\angle A, \angle B, \angle C$ meet at the Fermat's point of $\triangle ABC$. Moreover, this implies that the angle bisector of $\angle A$ and $\angle D$ and parallel. We restate the problem as follows. Quote: Let $\triangle ABC$ be a triangle with Fermat's point $P$. Construct points $D,E,F$ such that $$\measuredangle EAP = \measuredangle PAF = \measuredangle FBP = \measuredangle PBD = \measuredangle DCP = \measuredangle PCE=\theta$$then lines through $D,E,F$ parallel to $AP, BP, CP$ are concurrent. Now we use moving points. Note that $D$ has degree two as it is induced by projective pencils from $B,C$ thus move $D$ projectively. Since reflections across $BP, CP$ are projective, we get $E,F$ have degree $1$. Notice that $D=\infty_{AP}$ when $\angle PBD = \angle PCD = 120^{\circ}$. Thus line $D{\infty}_{AP}$ has degree $1$. Similarly lines $E{\infty}_{BP}$ and $F{\infty}_{CP}$ have degree $1$. The concurrency is degree $3$ hence it suffices to check four cases. When $\theta = 0$, we have $D=E=F=P$ so trivial. When $\theta = 90^{\circ}$, we have $\triangle DEF$ equilateral and the lines are concurrent at its center. When $\theta = 60^{\circ}$, we get $BPCD$ parallelogram. It's easy to see that $\triangle ABC$ and $\triangle DEF$ are homothetic. When $\theta = 30^{\circ}$, we get $D=\infty_{\perp AP}$ so all three lines are line at infinity.

i think about this 15 mineutes my first idea not work after 10 min i found that if we extend vertexes we have equilateral and then $PAOCY$ and... are cyclic (see v-enhance figure )and it solve problem but before it i tried to solve it with property ex-center and incenter . anyone have solution with this?

I think it's possible to solve using Jacobi theorem as it implies that $AD,BE,CF$ concur.

godly

[asy][asy] size(9cm); pair A=(15.178566974301635,11.828534760568738); pair B=(4.996868391022167,-8.09850389527818); pair C=(27.360242065011,-7.0076076184982385); pair D=(4.705962717214182,3.428633429363195); pair E=(25.869350486745077,10.628548856110804); pair M=(19.033067152257434,-2.025847954536509); pair F=(21.542128588851302,-11.407555934844); draw(A--B--C--A,purple); draw(D--E--F--D,orange); pair X2=extension(A,C,D,E); pair X1=extension(A,B,D,E); pair Y1=extension(A,B,D,F); pair Y2=extension(B,C,D,F); pair Z1=extension(B,C,F,E); pair Z2=extension(A,C,E,F); draw(circumcircle(D,B,M),green); draw(circumcircle(A,E,M),green); draw(circumcircle(C,F,M),green); draw(M--B,red+dotted); draw(M--D,red+dotted); draw(M--A,red+dotted); draw(M--E,red+dotted); draw(M--C,red+dotted); draw(M--F,red+dotted); dot("$A$",A,N); dot("$B$",B,SW); dot("$C$",C,SE); dot("$M$",M,SE); dot("$D$",D,NW); dot("$E$",E,NE); dot("$F$",F,S); dot("$X_1$",X2,NE); dot("$X_2$",X1,NW); dot("$Y_1$",Y1,W); dot("$Y_2$",Y2,SW); dot("$Z_2$",Z2,W); dot("$Z_1$",Z1,SE); [/asy][/asy] Nice problem though easy. Alter the labelling (refer diagram). Extend the sides to form $2$ triangles. We will claim that $\triangle{ABC}$ and $\triangle{DEF}$ are equilateral. Claim 1: Triangle $\triangle{ABC}$ and $\triangle{DEF}$ are equilateral. Proof: We will only prove that $\triangle{ABC}$ is equilateral since a similiar arguement can be used for $\triangle{DEF}$. Notice that $$\angle BAC=180^\circ-(360^\circ-\angle X_2+\angle X_1)=\angle X_2+\angle X_1-180^\circ=\angle Z_2+\angle Z_1-180^\circ=\angle ACB$$and applying this arguement cyclically we are done $\qquad \blacksquare$ We begin with a crucial claim. Claim 2: The angle bisectors of $\angle X_2,\angle Z_2,\angle Y_2$ are concurrent iff $\triangle{ABC}$ is congruent to $\triangle{DEF}$. Proof: For the first direction notice that if the triangles are concurrent then there obviously exists a rotation $\phi$ such that $\phi : \triangle{ABC} \mapsto \triangle{DEF}$. Let centre of rotation be $M$. Then notice that clearly distance of $M$ to $\overline{AB},\overline{DE}$ are equal and similiarly for the other sides. This implies the result. Now we state a subclaim for the second part. Subclaim: $(X_2DBY_2M)$ is cyclic and similiarly for others. Proof: Notice that $$\angle Y_2BX_2=60^\circ=\angle X_2DY_2$$hence $(X_2DBY_2)$ is cyclic. Now notice that $$\angle MX_2B+\angle MY_2B=90^\circ-\frac{\angle AX_2X_1}{2}+\frac{\angle Y_1Y_2B}{2}+90^\circ=180^\circ$$so we are done $\qquad \square$ Now for the other direction. Assume that the bisectors are concurrent. Then notice that there would exist a spiral similiarity at $M$ such that $\overline{BD} \mapsto \overline{AE}$. Now notice that $$\angle MDB=\angle MX_2B=\angle MX_2E=\angle MAE$$but by the spiral similiarity we have that $\angle MDB=\angle MEA$ hence we can conclude that $\overline{MA}=\overline{ME}$ and similiarly for the others. Now just taking a rotation centred at $M$ proves our claim $\qquad \blacksquare$ With this claim notice that we can conclude that $\triangle{ABC}$ and $\triangle{DEF}$ are congruent and so by symmetry we can conclude that the angle bisectors of $\angle X_1,\angle Y_1,\angle Z_1$ are concurrent $\qquad \blacksquare$

Posting for storage; solved with TheUltimate123 and eisirrational. Extend sides $AB$, $CD$, and $EF$ to form equilateral $\triangle XYZ$, as $\angle A + \angle B = \angle C + \angle D = \angle E + \angle F = 240^\circ$. Similarly, extend sides $BC$, $DE$, and $FA$ to form equilateral $\triangle PQR$. [asy][asy] import olympiad; defaultpen(fontsize(11pt)); size(5cm); pair U = (-0.2,-0.1); path tri = dir(90)--dir(210)--dir(330)--cycle; path fri = rotate(36,U)*tri; pair A = intersectionpoints(tri,fri)[1]; pair B = intersectionpoints(tri,fri)[2]; pair C = intersectionpoints(tri,fri)[3]; pair D = intersectionpoints(tri,fri)[4]; pair E = intersectionpoints(tri,fri)[5]; pair F = intersectionpoints(tri,fri)[0]; pair P = dir(210); pair Q = dir(330); pair R = dir(90); pair X = rotate(36,(-0.2,-0.1))*P; pair Y = rotate(36,(-0.2,-0.1))*Q; pair Z = rotate(36,(-0.2,-0.1))*R; filldraw(tri,red+white+white+white+white,dashed+red); filldraw(fri,blue+white+white+white+white,dashed+blue); filldraw(A--B--C--D--E--F--cycle, red+blue+white+white+white); draw(A--B,blue); draw(C--D,blue); draw(E--F,blue); draw(B--C,red); draw(D--E,red); draw(F--A,red); draw(circumcircle(U,A,E),red+blue); draw(circumcircle(U,A,C),red+blue); draw(circumcircle(U,E,C),red+blue); dot("$F$",F,dir(90)); dot("$A$",A,dir(180)); dot("$B$",B,dir(225)); dot("$C$",C,dir(315)); dot("$D$",D,dir(0)); dot("$E$",E,dir(45)); dot("$P$",P,dir(135)); dot("$Q$",Q,dir(45)); dot("$R$",R,dir(45)); dot("$X$",X,dir(225)); dot("$Y$",Y,dir(45)); dot("$Z$",Z,dir(135)); dot("$U$",U,dir(45)); [/asy][/asy] Evidently we have the cyclic quadrilaterals $ACXP$, $CEYQ$, and $EAZR$. By Miquel's theorem, the three circumcircles concur at a point $U$, which by definition is the spiral center sending $\triangle XYZ$ to $\triangle PQR$. The pith of the problem is the following claim. Claim. The angle bisectors of $\angle A$, $\angle C$, and $\angle E$ concur at $U$ if and only if $\triangle PQR \cong \triangle XYZ$. Proof. Note that \[\overline{AU} \text{ bisects } \angle XAF \iff UP = UX \iff \triangle PQR \cong \triangle XYZ,\]where the second result implies the spiral similarity is actually a rotation. Similarly, the angle bisectors of $\angle B$, $\angle D$, and $\angle F$ concur at the spiral center $V$ sending $\angle XYZ$ to $\triangle RQP$ if and only if $\triangle PQR \cong \triangle XYZ$. Thus, the concurrencies are both equivalent to the congruence. Hooray!

Apparently no one posted this yet so... Let $K$ be the point of intersection from the statement and let $FA\cap BC=X$, $BC\cap DE=Y$, $DE\cap FA=Z$, $AB\cap CD=U$, $CD\cap EF=V$, $EF\cap AB=W$. By angle condition, $XYZ$ and $UVW$ are equilateral, and since there is a rotation centered at $K$ taking $FA\rightarrow AB$, $BC\rightarrow CD$, $DE\rightarrow EF$, they're in fact congruent. Let $L$ be the center of rotation taking $X\rightarrow U$, $Y\rightarrow V$, $Z\rightarrow W$. We see that $L$ is equidistant from pairs of lines $(AB,BC)$, $(CD,DE)$, $(EF,FA)$, so $L$ is the desired intersection point.

Let's construct the problem in a different but equivalent way. Take a point $O$ and points $A, C, E$ such that $\angle AOC = \angle AOE = \angle COE = 120^\circ$. Say $AO = x, CO = y, EO = z$. Choose $B, D, F$ such that $\angle BAO = \angle BCO = \angle DCO = \angle DEO = \angle FEO = \angle FAO$. Let angle bisectors of $\angle D, \angle F$ intersect at point $P$ Claim: While $E$ is moving on line $EO$ with constant speed, $P$ moves on a certain line. Proof: It is clear that angle bisectors of $D$ and $F$ are now lines that move with constant speed. Therefore, their intersection also moves on a certain line. Now, choose two spesific examples for $z$, namely $z = x$ and $z = y$. It can be easily observed that $P$ lies on angle bisectors of $\angle B$ in both cases, so this certain line that $P$ is moving on is angle bisector of $\angle B$.

Here is another solution which is not using any rotation argument. Let $AB \cap CD = X$, $CD\cap EF = Y$ and $EF\cap AB = Z$. Since $\angle A = \angle C = \angle E$ and $\angle B = \angle D = \angle F$ triangle $XYZ$ is equilateral. Let $O_1, O_2, O_3$ be excircles of triangles $XBC$, $YDE$, $ZFA$ with respect to sides $BC$, $DE$ , $FA$, respectively. Obviously $XO_1, YO_2, ZO_3$ are all passing through incenter $I$ of triangle $XYZ$. If angle bisectors of $\angle A$, $\angle C$, $\angle E$ intersect at $T$, then we get $$\angle IO_1T = \angle TCD - \angle O_1XC = \frac{\angle C }{2} - 30^{\circ}$$By writing the same equation for $\angle IO_2T$ and $\angle IO_3T$ and using the fact that $\angle A = \angle C = \angle E$, we see that points $I, T, O_1, O_2, O_3$ all lie on the same circle $\Omega$. Let angle bisectors $B$ and $D$ intersect at $S$. Then using the same idea as above we get $$\angle IO_1S = \frac{\angle B }{2} - 30^{\circ} = \frac{\angle D }{2} - 30^{\circ} = \angle IO_2S$$which means $S$ also lie on the same circle. So, angle bisector of $\angle D$ passes through the second intersection point (or touching point if $S=O_1$) of the line $BO_1$ with circle $\Omega$. Similarly, angle bisector of $\angle F$ also passes through this point. Thus angle bisectors of $\angle B$, $\angle D$ and $\angle F$ concur at a point lying on $\Omega$.

v_Enhance wrote: In general, if hexagon $ABCDEF$ has $\angle A = \angle C = \angle E$ and $\angle B = \angle D = \angle F$, then its sides can be extended to form two equilateral triangles $PQR$ and $XYZ$, as shown. [asy][asy] size(5cm); pair P = dir(90); pair Q = dir(210); pair R = dir(330); pair O = (0.1,0.2); pair z = dir(-40); pair Y = O+z*(P-O); pair X = O+z*(R-O); pair Z = O+z*(Q-O); filldraw(P--Q--R--cycle, invisible, blue); filldraw(X--Y--Z--cycle, invisible, deepcyan); pair A = extension(Q, P, Y, Z); pair B = extension(P, R, Y, Z); pair C = extension(P, R, X, Y); pair D = extension(R, Q, X, Y); pair E = extension(R, Q, X, Z); pair F = extension(Q, P, X, Z); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$R$", R, dir(R)); dot("$Y$", Y, dir(Y)); dot("$X$", X, dir(X)); dot("$Z$", Z, dir(Z)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); [/asy][/asy] The problem is solved upon proving the following claim. Claim: The angle bisectors of $\angle A$, $\angle C$, $\angle E$ are concurrent if and only if the unique spiral similarity sending $PQR$ to $YZX$ is a rotation; equivalently, the two triangles are congruent. [asy][asy]size(10cm); pair P = dir(90); pair Q = dir(210); pair R = dir(330); pair O = (0.1,0.2); pair z = dir(-40); pair Y = O+z*(P-O); pair X = O+z*(R-O); pair Z = O+z*(Q-O); filldraw(P--Q--R--cycle, invisible, blue); filldraw(X--Y--Z--cycle, invisible, deepcyan); pair A = extension(Q, P, Y, Z); pair B = extension(P, R, Y, Z); pair C = extension(P, R, X, Y); pair D = extension(R, Q, X, Y); pair E = extension(R, Q, X, Z); pair F = extension(Q, P, X, Z); draw(circumcircle(O, A, C), deepgreen); draw(circumcircle(O, C, E), deepgreen); draw(circumcircle(O, E, A), deepgreen); draw(A--O, red); draw(C--O, red); draw(E--O, red); label("$2\theta$", A, 5*dir(40), orange); label("$2\theta$", C, 5*dir(280), orange); label("$2\theta$", E, 5*dir(160), orange); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$R$", R, dir(R)); dot("$O$", O, dir(O)); dot("$Y$", Y, dir(Y)); dot("$X$", X, dir(X)); dot("$Z$", Z, dir(Z)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); /* TSQ Source: P = dir 90 Q = dir 210 R = dir 330 O = (0.1,0.2) z := dir -40 Y = O+z*(P-O) X = O+z*(R-O) Z = O+z*(Q-O) P--Q--R--cycle 0.1 lightblue / blue X--Y--Z--cycle 0.1 lightcyan / deepcyan A = extension Q P Y Z B = extension P R Y Z C = extension P R X Y D = extension R Q X Y E = extension R Q X Z F = extension Q P X Z circumcircle O A C deepgreen circumcircle O C E deepgreen circumcircle O E A deepgreen A--O red C--O red E--O red !label("$2\theta$", A, 5*dir(40), orange); !label("$2\theta$", C, 5*dir(280), orange); !label("$2\theta$", E, 5*dir(160), orange); */ [/asy][/asy] Proof. [Proof of ``if'' direction] Let $O$ be the center of the rotation. Then $O$ is equidistant form $\overline{YZ}$ and $\overline{PQ}$ by rotation; similarly for the other pairs of sides. So the bisectors meet at $O$. $\blacksquare$ Proof. [Proof of ``only if'' direction] Let $O$ be the concurrency point. Let $\angle PAB = \angle RCD = \angle QEF = 2\theta$ Since $\angle APC = \angle AYC = 60^{\circ}$ the quadrilateral $PACY$ is cyclic. But $\angle PAO + \angle PCO = (90^{\circ}+\theta) + (90^{\circ}-\theta) = 180^{\circ}$, so $PAOCY$ is cyclic. Now $\angle PAO = \angle YCO \implies OP = OY$, and $\angle POY = 2\theta$. So $O$ is the center of rotation from $\triangle PQR$ to $\triangle YXZ$ with angle $2\theta$. $\blacksquare$ Why is d(O,YZ)=d(O,PQ)?

Because $O$ lies on the $A$-angle bisector.

Here is a different way to finish the proof, once we see that $XYZ$ and $PQR$ are equilateral and congruent, let $S$ be the center of rotation sending $PQR$ to $ZXY$ We have $\widehat{PSZ}=\widehat{QSX}$ and $PS=ZS$ ; $QS=XS$ So $PSZ$ and $QSX$ are similar which implies that $PSQ$ and $ZSX$ are similar and $F=(ZX)\cap (PQ)$ lies on the circles $(SPZ)$ and $(SQX)$ we also have $\widehat{PBA}=\widehat{PFZ}$, hence the points $S,B,F,P,Z$ are cyclic and so are the points $F,S,Q,X,D$ And as $SP=SZ$ we have $\widehat{ZBS}=180-\widehat{PBS}$ and hence $S$ lies on the angle bisector of $\widehat{B}$ which finishes the proof by symmetry

Let $X=BC\cap AF$, $Y=AF\cap DE$, $Z=DE\cap BC$, $U=CD\cap EF$, $V=AB\cap CD$, $W=EF\cap AB$. Then $\triangle XYZ,\triangle UVW$ are both equilateral. Let $P$ be the common intersection of the bisectors of $\angle A,\angle C,\angle E$. Angle chasing gives $P$ is a Miquel point of both $XYZ$ and $UVW$. Let $\measuredangle XAP = \measuredangle WAP = \alpha$, then by area $$(PA+PC+PE)\cos\alpha = \frac{\sqrt{3}}{2}UV = \frac{\sqrt{3}}{2}XY,$$so $UV=XY$. Let $Q$ be the common intersection of circles $(XBFW),(VBDZ),(UDFY)$ (exists by angle chasing). Then $Q$ is also a Miquel point of both $XYZ$ and $UVW$. Let $\measuredangle XBQ = \beta_1$, $\measuredangle VBQ = \beta_2$. Because $UV=XY$, $$(PB+PD+PF)\cos\beta_1 = \frac{\sqrt{3}}{2}XY = \frac{\sqrt{3}}{2}UV = (PB+PD+PF)\cos\beta_2,$$and since $\beta_1+\beta_2<\pi$, we know $\beta_1 = \beta_2$. Therefore $Q$ is the desired intersection.

Nice. Let $AB\cap CD=M,CD\cap EF=N,EF\cap AB=P$ and $AF\cap BC=X,BC\cap ED=Y,ED\cap AF=Z$. It is clear that $MNP$ and $XYZ$ are equilateral. Let $K$ be the intersection of the angular bisectors of $\angle A$ and $\angle C$. Then $K$ lies on $(MAC)$ because \[\angle AKC=360^\circ-\angle B-\angle A/2-\angle C/2=120^\circ=180^\circ-\angle AMC.\]Similarly, $K$ lies on $(XAC)$. Thus if $K$ is also on the angular bisector of $\angle F$, then it is the center of the spiral similarity mapping $PM\to ZX$, or the spiral sim mapping $\triangle MNP\to \triangle XYZ$. Additionally, remark that since $K$ lies on the exterior angle bisector of $\angle XCM$, then $KX=KM$. Thus the spiral similarity is a spiral congruence and $\triangle XYZ\cong\triangle MNP$. Moreover, if the two triangles are congruent then $K$ lies on the angle bisector of $\angle C$ etc so the lines concur iff $\triangle XYZ\cong\triangle MNP$. Similarly the angle bisectors of $\angle B,\angle D,\angle F$ concur iff $\triangle PMN\cong\triangle XYZ$, so done.

alifenix- wrote: Let $ABCDEF$ be a convex hexagon such that $\angle A = \angle C = \angle E$ and $\angle B = \angle D = \angle F$ and the (interior) angle bisectors of $\angle A, ~\angle C,$ and $\angle E$ are concurrent. Prove that the (interior) angle bisectors of $\angle B, ~\angle D, $ and $\angle F$ must also be concurrent. Note that $\angle A = \angle FAB$. The other interior angles of the hexagon are similarly described. Nice problem. Copying v_enhance's diagram for convenience [asy][asy]size(10cm); pair P = dir(90); pair Q = dir(210); pair R = dir(330); pair O = (0.1,0.2); pair z = dir(-40); pair Y = O+z*(P-O); pair X = O+z*(R-O); pair Z = O+z*(Q-O); filldraw(P--Q--R--cycle, invisible, blue); filldraw(X--Y--Z--cycle, invisible, deepcyan); pair A = extension(Q, P, Y, Z); pair B = extension(P, R, Y, Z); pair C = extension(P, R, X, Y); pair D = extension(R, Q, X, Y); pair E = extension(R, Q, X, Z); pair F = extension(Q, P, X, Z); draw(circumcircle(O, A, C), deepgreen); draw(circumcircle(O, C, E), deepgreen); draw(circumcircle(O, E, A), deepgreen); draw(A--O, red); draw(C--O, red); draw(E--O, red); label("$2\theta$", A, 5*dir(40), orange); label("$2\theta$", C, 5*dir(280), orange); label("$2\theta$", E, 5*dir(160), orange); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$R$", R, dir(R)); dot("$O$", O, dir(O)); dot("$Y$", Y, dir(Y)); dot("$X$", X, dir(X)); dot("$Z$", Z, dir(Z)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); /* TSQ Source: P = dir 90 Q = dir 210 R = dir 330 O = (0.1,0.2) z := dir -40 Y = O+z*(P-O) X = O+z*(R-O) Z = O+z*(Q-O) P--Q--R--cycle 0.1 lightblue / blue X--Y--Z--cycle 0.1 lightcyan / deepcyan A = extension Q P Y Z B = extension P R Y Z C = extension P R X Y D = extension R Q X Y E = extension R Q X Z F = extension Q P X Z circumcircle O A C deepgreen circumcircle O C E deepgreen circumcircle O E A deepgreen A--O red C--O red E--O red !label("$2\theta$", A, 5*dir(40), orange); !label("$2\theta$", C, 5*dir(280), orange); !label("$2\theta$", E, 5*dir(160), orange); */ [/asy][/asy] It can be seen that due to the conditions of the problem, the lines $\overline{AB}, \overline{CD}, \overline{EF}$ form an equilateral triangle $\triangle KLM$ and the lines $\overline{AF}, \overline{BC}, \overline{DE}$ form an equilateral triangle $\triangle XYZ$. Let angle bisectors of $\angle A, \angle C, \angle E$ intersect at a point $O$. We mainly utilize the fact that $d(O, PR) = d(O, XY), d(O, PQ) = d(O, YZ), d(O, QR) = d(O, XZ)$ We see that $$A(\triangle XYZ) = A(\triangle OYZ) + A(\triangle XOZ) + A(\triangle XYO) = \dfrac{XY \cdot (d(O, YZ) + d(O, XZ) + d(O, XY))}{2} = \dfrac{PQ \cdot (d(O, PQ) + d(O, QR) + d(O, PR))}{2} = A(\triangle OPQ) + A(\triangle QOR) + A(\triangle PRO) = A(\triangle PQR)$$and all these arguments mainly work only because $\triangle PQR$ and $\triangle XYZ$ are both equilateral triangles and therefore $\triangle XYZ \cong \triangle PQR$ Now, if the angle bisectors of $\angle B$ and $\angle D$ intersect at a point $K$, then $$A(\triangle XYZ) = A(\triangle KYZ) + A(\triangle XKZ) + A(\triangle XYK) = \dfrac{XZ \cdot (d(K, XY) + d(K, XZ) + d(K, YZ))}{2} = \dfrac{XZ \cdot (d(K, QR) + d(K, XZ) + d(K,PR))}{2} = \dfrac{QR \cdot (d(K, QR) + d(K, PQ) + d(K,PR))}{2}$$and we yield that $d(K, PQ) = d(K, XZ)$ which implies that $\overline{FK}$ bisects $\angle F$ which is the desired conclusion. After scrolling a bit, I see that this solution is same as CantonMathGuy solution but I am proud of this solution so posting here

Here is a solution that doesn't require much insight. We will consider the following version of the problem first, and later show that this implies the EGMO problem. Rephrased Version: Let $I$ be the point inside $\triangle ACE$ such that $\angle AIC=\angle CIE=\angle EIA$. Let $B,D$ and $F$ be points such that $\measuredangle FAI=\measuredangle IAB=\measuredangle BCI=\measuredangle ICD=\measuredangle DEI=\measuredangle IEF$. Let $l_b$ be the line through $B$ parallel to $IE$, and define $l_d$ and $l_f$ similarly. Then $l_b,l_d,l_f$ are concurrent. Suppose that the lengths of segments $IA$ and $IB$ are distinct, as otherwise the concurrency follows by symmetry. Let's move $C$ uniformly in the direction of ray $IC$. Then $B$ and $D$ also move uniformly along lines $AB$ and $ED$ respectively. This implies that $l_b$ and $l_d$ move uniformly too. Since $l_f$ is fixed, it suffices to check for two cases that the concurrency is true. (1) When $IC=IA$, the hexagon is symmetric about line $IE$, so the concurrency holds. (2) When $IC=IB$, the hexagon is symmetric about line $IA$, and the concurrency holds as well, so the proof is complete. $\blacksquare$ It is easy to see that this is equivalent to the original problem too. Suppose that the internal angle bisectors meet at $I$. Then by some angle chasing, we have $\angle AIC=\angle CIE=\angle EIA$. Moreover, $B$, $D$ and $F$ obviously satisfy the angle conditions. Now let $l_b$ be the angle bisector of $\angle B$ which meets $IC$ at $X$. Then \[2\angle BCI+2\angle XBC=240^\circ\Longrightarrow \angle BXI=120^\circ=\angle EIC,\]so it follows that $l_b$ is parallel to $IE$, and we're done.

Did no one come up with the solution involving Viviani's Theorem? That one is the most intuitive to me.

math31415926535 wrote: Did no one come up with the solution involving Viviani's Theorem? That one is the most intuitive to me. As in... post 2?

khina wrote: math31415926535 wrote: Did no one come up with the solution involving Viviani's Theorem? That one is the most intuitive to me. As in... post 2? o didn't see lol

We rewrite the polygon as $ADBECF$ where the angle bisectors of $\angle A, \angle B, \angle C$ concur. Notice that extending each side yields two equilateral triangles. Let them be $\triangle A_1B_1C_1$ and $\triangle A_2B_2C_2$ where $A$ lies between $A_1$ and $A_2$ and so on. First, we claim $\triangle A_1B_1C_1$ and $\triangle A_2B_2C_2$ are congruent. Indeed, If we let the point of concurrency be $P$ we have the sum of the lengths of the feet of the perpendicular to $A_1B_1$, $B_1C_1$, and $C_1A_1$ is equal to the sum of the feet to $A_2B_2$, $B_2C_2$, and $C_2A_2$ are equal since $P$ lies on three angle bisectors. But, those sums are the height of $\triangle A_1B_1C_1$ and $\triangle A_2B_2C_2$ which are equal. So, they are congruent as desired. Now, let $(DA_1A_2E),(EB_1B_2F),(FB_1B_2D)$ concur at $Q$ (they do this because Miquel). We claim $Q$ is the second concurrency point. Indeed, Notice that $A_2B_1=A_1B_2$. So, by spiral similarity $\triangle QA_1B_1$ and $\triangle QA_2B_2$ are congruent (the rest fallows cyclically). Thus $QA_1=QA_2$ etc... Finally, we have the following angle chase \[ \angle A_2DP = \angle PFC_2 = \angle C_2C_1P = \angle C_1C_2P = \angle PEC_1 = \angle PDB_1 \]as desired.

alifenix- wrote: Let $ABCDEF$ be a convex hexagon such that $\angle A = \angle C = \angle E$ and $\angle B = \angle D = \angle F$ and the (interior) angle bisectors of $\angle A, ~\angle C,$ and $\angle E$ are concurrent. Prove that the (interior) angle bisectors of $\angle B, ~\angle D, $ and $\angle F$ must also be concurrent. Note that $\angle A = \angle FAB$. The other interior angles of the hexagon are similarly described. Extraordinary problem! Solved with MathsCrazy Complete the hexagon to get triangles \(L_1M_1N_1\) and \(L_2M_2N_2\). It is trivial to notice both are equilateral, because \(\measuredangle M_1CD=180-\measuredangle C\) and \(\measuredangle M_1DC=180-\measuredangle D\). Therefore, \(\measuredangle M_1=180-(\measuredangle C+\measuredangle D)=180-120=60\), similarly for other angles and we're done. Now, we prove a claim. Claim 1. Let the \(B\) ex-bisector meet \(\odot(ABC)\) for a second time at \(P\). Then \(AP=PC\). Proof. Let \(X\) be a point on line \(AB\). Then, \[\measuredangle PAC=\measuredangle PBC=\measuredangle XBP=\measuredangle PCA\]as claimed. Also note that the converse holds true too. $\blacksquare$ We prove another claim which kills the problem. Claim 2. \(P,C,M_2,L_1,A\) are concyclic. Proof. As \(\measuredangle CM_2A=\measuredangle CL_1A=60\), \(CM_2L_1A\) is cyclic, now to prove that \(P\) lies on this circle, it suffices to prove that \(\measuredangle CPA=120\). But then, \(\measuredangle CPA=360-(\measuredangle CBA+\measuredangle PCB+\measuredangle PAB)=360-(\measuredangle ABC+\measuredangle BCD)=120 \), as claimed. $\blacksquare$ By claim 1 on triangle \(M_2L_1A\), we see that \(PL_1=PM_2\). Similarly, \(PM_1=PN_2\) and \(PN_1=PL_2\), implying that \(P\) is the center of rotation of triangles \(L_1M_1N_1\) and \(L_2M_2N_2\), implying that they are actually congruent. Since these steps are reversible, we see that \(L_1M_1N_1\) and \(L_2M_2N_2\) are congruent if and only if the angle bisectors of angles \(A,C,E\) concur. By symmetry, this also implies that the angle bisectors of \(B,D,F\) concur, hence proving the problem. $\blacksquare$

Let angle bisectors of $\angle A, ~\angle C,$ and $\angle E$ meet at $S$. Let $AB$ meet $CD$ and $EF$ at $X,Y$ and Let $EF$ and $CD$ meet at $Z$. Let $AF$ meet $BC$ and $DE$at $X'$ and $Y'$ and $ED$ and $BC$ meet at $Z'$. By some simple angle chasing $XYZ$ and $X'Y'Z'$ are equilateral so we only need to prove $S$ is the point sending $XYZ$ to $X'Y'Z'$. Claim $: X'ASCX$ is cyclic. Proof $:$ Note that $\angle AX'C = \angle AXC = \angle 60$ so $AX'XC$ is cyclic and $\angle ASC = \angle 360 - \angle B - \angle BAS - \angle BSC = \angle 360 - \angle 240 = \angle 120$ so $ASCXX'$ is cyclic. Note that we have the same approach for $YY'ESA$ is cyclic and $ZZ'CSE$ is cyclic so $\angle X'SX = \angle X'AX = \angle YAY' = \angle YSY' = \angle YEY' = \angle ZEZ' = \angle ZSZ'$ so $\angle X'SX = \angle Y'SY = \angle Z'SZ$ so $S$ is the point sending $XYZ$ to $X'Y'Z'$.

Great problem with a clean and memorable statement! [asy][asy] size(7cm); defaultpen(0.8); defaultpen(fontsize(9pt)); dotfactor*=0.6; pen redfill,orangefill,yellowfill,greenfill,turquoisefill,lightbluefill,bluefill,purplefill,pinkfill,grayfill,reddraw,orangedraw,yellowdraw,greendraw,turquoisedraw,lightbluedraw,bluedraw,purpledraw,pinkdraw,graydraw; redfill = RGB(255,204,204); orangefill = RGB(255,221,187); yellowfill = RGB(255,255,170); greenfill = RGB(204,255,204); turquoisefill = RGB(136,255,221); lightbluefill = RGB(187,255,255); bluefill = RGB(204,233,255); purplefill = RGB(238,204,255); pinkfill = RGB(255,204,255); grayfill = RGB(221,221,221); reddraw = RGB(255,85,85); orangedraw = RGB(255,136,0); yellowdraw = RGB(238,204,0); greendraw = RGB(0,187,0); turquoisedraw = RGB(0,170,153); lightbluedraw = RGB(68,204,255); bluedraw = RGB(0,102,255); purpledraw = RGB(170,34,255); pinkdraw = RGB(255,17,255); graydraw = RGB(119,119,119); pair P,Q,R,X,Y,Z,A,B,C,D,E,F,K,L; P = dir(90); Q = dir(210); R = dir(330); X = (-0.25,0)+dir(290); Y = (-0.25,0)+dir(50); Z = (-0.25,0)+dir(170); A = extension(Q,P,Z,Y); B = extension(P,R,Z,Y); C = extension(P,R,Y,X); D = extension(R,Q,Y,X); E = extension(R,Q,X,Z); F = extension(Q,P,X,Z); K = intersectionpoints(circumcircle(A,E,Q),circumcircle(C,E,R))[0]; L = intersectionpoints(circumcircle(B,D,Y),circumcircle(B,F,Z))[1]; fill(P--B--A--cycle,redfill); fill(Q--F--E--cycle,redfill); fill(R--D--C--cycle,redfill); fill(X--E--D--cycle,bluefill); fill(Y--C--B--cycle,bluefill); fill(Z--A--F--cycle,bluefill); fill(A--B--C--D--E--F--cycle,redfill+bluefill); draw(P--Q--R--cycle,reddraw); draw(X--Y--Z--cycle,bluedraw); draw(A--K,orangedraw); draw(C--K,orangedraw); draw(E--K,orangedraw); draw(B--L,greendraw); draw(D--L,greendraw); draw(F--L,greendraw); dot("$P$",P,dir(90)); dot("$Q$",Q,dir(210)); dot("$R$",R,dir(330)); dot("$X$",X,dir(290)); dot("$Y$",Y,dir(50)); dot("$Z$",Z,dir(170)); dot("$A$",A,dir(120)); dot("$B$",B,dir(70)); dot("$C$",C,dir(30)); dot("$D$",D,dir(300)); dot("$E$",E,dir(240)); dot("$F$",F,dir(200)); dot("$K$",K,dir(200)); dot("$L$",L,dir(20)); [/asy][/asy] For points $A$, $B$, and $C$, define $d(A,B,C)=\tfrac{[ABC]}{BC}$ using the signed area of $\triangle ABC$. Let lines $BC$, $DE$, and $FA$ determine $\triangle PQR$, and let lines $AB$, $CD$, and $EF$ determine $\triangle XYZ$, as shown. Since $\angle A+\angle B=240^\circ$, we have $\angle QPR=60^\circ$, and similarly, we can find that triangles $PQR$ and $XYZ$ are equilateral. Let the angle bisectors of $\angle A$, $\angle C$, and $\angle E$ concur at $K$. Notice that \[d(K,A,F)+d(K,C,B)+d(K,E,D)=d(K,B,A)+d(K,D,C)+d(K,F,E),\]so by Viviani's theorem, the altitudes of triangles $PQR$ and $XYZ$ are of equal length, so $\triangle PQR \cong \triangle XYZ$. Let $L$ be the center of rotation sending $\triangle PQR$ to $\triangle ZXY$. We have $d(L,B,A)=d(L,C,B)$ and analogous results for $D$ and $F$, so the angle bisectors of $\angle B$, $\angle D$, and $\angle F$ concur at $L$. $\square$

good luck to future me if i ever try to read this writeup oops Assume $ABCDEF$ to be labelled clockwise (we will use this later when describing stuff). Extend $\overline{CD}$ and $\overline{EF}$ to meet at $X$; extend $\overline{AB}$ and $\overline{EF}$ to meet at $Y$; extend $\overline{AB}$ and $\overline{CD}$ to meet at $Z$. We have due to the angle conditions that $\triangle XYZ$ is equilateral. Let $\ell_X, \ell_Y$ and $\ell_Z$ denote the $X$, $Y$ and $Z$ altitudes of $\triangle XYZ$, respectively. Let $Q$ be the point that lies on all three of these lines. Let $m_A$, $m_B$, $\dots$ denote the internal angle bisectors of $\angle A$, $\angle B$, $\dots$ respectively. Let $P$ be the point that lies on $m_A$, $m_C$ and $m_E$. Let $I_Z$ be the $Z$-excenter of $\triangle ZBC$. By the definition of excenter, this point is the intersection of lines $\ell_Z$ and $m_C$. Furthermore, $m_B$ also passes through $I_Z$, and is a $60^{\circ}$ counterclockwise rotation of $m_C$ about $I_Z$. Define points $I_X$ and $I_Y$ similarly; these points satisfy similar properties to the properties described for $I_Z$. Note that we can rotate $m_A$, $m_C$ and $m_E$ and shift the three lines in some way to obtain the lines $\ell_X$, $\ell_Y$ and $\ell_Z$, respectively. We have that $\angle PI_XQ$, $\angle PI_YQ$ and $PI_ZQ$ are all equal to this angle of rotation, hence $PQI_XI_YI_Z$ is cyclic; in particular, it is easy to see that $\triangle I_X I_Y I_Z$ is equilateral (i think there are a lot of config issues in proving this lol). Call this circle $\omega$. At this point, we can forget about everything we've defined previously except the points on $\omega$. We will show that, in fact, $m_B$, $m_D$ and $m_E$ also concur on $\omega$.

Therefore, $m_D$, $m_B$ and $m_F$ concur at a point $R$.

Remark: Angle bisectors should often be understood as equal altitudes to sides, especially when doing MS_Kekas's problems . Let $AF \cap BC = P$, $AF \cap ED = Q$, and $ED \cap BC = R$, and let $EF \cap CD = X$, $AB \cap CD = Y$, and $AB \cap EF = Z$. Note that $\triangle PQR$ and $\triangle XYZ$ are both equilateral since the sum of any two consecutive angles in $ABCDEF$ are equal. Let $O$ denote the intersection of the angle bisectors of $\angle A$, $\angle C$, and $\angle E$. Note that $O$ has equal distances to the pairs $(AF, AB)$, $(CB, CD)$, and $(ED, EF)$. By Viviani's theorem, the sum of the distances of a fixed point to the sides of an equilateral triangle is proportional to its side length; thus, $\triangle PQR$ and $\triangle XYZ$ have equal side length. Reusing the equal distances result, $O$ is the center of the rotation taking $\triangle PQR$ to $\triangle XYZ$, and thus also the center of rotation taking $\triangle XYZ$ to $\triangle PQR$. Reversing the previous logic, we have that the angle bisectors of $\angle B$, $\angle D$, and $\angle F$ concur, as desired.

Consider triangles $\triangle PQR$ and $\triangle XYZ$(vertices written in clockwise order) formed by the extensions of the sides(the intersections of both triangles form hexagon $ABCDEF$), so that $P = AF \cap BC$ and $X = AB \cap CD$. Note that both of these triangles are equilateral since $\angle A + \angle B = 240^{\circ} \implies \angle APB = 60^{\circ}$. Similarly $\angle Q = \angle R = \angle X = \angle Y = \angle Z = 60^{\circ}$. Let the angle bisectors of $A$, $C$, and $E$ concur at $I$. Let ${RS}_I$ be the altitude of $I$ to any line $RS$. Then from our angle bisector condition we get $XY_I = PQ_I$, $XZ_I + PR_I$, and $YZ_I = QR_I$. Summing gives $XY_I + XZ_I + YZ_I = PQ_I + PR_I + QR_I$ and by Viviani's theorem, the sum of altitudes from an interior point to the sides of an equilateral triangle is constant wrt the side length. So $\triangle PQR \cong \triangle XYZ$. Then, consider the center $K$ of the spiral similarity sending $\triangle PQR \to \triangle XYZ$. Then since both triangles are congruent, we have $PQ_K = XZ_K$ which holds symmetrically true. So the angle bisectors of $\angle B$, $\angle D$, and $\angle F$ also concur. The opposite direction also is true by symmetry. sorry

Let $U,V,W,X,Y,Z=FA\cap BC,AB\cap CD,BC\cap DE,CD\cap EF,DE\cap FA,EF\cap AB$ respectively. The angles give $UWY$ and $VXZ$ are equilateral. Let $P$ be the concurrence of the bisectors of $\angle A,\angle C$ and $\angle E.$ The angles give $\angle APC=\angle CPE=\angle EPA=120^\circ,$ so $(APCVU),(CPEXW),(EPAZY)$ are cyclic. In particular $P$ is the center of spiral similarity taking $UV$ to $YZ,$ so it is also the center taking $UY$ to $VZ$ and thus $\triangle UWY$ to $\triangle VXZ$ in that order. But now $AP$ is the external bisector of $\angle UAV,$ so $UP=PV$ and thus $\triangle UWY\cong\triangle VXZ.$ Let $Q$ be the spiral center taking $\triangle UWY$ to $\triangle ZVX$ in that order. Similarly to before we get $(BQDWV),(DQFYX),(FQBUZ)$ cyclic. But now $\triangle UWY\cong\triangle ZVX$ implies $QU=QZ,$ so $Q$ lies on the external bisectors of $\angle ZFU$ and $\angle ZBU,$ which are the internal bisectors of $\angle F$ and $\angle B.$ Similarly $Q$ lies on the bisector of $\angle D,$ so $Q$ is the desired concurrency point.

Let $AF$ and $BC$ meet at $P_{ab}$, and define $P_{bc},P_{cd},P_{de},P_{ef},P_{fa}$ similarly. Claim: $\triangle P_{ab}P_{cd}P_{ef}$ and $\triangle P_{bc}P_{de}P_{fa}$ are equilateral. We have $$\angle AP_{ab}B=\angle FAB+\angle ABC-180=\frac{720}{3}-180=60,$$and similar for the other ones. Redefine the hexagon as the intersection of two equilateral triangles. Note that all such hexagons satisfy $\angle A=\angle C=\angle E$ and $\angle B=\angle D=\angle F$. Now, the idea is to characterize all such configurations that work. Claim: The angle bisectors of $\angle A$, $\angle C$, $\angle E$ are concurrent if and only if $\triangle P_{ab}P_{cd}P_{ef}$ and $\triangle P_{bc}P_{de}P_{fa}$ are the same size. We will first show angle bisectors concurrent implies same size. Let the angle bisectors concur at $P$. Note that $\angle APC=\angle CPE=\angle EPA=120$. Note that $PAP_{ab}P_{bc}C$ is cyclic as $\angle AP_{ab}C=\angle AP_{bc}C=60$ and $\angle APC=120$. Furthermore, $\angle P_{ab}AP=\angle P_{bc}CP$ due to the angle bisector condition, so $P_{ab}P=P_{bc}P$. Similarly, $P_{cd}P=P_{de}P$ and $P_{ef}P=P_{fa}P$. Furthermore $$\angle P_{ab}PP_{bc}=\angle P_{ab}CP_{bc}=\angle P_{cd}CP_{de}=\angle P_{cd}PP_{de},$$so $P$ is the center of rotation between the two triangles, so they have the same size. Now, if the triangles have the same size, let $P'$ denote the center of rotation between $\triangle P_{ab}P_{cd}P_{ef}$ and $\triangle P_{bc}P_{de}P_{fa}$. Since the rotation at $P'$ takes line $P_{ef}P_{ab}$ to $P_{fa}P_{bc}$, $P'$ is equidistant from the two lines, and thus lies on the $A$-bisector, and similarly for the others. Thus, by symmetry, the $A,C,E$ bisectors concurring and the $B,D,F$ ones concurring are both equivalent to the triangles being the same size, so they are equivalent to each other, done. Remark: This has similar vibes to 2005 IMO/1 (where you also extend sides of a hexagon to form equilateral triangles) and 2011 USAMO/3 (where a hexagon satisfying strange conditions is explicitly characterized).