Nice problem! What number was it? Unfourtunately, I have a very long and ugly proof for this, but I hope it is worth mentioning. A shorter proof would definitely be more appropriate and appreciated:

This is my 500th post !

This was 5... so as long as you got the (apparently much easier) 1,2,4, you made the team by solving this one. I'm told the intended solution uses symmedians.

MellowMelon wrote: This was 5... so as long as you got the (apparently much easier) 1,2,4, you made the team by solving this one. I'm told the intended solution uses symmedians. Pretty much...I remember how distressed I was to find two geometry problems among the six total. Ah..here we go...here's their solution:

A very nice problem. This solution isn't nearly as nice as the official solution, it's a completely different approach: Lemma $ 1$: Consider a circle centre $ O$, radius $ r$, and a point $ A$ outside the circle. Let a line $ l$, parallel to $ AO$, intersect the circle at $ D, E$. If the perpendicular from $ A$ to $ l$ meets $ l$ at a point $ B$, choose $ C$ so that $ A$ is the midpoint of $ BC$. Let $ AD$ meet the circle again at $ G$, and $ AE$ meet the circle again at $ F$. Then $ G, F, C$ are collinear. Proof: Let's use inversion about a circle centre $ A$, radius $ \sqrt {AD.AG}$. Clearly, this inversion switches $ D,G$; $ F,E$, suppose $ C$ is taken to a point $ C'$. Proving that $ G, F,C$ are collinear reduces to proving that their images lie on a circle through $ A$, i.e. that $ DECC'$ is cyclic, or equivalantly $ BA*BC' = BD*BE$. But we note $ BD*BE = BO^2 - r^2 = BA^2 + AO^2 - r^2 = BA^2 + AC*AC' = BA^2 + BA*AC' = BA*(BA + AC') = BA*BC'$, as required. Here we use the fact that $ AC*AC' = AO^2 - r^2$, since both expressions are equal to the power of $ A$ with respect to the circle. Now let's solve the problem: Let the circle centre $ T$, radius $ TB = TC$ intersect $ AB$ again at $ B_{1}$ and $ AC$ again at $ C_{1}$. A quick angle chase tells us that $ B_{1}, C_{1}, T$ are collinear, and moreover, $ AB_{1}C_{1}$ is similar to triangle $ ACB$. Let $ B_{3}, C_{3}$ be the reflections of $ B_{1}, C_{1}$ about the line $ AT$ - by symmetry, we now have that $ AC_{3}B_{3}$ is similar to $ ABC$, $ T$ is the midpoint of $ B_{3}C_{3}$, and $ B_{1}C_{3}$, $ B_{3}C_{1}, AT$ are parallel. Let $ AB_{3}, AC_{3}$ intersect our circle (the circle centre $ T$, radius $ BT$), again at $ B'$ and $ C'$ respectively. Let the perpendicular from $ A$ to $ C_{1}B_{3}$ intersect $ C_{3}B_{1}$ at a point $ X$. From Lemma $ 1$ applied to this configuration, we have: $ B', C, X$ are collinear. But from Pascal applied to the $ 6$ points $ B_{3}, C_{3}, B, C, B_ {1}, B'$, we have that the $ 3$ points, $ BC \cap B_{3}C_{3}$, $ BB_{1} \cap B_{3}B'$, $ CB' \cap C_{3}B_{1}$, are collinear - i.e. $ A, X$ and $ B_{3}C_{3} \cap BC$ are collinear. If $ S'=B_{3}C_{3} \cap BC$ , we then have $ \angle TAS' = 90$. Since $ S'$ lies on $ BC$, going back to the problem statement, we have that, $ S' = S$. We also, the $ B_{3}$ in this diagram is $ B_{1}$ in the problem statement (not the $ B_{1}$ in my figure, sorry for the confusion.), and $ C_{3}$ in this diagram, is $ C_{1}$ in the problem statement. Since $ AB_{3}C_{3}$ is similar to $ ABC$ (already proven), the conclusion follows.

Nice problem! Could anyone tell me the author of this problem? Some more properties: Let $ SA \cap B_1C={K}$.Let K,A,B,C lie on a circle. And a similar one: Let $ BB_1 \cap CC_1={L}$.Then L is lie on the symmedian wrt triangle ABC and on the circumcircle of triangle ABC.

Let $\Gamma$ be the circle centered at $T$ with radius $TB$. Extend $AB$ and $AC$ to intersect $\Gamma$ at $B_2$ and $C_2$, respectively. It is clear that $\triangle ABC$ and $\triangle AC_2B_2$ are directly similar. Since $BC$ and $B_2C_2$ are antiparallel with respect to $AB$ and $AC$ and $AT$ is a symmedian of $\triangle ABC$, $AT$ is a median of $\triangle AC_2B_2$, i.e., $T$ is the midpoint of $B_2 C_2$. Reflect $B_2$ and $C_2$ across $AT$ to get $C_1'$ and $B_1'$, respectively. Since this reflection maps $\Gamma$ to itself, $B_1'$ and $C_1'$ both lie on $\Gamma$, and $B_1'$, $T$, and $C_1'$ are collinear. Since $\triangle AC_2 B_2$ and $\triangle AB_1' C_1'$ are directly similar, $\triangle ABC$ and $\triangle AB_1' C_1'$ are directly similar. Let $BC$ intersect $B_1' C_1'$ at $S'$ (which may be infinite.) Because $S' = BC \cap B_1' C_1'$, $T$ is the center of $BCC_1' B_1'$, and $A$ is the Miquel point of $BCC_1'B_1'$, we have $TA \perp AS'$. Hence, $S' = S$, whence $B_1 = B_1'$ and $C_1 = C_1$', so $\triangle ABC \sim \triangle AB_1C_1$, as desired.

Let $D$ and $E$ be on $AB$ and $AC$ such that $BCED$ is cyclic and $DE$ passes through $T$. By tangent angle theorem, $\angle{BCT}=\angle{CBT}=\angle{BAC}$. Hence $\angle{BDT}=\angle{ACB}=180^\circ - \angle{CBT} - \angle{ABC}=\angle{TBD}$. Hence $TD=TB=TC$. By the same argument, it follows that $TD=TB=TC=TE$. Let $M$ be the midpoint of $BC$. Since $BCED$ is cyclic, triangles $ABC$ and $AED$ are similar. Since $T$ is the midpoint of $DE$, it follows that $TB/TA=TD/TA=MC/MA$. Since $BTC$ is isosceles, $TM \perp BC$ which implies that $TMAS$ is cyclic. Hence $\angle{AMS}=\angle{ATS}$ which since $C_1 T/TA=BT/TA=MC/MA$ implies that triangles $ATC_1$ and $AMC$ are similar. By the same argument, triangles $ATB_1$ and $AMB$ are similar. This implies that $ABC$ and $AB_1 C_1$ are similar.

Dear Mathlinkers, the circle centered at T passing through B1, C1, B and C leads to the Boutin's theorem... then we can think to the Liquel situation... in order to have a synthetic proof. Sincerely Jean-Louis

Not so easy!

Zhero wrote: Let $\Gamma$ be the circle centered at $T$ with radius $TB$. Extend $AB$ and $AC$ to intersect $\Gamma$ at $B_2$ and $C_2$, respectively. It is clear that $\triangle ABC$ and $\triangle AC_2B_2$ are directly similar. Since $BC$ and $B_2C_2$ are antiparallel with respect to $AB$ and $AC$ and $AT$ is a symmedian of $\triangle ABC$, $AT$ is a median of $\triangle AC_2B_2$, i.e., $T$ is the midpoint of $B_2 C_2$. Reflect $B_2$ and $C_2$ across $AT$ to get $C_1'$ and $B_1'$, respectively. Since this reflection maps $\Gamma$ to itself, $B_1'$ and $C_1'$ both lie on $\Gamma$, and $B_1'$, $T$, and $C_1'$ are collinear. Since $\triangle AC_2 B_2$ and $\triangle AB_1' C_1'$ are directly similar, $\triangle ABC$ and $\triangle AB_1' C_1'$ are directly similar. Let $BC$ intersect $B_1' C_1'$ at $S'$ (which may be infinite.) Because $S' = BC \cap B_1' C_1'$, $T$ is the center of $BCC_1' B_1'$, and $A$ is the Miquel point of $BCC_1'B_1'$, we have $TA \perp AS'$. Hence, $S' = S$, whence $B_1 = B_1'$ and $C_1 = C_1$', so $\triangle ABC \sim \triangle AB_1C_1$, as desired. how you prove $A$ is Miquel of $BCC_1B_1$ very nice.

If $K = \overline{BB_1} \cap \overline{CC_1}$ then $\angle BKC = \tfrac{\widehat{B_1C_1} - \widehat{BC}}{2} = \tfrac{1}{2} \left( 180 - \angle BTC \right) = \angle BAC$, so $B,K,A,C$ are concyclic. So $A$ is the unique point with $AS \perp AT$ and $A \in (KBC)$; therefore $A$ is the Miquel point of $B_1BCC_1$, i.e. the center of the spiral similarity giving $\triangle ABC \sim \triangle AB_1C_1$. [asy][asy]/* DRAGON 0.0.9.6 Homemade Script by v_Enhance. */ import olympiad; import cse5; size(12cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(10pt)); /* Initialize Objects */ pair B_1 = (3.0, -3.0); pair C_1 = (11.0, -3.0); pair T = midpoint(B_1--C_1); pair B = (6.451128850753037, 0.9621636086265188); pair C = (10.082123836968163, -0.45038970554465996); pair S = IntersectionPoint(Line(B,C,lisf),Line(B_1,C_1,lisf)); pair K = IntersectionPoint(Line(B,B_1,lisf),Line(C,C_1,lisf)); pair A = foot(T,K,S); /* Draw objects */ draw(CirclebyPoint(T,B_1), rgb(0.0,0.6,0.6)); draw(circumcircle(A,B,C), rgb(0.0,0.6,0.8) + linetype("4 4")); draw(T--B, rgb(0.0,0.8,0.8)); draw(T--C, rgb(0.0,0.8,0.8)); draw(K--S, rgb(0.0,0.6,0.8)); draw(circumcircle(K,B_1,C_1), rgb(0.0,0.8,0.8) + linewidth(1.0) + dotted); draw(T--A, rgb(0.0,0.8,0.8) + linetype("4 4")); draw((abs(dot(unit(T-A),unit(S-A))) < 1/2011) ? rightanglemark(T,A,S) : anglemark(T,A,S), rgb(0.0,0.8,0.8)); draw(B--B_1, rgb(1.0,0.6,0.0) + linewidth(1.0)); draw(B_1--C_1, rgb(1.0,0.6,0.0) + linewidth(1.0)); draw(C_1--C, rgb(1.0,0.6,0.0) + linewidth(1.0)); draw(C--B, rgb(1.0,0.6,0.0) + linewidth(1.0)); draw(K--B, rgb(0.0,0.8,0.8) + dashed); draw(K--C, rgb(0.0,0.8,0.8) + dashed); draw(C--S, rgb(0.0,0.8,0.8) + dashed); draw(C_1--S, rgb(0.0,0.8,0.8) + dashed); /* Place dots on each point */ dot(B_1); dot(C_1); dot(T); dot(B); dot(C); dot(S); dot(K); dot(A); /* Label points */ label("$B_1$", B_1, lsf * dir(225)); label("$C_1$", C_1, lsf * dir(-45)); label("$T$", T, lsf * dir(-90)); label("$B$", B, lsf * dir(140)); label("$C$", C, lsf * dir(70) * 1.414); label("$S$", S, lsf * dir(-45)); label("$K$", K, lsf * dir(60)); label("$A$", A, lsf * dir(45)); [/asy][/asy]

But how do you know $K, A$ and $S$ are collinear?

Sorry, $TA \perp AS$ was meant rather than $TA \perp KS$. We don't need this collinearity at all since Miquel points does this all for us. I will edit my post when I get home, seems I am unable to do so on my phone.

Let $X$ be the intersection of $\omega$ and $AT$ and let $BX$ intersect $ST$ at $Y$ and $CX$ intersect $ST$ at $Z$ and let the foot of perpendicular from $X$ on $BC$ be $R$.Not hard to observe that $\angle ABX =\angle ARC,, .ARXS$ is cyclic $\Rightarrow \angle ABX=\angle ATY \Rightarrow ABTY$ is cyclic.$\Rightarrow \angle XBT=\angle XAB=\angle BYT \Rightarrow Y \equiv C_1$ .Similarly $Z \equiv B_1.,, \Rightarrow \angle C_1BB_1=90 ,\angle SBB_1=90+\angle C_1BS=90+\angle XB_1C_1=90+\angle XAB_1=\angle B_1AS \Rightarrow ABB_1S$is cyclic.Similarly $ACC_1S$ is cyclic. $\Rightarrow \angle ABC=\angle AB_1C_1$ and $\angle ACB=\angle AC_1B_1$ and we are done.

v_Enhance wrote: So $A$ is the unique point with $AS \perp AT$ and $A \in (KBC)$ Well technically, there are two points satisfying this (the intersections of $(KBC)$ and the circle with diameter $(ST)$.) I'm not sure exactly how to prove it's not the other intersection...

pi37 wrote: I'm not sure exactly how to prove it's not the other intersection... Switch $B$ and $C$. To elaborate, I inadvertently assumed that $A$ was on the major arc of $\widehat{BC}$. If in fact $\angle A \ge 90^{\circ}$, just switch the roles of $B$ and $C$ (but not $B_1$ and $C_1$), and everything works out as before. Note that in this case we get something like $\triangle ABC \sim \triangle AC_1B_1$.

this diagram seems to give a contradiction? I think that we need to assume wLOG that AB>AC, or else B_1 and C_1 must be exchanged...

Konigsberg wrote: I think that we need to assume wLOG that AB>AC ... This follows from the fact that $S$ lies on ray $BC$, meaning it is not the case that $B$ lies between $S$ and $C$.

Apologies for any typos. Lemma: If $(BB_1CC_1)\cap (AST) =D,E$ and $O$ is the circumcenter of $\triangle ABC$, then $O\in DE$. This follows since $OA$ is tangent to $(AST)$ and $OC$ is tangent to $(BB_1CC_1)$. Now if $P=DE\cap (ABC)$ on the same side of $BC$ as $A$, then $AS$ also passes through $P$. If $Q$ is diametrically opposite $P$, then $\angle C_1BB_1=\tfrac{\pi}{2}=\angle QBB_1$ so $Q\in BC_1$. So $BB_1$ also passes through $P$. By converse of radical axis on $(BB_1CC_1), (AST), ABB_1S$ then $ABB_1S$ is cyclic. Hence $\angle AB_1C_1=\angle ABC$. Similarly $ACC_1S$ is cyclic, so $\angle AC_1B_1=\angle ACB$, so $\triangle ABC\sim \triangle AB_1C_1$.

Notice that $BT=B_1T=CT=C_1T$ which implies $T$ is the circumcentre of $BCC_1B_1$ Let $BB_1 \cap CC_1=X$ and we know that $BC \cap B_1C_1=S$ By angle chasing, Let $\angle BB_1T=x$ and $\angle CC_1T=y$ $$\angle BTC=2x+2y-180 \implies \angle CBT=\angle BCT=180-(x+y)=\angle BAC=\angle BXC$$where the last part comes from the cyclic quadrilateral. and therefore $X \in \odot(ABC)$ Clubbing all the information it follows that, $A$ is the Miquel point of cyclic quadrilateral $BCC_1B_1$ and now taking spiral symmetry we get that $\triangle ABC \sim \triangle A_1B_1C_1$ as desired.

Easy one Let $BC_1$ and $CB_1$ meet at $K$. $\angle BKC = \angle 180 - \angle BC_1T - \angle CB_1T = \angle TBC = \angle BAC$ so $BAKC$ is cyclic so $A$ is Miquel point of $BCB_1C_1$. Now the problem is almost Done the rest is just angle chasing...

Let $F=SA\cap(ABC)$, $L=AT\cap(ABC)$, $Q=FL\cap BC$ and $M=SL\cap (ABC)$ Claim 1. M-Q-A Pf. Notice that $$-1=(A,L;B,C)\stackrel{F}{=}(S,Q;B,C)\stackrel{L}{=}(M,F;B,C)\stackrel{A}{=}(MA\cap BC,S;B,C)$$which implies that $MA\cap BC=Q$ See how $(M,F;B,C)=-1$ also shows that $T-M-F$, and now by looking at cyclic quadrilateral $(FALM)$ by Brocard we get that $ST$ is the polar of $Q$. Thus, if we let $O$ be the circumcenter of this circumference, then $OQ\perp ST$, but as $\angle LAF= 90$ we get that $FL\perp ST$. Deffine $N=FL\cap ST$, $C_1*=FC\cap ST$, and $B_1*=C_1*L\cap ST$. By Brocard on $(BFCL)$ we conclude $B_1*-B-F$, and thus $L$ is the orthocenter of $FC_1*B_1*$, which implies $C_1*=C_1$ and $B_1*=B_1$. Now we're done as $L$ and $A$ are inverses with respect to $(C_1BCB_1)$, and now it is well-known that $A$ is the Miquel point that sends $BC$ to $B_1C_1$ which implies the desired result.

A quick angle chasing gives that $M=CC_1\cap BB_1$ lies on $(ABC)$. We will prove that if $N=CB_1\cap BC_1$, then $A, T, N$ are collinear, implying that $A$ is a Miquel point of $C_1CBB_1$. Let $AB\cap ST=P, AT\cap (ABC)=N_1$ and $BN_1\cap ST=C_0$. Then $-1=(B,C;A,N_1)\stackrel{B}{=} (T,S;P,C_0)$ and $\angle SAT=90^\circ$,so $AT$ is a bisector of $\angle BAC_0$. Then $\angle C_0AT=\angle TAB=\angle N_1BT=\angle C_0BT$, so $AC_0TB$ is cyclic, implying that $TC_0=TB$, so $C_0\equiv C_1$.Similarly, $CB_1\cap AT\in (ABC)$, so $N=N_1$ and $N\in AT, N\in (ABC)$, so we are done.

Not hard at all. Suppose that B2 and C2 are the intersection points of ST and (SAB) and (SAC). Let M be the midpoint of BC. Since that TM is perpendicular to BC, we have that ASTM is cyclic. It's well-known that AT is a symmedian of triangle ABC. After that it's easy using just anglechase prove that TC2=BT=TB2.

[asy][asy] /* coordinates from geogebra */ import geometry; size(10cm); pair A,B,C,T,S,B_1,C_1,D; A=(-0.89,1.11); B=(1.96,-1.44); C=(-1.44,-1.06); T=(-0.1,-4.5); S=(-9.76,-0.13); B_1=(3.26,-6.02); C_1=(-3.47,-2.98); D=(1.16,1.4); fill(A--B--C--cycle,white+white); fill(B--C--C_1--B_1--cycle,white+white+white+white); draw(circle(A,B,C)); draw(circle(B_1,C_1)); draw(circle(A,C_1,B_1));draw(C_1--D--B_1--cycle,green); draw(A--B--C--cycle,red); draw(A--B_1--C_1--cycle,red); draw(D--S); draw(S--C); draw(D--B); draw(D--C); draw(B--T); draw(C--T); draw(A--T); draw(S--C);draw(S--C_1);dot(A^^B^^C^^T^^S^^B_1^^C_1^^D); label("$A$",A,NE); label("$B$",B,2W); label("$C$",C,NE); label("$T$",T,SW); label("$S$",S,2E); label("$B_1$",B_1,SW); label("$C_1$",C_1,SE); label("$D$",D,2N); [/asy][/asy] $B_1B \cap C_1C= D$ . Chase, $\angle BDC= 180^{\circ}-(\angle DB_1T +DC_1T) = \frac{180^{\circ}- \angle BTC}{2}=\angle BCT$.Thus, $ACBD$ is cyclic. As $T$ is the center$(B_1C_1CB)$ , and as $A,D,S$ are collinear by radical axes ,we use IMO 1985/5 , which proves $A$ is the miquel point of the complete quad $B_1C_1CBDS $. The property that the miquel point is the centre of the spiral sim. which sends $BC$ to $B_1C_1$ finishes the proof.

Thanks to SatisfiedMagma for teaching me how to make good asy diagrams. [asy][asy] import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2, xmax = 4, ymin = -4.7, ymax = 2; /* image dimensions */ draw((0.33709926477356134,0.6527195501031977)--(0.2983270446038727,0.5051548519589105)--(0.44589174274816,0.4663826317892219)--(0.4846639629178486,0.6139473299335092)--cycle, linewidth(0.75) + blue); /* draw figures */ draw((0.4846639629178486,0.6139473299335092)--(-0.6435760050714912,-0.10350482168263458), linewidth(0.5)); draw((-0.6435760050714912,-0.10350482168263458)--(0.5736726499363176,-0.10910136722290036), linewidth(0.5)); draw((0.5736726499363176,-0.10910136722290036)--(0.4846639629178486,0.6139473299335092), linewidth(0.5)); draw(circle((-0.033620893008616766,0.18314254712320893), 0.673952485546976), linewidth(0.5)); draw((-0.6435760050714912,-0.10350482168263458)--(-0.04083567317250101,-1.386072138521609), linewidth(0.5)); draw((-0.04083567317250101,-1.386072138521609)--(0.5736726499363176,-0.10910136722290036), linewidth(0.5)); draw((-0.04083567317250101,-1.386072138521609)--(0.4846639629178486,0.6139473299335092), linewidth(0.5)); draw((-0.2732001930556983,0.8130740606736085)--(3.283969388430581,-0.12156250165275902), linewidth(0.5)); draw((3.283969388430581,-0.12156250165275902)--(0.5736726499363176,-0.10910136722290036), linewidth(0.5)); draw(circle((-0.04083567317250101,-1.386072138521609), 1.4171361366930244), linewidth(0.5)); draw((-0.2732001930556983,0.8130740606736085)--(-1.3654078509418188,-1.8898413076860952), linewidth(0.5)); draw((1.2837365045968172,-0.8823029693571232)--(-0.2732001930556983,0.8130740606736085), linewidth(0.5)); draw((-1.3654078509418188,-1.8898413076860952)--(3.283969388430581,-0.12156250165275902), linewidth(0.5)); draw((-0.2732001930556983,0.8130740606736085)--(0.48686186746905835,-1.1853750711357969), linewidth(0.5)); draw((-0.6435760050714912,-0.10350482168263458)--(1.2837365045968172,-0.8823029693571232), linewidth(0.5) + linetype("4 4")); draw((0.5736726499363176,-0.10910136722290036)--(-1.3654078509418188,-1.8898413076860952), linewidth(0.5) + linetype("4 4")); draw(circle((1.9313054361793436,0.42502876216911906), 1.4589248703776139), linewidth(0.5) + blue); draw(circle((1.3118282790787361,-1.9326634023434865), 2.677578575444292), linewidth(0.5) + blue); /* dots and labels */ dot((0.4846639629178486,0.6139473299335092),linewidth(3pt) + dotstyle); label("$A$", (0.5270237491555813,0.7436458119447124), NE * labelscalefactor); dot((-0.6435760050714912,-0.10350482168263458),linewidth(3pt) + dotstyle); label("$B$", (-0.961798164673135,-0.08347747351568333), NE * labelscalefactor); dot((0.5736726499363176,-0.10910136722290036),linewidth(3pt) + dotstyle); label("$C$", (0.6564865242711218,-0.04032321514383659), NE * labelscalefactor); dot((-0.04083567317250101,-1.386072138521609),linewidth(3pt) + dotstyle); label("$T$", (-0.10590537363150583,-1.6298383985068579), NE * labelscalefactor); dot((-0.2732001930556983,0.8130740606736085),linewidth(3pt) + dotstyle); label("$Q$", (-0.4511394406062806,0.9090704690367915), NE * labelscalefactor); dot((3.283969388430581,-0.12156250165275902),linewidth(3pt) + dotstyle); label("$S$", (3.3680124253021653,-0.07628509712037554), NE * labelscalefactor); dot((-0.033620893008616766,0.18314254712320893),linewidth(3pt) + dotstyle); label("$O$", (-0.005212104097196526,0.18264045311070484), NE * labelscalefactor); dot((0.20595840703846502,-0.4467889664271907),linewidth(3pt) + dotstyle); label("$P$", (0.23213631694796114,-0.3639801529326871), NE * labelscalefactor); dot((-1.3654078509418188,-1.8898413076860952),linewidth(3pt) + dotstyle); label("$B_1$", (-1.7529595681569938,-1.9822648418769395), NE * labelscalefactor); dot((1.2837365045968172,-0.8823029693571232),linewidth(3pt) + dotstyle); label("$C_1$", (1.4044936693831338,-1.0616406632775426), NE * labelscalefactor); dot((0.48686186746905835,-1.1853750711357969),linewidth(3pt) + dotstyle); label("$S^*$", (0.48386949078373437,-1.4068747302523164), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $Q$ denote the $P$ antipode in $\odot(ABC)$. Now firstly we thus have that $\overline{Q-A-S}$ are collinear. Now Pascal on $BBPCCQ$ gives that $\overline{T-BP\cap CQ-CP\cap BQ}$ are collinear and then Pascal on $QAPBCC$ gives that $\overline{S-T-BP\cap QC}$ are collinear and thus we conclude that $BP\cap QC$ and $CP\cap BQ$ lie on $ST$. We redefine $B_1=QB\cap CP$ and $C_1=BP\cap QC$. Now note that, $\measuredangle B_1CC_1=\measuredangle PCC_1=\measuredangle PCQ=90^\circ$ and similarly that $\measuredangle B_1BC_1=90^\circ$. So we have that the center of $\odot(BCC_1B_1)$ is the midpoint of $B_1C_1$ and that it lies on the perpendicular bisector of $BC$ and thus $T$ becomes the center. This indeed matches the definitions of the points $B_1$ and $C_1$ as in the problem statement. Now, we have, $\measuredangle CAS=\measuredangle CAQ=\measuredangle CBQ=\measuredangle CBB_1=\measuredangle CC_1B_1=\measuredangle CC_1S\implies ACC_1S$ is cyclic. Similarly, $ABB_1S$ is also cyclic. Now note that the circles $\left\{\odot(BCB_1C_1),\odot(ABC)\right\}$ are orthogonal to each other. So upon performing an Inversion $\mathbf I(\odot(BCC_1B_1))$, we get that $A\leftrightarrow P$. Now $S^*$ denote the inverse of $S$ after the Inversion. Note that then as we have $ACC_1S$ is cyclic, we thus have $PCC_1S^*$ is also cyclic, and similarly $PBB_1S^*$ is also cyclic. Now a Radax on $\left\{\odot(PBB_1S^*),\odot(PCC_1S^*),\odot(BCC_1B_1)\right\}$ gives that $\overline{Q-P-S^*}$ are collinear. We finally have $SA\cdot SQ=SC\cdot SB=SC_1\cdot SB_1\implies AQB_1C_1$ is also cyclic. Thus to finish, we have that as $Q=BB_1\cap CC_1$, thus as $A=\odot(QBC)\cap\odot(QB_1C_1)$, it becomes the miquel point mapping $\overline{BC}\mapsto\overline{B_1C_1}$ and thus we are done.

Let $D = \overline{BB_1} \cap \overline{CC_1}$ and $E = \overline{B_1C} \cap \overline{BC_1}$. Note that $T$ is the center of $(BB_1C_1C)$ so $\overline{B_1B} \perp \overline{BC_1}$ and $\overline{B_1C} \perp \overline{CC_1}$. Hence $E$ is the orthocenter of $\triangle KB_1C_1$, and $\overline{TB}$ and $\overline{TC}$ are tangent to $(DBEC)$ by three tangents lemma, which is enough to imply that $(DBEC)$ and $(ABC)$ coincide. Thus, we now know that $A$ is one of the intersections of $(ST)$ with $(DBEC)$; denote these two intersections as $A_1$ and $A_2$. Then it's well-known that $A_1$ and $A_2$ are just the $K$-Humpty and Queue points of $\triangle DB_1C_1$ (as $S$ is the $D$-ex point); WLOG assume that $A_1$ is the Humpty point and $A_2$ is the Queue point. If $A = A_1$, then $A$ lies on $(EB_1C_1)$ so $\triangle ABC \sim \triangle AC_1B_1$. If $A = A_2$, then $A$ lies on $(DB_1C_1)$ so $\triangle ABC \sim \triangle AB_1C_1$.

hello Let $G=BB_1\cap CC_1$, notice that we have the good old orthocenter configuration hence $G\in \omega$. Let $A'$ be the foot of the altitude from $T$ to $SG$, it's well-known that $A'$ is the Miquel Point of $BCC_1B_1$. Hence $A'=\omega\cap (ST)$ which is also just $A$ so we're done?? Wait I think I fakesolved

Note that $BCB_1C_1$ is cyclic with circumcenter $T$. let $P$ be the intersection of $BC_1$ and $CB_1$. It is well known that $TB$ and $TC$ are tangent to $(PBC)$ (this can be shown by angle chasing and showing that $\angle BTC=2\angle B_1+2\angle C_1-180$). Hence, $P$ lies on $(ABC)$. Thus, $A$ lies on the circle with diameter $ST$ as well as the $(PBC)$. Thus, it is the Miquel point, which establishes the spiral similarity, as desired.

We first make the following claims: Claim 1: $B_1BCC_1$ is cyclic with center $T$. Proof: Obviously $T$ is the center if we indeed somehow have $B_1BCC_1$ is cyclic. Now observe that $\angle CTB_1 = 2 \angle CC_1T$. Hence we have that \begin{align*} \angle CTB_1 &= (180^{\circ} - 2 \angle TB_1B) + (180 - 2 \angle A) \\ \implies \angle CC_1T &= (180^{\circ} - 2 \angle TB_1B) + (180^{\circ} - 2 \angle A) \\ \implies \angle CC_1T &= 180^{\circ} - \angle TB_1B - \angle A \\ &= 180^{\circ} - \angle TBB_1 - \angle TBC \\ &= 180^{\circ} - \angle B_1BC. \phantom{c} \square \end{align*} Claim 2: Let $V = B_1B \cap C_1C$. Then $V$ lies on $(ABC)$. Proof: \begin{align*} \angle BVC &= 180^{\circ} - (\angle BB_1T + \angle CC_1T) \\ &= \angle A, \end{align*}from the first claim. $\square$ Make the crucial observation that as now $B_1BCC_1$ is cyclic, it must be that $A$ is its Miquel point, as $SV$ meets $(ABC)$ at one other point other than $V$, and must also have the property that the line that point forms with the center of this quadrilateral is perpendicular to $SV$. Now a trivial spiral similarity at $A$ sends $B_1C_1$ to $BC$, providing us with the desired similarity. $\blacksquare$

The length conditions can be interpreted by drawing a circle centered at $T$ passing through $B$. We show that $A$ is the Miquel point of $BB_1C_1C$. Let $Q = \overline{BB_1} \cap \overline{CC_1}$. Then, by the three tangents configuration, $Q$ lies on $(ABC)$. Fixing $BB_1C_1C$, only one point $A$ satisfies $\angle BAC$ being acute, $A$ being on $(QBC)$, and $\angle TAS = 90$. The Miquel point also satisfies these criteria whenever $BB_1C_1C$ is a quadrilateral following from the problem statement (based on various properties listed in EGMO), so $A$ must be the Miquel point. Note: for me, the motivation for $A$ being a Miquel point came from the fact that $E = \overline{B_1C} \cap \overline{C_1B}$ lies on $(ABC)$ when drawn and looks like the inverse of $A$ (orthogonal circles).

Suppose that $AS, AT$ intersect $\omega$ at $U, V$. Redefine $B_1$ is intersection of $UB$ and $CV;$ $C_1$ is intersection of $UC$ and $BV$. Applying Pascal theorem for $\left( \begin{array}{l} B\ C\ V \\ C\ B\ U \end{array} \right),$ we have $B_1, T, C_1$ are collinear. But $\angle{B_1BC_1} = \angle{B_1CC_1} = 90^{\circ}$ then $T$ is midpoint of $B_1C_1$. It's easy to see that $V$ is orthocenter of $\triangle UB_1C_1,$ hence $A \in (UB_1C_1)$. From this, we have $\angle{ABC} = 180^{\circ} - \angle{AUC} = \angle{AB_1C_1}$ and $\angle{ACB} = \angle{AUB} = \angle{AC_1B_1}$. Therefore $\triangle ABC \sim \triangle AB_1C_1$. Since $\overline{SA} \cdot \overline{SU} = \overline{SB} \cdot \overline{SC}$ we have $S$ lies on radical axis of $(B_1BCC_1)$ and $(UB_1C_1)$ or $S \in B_1C_1$

Let $BB_1 \cap CC_1 = D$ and let $H$ be the orthocentre of triangle $DB_1C_1$. Then by Brokard's triangle $DHS$ is self-polar. From here we have that $A$ is either the miquel point of quadrilateral $BCC_1B_1$, or the $D$ humpty point in triangle $DB_1C_1$, and we can check the conditions to hold in each case.