Let $D=B_1B_2\cap C_1C_2, E=B_1C_2\cap B_2C_1$. Then $D, E$ lie on $\omega$. By Miquel, $(C_1DB_2)$ and $(B_1DC_2)$ concur on $K\in BC$. $\measuredangle B_2KC_1=\measuredangle B_2DC_1=\measuredangle B_1DC_1=\measuredangle BAC$ so $ABC_1K$ is cyclic, likewise $AB_1CK$ is cyclic. ELMO SL 2013 G3.

tastymath75025 wrote: Given a triangle $ABC$ for which $\angle BAC \neq 90^{\circ}$, let $B_1, C_1$ be variable points on $AB,AC$, respectively. Let $B_2,C_2$ be the points on line $BC$ such that a spiral similarity centered at $A$ maps $B_1C_1$ to $C_2B_2$. Denote the circumcircle of $AB_1C_1$ by $\omega$. Show that if $B_1B_2$ and $C_1C_2$ concur on $\omega$ at a point distinct from $B_1$ and $C_1$, then $\omega$ passes through a fixed point other than $A$. Proposed by Max Jiang Solution. We claim that the fixed point is the A-Humpty Point of $\triangle ABC$, call it $A_{HM}$. We have $B_1B_2\cap C_1C_2 = D$ and $B_1C_2\cap C_1B_2 = E$ both intersect on $\omega$. Therefore from Brokard, polar of $B_1C_1\cap DE$ is line $BC$ which gives that tangents at $B_1$ and $C_1$ to $\omega$ meet at $BC$. Suppose $\omega$ passes through $A_{HM}$ and tangents at $B_1$ to $\omega$ intersects $BC$ at $Y$. This gives us \[\angle YB_1A_{HM} = \angle B_1AA_{HM} = \angle YBA_{HM}\]so $Y,B,B_1,A_{HM}$ are concyclic. This gives $\angle A_{HM}YB = \angle A_{HM}B_1A = \angle A_{HM}C_1C$ giving $A_{HM},Y,C,C_1$ are concyclic which gives $YC_1$ is tangent to $\omega$. Therefore tangents at $B_1$ and $C_1$ to $\omega$ meet at $BC$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(6cm); 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 = -5.680465499906917, xmax = 12.515849106577052, ymin = -1.671769681258384, ymax = 8.037903112857242; /* image dimensions */ pair A = (0.,6.), B = (-2.,0.), C = (7.,0.), M = (2.5,0.), B_1 = (-0.8993043727972788,3.3020868816081634), C_1 = (2.1423222095103633,4.163723820419689), Y = (1.6789724341706864,0.); draw(A--B--C--cycle, linewidth(2.)); /* draw figures */ draw(A--M, linewidth(1.)); draw(circle((0.4459902826727137,4.352495951113631), 1.7068031296360655), linewidth(1.)); draw(C_1--Y, linewidth(1.)); draw(B_1--Y, linewidth(1.)); draw((1.301775147928994,2.8757396449704142)--B, linewidth(1.)); draw((1.301775147928994,2.8757396449704142)--B_1, linewidth(1.)); draw((1.301775147928994,2.8757396449704142)--C_1, linewidth(1.)); draw((1.301775147928994,2.8757396449704142)--C, linewidth(1.)); /* dots and labels */ dot(A,dotstyle); label("$A$", (0.06181410951630287,6.143696592346206), NE * labelscalefactor); dot(B,dotstyle); label("$B$", (-2.220182722280457,-0.6), NE * labelscalefactor); dot(C,dotstyle); label("$C$", (7.101699760549445,-0.6), NE * labelscalefactor); dot(M,linewidth(4.pt) + dotstyle); label("$M$", (2.2990659053954796,-0.6), NE * labelscalefactor); dot((1.301775147928994,2.8757396449704142),linewidth(4.pt) + dotstyle); dot(B_1,dotstyle); dot(C_1,linewidth(4.pt) + dotstyle); label("$C_1$", (2.1946608215877843,4.279320095780226), NE * labelscalefactor); dot(Y,linewidth(4.pt) + dotstyle); label("$Y$", (1.5980603426866709,-0.6), NE * labelscalefactor); label("$A_{HM}$", (1.4041651870438088,2.6), NE * labelscalefactor); label("$B_1$", (-1.7,3), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Now suppose $\omega$ does not pass through $A_{HM}$. Therefore consider a homothety centered at $A$ mapping $\omega\cap AA_{HM}$ to $A_{HM}$. Let tangents at $B_1,C_1$ meet at $Y$ which lies on $BC$. Further let $B',C'$ be images of $B_1,C_1$ under the homothety. Let tangents at $B',C'$ to $\odot(AB'C')$ meet at $Y'$. As $A_{HM}\in \odot(AB'C')$, $Y'$ lies on $BC$. We have from the homothety $B_1Y||B'Y'$ and $C_1Y||C'Y'$, so the homothety maps $Y$ to $Y'$. But $Y$ and $Y'$ both lie on $BC$, so $\omega\cap AA_{HM} = A_{HM}$, so $\omega$ passes through $A_{HM}$ and we are done. $~\square$ Edit: 3000th post

Can anyone tell why this is g(5) and not a g(1) ?

Let $B_1C \cap BC_1 = \{ T \}$. Because of spiral similarity proprieties we have that $C_2B_1 \cap B_2C_1 = \{ Y \} \in (AB_1C_1)$. Tangents at $B_1$ and $C_1$ to $\omega$ intersect on $BC$. Because of the converse of Pascal theorem applied on $AC_1C_1TB_1B_1$ we get that $T \in \omega$. Now let $R$ be the second intersection point of $\omega$ and $(BCT)$. By easy angle chasing we get that $\angle BAR = \angle CBR$, and $\angle CAR = \angle BCR$, so $R$ is the $A$ - humpty point of $ABC$.