Really nice problem which has really short solution.

This was also south korean TST #2.

The construction in this problem is really weird, but beside that it felt easy for G5 Let $B'B$ cut $\omega$ again at $X$ and let $XP$ cut $\omega$ again at $E'$. Let $CP$ and $AP$ cut $AX$ and $X$ at $R$ and $S$. Notice that, because $AA'C'C$ is an isosceles trapezoid, the internal angle bisector of angle $APC$ is parallel to $AA'$ and $CC'$, and thus is also parallel to $BB'$. Notice that $XB$ is the internal angle bisector of $\angle AXB$ because $B$ is the midpoint of arc $AC$ in $\omega$. Thus, the internal angle bisectors of angles $APC$ and $AXC$ are parallel. By considering the feet of these bisector onto side $AC$, it is easy by angle chasing to show that this implies that quadrilateral $ARSC$ is cyclic. Now, by the Law of Sines in triangle $AEC$ $$\frac{AE'}{E'C} = \frac{\sin(\angle ACE')}{\sin(\angle CAE')} = \frac{\sin(\angle AXP)}{\sin(\angle CXP)}$$ And from Law of Sines in triangles $APX$ and $CPX$ we have $$\sin(\angle AXP) = \frac{AP \cdot \sin(\angle XAP)}{PX} \qquad \sin(\angle CXP) = \frac{CP \cdot \sin(\angle XCP)}{PX}$$ However, we have $\angle XAP = \angle XCP$ because $ARSC$ is cyclic. So the previous equality implies $$\frac{AE'}{E'C} = \frac{\sin(\angle AXP)}{\sin(\angle CXP)} = \frac{AP}{CP}$$ This equality implies that the external angle bisectors of $\angle APC$ and $\angle AE'C$ cut line $AC$ at the same point. By simmetry, the external angle bisector of $\angle APC$ cuts $AC$ at the point $Y = AC \cap A'C'$. This implies that $Y, E', B$ are collinear because $B$ is the midpoint of arc $AEC$. Finally, by Power of a Point from $Y$ we have $$YE' \cdot YB = YA \cdot YC = YA' \cdot YC'$$ Implying that $A', C', B$ and $E'$ are collinear. Thus $E = E'$ and $EP$ and $BB'$ are concurrent at point $X$ on $\omega$.

Let $\ell$ denote this perpendicular bisector, and let $V = \overline{AC} \cap \overline{A_1 C_1} \cap \ell$. Also, let $X = \overline{BB_1} \cap \omega \ne B$. We contend $E = \overline{BV} \cap \omega \ne B$; indeed, this gives $VE \cdot VB = VA \cdot VC = VA_1 \cdot VC_1$. Now re-define $D = \overline{XE} \cap \ell$, so that $\overline{VD} \perp \overline{XB}$. Then, if we can show that $\ell$ is the external angle bisector of $\angle ADC$, we would conclude $D = \overline{CA_1} \cap \overline{AC_1}$ and solve the problem. We present two approaches now. [asy][asy] size(12cm); pair X = dir(138); pair A = dir(210); pair C = dir(330); pair M = dir(90); pair B = dir(-90); pair E = dir(235); pair S = extension(M, E, A, C); pair L = extension(E, M, X, B); pair V = extension(E, B, A, C); pair K = extension(X, M, V, B); pair D = extension(S, S+X-L, X, E); draw(X--A--C--cycle, red); draw(unitcircle, red); draw(M--K, grey); draw(X--B, lightcyan); draw(D--S, lightcyan+dashed); draw(M--E, red); draw(K--B, red); draw(V--A, red); draw(X--E, red); pair H_A = foot(A, V, D); pair H_C = foot(C, V, D); pair A_1 = 2*H_A-A; pair C_1 = 2*H_C-C; draw(M--(2*M-X), grey); draw(V--(3*D-2*V), grey); draw(A--A_1, lightcyan); draw(C--C_1, lightcyan); draw(V--C_1, dotted+red); draw(B--(1.5*B-0.5*L), lightcyan); draw(K--(1.8*L-0.8*K), orange+dashed); label("$\infty_2$", 3*D-2*V, dir(D-V)); label("$\infty_2$", 2*M-X, dir(D-V)); label("$\infty_1$", 1.5*B-0.5*L, dir(B-L), lightblue); label("$\infty$", 1.8*L-0.8*K, dir(0), red); pair N = extension(V, D, M, B); draw(M--B, orange); dot("$X$", X, dir(X)); dot("$A$", A, dir(A)); dot("$C$", C, dir(C)); dot("$M$", M, dir(M)); dot("$B$", B, dir(B)); dot("$E$", E, dir(E)); dot("$S$", S, dir(315)); dot("$L$", L, dir(10)); dot("$V$", V, dir(225)); dot("$K$", K, dir(K)); dot("$D$", D, dir(50)); dot("$A_1$", A_1, dir(A_1)); dot("$C_1$", C_1, dir(C_1)); dot("$N$", N, dir(325)); /* TSQ Source: !size(12cm); X = dir 138 A = dir 210 C = dir 330 M = dir 90 B = dir -90 E = dir 235 S = extension M E A C R315 L = extension E M X B R10 V = extension E B A C R225 K = extension X M V B D = extension S S+X-L X E R50 X--A--C--cycle 0.1 lightred / red unitcircle 0.1 yellow / red M--K grey X--B lightcyan D--S lightcyan dashed M--E red K--B red V--A red X--E red H_A := foot A V D H_C := foot C V D A_1 = 2*H_A-A C_1 = 2*H_C-C M--(2*M-X) grey V--(3*D-2*V) grey A--A_1 lightcyan C--C_1 lightcyan V--C_1 dotted red B--(1.5*B-0.5*L) lightcyan K--(1.8*L-0.8*K) orange dashed !label("$\infty_2$", 3*D-2*V, dir(D-V)); !label("$\infty_2$", 2*M-X, dir(D-V)); !label("$\infty_1$", 1.5*B-0.5*L, dir(B-L), lightblue); !label("$\infty$", 1.8*L-0.8*K, dir(0), red); N = extension V D M B R325 M--B orange */ [/asy][/asy] First completion (projective, Evan Chen) Define $M$ to be the $B$-antipode of $\omega$, and let $S = \overline{EM} \cap \overline{AC}$. Claim: We have $\overline{DS} \parallel \overline{XB}$. Proof. This is totally projective. Let \[ K = \overline{XM} \cap \overline{BEV}, \quad L = \overline{EM} \cap \overline{BX}, \quad \infty = \overline{KL} \cap \overline{MM} \cap \overline{BB}, \]\[ \infty_1 = \overline{XB} \cap \overline{AA_1} \cap \overline{CC_1}, \quad \text{and}\quad \infty_2 = \ell \cap \overline{XKM}. \]with the latter three points at infinity. We aim to show $D$, $S$, $\infty_1$ are collinear. Note that \[ \triangle X\infty_1\infty_2 \quad\text{and}\quad \triangle ESV \]are perspective from a line, since $\overline{X\infty_1} \cap \overline{ES} = L$, $\overline{\infty_1 \infty_2} \cap \overline{SV} = \infty$, and $\overline{\infty_2 X} \cap \overline{VE} = K$, and $\infty \in \overline{KL}$ by Brokard's theorem. Thus by Desargue's theorem, they are perspective from a line, meaning $\overline{XE}$, $\overline{\infty_1 S}$, $\overline{\infty_2 V}$ concur --- at $D$. $\blacksquare$ Now $-1 = (MB;AC) \overset{E}{=} (SV;AC)$, and so $\overline{SD}$ and $\overline{SV}$ are the internal and external angle bisectors, as desired. Remark: The key projective step seems to be essentially the following lemma (once the polar of $\overline{BM}$ is eliminated via Brokard theorem). Let $ABCD$ be a (complete) quadrilateral. Pick $W \in \overline{AB}$, $X \in \overline{AD}$, $Y \in \overline{BC}$, $Z \in \overline{CD}$. Then $\overline{WX}$, $\overline{YZ}$, $\overline{BD}$ are concurrent, if and only if $\overline{WY}$, $\overline{XZ}$, $\overline{AC}$ are concurrent. The proof is by the same application of Desargue's theorem. For example, if we are given that $\overline{WX}$, $\overline{YZ}$, $\overline{BD}$ are concurrent, then $\triangle AWX$ and $\triangle CYZ$ are perspective from a line, hence $\overline{AC}$, $\overline{WY}$, $\overline{XZ}$ are concurrent. Second completion (synthetic, from the official shortlist) Define $M$ to be the $B$-antipode of $\omega$, as before. Let $N = \ell \cap \overline{MB}$. Then \[ \measuredangle DNB = \measuredangle XMB = \measuredangle XEB \]so $DNEB$ is cyclic. Now, as $VD \cdot VN = VE \cdot VB = VA \cdot VC$, we see $ADNC$ is cyclic too. But as $NA = NC$, it follows that $\ell \equiv \overline{VDN}$ must be the external angle bisector of $\angle ADC$ as desired. Third solution (synthetic, Ankan Bhattacharya) We present an independent variation on the second solution. Again let lines $AC$ and $A_1C_1$ meet at $V$ (on the common perpendicular bisector $\ell$) and let $N$ be the point on $\ell$ with $AN = CN$. [asy][asy] unitsize(50); pair A, B, C, A1, B1, C1, D, E, V, N, Z; A = (-1, 1); B = (0.2, 2.2); C = rotate(120, B) * A; A1 = yscale(-1) * A; B1 = yscale(-1) * B; C1 = yscale(-1) * C; D = extension(A, C1, C, A1); V = extension(A, C, A1, C1); E = 2 * foot(circumcenter(A, B, C), V, B) - B; N = circumcenter(A, C, A1); Z = 2 * foot(circumcenter(A, B, C), B, B1) - B; draw(circumcircle(A, B, C)); draw(circumcircle(A, N, C)); draw(circumcircle(B, D, N)); draw(E--Z, dashed); draw(B--Z); draw(V--B^^V--C^^V--N^^B--N); draw(A--B--C--N--cycle); dot(A^^B^^C^^D^^E^^V^^N^^Z); label("$A$", A, dir(140)); label("$B$", B, dir(100)); label("$C$", C, dir(40)); label("$D$", D, dir(220)); label("$E$", E, dir(130)); label("$V$", V, dir(190)); label("$N$", N, dir(290)); label("$Z$", Z, dir(250)); [/asy][/asy] We make three observations: The quadrilateral $ACC_1A_1$ is an isosceles trapezoid and is cyclic, and by radical axis theorem we learn $B$, $E$, $V$ are collinear. Quadrilateral $ABCN$ is also a kite, and thus $A$ and $C$ are reflections about $\overline{BN}$. Finally $\ell$ is the external bisector of $\angle ADC$ and so $A$, $D$, $N$, $C$ are concyclic. Now we may delete points $A_1$, $B_1$, $C_1$; this is a problem about $\triangle VBN$. Points $A$ and $C$ belong to its $V$-altitude and are reflections about $\overline{BN}$, and the circumcircles $\triangle BAC$ and $\triangle NAC$ meet $\overline{VB}$, $\overline{VN}$ respectively at $E$, $D$. The task is to show that the $B$-altitude meets $\overline{DE}$ on circle $\omega$. Define $Z$ as the second intersection of the $B$-altitude and $\omega$, so we show $E$, $D$, $Z$ collinear. $VE \cdot VB = VA \cdot VC = VD \cdot VN$, so $\overline{DE}$ and $\overline{BN}$ are antiparallel. As $BA = BC$ the center of $\omega$ lies on $\overline{BN}$. Thus $\omega$ is tangent to circle with diameter $\overline{BN}$. The homothety taking $\omega$ to this circle takes $\overline{ZE}$ to the segment joining the $B$- and $N$-altitude feet, so $\overline{EZ}$ and $\overline{BN}$ are antiparallel. Thus $E$, $D$, $Z$ are collinear as desired.

Let $T=\overleftrightarrow{A_1C_1}\cap \overleftrightarrow{AC},$ and let $R$ be the center of the circumcircle of the isosceles trapezoid $\odot (A_1C_1CA).$ The following lemma allows us to erase $A_1$ and $C_1$. Lemma: $EDRB$ is cyclic. Proof: Clearly $T,E,B$ are collinear by considering the radical center of $\odot (ABC),\odot (A_1BC_1),\odot(A_1C_1CA).$ Now $$2\angle ARC=\angle ARC+\angle A_1RC_1=\angle ADC+\angle A_1DC_1=2\angle ADC,$$so $R\in \odot (ADC).$ By radical axis once more, we see that $DRBE$ is cyclic. Now, let $X=\overleftrightarrow{DE}\cap \odot (ABC).$ Note that it suffices to show $\overline{BX}\perp \overline{TDR}$ from the perpendicular bisector conditions. To finish, let $B'$ be the antipode of $B$ in $\odot (ABC),$ and let $\overline {BX}\cap \overline{TDR}=U.$ Then $$\angle (\overline{XB},\overline{DR})=\angle RUB=180^{\circ}-\angle DRB-\angle UBR=\angle XEB-\angle XBR=180-\angle XB'B-\angle XBR=90^{\circ},$$finishing the problem.

Nice problem! Quite a rich configuration(although, surprisingly, quite easy for a G5). My solution: In the upcoming solution, WLOG I assume $AA_1>CC_1$. Let $\ell$ denote the common perpendicular bisector of $AA_1, BB_1, CC_1$. Note that $\triangle A_1B_1C_1$ is the reflecton of $\triangle ABC$ about $\ell$. Let $BB_1 \cap \omega = P, BB_1 \cap AC = T, AC \cap A_1C_1 = X, BX \cap \omega = E', \odot (DEX) \cap AC = F$. Due to symmetry, $D$ and $X$ lie on $\ell$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(15); defaultpen(dps); /* default pen style */ pen dotstyle = lightcyan; /* point style */ real xmin = -12.091253394599342, xmax = 13.089144910773957, ymin = -6.60064145094755, ymax = 4.9708087626666355; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-1.0195668114733758,3.3754094807231514)--(-4.633242293465882,1.9658197537048119), linewidth(2) + wrwrwr); draw((1.681337782237601,0.591400130282612)--(-1.0195668114733758,3.3754094807231514), linewidth(2) + wrwrwr); draw((xmin, 0.16282894736842105*xmin-1.0835197368421052)--(xmax, 0.16282894736842105*xmax-1.0835197368421052), linewidth(2) + wrwrwr); /* line */ draw((2.1258477372440456,-2.1385195933933305)--(0.44768124451565317,-5.635568479290281), linewidth(2) + wrwrwr); draw((1.681337782237601,0.591400130282612)--(2.1258477372440456,-2.1385195933933305), linewidth(2) + wrwrwr); draw((-4.633242293465882,1.9658197537048119)--(-3.4265104455575948,-5.445220281933963), linewidth(2) + wrwrwr); draw((-1.0195668114733758,3.3754094807231514)--(0.44768124451565317,-5.635568479290281), linewidth(2) + wrwrwr); draw((-3.4265104455575948,-5.445220281933963)--(0.44768124451565317,-5.635568479290281), linewidth(2) + wrwrwr); draw((-4.633242293465882,1.9658197537048119)--(2.1258477372440456,-2.1385195933933305), linewidth(2) + wrwrwr); draw((1.681337782237601,0.591400130282612)--(-3.4265104455575948,-5.445220281933963), linewidth(2) + wrwrwr); draw(circle((-1.765148213852266,-0.05006072360510188), 3.505672224882686), linewidth(2) + wrwrwr); draw((0.02863768282365131,-3.0620484236464707)--(0.7721296488121127,2.369021498219366), linewidth(2) + wrwrwr); draw((-4.633242293465882,1.9658197537048119)--(5.363852980862329,-0.21012920212932454), linewidth(2) + wrwrwr); draw((-3.4265104455575948,-5.445220281933963)--(5.363852980862329,-0.21012920212932454), linewidth(2) + wrwrwr); draw((-1.0195668114733758,3.3754094807231514)--(5.363852980862329,-0.21012920212932454), linewidth(2) + wrwrwr); draw(circle((2.1337507479366074,-0.5838045523523681), 3.251644768807941), linewidth(2) + wrwrwr); draw((0.7721296488121127,2.369021498219366)--(1.681337782237601,0.591400130282612), linewidth(2) + wrwrwr); draw((-0.01853058372847902,0.9613902818697114)--(0.7721296488121127,2.369021498219366), linewidth(2) + wrwrwr); draw((-0.01853058372847902,0.9613902818697114)--(0.3063184592594781,-1.033642224561368), linewidth(2) + wrwrwr); /* dots and labels */ dot((-1.0195668114733758,3.3754094807231514),dotstyle); label("B", (-0.9426800990905536,3.5676262616801977), NE * labelscalefactor); dot((-4.633242293465882,1.9658197537048119),dotstyle); label("A", (-5.171449280145612,2.2413304730765784), NE * labelscalefactor); dot((1.681337782237601,0.591400130282612),dotstyle); label("$C$", (1.7867981904995291,0.7804829378030267), NE * labelscalefactor); dot((-3.4265104455575948,-5.445220281933963),dotstyle); label("$A_1$", (-4.133478662977551,-5.754887614736546), NE * labelscalefactor); dot((2.1258477372440456,-2.1385195933933305),dotstyle); label("$C_1$", (2.1712317524136253,-1.6222268241600517), NE * labelscalefactor); dot((0.44768124451565317,-5.635568479290281),dotstyle); label("$B_1$", (0.787270929522879,-5.812552649023661), NE * labelscalefactor); dot((0.3063184592594781,-1.033642224561368),linewidth(4pt) + dotstyle); label("$D$", (0.5181674361830118,-1.7760002489256888), NE * labelscalefactor); dot((0.7721296488121127,2.369021498219366),linewidth(4pt) + dotstyle); label("$E$", (0.6527191828529454,2.625764034990671), NE * labelscalefactor); dot((0.02863768282365131,-3.0620484236464707),linewidth(4pt) + dotstyle); label("$P$", (-0.8273500305163248,-3.0254093251464895), NE * labelscalefactor); dot((5.363852980862329,-0.21012920212932454),linewidth(4pt) + dotstyle); label("$X$", (5.6695771658319005,-0.6995862755662295), NE * labelscalefactor); dot((-0.6488331877428701,1.0985807612469138),linewidth(4pt) + dotstyle); label("$T$", (-1.307891982908945,1.5301283835355073), NE * labelscalefactor); dot((-0.01853058372847902,0.9613902818697114),linewidth(4pt) + dotstyle); label("$F$", (-0.32758640002799977,0.5306011225588666), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Now, $XB \cdot XE' = XA \cdot XC = XA_1 \cdot XC_1 \Rightarrow A_1BC_1E'$ is cyclic $\Rightarrow E' = E$ Also, $\angle XEC = \angle BAC = \angle BCA \Rightarrow \angle ACE = \angle XEC + \angle CXE = \angle BCA + \angle CXE$ $\Rightarrow \angle BCE = \angle CXE \Rightarrow BC$ is tangent to $\odot (CXE) \Rightarrow BC^2 = BE \cdot BX$ But, As $B$ is the midpoint of $\overarc {AC}$, by the Shooting Lemma, $BC^2 = BT \cdot BP$ $\Rightarrow BT \cdot BP = BE \cdot BX \Rightarrow PTEX$ is cyclic. Now, $XE$ is the external angle bisector of $\triangle AEC \Rightarrow \frac{XC}{XA} = \frac{EC}{EA}$ Also, $XD$ is the external angle bisector of $\triangle ADC \Rightarrow \frac{DC}{DA} = \frac{XC}{XA} = \frac{EC}{EA}$ $\Rightarrow \odot (DEX)$ is the $E$-Appolonius Circle of $\triangle EAC \Rightarrow FX$ is a diameter of $\odot (DEX)$ $\Rightarrow \angle FDX = 90^{\circ} \Rightarrow FD \perp \ell \Rightarrow FD \parallel TP$ $\Rightarrow \angle XFD = \angle XTP \Rightarrow \angle XED = \angle XEP \Rightarrow E, D, P$ are collinear. $\blacksquare$

fastlikearabbit wrote: Let $ABCC_1B_1A_1$ be a convex hexagon such that $AB=BC$, and suppose that the line segments $AA_1, BB_1$, and $CC_1$ have the same perpendicular bisector. Let the diagonals $AC_1$ and $A_1C$ meet at $D$, and denote by $\omega$ the circle $ABC$. Let $\omega$ intersect the circle $A_1BC_1$ again at $E \neq B$. Prove that the lines $BB_1$ and $DE$ intersect on $\omega$. Let $P=\overline{BE} \cap \overline{AC}$ and $\ell$ be the perpendicular bisector of $\overline{BB_1}$. By radical axis theorem on $\odot(AA_1CC_1), \odot(ABC), \odot(A_1BC_1)$ we conclude that $P \in \overline{A_1C_1} \implies P \in \ell$. Note that $\ell=\overline{PD}$ as $AA_1C_1C$ is an isosceles trapezoid. Let $U=\overline{AA_1} \cap \overline{PD}$ and $V=\overline{BB_1} \cap \overline{PD}$; then $(\overline{UV}; \overline{DP})=-1$. Project on line $\overline{AC}$; if $T \in \overline{AC}$ with $\angle PDT=90^{\circ}$ then $(\overline{AC}; \overline{TP})=-1$. Now $\overline{EP}$ bisects angle $AEC$ externally, so $\overline{ET}$ bisects angle $AEC$ internally. Let $\overline{BB'}$ be a diameter of $\odot(ABC)$; then $E,T,B'$ are collinear. Note also that $P,D,T,E$ are concyclic. So $\measuredangle DET=\measuredangle DPT$. Now let $X=\overline{BB_1} \cap \odot(ABC)$ with $X \ne B$. Then $\overline{BX} \perp \ell$ and $\overline{BB_1} \perp \overline{AC}$ so $\measuredangle XEB'=\measuredangle XBB'=\measuredangle DPT$ and we're done. $\blacksquare$

i have the same proof as @ above a part from i ve proved that $PETD$ is cyclic whereas he has takes it for obvious which i don't estimate so then if necessary i will post the proof at least the fore-mentionned part . RH HAS

PROF65 wrote: i have the same proof as @ above a part from i ve proved that $PETD$ is cyclic whereas he has takes it for obvious which i don't estimate so then if necessary i will post the proof at least the fore-mentionned part . RH HAS I have shown that $PETD$ is cyclic in my proof in post#7(In my solution $P$ has been replaced by $X$ and $T$ has been replaced by $F$)

math_pi_rate wrote: ...Also, $XD$ is the external angle bisector of $\triangle A\color{red}{D}$$C $ $\blacksquare$ you haven't proved this result which is not simple!

PROF65 wrote: math_pi_rate wrote: ...Also, $XD$ is the external angle bisector of $\triangle A\color{red}{D}$$C $ $\blacksquare$ you haven't proved this result which is not simple! I don't get it. Isn't it obvious. $XD$ is the perpendicular bisector of $CC_1$ in $\triangle CDC_1$, which is isosceles, and so $XD$ is the internal angle bisector of $\triangle CDC_1$, i.e. $XD$ is the external angle bisector of $\triangle ADC$.

Since $\overline{AA_1},\ \overline{BB_1}$ and $\overline{CC_1}$ have the same perpendicular bisector, quadrilaterals $AA_1B_1B,\ AA_1C_1C$ and $BB_1C_1C$ must be isosceles trapezoids. Let $F=\overline{BB_1}\cap \omega,\ P=\overline{A_1C}\cap \omega,\ Q=\overline{AC_1}\cap \omega,\ P\neq C,\ Q\neq A$. It suffices to show that $EF,\ PC$ and $AQ$ concur at $D$. By Ceva's theorem on circumference $\omega$, it suffices to prove that, \begin{eqnarray} \dfrac{EA}{AP}\dfrac{PF}{FQ}\dfrac{QC}{CE}=\frac{EA}{CE}\dfrac{PF}{QF}\dfrac{QC}{AP}=1 \end{eqnarray}By the radical axes theorem applied to $\omega,\ (A_1BC_1),\ (AA_1C_1C)$ we conclude that $BE,\ AC,\ A_1C_1$ concur at a point, say $K$. Since $B$ is the midpoint of $\widehat{ABC}$, $KB$ is the external bisector of $\angle AEC$, so, \begin{eqnarray} \dfrac{EA}{CE}=\dfrac{KA}{KC}=\dfrac{AA_1}{CC_1} \end{eqnarray}On the other hand, notice that $\angle A_1AP=\angle A_1C_1-\angle PAQ=\angle A_1CC_1-\angle PCQ=\angle QCC_1$ and we have $\angle PA_1A=\angle QC_1C$, so $\bigtriangleup A_1AP\sim \bigtriangleup C_1CQ$, thus, \begin{eqnarray} \dfrac{CQ}{AP}=\dfrac{CC_1}{AA_1} \end{eqnarray}Let $J=\overline{A_1C}\cap \overline{BB_1},\ L=\overline{AC_1}\cap \overline{BB_1}$. Since $BB_1\parallel AA_1\parallel CC_1$, we get $\angle FLQ=\angle AC_1C=\angle AA_1C=\angle CJB$ then $$\dfrac{1}{2}\left(\widehat{QF}+\widehat{AB}\right)=\dfrac{1}{2}\left(\widehat{BC}+\widehat{PF}\right)$$whence $\widehat{QF}=\widehat{PF}$, so $PF=CF$. Substituting this fact together with (2) and (3) in (1) we conclude the required equality.

math_pi_rate wrote: PROF65 wrote: math_pi_rate wrote: ...Also, $XD$ is the external angle bisector of $\triangle A\color{red}{D}$$C $ $\blacksquare$ you haven't proved this result which is not simple! I don't get it. Isn't it obvious. $XD$ is the perpendicular bisector of $CC_1$ in $\triangle CDC_1$, which is isosceles, and so $XD$ is the internal angle bisector of $\triangle CDC_1$, i.e. $XD$ is the external angle bisector of $\triangle ADC$. since you neither have $(A,C;F,X)=-1 $ nor $XD\perp DF$ you can't conclude it

PROF65 wrote: since you neither have $(A,C;F,X)=-1 $ nor $XD\perp DF$ you can't conclude it I guess you didn't notice but I have taken $F$ as the point where $\odot (DEX)$ meets $AC$, and so $F$ is the foot of internal angle bisector of $\angle AEC$, as it lies on the $E$-Appolonius circle of $\triangle AEC$. Thus $FX$ is a diameter of $\odot (DEX)$, and so $FD \perp \ell$.

math_pi_rate wrote: PROF65 wrote: since you neither have $(A,C;F,X)=-1 $ nor $XD\perp DF$ you can't conclude it I guess you didn't notice but I have taken $F$ as the point where $\odot (DEX)$ meets $AC$, and so $F$ is the foot of internal angle bisector of $\angle AEC$, as it lies on the $E$-Appolonius circle of $\triangle AEC$. Thus $FX$ is a diameter of $\odot (DEX)$, and so $FD \perp \ell$. you have used it to ''prove''that $\odot (DEX)$ is $E$-Appolonius circle thus you keep in vicious circle

PROF65 wrote: you have used it to ''prove''that $\odot (DEX)$ is $E$-Appolonius circle thus you keep in vicious circle I am still not getting your point. Can you please explain it again? I have nowhere used $F$ while showing that $\odot (DEX)$ is the $E$-Appolonius circle in $\triangle AEC$. And it is well known that the point where the $E$-Appolonius circle meets the opposite side $AC$ is the foot of angle bisector of $\angle AEC$ $\Rightarrow$ $F$ is the foot of internal angle bisector of $\angle AEC$ and $FX$ is a diameter of $\odot (DEX)$.

Let the common perp. bisector be $\ell$, let $X=\ell\cap AC$, and let $BB_1\cap \omega=P$. Consider the circles $\omega, (A_1BC_1), (AA_1C_1C)$. The radical axis theorem tells us $X,E,B$ are collinear, so it suffices to prove that $P,D,E$ are collinear. But looking at the isosceles trapezoid $AA_1C_1C$ easily gives $AD/DC = XA/XC$ and the similarities $\Delta XAE \sim \Delta XBC, \Delta XAB \sim \Delta XEC$ easily give $XA/XC = (XA/XC)(BC/BA) = AE/EC$. Hence, $E,D,X$ lie on the same Apollonius circle of $AC$. If this circle intersects $AC$ again at $F$ then clearly $\angle XDF=\angle XEF=90$, and $FD \parallel BP$. Thus, proving $P,D,E$ are collinear is equivalent to proving $\angle EDF = \overarc{EB}/2$. But since $XEFD$ is cyclic, the problem reduces to proving $\angle BXC = \overarc{EB}/2$. Noticing $\angle BXC=(\overarc{BC}-\overarc{AE})/2$ and $\overarc{AB}=\overarc{BC}$ finishes the problem. $\square$

[asy][asy] unitsize(0.3inches); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -23.773547968982495, xmax = 16.197327065916014, ymin = -23.31059813999672, ymax = 25.487178465108443; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-9.531925194048181,7.208164221024108)--(-12.7,-0.49), linewidth(2) + wrwrwr); draw((-12.7,-0.49)--(-4.386987223255936,-0.05153800204314446), linewidth(2) + wrwrwr); draw((-9.531925194048181,7.208164221024108)--(-4.386987223255936,-0.05153800204314446), linewidth(2) + wrwrwr); draw((-9.531925194048181,7.208164221024108)--(0.5,1.07), linewidth(2) + wrwrwr); draw((1.964695434005493,9.846578511955038)--(-4.386987223255936,-0.05153800204314446), linewidth(2) + wrwrwr); draw(circle((-8.682117392623049,2.357466211933077), 4.924575562905164), linewidth(2) + wrwrwr); draw(circle((-6.870612422263731,6.810566649923873), 9.342378334313304), linewidth(2) + wrwrwr); draw(circle((-3.2509867603635816,6.206498769419799), 6.360308269626089), linewidth(2) + wrwrwr); draw((-0.5829568651644716,-5.4191688275318946)--(-7.7929475260872385,-2.4861709981815916), linewidth(2) + wrwrwr); draw((-7.7929475260872385,-2.4861709981815916)--(-12.7,-0.49), linewidth(2) + wrwrwr); draw((-4.386987223255936,-0.05153800204314446)--(-0.5829568651644716,-5.4191688275318946), linewidth(2) + wrwrwr); draw((-0.5829568651644716,-5.4191688275318946)--(0.5,1.07), linewidth(2) + wrwrwr); draw((0.5,1.07)--(1.964695434005493,9.846578511955038), linewidth(2) + wrwrwr); draw((-7.7929475260872385,-2.4861709981815916)--(-3.613759651117511,7.787244565664228), linewidth(2) + wrwrwr); draw((-12.7,-0.49)--(-2.2638261512790203,1.9050473026756387), linewidth(2) + wrwrwr); draw((-0.5829568651644716,-5.4191688275318946)--(-3.613759651117511,7.787244565664228), linewidth(2) + wrwrwr); /* dots and labels */ dot((-9.531925194048181,7.208164221024108),dotstyle); label("$A$", (-9.384032956419029,7.533593761933233), NE * labelscalefactor); dot((-12.7,-0.49),dotstyle); label("$B$", (-13.447738584967045,-1.459853120918913), NE * labelscalefactor); dot((-4.386987223255936,-0.05153800204314446),dotstyle); label("$C$", (-4.254437326940387,0.2722181305933522), NE * labelscalefactor); dot((0.5,1.07),dotstyle); label("$C_1$", (0.6419948648346803,1.404726256582141), NE * labelscalefactor); dot((1.964695434005493,9.846578511955038),dotstyle); label("$A_1$", (2.1075936161142925,10.165009701730712), NE * labelscalefactor); dot((-7.7929475260872385,-2.4861709981815916),linewidth(4pt) + dotstyle); label("$E$", (-7.818507017552171,-3.258542497489342), NE * labelscalefactor); dot((-2.4923744936347423,2.900923346959488),linewidth(4pt) + dotstyle); label("$D$", (-2.3558207627827077,3.170106570623488), NE * labelscalefactor); dot((-0.5829568651644716,-5.4191688275318946),linewidth(4pt) + dotstyle); label("$T$", (-0.12411357333420778,-5.423631561879674), NE * labelscalefactor); dot((-2.2638261512790203,1.9050473026756387),linewidth(4pt) + dotstyle); label("$F$", (-1.7895666997883124,1.571271569227551), NE * labelscalefactor); dot((-3.613759651117511,7.787244565664228),linewidth(4pt) + dotstyle); label("$F'$", (-3.488328888771499,8.066538762398546), NE * labelscalefactor); dot((-3.2509867603635816,6.206498769419799),linewidth(4pt) + dotstyle); label("$O$", (-3.121929200951596,6.467703761002608), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $\ell$ be the common perpendicular bisector. We see that $AC,A_1C_1,\ell$ concur by symmetry ($A_1C_1$ is the reflection of $AC$ over $\ell$), and let that concurrence be $T$. We also have that $AC_1,A_1C,\ell$ concur at $D$ by the same logic. Note that by radical axis on $(ACA_1C_1),(ABC),(A_1BC_1)$, we have that $T,E,B$ collinear. Note that $B_1$ is not doing anything except providing line $BB_1$, so defining $F$ to be the foot from $B$ to $\ell$, we see that $BF=BB_1$, thus eliminating $B_1$ from the picture. Let $O$ be the center of $(ACA_1C_1)$ and let $F'$ be the point on $\ell$ such that $F'E\perp BE$. Now, let $\phi$ be the inversion at $T$ with power $TE\cdot TB$. We will invert the problem statement, and show that. We first need a lemma. Lemma: $\phi(D)=O$. Proof of Lemma: [asy][asy] unitsize(0.2inches); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12, xmax = 12, ymin = -14.65, ymax = 14.65; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-5.813244360660654,4.410156380540263)--(0.6509885240649234,-3.388704239604645), linewidth(2) + wrwrwr); draw((-5.706755639339346,-0.19015638054026374)--(0.3896884108764894,7.899460650135696), linewidth(2) + wrwrwr); draw((-10.08,2.01)--(3.5254940804020505,2.324941992601899), linewidth(2) + wrwrwr); draw((-10.08,2.01)--(0.3896884108764894,7.899460650135696), linewidth(2) + wrwrwr); draw((0.6509885240649234,-3.388704239604645)--(-10.08,2.01), linewidth(2) + wrwrwr); draw((-5.813244360660654,4.410156380540263)--(-4.535296440986093,2.1383496194216183), linewidth(2) + wrwrwr); draw((0.3896884108764894,7.899460650135696)--(3.5254940804020505,2.324941992601899), linewidth(2) + wrwrwr); draw((0.3896884108764894,7.899460650135696)--(0.6509885240649234,-3.388704239604645), linewidth(2) + wrwrwr); draw((-5.813244360660654,4.410156380540263)--(-5.706755639339346,-0.19015638054026374), linewidth(2) + wrwrwr); /* dots and labels */ dot((-10.08,2.01),dotstyle); label("$T$", (-10,2.21), NE * labelscalefactor); dot((-5.76,2.11),dotstyle); label("$X$", (-5.68,2.31), NW * labelscalefactor); dot((0.5203384674707063,2.255378205265526),dotstyle); label("$Y$", (0.6,2.45), NE * labelscalefactor); dot((-5.813244360660654,4.410156380540263),dotstyle); label("$A$", (-5.74,4.61), NE * labelscalefactor); dot((0.3896884108764894,7.899460650135696),linewidth(4pt) + dotstyle); label("$C$", (0.46,8.05), NE * labelscalefactor); dot((-5.706755639339346,-0.19015638054026374),dotstyle); label("$A_1$", (-6.14,-0.73), NE * labelscalefactor); dot((0.6509885240649234,-3.388704239604645),linewidth(4pt) + dotstyle); label("$C_1$", (0.74,-3.23), NE * labelscalefactor); dot((-3.941605448252763,2.1520924664756302),linewidth(4pt) + dotstyle); label("$D$", (-4,1.47), NE * labelscalefactor); dot((-4.535296440986093,2.1383496194216183),linewidth(4pt) + dotstyle); label("$Y'$", (-5.12,2.31), S * labelscalefactor); dot((3.5254940804020505,2.324941992601899),linewidth(4pt) + dotstyle); label("$X'$", (3.6,2.49), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] I'm sure there are far easier proofs for this, but here is one that I came up with. Let $X$ and $Y$ be the feet from $A$ and $C$ to $\ell$, and let $X'$ and $Y'$ be on $\ell$ such that $AY',AX'\perp AC$. Note that $DX/DY=A_1X/CY=AX/CY=TX/TY$ by similar triangles, so $(TD;XY)=-1$. Inverting this with $\phi$, we see that $\phi(D)$ is the midpoint of $\phi(X)\phi(Y)$. But $\phi(AX)=(TCX')$, so in fact $\phi(X)=X'$ and $\phi(Y)=Y'$. Therefore, $\phi(D)$ is the midpoint of $X'Y'$. But this is clearly the intersection of the perpendicular bisector of $AC$ with $\ell$, or $O$. $\blacksquare$ Note that $\phi(BF)$ is a circle passing through $T,E$ with center on $\ell$ (since $BF\perp\ell)$, or $(TEF')$. Also $\phi(DE)=(OBT)$, and $\phi(ABC)=(ABC)$. Therefore, the inverted problem asks us to show that $(ABC),(TEF'),(OBT)$ all pass through a common point. Here is a diagram with only the relevant points: [asy][asy] unitsize(0.3inches); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -16, xmax = 8, ymin = -15.79, ymax = 13.51; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-4.58,2.59)--(-0.9466390632076092,3.6891680144918153), linewidth(2) + wrwrwr); draw(circle((-3.403200802169263,5.254747096059504), 2.9130282247126664), linewidth(2) + wrwrwr); draw((-4.58,2.59)--(-6.316202153075583,5.242234398884779), linewidth(2) + wrwrwr); draw((-6.316202153075583,5.242234398884779)--(-0.9466390632076092,3.6891680144918153), linewidth(2) + wrwrwr); draw((-9.34,1.15)--(-0.6593835340669009,1.0623477420349394), linewidth(2) + wrwrwr); draw((-9.34,1.15)--(-4.58,2.59), linewidth(2) + wrwrwr); draw((-9.34,1.15)--(-4.246699836810639,8.042980016124057), linewidth(2) + wrwrwr); draw((-6.316202153075583,5.242234398884779)--(-0.6593835340669009,1.0623477420349394), linewidth(2) + wrwrwr); draw((-4.246699836810639,8.042980016124057)--(-2.1394330240819484,1.077292496270883), linewidth(2) + wrwrwr); /* dots and labels */ dot((-9.34,1.15),dotstyle); label("$T$", (-9.26,1.35), NE * labelscalefactor); dot((-4.58,2.59),dotstyle); label("$A$", (-4.5,2.79), NE * labelscalefactor); dot((-0.9466390632076092,3.6891680144918153),dotstyle); label("$C$", (-0.86,3.89), NE * labelscalefactor); dot((-4.246699836810639,8.042980016124057),dotstyle); label("$B$", (-4.16,8.25), NE * labelscalefactor); dot((-2.1394330240819484,1.077292496270883),dotstyle); label("$O$", (-2.06,1.27), NE * labelscalefactor); dot((-6.316202153075583,5.242234398884779),linewidth(4pt) + dotstyle); label("$E$", (-6.7,5.45), NE * labelscalefactor); dot((-0.6593835340669009,1.0623477420349394),linewidth(4pt) + dotstyle); label("$F'$", (-0.58,1.23), NE * labelscalefactor); dot((-2.5597017675278857,2.466514175994953),linewidth(4pt) + dotstyle); label("$M$", (-2.48,2.63), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $M$ be the midpoint of arc $AC$. Clearly, $BO$ passes through it, since $BO$ is the perpendicular bisector of $AC$. To show that $E,M,F'$ collinear, it suffices to show that $EF'$ bisects $\angle AEC$. We see that \[\angle BEC=\angle BAC=\angle BCA=\angle TEA,\]which combined with the fact that $EF'\perp BT$ proves that $EF'$ bisects $\angle AEC$, or that $M\in EF'$. Now, by Miquel's theorem on $EMOT$, we see that $(BEM),(BOT),(F'OM),(F'ET)$ concur. Inverting back, we see then that $(ABC),DE,BF$ are concurrent, as desired. $\blacksquare$

Case $1 : AA_1 = CC_1$. Note that in this case both centers of $A_1BC_1$ and $ABC$ lie on $BB_1$ so circles are tangent to each other at $B$ so there exists no $E$. Case $2 : AA_1 \neq CC_1$. WLOG assume $AA_1 < CC_1$. Let $S$ be intersection of $AC$ and $A_1C_1$ and perpendicular bisector of $AA_1$. Note that by Radical Axis on $ABC$ and $A_1BC_1$ and $AA_1C_1C$ we have $S,E,B$ are collinear. Let $O$ be center of $AA_1C_1C$. Claim $: ADOC$ is cyclic. Proof $:$ Note that $\angle AOC = 2\angle AC_1C = \angle DC_1C + \angle DCC_1 = \angle ADC$. Let $BB_1$ meet $\omega$ at $K$ and Let $KE$ meet perpendicular bisector of $AA_1$ at $D'$ and Let $BO$ meet $\omega$ at $T$. Claim $: BED'O$ is cyclic. Proof $:$ Note that $\angle SED' = \angle SEK = \angle KTB$ and note that $KT \perp KB \perp SD'$ so $\angle KTB = \angle D'OB$. Claim $: AD'OC$ is cyclic. Proof $:$ Note that $SD'.SO = SE.SB = SA.SC$. Now Note that $AD'DOC$ meets $SD'$ at $3$ points so either $D'$ is $O$ or $D$. Note that $D'$ can't be $O$ cause $\angle KOE < \angle KOB < B_1OB < 180$ so $D'$ is $D$ and we're Done.

Let $AC_1,BB_1,CA_1$ meet $\omega$ again at $F,G,H$ respectively. By radical axis for $\omega, \odot (AB_1C_1),\odot (ACC_1A_1)$ lines $BE,AC,A_1C_1$ concur at $S.$ Observe that $$\measuredangle CFC_1=-\measuredangle AHA_1,\measuredangle CC_1F=-\measuredangle AA_1H\implies CC_1F\stackrel{-}{\sim} AA_1H\implies |FG|=|GH|.$$Furthermore $S$ is the foot of $E-\text{external bisector}$ in $AEC,$ therefore $$\frac{|AE|}{|EC|}\cdot \frac{|CF|}{|FG|}\cdot \frac{|GH|}{|HA|}=\frac{|AS|}{|SE|} \cdot \frac{|CC_1|}{|AA_1|}=1,$$which yields that $AC_1,CA_1,EG$ concur at $D$ as desired.

Let $F = \overline{BB'} \cap \overline{DE}$, $S = \overline{AC} \cap \overline{A_1C_1}$, $\Omega$ be the circumcircle of $ACC_1A_1$, and $O$ be its center. Then by radical axes, $\overline{BE}$ passes through $S$. Moreover, since $ADOC$ is also cyclice, we have $SD \cdot SO = SA \cdot SC = SE \cdot SB$, so $BEDO$ is cyclic, and \[ \measuredangle EFB = \measuredangle FES + \measuredangle EBF = \measuredangle SOB + \measuredangle SBB' = 90^\circ + \measuredangle OSC + 90^\circ + \measuredangle BSO = \measuredangle BSC. \]Finally, $\triangle BCS \sim \triangle BEC$, so $\measuredangle BSC = -\measuredangle BCE = \measuredangle ECB$ and so $F$ lies on $\omega$.

so i failed to solve 2015 G7 and then 2016 G4 (both were extremely easy by like one construction), and i went to sleep basically depressed lmao. well we're coming back with a vengeance i hope im starting to get the hang of this Let $BB_1\cap AC=P$ and let $BB_1\cap \omega$ be $L$. Also define $\omega_1=(A_1BC_1)$. Finally, let $AC\cap A_1C_1=F$. Wait, we also define $N\in AC$ such that $\angle NDF=90^{\circ}$. In particular, $N$ is the harmonic conjugate of $F$ w.r.t $AC$. Since \[\text{Pow}_F(\omega)=FA\cdot FC=FA_1\cdot FC_1=\text{Pow}_F(\omega_1)\]it follows that $F\in EB$, or that $E=BF\cap \omega$. Let $J$ be the midpoint of $AC$. Verify $FA\cdot FC=FN\cdot FJ=FE\cdot FB$ hence $EBNJ$ cyclic, hence $E\in (NF)$. By Shooting Lemma $BE\cdot BF=BP\cdot BL=BA^2=BC^2$, hence $FEPL$ cyclic. We have $FEND$ cyclic, and $ND\parallel PL$. Hence by Reim's $L\in ED$. Done!

Cool problem! First, observe that $AA_1C_1C$ is an isosceles trapezium, and $AA_1 // BB_1 // CC_1$. Let $X= BB_1 \cap \omega$, $F = AC \cap A_1C_1$, and let $T$ be the point on $AC$ such that $DT // AA_1$. From radical axis on $(ACC_1A_1)$, $(ABC)$ and $(A_1BC_1)$, we have that $B,E,F$ are collinear. Claim: $DTEF$ is cyclic Proof: Note that since $B,E,F$ are collinear, $F$ is the foot of the external angle bisector of $E$ onto $AC$. Thus, \[\frac{TA}{TC} = \frac{DA}{DC_1} = \frac{AA_1}{CC_1} = \frac{FA}{FC} \]which means that $(A,C;T,F)=-1$. Since $EF$ is an external angle bisector, $ET$ must be the internal angle bisector, so $ET \perp EF$. Furthermore, $DF$ is the perpendicular bisector of $AA_1$, so $DT // AA_1 \implies DT \perp DF$. Thus, $DTEF$ is cyclic with diameter $TF$. Hence, \begin{align*} \angle XEF = \angle BAX = 180^{\circ} - \angle ABX - \angle AXB &= \angle BAA_1 - \angle ACB \\ &=\angle BAC + \angle CAA_1 - \angle ACB \\ &= \angle CAA_1 \\ &= \angle DTF \\ &= \angle DEF \end{align*}implying $D,X,E$ are collinear as desired.

Note that $ABEC$, $A'BEC'$, and $AA'C'C$ are all cyclic. Thus, $AC$, $A'C'$, $BE$ concur at a point, $P$ on the perpendicular bisector. Let $D'$ be the point on the perpendicular bisector such that $ACD'Y$ is cyclic. Let the angle bisector of $\angle ABC$, which is also the diameter of $\omega$ from $B$, intersect $\omega$ at $X$ and the perpendicular bisector at $Y$. We have \[\measuredangle CD'P=\measuredangle CD'Y=\measuredangle CAY=\measuredangle YCA=\measuredangle YDA\]By symmetry, $\measuredangle CD'C=\measuredangle A'D'A$ so $D'=D$. Thus, we have $PD\cdot PY=PA\cdot PC=PE\cdot PB$, so $BEDY$ is cyclic. Indeed, let $BB'$ intersect $\omega$ at $F$ then $BF\perp FX$ because $BX$ is a diameter, and $BF\perp DY$. Thus, $DY\parallel FX$. \[\measuredangle BED=\measuredangle BYD=\measuredangle BXF=\measuredangle BEF\]so $F$ is on $ED$ as desired.

decent problem

Let $D'=AC\cap A_1C_1$. By radical axis, $B,E,D'$ are collinear. Moreover, $E$ lies on the inverse of the perpendicular bisector of $AC$ under inversion at $D'$ fixing $(ACA_1C_1)$. Call this inverse $\Omega$. Let $D_1$ be the point on $\Omega$ with $D'D_1\parallel BB_1$. Also, let $\omega$ and $\Omega$ intersect at $K \neq E$. By Reim, $T,K,D_1$ are collinear. Claim: $TT \parallel A_1C_1D'$. Proof. Let $X=CT\cap A_1C_1$, as $\angle XC_1B_1 = \angle BAC=\angle BTC$ we find $TXC_1B_1$ is cyclic. Then, \[\measuredangle (TT,CT)=\measuredangle TBC=\measuredangle C_1B_1T=\measuredangle C_1XT\]and we conclude the claim. $\blacksquare$ Claim: $DD_1 \parallel A_1C_1D'$ Proof. Invert around $D'$ and $D$ is sent to $O$, the circumcenter of $ACA_1C_1$. It suffices to show some circle tangency, or and angle equality: $\angle C_1D'O$ is equal to the angle between the perpendicular bisector of $AC$ and line $BB_1$. I This is not hard to show since if $\angle CC_1D'=x$ then both are $90^{\circ} - x$ after some light angle chase by dropping feet from $O$ to $AC$ and $A_1C_1$ $\blacksquare$ Thus $DD_1 \parallel TT$ are we finish by Reim.

Let $K = BB_1 \cap \omega$, redefine $E = DK \cap \omega$. We'll prove that $A_1, B, E, C_1$ are concyclic. Let $Z = AC \cap A_1C_1$, and $F \in AC$ so that $BK \parallel FD$. [asy][asy] pair A = (-5.2,4.52); pair B = (-7,0.64); pair C = (-3.50172,-1.82099); pair P = (1.36,4.52); pair Q = (3.16,0.64); pair R = (-0.33827,-1.82099); pair K = (-0.55640,0.64); pair E = (-6.03000,-0.95759); pair D = (-1.92,0.24200); pair Z = (-1.92,-7.72683); pair F = (-4.05424,0.24200); import graph; size(12cm); pen ttqqtt = rgb(0.2,0,0.2); pen zzccff = rgb(0.6,0.8,1); pen ffttww = rgb(1,0.2,0.4); pen ffwwzz = rgb(1,0.4,0.6); pen ttffcc = rgb(0.2,1,0.8); pen zzwwff = rgb(0.6,0.4,1); pen qqwuqq = rgb(0,0.39215,0); pen ffzzqq = rgb(1,0.6,0); pen ttttff = rgb(0.2,0.2,1); pen ttzzqq = rgb(0.2,0.6,0); draw(circle((-3.77820,1.50287), 3.33534), linewidth(0.7) + zzccff); draw(circle((-2.64064,2.19355), 4.62790), linewidth(0.7) + linetype("4 4") + ffttww); draw(K--E, linewidth(0.7) + ffwwzz); draw(A--B, linewidth(0.7) + ttffcc); draw(B--C, linewidth(0.7) + ttffcc); draw(C--R, linewidth(0.7) + ttffcc); draw(R--Q, linewidth(0.7) + ttffcc); draw(Q--P, linewidth(0.7) + ttffcc); draw(P--A, linewidth(0.7) + ttffcc); draw(A--K, linewidth(0.7) + zzwwff); draw(K--C, linewidth(0.7) + zzwwff); draw(B--Q, linewidth(0.7) + qqwuqq); draw(A--Z, linewidth(0.7) + ffzzqq); draw(A--D, linewidth(0.7) + ttttff); draw(D--C, linewidth(0.7) + ttttff); draw(F--D, linewidth(0.7) + ttzzqq); dot("$A$", A, dir((-10, 10))); dot("$B$", B, dir((-10, -10))); dot("$C$", C, dir((-5, -10))); dot("$A_1$", P, dir((10, 10))); dot("$B_1$", Q, dir((10, -10))); dot("$C_1$", R, dir((5, -10))); dot("$K$", K, dir((10, -10))); dot("$E$", E, dir((-10,-10))); dot("$D$", D, dir((5, -10))); dot("$Z$", Z, dir((10, 10))); dot("$F$", F, dir((-10, 0))); [/asy][/asy] As $\angle CDZ = \angle C_1DZ$, we have $\angle ADF = \angle CDF$, so $(A, C; F, Z) = -1$. Now, we get $\angle KAD = \angle KCD$ since $$\measuredangle KAD = \measuredangle FDA - \measuredangle BKA = \measuredangle CDF - \measuredangle CKB = \measuredangle DCK.$$This implies the radius of $(KAD)$ and $(KCD)$ are equal. By the law of sines, $$\frac{AE}{EC} = \frac{\sin \angle AKD}{\sin \angle CKD} = \frac{AD}{DC} = \frac{AF}{FC}$$, and so $EF$ bisects $\angle AEC$. This means $BE \perp EF, ZE \perp EF$. (Remark that $AB = BC$, $(A, C; F, Z) = -1$.) Therefore, $ZB \cdot ZE = ZA \cdot ZC = ZA_1 \cdot ZC_1$, we have $A_1, B, E, C_1$ are concyclic. $\blacksquare$

Letting $F=AC\cap A_1C_1$ we get by Radical Axis theorem that $F,B,E$ collinear. Next, by observing that $FD$ is the external bisector of $\angle ADC,$ we can delete points $A_1,B_1,C_1.$ Our new problem is: (with some points renamed) In $\triangle ABC$ let $D$ be the intersection of the external $\angle A$ bisector with $BC.$ Let $E$ be a point with $BE=CE$ and let $DE$ meet $(BCE)$ at $F.$ If $AF$ meets $(BCE)$ at $X,$ show that $EX\perp AD.$ Letting $(BCE)$ meet $AB$ and $AC$ at points $P,Q$ we get that the conclusion is equivalent to $X$ being the midpoint of minor arc $PQ$ by some angle chasing. Now notice that $E$ is an arc midpoint, so $DE$ is the external bisector of $\angle BFC.$ Thus this implies $AB\cdot CF=AC\cdot BF.$ Finally, invert at $F.$ Our new problem is: (with some points renamed) In $\triangle ABC$ let $P$ satisfy $BP=CP$ and let $X=AP\cap BC.$ Let $(ABP),(ACP)$ meet $BC$ again at $D,E$ respectively. Show that $AP$ bisects $\angle DAE.$ Now this is because $\angle DAP=\angle DBP=\angle CBP=\angle BCP=\angle ECP=\angle EAP$ so we are done.

Let lines $BB_1$ and $DE$ intersect at $K$. We would like to show that $A, B, C, E, K$ are concyclic. A little bit of observation yields the following claim, which helps us make use of the weird circle conditions: Claim 1: $BE, AC, A_1C_1$ and the common perpendicular bisector of $AA_1, BB_1, \text{and } CC_1$ concur. Proof. Radical Axis Concurrence Theorem on $(ABEC), (AA_1C_1C), \text{and } (A_1BEC_1)$. Let us call this point of concurrence $F$. An obvious observation after constructing $F$ is the following claim: Let $G$ be the intersection of $BB_1$ and $AC$. Claim 2: $G, E, F, K$ are concyclic. Now note that $\Delta{BCE} \sim \Delta{BFC}$, so the claim would imply $\measuredangle{BKE} = \measuredangle{GKE} = \measuredangle{GFE} = \measuredangle{CFB} = \measuredangle{BCE}$ as desired. Hence it would suffice to prove Claim 2. We have to find a way to use the $AB = BC$ condition, so a natural next step is to add in the perpendicular bisector of $AC$. Alternatively, we have that $FB \cdot FE = FC \cdot FA = FC' \cdot FA' = \text{Pow}(F)$, so it is also natural to think about the point on the common perpendicular bisector such that $FD \cdot F \text{point} = \text{Pow}(F)$. Either way, we are led to the following claim: Let $I$ be the intersection of the perpendicular bisector of $AC$ with the common perpendicular bisector of $AA_1, BB_1, \text{and } CC_1$. Claim 3: $A, C, D, I$ are concyclic. Proof. Trivial This also implies that $B, I, D, E$ are concyclic (by power of a point). Some angle chasing now yields Claim 2: Let $M$ be the midpoint of $AC$ and $J$ be the intersection of $BB_1$ and the common perpendicular bisector of $AA_1, BB_1, \text{and } CC_1$, then we have that $G, M, I, J$ are concyclic, implying that $\measuredangle{KGF} = \measuredangle{BGA} = \measuredangle{BGM} = \measuredangle{JIM} = \measuredangle{DIB} = \measuredangle{DEF} = \measuredangle{KEF}$ implying that $G, E, F, K$ are concyclic as desired, which finishes the problem.

[asy][asy] size(12cm); import geometry; pair A = dir(0); pair B = dir(105); pair C = dir(210); pen dps = linewidth(0.7); defaultpen(dps); line l = line( (0, -0.8), (1, -0.8)); pair A_1 = reflect(l) * A; pair C_1 = reflect(l) * C; pair B_1 = reflect(l) * B; point D = intersectionpoint(line(A, C_1), line(A_1, C)); point X = intersectionpoints(line(B, B_1), circle(A, B, C))[0]; point T = intersectionpoint(line(A, C), line(A_1, C_1)); point E = intersectionpoint(line(B, T), line(D, X)); point F = intersectionpoint(line(B, X), line(A, C)); point M = midpoint(segment(A, C)); point O = intersectionpoint(line(B, M), line(T, D)); point N = intersectionpoint(line(B, X), line(T, D)); draw(circle(A, B, C)); draw(circle(A_1, B, C_1)); draw(A--B--C--cycle); draw(T--B, red); draw(T--O, red); draw(T--A, red); draw(circle(B,E,D), cyan+dashed); draw(E--X, blue+dashed); draw(B--O); draw(B--X); draw(circle(A,C,O), green+dashed); draw(circle(F,M,N), red+dashed); draw(circle(A,C,C_1)); draw(T--A_1); draw(A--A_1); draw(C--C_1); dot("$A$", A, dir(140)); dot("$B$", B, dir(100)); dot("$C$", C, dir(40)); dot("$D$", D, dir(220)); dot("$E$", E, dir(130)); dot("$T$", T, dir(190)); dot("$O$", O, dir(290)); dot("$X$", X, dir(250)); dot("$F$", F, dir(140)); dot("$M$", M, dir(45)); dot("$N$", N, dir(45)); dot("$A_1$", A_1, dir(315)); dot("$C_1$", C_1, dir(180)); [/asy][/asy] First, notice that by Radical Axis on $(ABC)$, $(A_1BC_1)$ and $(ACA_1C_1)$, then we find that $BE$, $AC$ and $A_1C_1$ concur. Now, define $X$ as the intersction of $BB_1$ with $(ABC)$ such that $X \neq B$, and define $F$ as the intersection of $BX$ with $AC$. Notice that by Shooting Lemma, we have \[ BE \cdot BT = BC^2 = BF \cdot BX \]Hence $\boxed{TEFX \text{ is cyclic}}$. Let $M$ be the midpoint of $AC$. Let line $TD$ (which is the perpendicular bisector of $CC_1$ and $AA_1$) intersect $BM$ at $O$, so then $O$ is the center of $(ACC_1A_1)$. Notice that \[ \angle AOC = 2\angle AC_1C = 2(90 - \angle TDC) = 180 - \angle TDC - \angle ADO = \angle ADC \]Hence we find that $ACDO$ is cyclic; but then by radical axis, \[ TN \cdot TO = TC \cdot TA = TE \cdot TB \implies \boxed{BEDO \text{ is cyclic}} \]Now, the rest is just angle chasing. Define $N$ to be the intersection of $BX$ with $TO$. Then $\angle BNO = 90 = \angle BMT$ which implies $MNOF$ cyclic. Then \[ \measuredangle TED = -\measuredangle TOB = \measuredangle MON = \measuredangle MFN = \measuredangle TFX = \measuredangle TEX \]as desired. $\blacksquare$

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