GOTEEM #5: Circumcircle passes through fixed point

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GOTEEM #5: Circumcircle passes through fixed point

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Source: GOTEEM: Mock Geometry Contest

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tworigami

844 posts

tworigami

#1 h Jan 2, 2020, 8:52 PM • 9 Y

Y by rocketscience, NJOY, amar_04, GeoMetrix, mijail, CyclicISLscelesTrapezoid, Adventure10, Mango247, Rounak_iitr

Let $ABC$ be a triangle and let $B_1$ and $C_1$ be variable points on sides $\overline{BA}$ and $\overline{CA}$ , respectively, such that $BB_1 = CC_1$ . Let $B_2 \neq B_1$ denote the point on $\odot(ACB_1)$ such that $BC_1$ is parallel to $B_1B_2$ , and let $C_2 \neq C_1$ denote the point on $\odot(ABC_1)$ such that $CB_1$ is parallel to $C_1C_2$ . Prove that as $B_1, C_1$ vary, the circumcircle of $\triangle AB_2C_2$ passes through a fixed point, other than $A$ .

Proposed by tworigami

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rocketscience

466 posts

rocketscience

#2 h Jan 3, 2020, 2:59 AM • 3 Y

Y by tworigami, mijail, Adventure10

My submission, with typos corrected (I think). Also my favorite problem of the test.

The following solution will make use of the spiral similarity lemma: if $P: XY \mapsto WZ$ is a spiral similarity centered at $P$ mapping $\overline{XY}$ to $\overline{WZ}$ , then $(PXW)$ and $(PYZ)$ meet a second time at $XY \cap WZ$ .

Let $M$ be the midpoint of arc $\widehat{BAC}$ on $(ABC)$ . It's easy to see that $M$ is the center of spiral congruence taking $BB_1 \mapsto CC_1$ , so $MB_1 = MC_1.$

Define $P = (ABC_1) \cap (AC_1B)$ to be the second intersection of those circles; note that $P$ is the center of the spiral congruence $BB_1 \mapsto C_1C$ . Thus $\mathrm{dist}(P, AB)=\mathrm{dist}(P, AC)$ , which implies that $P$ lies on the internal $A$ -angle bisector. Next we claim that $P = C_1B_2 \cap B_1C_2$ . Indeed, the given parallel lines imply $\triangle AC_1C_2 \sim \triangle AB_1B_2$ , i.e. $A: C_2B_1 \mapsto C_1B_2$ , meaning $P = (ABC_1) \cap (AC_1B)$ is the intersection of $C_1B_2$ and $B_1C_2$ , as desired. This implies that $B_1C_1B_2C_2$ is cyclic: $\angle B_1C_2C_1 = \angle BAP = \angle CAP = \angle C_1B_2B_1 = \frac 12 \angle A$ .

I'm pretty sure the rest is angle-chaseable, but I succumb to my projective/inversive tendencies. Plus, this is cool. Let $O$ be the circumcenter of $\Gamma = B_1C_1B_2C_2$ . Radical center on $\Gamma, (ABC_1), (ACB_1)$ yields a point $Q = C_1C_2 \cap B_1B_2 \cap AP$ . Consider the inversion at $P$ which fixes $\Gamma$ . It swaps $(ABC_1) \leftrightarrow B_1B_2$ and $(ACB_1) \leftrightarrow C_1C_2$ , and thus $A \leftrightarrow Q$ . Brokard implies $Q$ lies on the polar of $P$ wrt $\Gamma$ , and since the image of the polar of $P$ under our inversion is $(PO)$ , we have $A \in (PO)$ , i.e. $\angle PAO = 90^\circ$ .

We now have two characterizations that uniquely determine $O$ : first, it lies on the perpendicular bisector of $\overline{B_1C_1}$ , and second it lies on the line through $A$ perpendicular to $AP$ . Surprise, we claim $O = M$ . Indeed, we verified earlier that $MB_1 = MC_1$ , and also $\angle MAP = 90^\circ$ since $AP$ is the internal angle bisector.

Armed with this, we can finally slay the problem. The required fixed point is $M$ ; observe that $AP$ is the internal angle bisector of $\angle B_2AC_2$ by some earlier angle work, so $AM$ is its external angle bisector; also, $M$ is the center of $\Gamma$ so $MB_2 = MC_2$ , and together these imply that $AMB_2C_2$ is cyclic, as desired.

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Sugiyem

115 posts

Sugiyem

#3 h Jan 3, 2020, 7:34 AM • 3 Y

Y by Adventure10, Mango247, TheHimMan

Let $D$ be the midpoint of arc $\widehat{BAC}$ of $(ABC)$ , $E$ the second intersection of $(ABC_{1})$ and $(ACB_{1})$ , $F$ the intersection of $C_{1}C_{2}$ and $B_{1}B_{2}$ .

It's easy to check by spiral similarity that $D$ is the second intersection of $(AB_{1}C_{1})$ and $(ABC)$ .

By spiral similarity also we can get that $\triangle EBB_{1}\thicksim \triangle EC_{1}C$ .
Thus $\frac{dist (E,AB)}{dist (E,AC)}=\frac{BB_{1}}{CC_{1}}=1$ .
So $E$ lies on the internal bisector of $\angle BAC.$

$\textbf{Claim 1}$ : $E$ is the intersection of $B_{1}C_{2}$ and $C_{1}B_{2}$
$\textbf{Proof}$ :
Notice that $\angle AB_{1}B_{2}=\angle ABC_{1}=\angle AC_{2}C_{1}$ . Moreover $\angle AC_{1}C_{2}=\angle ACB_{1}=\angle AB_{2}B_{1}$
So $\triangle AC_{2}C_{1}\thicksim \triangle AB_{1}B_{2}$
Which is equivalent to $\triangle AB_{1}C_{2}\thicksim \triangle AB_{2}C_{1}$ .
$\angle AB_{1}C_{2} +\angle AB_{1}E=\angle AB_{2}C_{1} +\angle AB_{1}E=180^{\circ}$ because $AB_{1}EB_{2}$ is cyclic.
So $E,B_{1},C_{2}$ are collinear. Analagously we can prove that $E,C_{1},B_{2}$ are collinear. So the claim 1 is proven.

Now let's move on to the second claim.

$\textbf{Claim 2}$ : $B_{1}C_{1}B_{2}C_{2}$ lies on a circle centered at $D$ .
$\textbf{Proof}$ :
Note that $\angle B_{1}C_{2}C_{1}=\angle EC_{2}C_{1}=\angle EAC_{1}=\frac{\angle B_{1}AC_{1}}{2}=\frac{\angle B_{1}DC_{1}}{2}$ .
With the same way, we can get that $\angle B_{1}B_{2}C_{1}=\frac{\angle B_{1}DC_{1}}{2}$
So the claim 2 is done.

$\textbf{Claim 3}$ : $F$ lies on the internal bisector of $\angle BAC$
$\textbf{Proof}$ : This is equivalent to show that $A,F,E$ are collinear.
Because $A$ is the center of spiral similarity which sends $B_{1}C_{2}$ to $B_{2}C_{1}$ , we have that $AFB_{1}C_{2}$ and $AFC_{1}B_{2}$ are both cyclic.
Thus by radical center at $(B_{1}C_{1}B_{2}C_{2})$ , $(AFB_{1}C_{2})$ and $(AFC_{1}B_{2})$ , we have $A,F,E$ collinear.
Therefore the claim 3 is done too.

Now we'll prove that point $D$ is the fixed point that the problem wants,
Define $X$ to be the intersection of $B_{1}C_{1}$ and $B_{2}C_{2}$
By Brokard's at $(B_{1}C_{1}B_{2}C_{2})$ , we have $EF$ is the polar of $X$ WRT $(B_{1}C_{1}B_{2}C_{2})$ .
Hence, $DX\perp EF$ , but $AD$ also perpendicular to $EF$ because $AD$ is the external bisector of $\angle BAC$ and $EF$ is the internal bisector of $\angle BAC$ .
So, $D,A,X$ collinear.
Therefore, because $ADB_{1}C_{1}$ and $B_{1}C_{1}B_{2}C_{2}$ are both cyclis, we have:
$XA\times XD=XC_{1}\times XB_{1}=XB_{2}\times XC_{2}$ .
So, $D$ lies on $(AB_{2}C_{2})$ as we want.

This post has been edited 2 times. Last edited by Sugiyem, Jan 3, 2020, 7:36 AM

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Th3Numb3rThr33

1247 posts

Th3Numb3rThr33

#4 h Jan 6, 2020, 1:04 AM • 2 Y

Y by Adventure10, Mango247

The inversion is probably unnecessary, but I don't like circles so...

The fixed point is the midpoint $T$ of major arc $\widehat{BAC}$ .

Let $T'$ denote the Miquel point of $BB_1C_1C$ , which is on $(ABC)$ and sends $\overline{BB_1}$ to $\overline{CC_1}$ . But $BB_1 = CC_1$ by definition, so $T'B = T'C$ and thus $T = T'$ . So $T$ lies on $(AB_1C_1)$ .

Now take a $\sqrt{bc}$ -inversion at $A$ and reflect about the angle bisector of $\angle BAC$ , yielding the following problem (with stars dropped).

Quote:

Let $ABC$ be a triangle, and $T$ be the foot of the $A$ -external bisector onto $\overline{BC}$ . A variable line $\ell$ through $T$ meets $\overline{AB}$ and $\overline{AC}$ at $B_1$ and $C_1$ , respectively. Let $B_2 \neq B_1$ denote the point on $CB_1$ such that $\measuredangle AB_2B_1 = \measuredangle AC_1B$ , and let $C_2 \neq C_1$ denote the point on $BC_1$ such that $\measuredangle AC_2C_1 = \measuredangle AB_1C$ . Prove that $T,B_2,C_2$ are collinear.

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pinetree1

1207 posts

pinetree1

#5 h Jan 8, 2020, 9:10 PM • 3 Y

Y by tworigami, Adventure10, Rounak_iitr

Here's my solution with a different finish (which might actually be essentially the same, but phrased differently).

Let $L$ be the midpoint of arc $BAC$ and $R=\overline{AT}\cap\overline{B_1C_1}$ , and denote by $\omega_B$ and $\omega_C$ the circumcircles of $\triangle ABC_1$ and $\triangle ACB_1$ . Finally, let $T = \omega_B\cap \omega_C\neq A$ . We will show that $L$ is the desired fixed point.
$[asy] defaultpen(fontsize(10pt)); size(350); pair A, B, C, B1, C1, B2, C2, L, R, S, T; A = dir(120); B = dir(210); C = dir(330); real r = 0.7; B1 = intersectionpoint(Circle(B, r), A--B); C1 = intersectionpoint(Circle(C, r), A--C); B2 = intersectionpoints(Line(B1, B1+C1-B, 20), circumcircle(A, C, B1))[1]; C2 = intersectionpoints(Line(C1, C1+B1-C, 20), circumcircle(A, B, C1))[1]; L = dir(90); S = extension(B1, C1, B2, C2); T = extension(B1, C2, C1, B2); R = extension(A, T, B1, C1); draw(A--B--C--cycle, orange); draw(circumcircle(A, B, C), red); draw(circumcircle(A, C, B1), lightblue); draw(circumcircle(A, B, C1), lightblue); draw(arc(circumcenter(A, B2, C2), circumradius(A, B2, C2), 55, 155), magenta+dotted); draw(circumcircle(A, B1, C1), heavygreen+dashed); draw(arc(L, circumradius(B1, C1, B2), 190, 365), heavycyan+dashed); draw(B--C1^^C--B1, purple+dotted); draw(B1--B2^^C1--C2, purple+dotted); draw(B2--T--C2, heavycyan); draw(L--S^^B2--S^^C1--S, deepgreen); draw(A--T, purple+dotted); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$B_2$", B2, dir(20)); dot("$C_2$", C2, dir(150)); dot("$B_1$", B1, dir(210)); dot("$C_1$", C1, dir(0)); dot("$L$", L, dir(90)); dot("$T$", T, dir(270)); dot("$S$", S, dir(180)); dot("$R$", R, dir(50)); dot(extension(B1, B2, C1, C2)); label("$\omega_B$", (-1.21, -0.1)); label("$\omega_C$", (1.33, 0.1)); [/asy]$

Claim: We have $T = \overline{B_1C_2}\cap \overline{C_1B_2}$ .

Proof. Redefine $B_2 = \overline{TC_1}\cap \omega_C$ and $C_2 = \overline{TB_1}\cap \omega_B$ ; we will show that $\overline{B_1B_2}\parallel \overline{BC_1}$ and $\overline{C_1C_2}\parallel \overline{CB_1}$ .

Indeed, we have
$\angle AB_1B_2 = \angle ATB_2 = \angle ATC_1 = \angle ABC_1,$ which implies $\overline{B_1B_2}\parallel \overline{BC_1}$ . The other pair of parallel lines follows similarly. $\blacksquare$

Now observe that:

Since $LB = LC$ , $BB_1 = CC_1$ , and $\angle B_1BL = \angle C_1CL$ , we have $\triangle LBB_1\cong \triangle LCC_1$ . Hence $L$ is the center of the spiral similarity (congruence) sending $\overline{BB_1}\to \overline{CC_1}$ ; in particular, $LAB_1C_1$ is cyclic.
By definition, $T$ is the center of the spiral similarity sending $\overline{BB_1}\to \overline{C_1C}$ , and since $BB_1 = CC_1$ , we actually have $\triangle TBB_1\cong \triangle TC_1C$ .

This analysis also shows that $\overline{ART}$ is $\angle A$ -bisector (since $BT = TC_1$ and $B_1T = TC$ ).

Claim: $B_1C_1B_2C_2$ is cyclic.

Proof. We have $\angle BTC_1 = \angle B_1TC = 180^\circ - \angle A$ , so
$\angle B_1C_2C_1 = \angle TB_1C = \angle A/2 = \angle TC_1B = \angle C_1B_2B_1,$ which is enough. $\blacksquare$

Claim: $\overline{AL}$ , $\overline{B_1C_1}$ , $\overline{B_2C_2}$ concur at a point $S$ .

Proof. Let $S = \overline{B_1C_1}\cap \overline{B_2C_2}$ ; we'll show that $S$ lies on $\overline{AL}$ .

By Radical axis on $\omega_B$ , $\omega_C$ , $(B_1C_1B_2C_2)$ , we know that $\overline{AT}$ , $\overline{B_1B_2}$ , $\overline{C_1C_2}$ concur, which forces $(S, R; B_1, C_1) = -1.$ On the other hand, $\overline{AT}$ and $\overline{AL}$ are the internal/external angle bisectors of $\angle B_1AC_1$ , so
$(\overline{AL}\cap \overline{B_1C_1}, R; B_1, C_1) = -1,$ which is sufficient. $\blacksquare$

Note that $S$ is the radical center of $(LAB_1C_1)$ , $(B_1B_2C_1C_2)$ , and $(AB_2C_2)$ . Thus $\overline{SAL}$ is the radical axis of $(AB_2C_2)$ and $(LAB_1C_1)$ , which forces $L$ to lie on $(AB_2C_2)$ , as desired.

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yayups

1614 posts

yayups

#6 h Jan 11, 2020, 2:20 AM • 3 Y

Y by amar_04, sameer_chahar12, Adventure10

By spiral similarity, the circle $(AB_1C_1)$ passes through the arc midpoint of $BAC$ . We'll now $\sqrt{bc}$ -invert the entire problem. Stated below is the inverted problem, which we'll solve.

Inverted Problem wrote:

Let $ABC$ be a triangle, and let $S$ be the intersection of the external angle bisector of $\angle BAC$ with $BC$ . Let $B_1\in AC$ and $C_1\in AB$ such that $S,B_1,C_1$ collinear. Let $B_2\in BB_1$ such that $(AB_1B_2)$ is tangent to $(AC_1C)$ , and define $C_2$ similarly. Show that $B_2C_2$ passes through a fixed point on $BC$ .

$[asy] unitsize(2.5inches); pair A=dir(125); pair B=dir(220); pair C=dir(-40); pair I=incenter(A,B,C); pair D=extension(A,I,B,C); pair E=extension(B,I,A,C); pair F=extension(C,I,A,B); pair S=extension(E,F,B,C); pair P=0.8*D+0.2*A; pair B1=extension(B,P,A,C); pair C1=extension(C,P,A,B); pair Y=extension(B,C1+B1-B,A,C); pair Z=extension(C,B1+C1-C,A,B); pair B2=2*foot(circumcenter(A,Z,B1),B,B1)-B1; pair C2=2*foot(circumcenter(A,Y,C1),C,C1)-C1; pair Q=extension(A,D,B2,C1); draw(circumcircle(A,B,C)); draw(A--B--C--cycle); draw(A--D); draw(C--C1); draw(B--B1); draw(B1--Z,dotted); draw(C1--Y,dotted); draw(B2--C1,dotted); draw(C2--B1,dotted); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$B_1$",B1,dir(B1)); dot("$C_1$",C1,dir(C1)); dot("$Y$",Y,dir(Y)); dot("$Z$",Z,dir(Z)); dot("$B_2$",B2,dir(B2)); dot("$C_2$",C2,dir(C2)); dot("$Q$",Q,dir(Q)); dot("$P$",P,dir(P)); dot("$D$",D,dir(D)); [/asy]$

As expected, we'll show that the fixed point is $S$ . Let $Y\in AC$ and $Z\in AB$ such that $C_1Y\parallel BB_1$ and $B_1Z\parallel CC_1$ . Then, we have that $B_2\in(AZB_1)$ and $C_2\in(AYC_1)$ . Let $P=BB_1\cap CC_1$ , and let $D$ be the foot of the angle bisector from $A$ to $BC$ . Since $B_1C_1$ passes through $S$ , we have by Ceva-Menalaus that $P\in AD$ .

We want to show that $B_1C_1$ , $B_2C_2$ , and $BC$ are concurrent. By the so called prism lemma, it suffices to show that $(BP;B_2B_1)=(CP;C_2C_1)$ , or that
$(AB,AD;AB_2,AC) = (AC,AD;AC_2,AB).$ By reflecting in $AD$ , we see then that it suffices to show $AB_2$ and $AC_2$ are isogonal.

Note that
$AY\cdot AB = \frac{AC_1}{AB}\cdot AB_1\cdot AB = AB_1\cdot AC_1,$ and similarly $AZ\cdot AC=AB_1\cdot AC_1$ . Thus, $Y\mapsto B$ and $Z\mapsto C$ under $\sqrt{bc}$ inversion in triangle $AB_1C_1$ . Thus, $(AB_1Z)$ is sent to $C_1C$ and $(AC_1Y)$ is sent to $B_1B$ . Since $C_1C$ and $B_1B$ concur at $P$ , which is on $AD$ , we see that $(AB_1Z)$ and $(AC_1Y)$ concur at $Q$ , which is also on $AD$ . Then, we have (all angles directed mod $\pi$ )
$\angle AQC_1=\angle AYC_1 = \angle AB_1B = \angle AQB_2,$ so $C_1,Q,B_2$ collinear. Similarly, $C_2,Q,B_1$ collinear.

Now, by DDIT on $C_1C_2B_2B_1$ with pencil point $A$ , we see that there is an involution swapping the pairs $\{AC_1,AB_1\}$ , $\{AC_2,AB_2\}$ , and $\{AQ,AP\}$ . The first and third imply that the involution is isogonality, so $AB_2$ and $AC_2$ isogonal, as desired.

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Physicsknight

635 posts

Physicsknight

#7 h Jan 14, 2020, 2:58 PM • 1 Y

Y by Adventure10

Let $D$ be the midpoint of arc $\widehat {BAC}$ of $\triangle ABC$ . Let $\odot(AB_1C)$ cuts the internal angle bisector of $\widehat {BAC}$ at $H\neq A$ , and $\odot (ABH)$ cuts $AC$ at $C_3\neq A$ . Note, $HB=HC_3$ , $HC=HB_1$ , $\widehat {BHC_3}=180^{\circ}-\widehat {BAC}=\widehat {B_1HC}\implies\triangle BHB_1=\triangle CHC_3\implies CC_3=BB_1=CC_1\implies C_3\equiv C_1$ .
But, $\widehat {HB_1C}=\widehat {HAC}=\widehat {HC_2C_1}\implies\overline {H,B_1,C_2}$ .
Similarly, $\overline {H,C_1,B_2}$ . Thus $\widehat {HC_2C_1}=\widehat {HAC}=\widehat {HAB}=\widehat {HB_2B_1}\implies B_1,B_2,C_1,C_2$ are concyclic.
$\triangle DBB_1=\triangle DCC_1\implies DB_1=DC_1$ , and $\widehat {B_1DC_1}=\widehat {BAC}=\widehat {BDC}=2\widehat {BAH}=2\widehat {B_1C_2C_1}=\widehat {B_2B_1C}+2\widehat {BAH}+\widehat{C_2C_1B}=\widehat{B_2AC}+\widehat {CAB}+\widehat {BAC_2}=\widehat {B_2AC_2}$ .
It is proved that $\odot (AB_2C_2)$ passes through $D$ , which is fixed point.
$\blacksquare$

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TheUltimate123

1740 posts

TheUltimate123

#8 h Jan 20, 2020, 8:25 AM • 2 Y

Y by CyclicISLscelesTrapezoid, Adventure10

$[asy] size(9cm); defaultpen(fontsize(10pt)); pair A,B,C,I,D,B1,C1,B2,C2,K,X,Y; A=dir(125); B=dir(210); C=dir(330); I=incenter(A,B,C); real t=0.7; D=(t+1)*I-t*A; B1=2*foot(circumcenter(A,D,C),A,B)-A; C1=2*foot(circumcenter(A,D,B),A,C)-A; B2=2*foot(circumcenter(A,D,C),B1,B1+C1-B)-B1; C2=2*foot(circumcenter(A,D,B),C1,C1+B1-C)-C1; K=dir(90); X=extension(B1,B2,C1,C2); Y=extension(B1,C1,B2,C2); //draw(B1--B2,Dotted); draw(C1--C2,Dotted); draw(B--C1,Dotted); draw(C--B1,Dotted); draw(A--D,dashed); draw(arc(circumcenter(B1,C1,B2),circumradius(B1,C1,B2),170,10,CCW),gray); draw(B2--D--C2,gray); draw(circumcircle(A,B1,C1),dashed); draw(circumcircle(A,D,C)); draw(circumcircle(A,D,B)); draw(A--B--C--A); draw(K--Y--C1); draw(Y--B2); dot("$A$",A,NW); dot("$B$",B,SW); dot("$C$",C,SE); dot("$D$",D,S); dot("$B_1$",B1,dir(215)); dot("$C_1$",C1,E); dot("$B_2$",B2,E); dot("$C_2$",C2,dir(140)); dot("$K$",K,N); dot("$X$",Y,W); [/asy]$

Let $K$ be the midpoint of arc $BAC$ on the circumcircle of $\triangle ABC$ , and let $(ABC_1)$ and $(ACB_1)$ intersect again at $D$ . Denote $X=\overline{B_1B_2}\cap\overline{C_1C_2}$ and $Y=\overline{B_1C_1}\cap\overline{B_2C_2}$ . I claim that $K$ is the fixed point.

Since $KB=KC$ , $BB_1=CC_1$ , and $\measuredangle KBB_1=\measuredangle KCC_1$ , $\triangle KBB_1\cong\triangle KCC_1$ , so $K$ is the center of spiral similarity sending $\overline{BB_1}$ to $\overline{CC_1}$ and $K$ lies on $(AB_1C_1)$ . Moreover, $D$ is the center of spiral similarity sending $\overline{BB_1}$ to $\overline{C_1C}$ . Since $BB_1=CC_1$ , we have $DB=DC_1$ , so $\overline{AD}$ bisects $\angle BAC$ .

Notice that $\measuredangle ADB_1=\measuredangle ACB_1=\measuredangle AC_1C_2=\measuredangle ADC_2$ and $\measuredangle ADC_1=\measuredangle ADB_2$ , so $D=\overline{B_1C_2}\cap\overline{B_2C_1}$ . Furthermore $\measuredangle B_1B_2C_1=\measuredangle B_1AD=\measuredangle DAC_1=\measuredangle B_1DC,$ so $B_1C_1B_2C_2$ is cyclic.

Finally $A$ is the Miquel point of $B_1B_2C_1C_2$ , so by a well-known property of complete quadrilaterals, $A$ is the foot from $X$ to $\overline{AD}$ , id est $\angle XAD=90^\circ$ . It follows that $K\in\overline{AX}$ , so $XB_2\cdot XC_2=XB_1\cdot XC_1=XA\cdot XK,$ and we are done.

This post has been edited 1 time. Last edited by TheUltimate123, Jan 20, 2020, 8:28 AM

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v_Enhance

6870 posts

v_Enhance

#9 h Feb 24, 2020, 3:20 AM • 7 Y

Y by Aryan-23, GeoMetrix, amar_04, sameer_chahar12, mijail, v4913, Mango247

Solution with Aditya Khurmi, Anant Mudgal, Anushka Aggarwal, Arindam, Dhrubajyoti Ghosh, Paul Hamrick, Pranjal Srivastava, Rishabh, Robin Son, Rohan Goyal, William Hu, Zifan Wang:

Let $D = \overline{B_1B_2} \cap \overline{C_1C_2}$ and $E = \overline{BC_1} \cap \overline{B_1C}$ . Moreover let $X$ be the arc midpoint of $\widehat{BAC}$ . We will show that $X$ is the desired fixed point.

$[asy] size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -15.08, xmax = 18.82, ymin = -6.34, ymax = 10.22; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); pen zzffff = rgb(0.6,1.,1.); pen ttffqq = rgb(0.2,1.,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen ffdxqq = rgb(1.,0.8431372549019608,0.); draw((-1.,8.)--(-5.,-2.)--(7.,-2.)--cycle, linewidth(1.) + zzttqq); /* draw figures */ draw((-1.,8.)--(-5.,-2.), linewidth(1.) + zzttqq); draw((-5.,-2.)--(7.,-2.), linewidth(1.) + zzttqq); draw((7.,-2.)--(-1.,8.), linewidth(1.) + zzttqq); draw(circle((3.,3.), 6.4031), linewidth(1.) + zzffff); draw(circle((-1.3780,2.3512), 5.6614), linewidth(1.) + zzffff); draw((-3.34482,2.1379)--(8.3309,6.5469), linewidth(1.)); draw((4.2159,1.4800)--(-6.0296,5.5783), linewidth(1.)); draw(circle((1.8537,-4.3609), 12.686), linewidth(1.) + linetype("2 2")); draw((-5.,-2.)--(4.2159,1.4800), linewidth(1.)); draw((7.,-2.)--(-3.3448,2.1379), linewidth(1.)); draw(circle((1.,1.4), 6.8963), linewidth(1.) + red); draw(circle((1.,8.2963), 7.5368), linewidth(1.) + ttffqq); draw(circle((0.6219,3.9512), 4.3615), linewidth(1.) + red); draw((0.6219,-2.9451)--(-6.0296,5.5783), linewidth(1.)); draw((0.6219,-2.9451)--(8.3309,6.5469), linewidth(1.)); draw((-1.,8.)--(1.,-5.4963), linewidth(1.)); draw(circle((-2.8092,5.3236), 3.2304), linewidth(1.) + ffdxqq); draw(circle((3.5076,6.2597), 4.8318), linewidth(1.) + ffdxqq); /* dots and labels */ dot("$A$", (-1.,8.), dir(90)); dot("$B$", (-5.,-2.), dir(225)); dot("$C$", (7.,-2.), dir(-45)); dot("$B_1$", (-3.3448,2.1379), dir(190)); dot("$C_1$", (4.2159,1.4800), dir(0)); dot("$B_2$", (8.3309,6.5469), dir(30)); dot("$C_2$", (-6.0296,5.5783), dir(160)); dot((1.,-5.4963)); dot("$X$", (1.,8.2963), dir(90)); dot("$D$", (-0.3016,3.2871), dir(75)); dot("$E$", (1.1645,0.3278), dir(-90)); dot("$Y$", (0.6219,-2.9451), dir(-45)); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

To begin, note that $B_1DC_1E$ is a parallelogram. Define $\theta = \measuredangle EB_1D = \measuredangle DC_1E$ ; this angle appears in several places. We now proceed in eight steps.

Since $\measuredangle C_2AB = \measuredangle C_2 C_1 B = \theta = \measuredangle CB_1 B_2 = \measuredangle CA_2 B_2.$ it follows $\overline{AC_2}$ and $\overline{AB_2}$ are isogonal with respect to $\angle A$ .
Let $D = \overline{B_1B_2} \cap \overline{C_1C_2}$ . Since $\measuredangle C_2AB_1 = \theta = \measuredangle C_2DB_1$ , we have $C_2ADB_1$ is cyclic; similarly $B_2ADC_2$ are cyclic.
Note that $A$ is Miquel point of self-intersecting quadrilateral $C_1C_2B_1B_2$ . It follows that $Y \overset{\text{def}}{=} \overline{C_1B_2} \cap \overline{C_2B_1}$ lies on both $(C_2AC_1)$ and $(B_2AB_1)$ .
The isogonal lemma on $\triangle AB_1C_1$ now gives $AD$ , $AY$ isogonal with respect to $\angle BAC$ . (This is equivalent to DDIT on $B_1B_2C_1C_2DY$ from $A$ .)
Since $Y$ is now the spiral center mapping congruent segments $BB_1$ and $C_1C$ , it follows $YB_1 = YC$ , $YB = YC_1$ , so $\overline{AY}$ is an angle bisector. Since $\overline{AD}$ and $\overline{AY}$ were isogonal, we conclude $ADY$ are collinear.
By power of a point from $Y$ , we find $B_1B_2C_1C_2$ is cyclic (with Miquel point $A$ ).
Then it follows by Miquel theory, and from $XB_1 = XC_1$ , that $X$ is the circumcenter of $B_1B_2C_1C_2$ .
By Miquel theory, $XAB_2C_2$ concyclic as needed.

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NJOY

495 posts

NJOY

#11 h Feb 24, 2020, 8:41 AM • 2 Y

Y by char2539, Purple_Planet

My $256\text{th}$ post.

Common remarks/Constructions. Let $T$ be the mid-point of the arc $\widehat{BAC}$ of $\odot(ABC)$ . Again, let $P$ be the point of intersection of $\odot(ABC_1)$ and $\odot(ACB_1)$ .

We prove that the desired fixed point is the point $T$ .

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(8); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.767963430995203, xmax = 11.435801392656167, ymin = -3.0611011252211666, ymax = 8.06902586020495; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen ffffww = rgb(1,1,0.4); pen ffzzcc = rgb(1,0.6,0.8); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); draw((0,6)--(-2.34,0.64)--(5.58,0.04)--cycle, linewidth(1) + rvwvcq); draw(circle((2.07585201821125,6.357246640388502), 5.306669925109475), linewidth(1) + orange); draw(circle((1.7492097630383627,2.0455688721063865), 4.3240328791953795), linewidth(1) + ffzzcc); draw(circle((1.9115843695020698,1.1020592167954548), 5.257754151489301), linewidth(1) + sexdts); /* draw figures */ draw((0,6)--(-2.34,0.64), linewidth(1) + rvwvcq); draw((-2.34,0.64)--(5.58,0.04), linewidth(1) + rvwvcq); draw((5.58,0.04)--(0,6), linewidth(1) + rvwvcq); draw((-1.539797717153027,2.472941981222126)--(5.58,0.04), linewidth(1)); draw((4.213095871870886,1.4999907891845023)--(-2.34,0.64), linewidth(1)); draw(circle((0.7513492591259361,2.4812020025457664), 3.5981196556086097), linewidth(1)); draw(circle((2.5440342362231454,2.7897166171350927), 4.096099314393395), linewidth(1)); draw((-2.5869508073840293,3.8236690308248895)--(4.213095871870886,1.4999907891845023), linewidth(1)); draw((-1.539797717153027,2.472941981222126)--(6.571316432489792,3.5373981703432285), linewidth(1)); draw((-2.5869508073840293,3.8236690308248895)--(0,6), linewidth(1)); draw((0,6)--(6.571316432489792,3.5373981703432285), linewidth(1)); draw((-2.5869508073840293,3.8236690308248895)--(6.571316432489792,3.5373981703432285), linewidth(1)); draw((2.075852018211249,6.357246640388503)--(6.571316432489792,3.5373981703432285), linewidth(1) + linetype("4 4") + sexdts); draw((0,6)--(2.075852018211249,6.357246640388503), linewidth(1)); draw((0,6)--(1.219531477137834,-1.0863280207076442), linewidth(1) + linetype("4 4")); draw((-2.34,0.64)--(2.075852018211249,6.357246640388503), linewidth(1) + dtsfsf); draw((2.075852018211249,6.357246640388503)--(5.58,0.04), linewidth(1) + dtsfsf); draw((-1.539797717153027,2.472941981222126)--(2.075852018211249,6.357246640388503), linewidth(1) + linetype("4 4") + sexdts); draw((2.075852018211249,6.357246640388503)--(4.213095871870886,1.4999907891845023), linewidth(1) + linetype("4 4") + sexdts); draw((-2.5869508073840293,3.8236690308248895)--(1.219531477137834,-1.0863280207076442), linewidth(1) + wvvxds); draw((1.219531477137834,-1.0863280207076442)--(6.571316432489792,3.5373981703432285), linewidth(1) + wvvxds); draw((-2.34,0.64)--(1.219531477137834,-1.0863280207076442), linewidth(1) + wvvxds); draw((1.219531477137834,-1.0863280207076442)--(5.58,0.04), linewidth(1) + wvvxds); draw((-2.5869508073840293,3.8236690308248895)--(2.075852018211249,6.357246640388503), linewidth(1) + linetype("4 4") + sexdts); draw((-1.539797717153027,2.472941981222126)--(4.213095871870886,1.4999907891845023), linewidth(1)); /* dots and labels */ dot((0,6),linewidth(3pt) + dotstyle); label("$A$", (-0.11,6.12988679114503), NE * labelscalefactor); dot((-2.34,0.64),linewidth(3pt) + dotstyle); label("$B$", (-2.7,0.40), NE * labelscalefactor); dot((5.58,0.04),linewidth(3pt) + dotstyle); label("$C$", (5.65,-0.19223784496813395), NE * labelscalefactor); dot((-1.539797717153027,2.472941981222126),linewidth(3pt) + dotstyle); label("$B_1$", (-2.1,2.33), NE * labelscalefactor); dot((4.213095871870886,1.4999907891845023),linewidth(3pt) + dotstyle); label("$C_1$", (4.36989738758852,1.3617297651689253), NE * labelscalefactor); dot((-2.5869508073840293,3.8236690308248895),linewidth(3pt) + dotstyle); label("$C_2$", (-3.02,3.8852669098359445), NE * labelscalefactor); dot((6.571316432489792,3.5373981703432285),linewidth(3pt) + dotstyle); label("$B_2$", (6.7074896900169145,3.473531902021852), NE * labelscalefactor); dot((2.075852018211249,6.357246640388503),linewidth(3pt) + dotstyle); label("$T$", (2.1252775062794367,6.435367603394195), NE * labelscalefactor); dot((1.219531477137834,-1.0863280207076442),linewidth(3pt) + dotstyle); label("$P$", (1.1,-1.43), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

First, we prove the following claim.

Claim. The lines $B_1C_2$ and $C_1B_2$ meet at point $P$ .
Proof. To prove this, let us redefine the point $B_2$ as the intersection of lines $PC_1$ and $\odot(ACB_1)$ . Redefine the point $C_2$ similarly. Then, we have $\angle AB_1B_2 = \angle APB_2 = \angle APC_1 = \angle ABC_1,$ which implies that the lines $B_1B_2$ and $BC_1$ are parallel. Similarly, $C_1C_2\parallel CB_1$ . Thus, we obtain that $B_2$ , $C_2$ are the points as defined earlier and hence, the lines $B_1C_2$ and $C_1B_2$ meet at point $P$ . $\square$

By the problem, we have that $P$ is the center of spiral similarity that maps $BB_1 \mapsto CC_1$ , and thus, we conclude that $AP$ is the internal angle bisector of $\angle BAC$ . It follows that $AT\perp AP$ . This further implies that the quadrilateral $AB_1C_1T$ is cyclic, which yields $\angle B_1AC_1=\angle B_1TC_1$ .

Next, we prove this crucial claim.

Claim. The quadrilateral $B_1B_2C_1C_2$ is cyclic with center $T$ .
Proof. Note that $\angle B_1C_2C_1=\angle PC_2C_1=\angle PAC_1=\frac{\angle B_1AC_1}2=\frac{\angle B_1TC_1}2.$ In the same way, we can prove that $\angle B_1B_2C_1=\frac{\angle B_1TC_1}2,$ which thus yields that the quadrilateral $B_1C_1B_2C_2$ is cyclic with center $T$ . $\square$

The above claim also yields that $TB_1=TC_1=TB_2=TC_2.$
Finally, observe that $\angle BAC_2 = \angle BC_1C_2 = \angle CB_1B_2 = \angle CAB_2,$ which thus implies that the lines $AB_2$ and $AC_2$ are isogonal with respect to $\angle BAC$ . It follows that $AP$ is also the angle bisector of $\angle B_2AC_2$ . Again, as $AT\perp AP$ , we obtain that $AT$ is the external angle bisector of $\angle B_2AC_2$ . Now, as $TB_2=TC_2$ , we conclude that $T$ is the mid-point of arc $\widehat{B_2AC_2}$ , which finally implies that the circumcircle of $\odot(AB_2C_2)$ pass through the point $T$ , as desired. $\blacksquare$

This post has been edited 2 times. Last edited by NJOY, Feb 24, 2020, 8:45 AM

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amar_04

1915 posts

amar_04

#13 h Feb 29, 2020, 11:20 AM • 7 Y

Y by mueller.25, GeoMetrix, lilavati_2005, Hexagrammum16, AmirKhusrau, Purple_Planet, Bumblebee60

Great Problem! Congrats to Tworigami.

Here's a bashy and a different solution which I got and also this is the first time I carried out a Menelaus and Trig Bash properly.

tworigami wrote:

Let $\odot(AB_1C_1)\cap\odot(ABC)=T$ . So there $\exists$ a unique Spiral Similarity $(\sigma)$ centered at $T$ such that $\sigma:\triangle TB_1B\mapsto \triangle TC_1C\implies TB=TC$ as $BB_1=CC_1$ . So, $T$ is the midpoint of $\widehat{BAC}$ . Now we will Invert around $A$ with radius $\sqrt{AB\cdot AC}$ followed by a reflection along the angle bisector of $\angle BAC$ . The Problem becomes equivalent to this one. (Diagram taken from #4)

Inverted Problem wrote:

$ABC$ be a triangle and the External Bisector of $\angle BAC$ intersects $BC$ at $T$ . A ray passing through $T$ intersects $\{AB,AC\}$ at $\{B_1,C_1\}$ respectively. Let $BC_1\cap CB_1=X$ . Let $\dot(AB_1X)\cap BC_1=C_2$ and $\odot(AC_1X)\cap CB_1=B_2$ . Then $\overline{T-B_2-C_2}.$

$[asy] defaultpen(fontsize(11pt)); size(10cm); real t = 0.5; pair A = dir(160); pair B = dir(225); pair C = dir(540-225); pair X = t*A + (1-t)*dir(270); pair C1 = extension(B,X,A,C); pair B1 = extension(C,X,A,B); path circ1 = circumcircle(A,B1,X); path circ2 = circumcircle(A,C1,X); pair B2 = intersectionpoints(C--2B1-C, circ2)[1]; pair C2 = intersectionpoints(B--2C1-B, circ1)[1]; pair T = extension(B,C,B1,C1); pair O = circumcenter(B1,C1,B2); filldraw(A--B--C--cycle,pink+white+white); draw(circ1,orange); draw(circ2,orange); draw(B--C2,red); draw(C--B2,red); draw(T--C2,gray(0.5)); draw(T--C1,gray(0.5)); draw(T--B); dot("$A$", A, dir(90)); dot("$B$", B, dir(270)); dot("$C$", C, dir(270)); dot("$X$", X, dir(270)); dot("$B_1$", B1, dir(10)); dot("$C_1$", C1, dir(180)); dot("$B_2$", B2, dir(180)); dot("$C_2$", C2, dir(0)); dot("$T$", T, dir(270)); draw(arc(O,abs(O-B1),225,360),dashed); [/asy]$

Notice that if $AX\cap BC=Y$ , then $(TY;BC)$ is harmonic $\implies AX$ is the bisector of $\angle A$ .

Claim:- $B_1B_2C_1C_2$ is a cyclic quadrilateral.
Proof:- $\angle C_1B_2B_1=\angle C_1B_2X=\angle C_1AX=\angle XAB_1=\angle B_1C_2X=\angle B_1C_2C_1$ . So, $B_1B_2C_1C_2$ is a cyclic quadrilateral.

Now we will use Menelaus Theorem in order to prove that $\overline{T-B_2-C_2}$ .

So let's compute $\frac{BC_2}{XC_2}$ and $\frac{XB_2}{CB_2}$ first. We will use PoP to compute them. $\begin{cases} \frac{BC_2\cdot BX}{XC_2\cdot XC_1}=\frac{BA\cdot BB_1}{XB_2\cdot XB_1}\implies\frac{BC_2}{XC_2}=\frac{BA\cdot BB_1\cdot XC_1}{XB_1\cdot XB_2\cdot XB} \\ \\ \frac{XB_2\cdot XB_1}{CB_2\cdot CX}=\frac{XB_1\cdot XB_2}{CC_1\cdot CA}\implies \frac{XB_2}{CB_2}=\frac{XB_1\cdot XB_2\cdot XC}{CC_1\cdot CA\cdot XB_1} \end{cases}$ Now to prove that $\overline{T-B_2-C_2}$ it suffices to prove this $\longrightarrow$

$\begin{align*} \iff\frac{CT}{TB}\cdot\frac{BC_2}{XC_2}\cdot\frac{XB_2}{CB_2}=1 \\ &\iff \frac{CA}{AB}\cdot\frac{BA\cdot BB_1\cdot XC_1}{XB_1\cdot XB_2\cdot XB}\cdot\frac{XB_1\cdot XB_2\cdot XC}{CC_1\cdot CA\cdot XB_1}=1\\ &\iff \frac{BB_1\cdot XC_1\cdot XC}{XB\cdot CC_1\cdot XB_1}=1\cdots\cdots\cdots (\star) \end{align*}$

Now we will prove the following Lemma.

Lemma:- In a $\triangle ABC$ , let $G$ be an arbitary point inside the plane of $\triangle ABC$ and $\triangle XEF$ be the $G-$ Cevian Triangle WRT $\triangle ABC$ such that $AX$ is the bisector of $\angle BAC$ . Then $\frac{GE\cdot GC}{GF\cdot GB}=\frac{CE}{BF}$ .

Proof:- We will make use of Sine Rule extensively. $\begin{cases}\frac{GE}{GF}=\frac{\sin\angle GFE}{\sin\angle GEF} \\ \frac{GC}{GB}=\frac{\sin\angle GBC}{\sin\angle GCB}\end{cases}$ Now by Trig Form of Ceva's Theorem we get that $\frac{\sin\angle BAX}{\sin\angle CAX}\cdot\frac{\sin\angle ACF}{\sin\angle BCF}\cdot\frac{\sin\angle EBC}{\sin\angle EBA}=1\implies\frac{\sin\angle GBC}{\sin\angle GCB}=\frac{\sin\angle EBF}{\sin\angle ECF}$
Hence putting the values back we get that,
$\frac{GE}{GF}\cdot\frac{GC}{GB}=\frac{\sin\angle GFE\cdot\sin\angle GBC}{\sin\angle GEF\cdot\sin\angle GCB}=\frac{\sin\angle CFE}{\sin\angle BEF}\cdot\frac{\sin\angle EBF}{\sin\angle ECF}=\frac{CE}{EF}\cdot\frac{EF}{BF}=\frac{CE}{BF}$

Back to the Inverted Problem:-

From (Lemma) indeed we get that $(\star)=1$ . So, by Converse of Menelaus Theorem we get that $\overline{T-B_2-C_2}$ . So, Inverting back we get that $\odot(AB_2C_2)$ passes through the midpoint of $\widehat{BAC}$ which is $T$ . $\blacksquare$

This post has been edited 1 time. Last edited by amar_04, Feb 29, 2020, 11:26 AM

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GeoMetrix

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GeoMetrix

#14 h Apr 24, 2020, 4:43 AM • 6 Y

Y by AmirKhusrau, mueller.25, amar_04, Purple_Planet, Mango247, Mango247

Really nice problem. Although my solution is similiar to the ones above but since this is too nice i cant resist to post.

$[asy] size(10cm); pair A=(4.089884929030667,7.097525729001119); pair B=(-0.12839547437282756,-4.869227188455621); pair C=(16.954144315296645,-4.629892130106486); pair B1=(1.9358694038884572,1.0243986233918232); pair C1=(12.317027559782165,-0.4714454912902692); pair B2=(17.552481961169484,6.559021847715566); pair C2=(-0.5173149441901711,4.345172557986069); pair D=(8.218414685553237,8.054865962397658); pair F=(5.316477103069982,2.1612401505502135); pair G=(7.560243275093117,-6.843741419835983); filldraw(A--B--C--cycle,orange+white+white+white,orange); draw(circumcircle(A,B,C),red); draw(C--B1,green+dotted); draw(B--C1,green+dotted); draw(circumcircle(A,B,C1),purple+dotted); draw(circumcircle(A,C,B1),purple+dotted); draw(C1--C2,green+dotted); draw(B1--B2,green+dotted); draw(A--G,red+dotted); draw(C2--B,green); draw(C--B2,green); draw(G--B2,cyan); draw(G--C2,cyan); draw(circumcircle(A,B1,C1),cyan+dashed); draw(arc(circumcenter(A, B2, C2), circumradius(A, B2, C2), 55, 150),pink); draw(arc(D,circumradius(B1,B2,C1),180,360),lightblue); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$G$",G,dir(G)); dot("$B_1$",B1,dir(B1)); dot("$C_1$",C1,dir(C1)); dot("$B_2$",B2,dir(B2)); dot("$C_2$",C2,dir(C2)); dot("$F$",F,dir(F)); [/asy]$
We proceed with several claims.

Claim 1: If $\odot(C_2BC_1) \cap (B_2CB_1)$ and $G \neq A$ then $\overline{GC_2}$ bisects $\angle BC_2C_1$ and $\overline{GB_2}$ bisects $\angle B_1B_2C$

Proof: Notice that there exists a spiral similiarity at $G$ such that $G: \overline{BB_1} \mapsto \overline{CC_1}$ . This implies that $\overline{GB}=\overline{GC_1}$ and $\overline{GB_1}=\overline{GC}$ and this proves the claim $\qquad \square$

Claim 2: $B_1\in \overline{GC_2}$ and also $C_1 \in \overline{GB_2}$

Proof: Notice that $\angle C_1C_2G=\angle C_1AG=\angle CAG=\angle CB_1G$ but since $\overline{CB_1}\parallel \overline{C_1C_2}$ hence we have that $B_1\in \overline{GC_2}$ . A similiar arguement shows that $C_1 \in \overline{GB_2}$ and we're done $\qquad \square$

Claim 3: $B_1C_1B_2C_2$ is cyclic

Proof: Notice that $\angle B_1B_2C_1=\angle BC_1G=\angle BC_2G=\angle GC_2C_1=\angle B_1C_2C_1$ and this proves the claim $\qquad \square$

Claim 4: If $F=\overline{B_1B_2} \cap \overline{C_1C_2}$ then $F \in \overline{AG}$ . Also $AG$ is the angle bisector OF $\angle BAC$

Proof: The first part is just radical axis on $\odot(ABC_1),\odot(ACB_1),\odot(B_1B_2C_1C_2)$ . For the second just notice that $\angle BAG=\angle BC_2G=\angle GC_2C_1=\angle GAC_1$ which proves the second part $\qquad \square$

Claim 5: If $D$ is the midpoint of $\widehat{BAC}$ then $D$ is the circumcenter of $\odot(B_1B_2C_1C_2)$

Proof: Notice that simple congruency stuff implies that $\triangle{DB_1B} \cong \triangle{DC_1C}$ so $\overline{DB_1}=\overline{DC_1}$ . Now just notice that this implies that $D\in \odot(AB_1C_1)$ . Hence we have that $\angle B_1DC_1=\angle B_1AC_1=\angle BAC_1=\angle BC_2C_1=2\angle B_1C_2C_1$ and this proves the claim $\qquad \square$

Now back to the main problem. We will claim that $D$ is infact the required fixed point. Notice that $\angle C_2DB_2=2\angle C_2C_1G=2\angle C_2AG$ But since $\angle C_2AB=\angle C_2GB=\angle B_2GC=\angle B_2AC$ hence $2\angle C_2AG=\angle C_2AG+\angle B_2AG=\angle C_2AB_2$ and with this we are done $\qquad \blacksquare$

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mathlogician

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mathlogician

#16 h Jul 12, 2020, 8:41 PM • 3 Y

Y by Mango247, Mango247, Mango247

There was a small hole in the previously posted solution, which has now been fixed.

Let $M$ be the second intersection of circles $(AB_1C_1)$ and $(ABC)$ .

We begin with some easy observations. Note that $M$ is the Miquel point of $B_1BCC_1$ , so $\triangle MB_1B \sim \triangle MC_1C$ . But since $B_1B = C_1C$ , this similarity is in fact a congruence, so $MB_1= MC_1$ and $MB=MC$ . Therefore, $M$ is invariant of $B_1$ and $C_1$ . I claim that $M$ is the desired fixed point.

Claim: $A$ is the Miquel Point of $C_1C_2B_1B_2$ .

Proof: Let $T = C_1C_2 \cap B_1B_2$ . Notice that $\measuredangle B_2TC_1 = \measuredangle B_2B_1C = \measuredangle B_2AC = \measuredangle B_2AC_1$ whence $B_2ATC_1$ is cyclic. Similarly, $C_2ATB_1$ is cyclic, proving the desired claim.

Let $K = B_1C_2 \cap C_1B_2$ . Note that by properties of Miquel point that $AC_1KC_2$ and $AB_1KB_2$ are cyclic.

Claim: $\angle BAK = \angle CAK$ .

Proof: Note that there exists a spiral similarity taking $BB_1$ to $C_1C$ . Since $B_1B= C_1C$ this similarity is a congruence, so $BK = C_1K$ and by Fact 5 we have our desired angle equality.

Claim: $B_1C_1B_2C_2$ is cyclic.

Proof:
$\measuredangle C_1C_2B_1 = \measuredangle CB_1K = \measuredangle CAK = \measuredangle KAB = \measuredangle KC_1B = \measuredangle C_1B_2B_1.$
Claim: $M$ is the circumcenter of $B_1C_1B_2C_2$ .

Proof: Let $M'$ be the circumcenter of $B_1C_1B_2C_2$ . Note that by Miquel, $M'AB_1C_1$ is cyclic, and $M'B_1 = M'C_1$ . But there is only one unique point that satisfies both conditions, namely $M$ , so $M$ is the circumcenter.

This last claim finishes the problem by a well-known property of Miquel points.[/hide]

Motivational Remarks

This post has been edited 2 times. Last edited by mathlogician, Jul 12, 2020, 8:42 PM

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MP8148

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MP8148

#17 h Feb 9, 2021, 5:55 AM

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Here is a slightly different inversion solution:

First, like everyone else, note that by spiral sim $(AB_1C_1)$ passes through the midpoint of $\widehat{BAC}$ . Now $\sqrt{bc}$ inversion gives the following (primes are omitted):

inverted problem wrote:

In $\triangle ABC$ , points $B_1$ , $C_1$ lie on sides $\overline{AB}$ , $\overline{AC}$ such that $K = \overline{B_1C_1} \cap \overline{BC}$ lies on the external $A$ -bisector. Let $B_2$ be the point on $\overline{CB_1}$ such that $(AB_1B_2)$ is tangent to $(ABC_1)$ , and similar for $C_2$ . Prove that $B_2$ , $C_2$ , $K$ are collinear.

$[asy] size(10cm); defaultpen(fontsize(10pt)+linewidth(0.4)); dotfactor *= 1.5; pair A = dir(160), B = dir(235), C = dir(305), K = extension(A,dir(90),B,C), B1 = A+dir(A--B)*abs(A-B)*0.6, C1 = extension(A,C,K,B1), S = extension(B,C1,C,B1), B3 = intersectionpoint(circumcircle(A,B,C1),B+dir(C--B1)*0.01--B+dir(C--B1)*100), B2 = extension(A,B3,C,B1), C2 = extension(B,C1,K,B2), C3 = extension(A,C2,C,C+dir(B--C1)); draw(A--B--C--A); draw(circumcircle(A,B,C1)^^circumcircle(A,C,B1)); draw(A--K--B^^C1--K); draw(C2--K, red+dashed); draw(B--C2^^C--B2, blue); draw(B--B3--A--C3--C, blue); draw(B1--C2^^C1--B2, purple); draw(A--S, purple); dot("$A$", A, dir(135)); dot("$B$", B, dir(270)); dot("$C$", C, dir(270)); dot("$C_1$", C1, dir(5)); dot("$B_1$", B1, dir(250)); dot("$B_2$", B2, dir(135)); dot("$B_3$", B3, dir(225)); dot("$C_2$", C2, dir(95)); dot("$C_3$", C3, dir(10)); dot("$K$", K, dir(210)); dot("$S$", S, dir(285)); [/asy]$
Let $B_3$ be the point on $(ABC_1)$ such that $\overline{BB_3} \parallel \overline{CB_1}$ ; by homothety we can redefine $B_2 = \overline{CB_1} \cap \overline{AB_3}$ . Do the same thing for $C_3$ and $C_2$ . Then $\measuredangle B_3AC_1 = \measuredangle(\overline{B_3B}, \overline{BC_1}) = \measuredangle(\overline{B_1C}, \overline{CC_3}) = \measuredangle B_1AC_3,$ which implies $\overline{AB_2}$ and $\overline{AC_2}$ are isogonal wrt $\angle BAC$ .

Furthermore, note that $S = \overline{BC_1} \cap \overline{CB_1}$ lies on the internal $A$ -bisector by Ceva-Menelaus, so by isogonal lemma $\overline{C_1B_2} \cap \overline{B_1C_2}$ lies on $\overline{AS}$ . Now note that $\overline{B_2C_2}$ meets both $\overline{AK}$ and $\overline{B_1C_1}$ at the harmonic conjugate of $\overline{AS} \cap \overline{B_2C_2}$ : the former by a well-known lemma (right angle and angle bisector), the the latter by Ceva-Menelaus. This implies the lines are concurrent, as desired. $\blacksquare$

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rcorreaa

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rcorreaa

#18 h May 19, 2021, 10:23 PM

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We claim that the desired point is the arc midpoint $X$ of $\widehat{BAC}$ . Let $Y= (ABC_1) \cap (ACB_1)$ , with $A \neq Y$ , $Z= B_1B_2 \cap C_1C_2$ and $W= B_1C \cap BC_1$ .

Observe that $Y$ is the center of the spiral similarity mapping $BB_1 \mapsto C_1C$ , so $\angle B_1YB= \angle B_1WB= \angle CB_1B_2$ (since $B_1B_2 \parallel BC_1$ ), and also $\angle CB_1B_2= \angle C_2C_1B$ , because $ZB_1WC_1$ is a parallelogram. Thus, $\angle CB_1B_2= \angle C_2C_1B= \angle C_2YB \implies \angle B_1YB= \angle C_2YB$ , so $C_2,B_1,Y$ are collinear. Similarly, $B_2,C_1,Y$ are collinear.

Now, notice that $\angle B_1C_2C_1= \angle YC_2C_1= \angle YBC_1$ , and since $Y$ is the center of the spiral similarity mapping $BB_1 \mapsto C_1C$ , we have that $YB_1B ~ YCC_1$ , then since $BB_1=CC_1 \implies YB=YC_1,YC=YB_1$ , hence $\angle YBC_1= \angle YC_1B= \angle YCB_1= \angle YB_2B_1= \angle C_1B_2B_1$ . Therefore, $\angle B_1C_2C_1= \angle C_1B_2B_1$ , so $B_1C_1B_2C_2$ is cyclic. Furthermore, observe that $\angle BAC= 180º- \angle BYC_1= 2 \angle YBC_1= 2 \angle B_1C_2C_1$ , so $A$ is the circumcenter of $B_1C_1B_2C_2$ . Let $\Omega= (B_1C_1B_2C_2)$ .

Consider the inversion $\Phi$ about $\Omega$ centered at $A$ . Clearly, since $A \in (C_1C_2B)$ , $\Phi(B)=B' \in C_1C_2$ . Similarly, $\Phi(C)=C' \in B_1B_2$ . Our goal is to prove that $B_1C_1,B'C', B_2C_2$ are concurrent (it's well-known that $X=(AB_1C_1) \cap (ABC)$ ). Now, notice that since $B_1C_1B_2C_2$ is cyclic $ZB_1.ZB_2=ZC_1.ZC_2$ , so $Z$ lies on the radical axis of $(ABC_1),(AB_1C)$ , which is $AY$ . Thus, since $A,Y,Z$ are collinear, by Desargues' Theorem, we have that $B_1B'C_2,C_1C'B_2$ are in perspective, so $B_1C_1,B'C', B_2C_2$ are concurrent, as desired.
$\blacksquare$

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jeteagle

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jeteagle

#19 h Dec 30, 2021, 7:35 PM

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Very fun problem.

I claim this point is the point $M$ that is the midpoint of arc $BAC$ . It is well known that $M$ is the Miquel Point of $BB_1C_1C$ and it lies on the perpendicular bisector of $B_1C_1$ . Let $N \ne A$ and be on both $(AC_1B)$ and $(AB_1C)$ and $F = B_1C \cap BC_1$ . This is the Miquel Point of complete quadrilateral $B_1CC_1B$ so $BB_1FN$ and $CC_1FN$ are cyclic quads. We also have $NBB_1 \sim NC_1C$ and $BB_1 = C_1C$ so $NB = NC_1$ and $NB_1 = NC$ .

Let $D$ and $E$ be the second points of intersection of $B_1C$ and $BC_1$ with $(ABC)$ respectively. Additionally we have $\measuredangle{BDC} = \measuredangle{BAC} = \measuredangle{BAC_1} = \measuredangle{BC_2C_1}$ so $B, D, C_2$ are collinear. Similarly $C, E, B_2$ are collinear. We also have $\measuredangle{NC_2C_1} = \measuredangle{NBC_1} = \measuredangle{NBF} = \measuredangle{NB_1F}$ so $N, B_1, C_2$ are collinear and so is $N, C_1, B_2$ . Now because $AC_1NC_2$ and $AB_1NB_2$ are cyclic quads and also $B_1C_2 \cap C_1B_2 = N$ , this means $A$ is the Miquel Point of $B_1B_2C_1C_2$ . Let $G = B_1B_2 \cap C_1C_2$ . Notice that $\measuredangle{B_1AG} = \measuredangle{BC_2G} = \measuredangle{NC_2C_1} = \measuredangle{NBC_1} = \measuredangle{BC_1N} = \measuredangle{BAN} = \measuredangle{B_1AN}$ so $A, G, N$ are collinear. This means $B_1B_2C_1C_2$ is a cyclic quad and let $M'$ be its center. It is well-known that $AB_1B_2M'$ and $M'$ lies on the perpendicular bisector of $B_1B_2$ . There are two points that satisfy this with one of them being $M$ . Clearly the other point doesn't work I'm too lazy to rigorously prove so $M = M'$ . Finally, it is again well-known that $(AC_1C_2)$ contains the center of this circle so it always passes through $M$ which is fixed. $\blacksquare$

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cursed_tangent1434

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cursed_tangent1434

#20 h Jul 5, 2024, 3:23 PM • 1 Y

Y by GeoKing

Very beautiful problem. I actually found this pretty straightforward. Doing certain other problems such as 2014 EGMO 2 with the given mysterious length condition helped. We let $M$ denote the arc midpoint of major arc $BC$ in $(ABC)$ (containing $A$ ). We know that the given condition translates to the following claim.

Claim : Points $A$ , $B_1$ , $C_1$ and $M$ are concyclic.
Proof : It is not hard to see that since $MC=MB$ , $CC_1=BB_1$ and
$\measuredangle C_1CM = \measuredangle ACM = \measuredangle ABM = \measuredangle B_1BM$ it follows that $\triangle MBB_1 \cong \triangle MCC_1$ . In particular, this implies that $MB_1=MC_1$ . Further, we can also note that there must exist a spiral similarity centered at $M$ mapping $BB_1 \mapsto CC_1$ and in turn, there must also exist a spiral similarity centered at $M$ mapping $B_1C_1 \mapsto BC$ . This gives us that $\triangle MB_1C_1 \sim \triangle MBC$ and thus,
$\measuredangle C_1MB_1 = \measuredangle CMB = \measuredangle CAB = \measuredangle C_1AB_1$ which most clearly implies the claim.

Let $\omega$ denote the circle centered at $M$ with radius $MB_1=MC_1$ . We can proceed with proving the following key claim.

Claim : Quadrilateral $B_1C_1B_2C_2$ is cyclic (with circumcenter $M$ ).
Proof : Let $R = (ABC_1) \cap (AB_1C)$ . Then, $BB_1=CC_1$ , $\measuredangle RB_1B = \measuredangle RCA$ and $\measuredangle RBA = \measuredangle CC_1R$ . Thus, $\triangle RB_1B \cong \triangle RCC_1$ . Now, this means there exists a spiral similarity centered at $R$ mapping $BB_1 \mapsto CC_1$ . Thus, there must also exist a spiral similarity centered at $R$ mapping $BC_1 \mapsto CB_1$ . This allows us to conclude that $\triangle RBC_1 \sim \triangle RCB_1$ . Let $C_2 ' = \overline{RB_1} \cap (AC_1B)$ and $B_2' = \overline{RC_1} \cap (AB_1C)$ . Then,
$\measuredangle CB_1R = \measuredangle C_1BR = \measuredangle C_1C_2'R$ and thus, $\overline{C_1C_2'} \parallel \overline{CB_1}$ which implies that $C_2'=C_2$ . Similarly we conclude that $B_2'=B_2$ . Thus, lines $\overline{B_1C_2}$ and $\overline{B_2C_1}$ intersect at $R$ . Clearly, $R$ is the arc midpoint of minor arc $BC_1$ (since $RB=RC_1$ due to the previously establish congruency). So, $\overline{C_2B_1}$ is the $\angle C_1C_2B$ -bisector. Now, we can note that,
$2\measuredangle C_1C_2B_1 = \measuredangle C_1CB = \measuredangle C_1AB_1 = \measuredangle C_1MB_1$ so it follows that $C_2$ lies on $\omega$ . A similar argument implies that $B_2$ also lies on $\omega$ which finishes the proof of the claim.

Now, the finish is clear. Simply note that,
$\begin{align*} \measuredangle B_2MC_2 &= \measuredangle B_2MC_1 + \measuredangle C_1MB_1 + \measuredangle B_1MC_2\\ &= 2\measuredangle B_2B_1C_1 + \measuredangle C_1AB + 2\measuredangle B_1C_1C_2\\ &= 2(\measuredangle B_2B_1C_1 + \measuredangle C_1B_1C) + \measuredangle C_1AB\\ &= 2\measuredangle B_2B_1C + \measuredangle C_1AB_1\\ &= \measuredangle B_2B_1C + \measuredangle BC_1C_2 + \measuredangle C_1AB_1\\ &= \measuredangle B_2AC + \measuredangle BAC_2 + \measuredangle CAB\\ &= \measuredangle B_2AC_2 \end{align*}$ and thus, $M$ lies on $(AB_2C_2)$ . Thus, as $B_2$ and $C_2$ vary along $\overline{AB}$ and $\overline{AC}$ , the circle $(AB_2C_2)$ passes through the fixed point $M$ , which is the midpoint of arc $BC$ not containing $A$ of $(ABC)$ which is what we wished to show.

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bronzetruck2016

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bronzetruck2016

#21 h Nov 14, 2024, 11:32 PM

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This is a very interesting problem. It took me a while but it was very fun.

Define $O$ as the circumcenter of $ABC$ , $O_1$ as the circumcenter of $AB_1C$ , $O_2$ as the circumcenter of $ABC_1$ , and $O_3$ as the circumcenter of $AB_2C_2$ . Let $D$ and $E$ be the midpoints of sides $AB$ and $AC$ , respectively.

Lemma 1: O_2O=O_1O
Proof of Lemma 1: Let $O'$ be circumcenter of $AB_1C_1$ . Then, the sides of quadrilateral $O'O_1OO_2$ are the perpendicular bisectors of segments $AC_1$ , $AC$ , $AB_1$ , and $AB$ , which are 2 pairs of parallel lines, with each pair separated by the same distance $\frac{1}{2}C_1C=\frac{1}{2}B_1B$ . Therefore, $O'O_1OO_2$ is a rhombus, so we are done.

Now, define $M$ and $N$ as the midpoints of $AB_2$ and $AC_2$ . I claim that $\measuredangle{OO_2O_3}=\measuredangle{O_3O_1O}$ .
Time to angle chase!
$\measuredangle{OO_2O_3}=\measuredangle{DO_2N}=\measuredangle{DO_2A}-\measuredangle{NO_2A}=\measuredangle{BC_1A}-\measuredangle{C_2C_1A}=\measuredangle{BC_1C_2}$ $=\measuredangle{B_2B_1C}=\measuredangle{CB_1A}+\measuredangle{AB_1B_2}=\measuredangle{AO_1D}+\measuredangle{MO_1A}=\measuredangle{MO_1D}=\measuredangle{O_3O_1O}$
Combining this result with Lemma 1 gets that triangle $O_1O_2O_3$ is isosceles and $O_1O_3O$ is isosceles. This means that $O_3O$ is the perpendicular bisector of $O_1O_2$ , which we will call $l$ . Since $O_1$ and $O_2$ always lie on the perpendicular bisectors of $AC$ and $AB$ , by Lemma 1, $l$ does not change as $O_1$ and $O_2$ vary. Therefore, $O_3$ always lies on a fixed line through $O$ , so our desired fixed point is simply the reflection of $A$ across $l$ .

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Saucepan_man02

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Saucepan_man02

#22 h Dec 4, 2024, 6:52 AM

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Let's G(e)o:

Let $D = C_1C_2 \cap B_1 B_2$ and $F = B_1 C_2 \cap C_1 B_2$ .

Claim: $D, F$ lie on the angle bisector on angle $\angle BAC$ .
Note that: $A$ is the Miquel point of $B_1 B_2 C_1 C_2$ . Thus: $AC_2 B_1 D, A B_2 C_1 D, A C_2 F C_2, A B_2 F B_1$ are cyclic. Thus; $C_2, B_2$ and $C_1, B_1$ are isogonal wrt $\angle BAC$ . Therefore, due to isogonal lemma, $D, F$ are isogonal wrt $\angle BAC$ .
Note that, $F$ is the spiral center sending $BB_1$ to $C C_1$ . Since $BB_1 = C C_1$ , then we have: $BF=C_1F$ and $B_1 F = CF$ .
Therefore; $F$ lies on the angle bisector of $\angle BAC$ .

Let $X$ be the midpoint of arc $BAC$ in $(ABC)$ .
Now, we finish the problem with Miquel Properties:
Notice that: $\triangle XBB_1 \cong \triangle XCC_1$ which implies $XB_1 = XC_1$ . Thus: $X$ maps $BB_1$ to $CC_1$ (spiral similarity) and by Miquel theorem on $BB_1 C_1 C$ , we have: $X$ to be its Miquel center with $AXB_1 C_1$ to be cyclic. Due to Miquel theorem on $C_1 B_1 C_1 B_2$ , we have $X$ to be its circumcenter.

By Miquel theorem on $C_1 B_1 C_1 B_2$ , we must have $AXB_2C_2$ to be cyclic. Therefore, $X$ is the claimed fixed point and we are done,

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HamstPan38825

8857 posts

HamstPan38825

#23 h Dec 22, 2024, 3:50 PM

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I thought this was fairly straightforward if one knows Miquel theory well. I claim the fixed point is the Miquel point $X = (AB_1C_1) \cap (ABC)$ of $B_1BCC_1$ .

Claim: $M = (ABC_1) \cap (AB_1C)$ is the arc midpoint of $\widehat{BC_1}$ and $\widehat{B_1C}$ , and it lies on $\overline{B_1C_2}$ and $\overline{C_1B_2}$ .

Proof: A spiral similarity at $M$ takes $\overline{BB_1} \to \overline{CC_1}$ , so in fact it must be a spiral congruence. Thus $MB = MC_1$ and $MB_1 = MC$ , and the first part follows.

For the second part, define $C_2' = \overline{MB_1} \cap (ABC_1)$ and $B_2'$ similarly; I will show $\overline{C_1C_2'} \parallel \overline{B_1C}$ . This is true because $\measuredangle C_1C_2M = \measuredangle C_1AM = \measuredangle CB_1M$ , as needed. $\blacksquare$

Claim: $X$ is the circumcenter of $C_2B_1C_1B_2$ .

Proof: We will show that $X$ is the circumcenter of both triangles $C_2B_1C_1$ and $B_2C_1B_1$ . In fact, as $XB_1 = XC_1$ by the same Miquel point properties, it suffices to show that $\measuredangle B_1XC_1 = \measuredangle B_1AC_1 = 2\measuredangle MAC_1 = 2\measuredangle B_1C_2C_1. \ \blacksquare$
Now note that $XB=XC$ , so $\overline{AX}$ bisects the external $\angle BAC$ . But $\measuredangle C_2AB = \measuredangle C_2C_1B = \measuredangle CB_1B_2 = \measuredangle CAB_2$ by parallelogram properties, so the external bisectors of $\angle BAC$ and $\angle C_2AB_2$ coincide; so $\overline{AX}$ bisects the external $\angle C_2AB_2$ and as $XC_2 = XB_2$ , $X$ lies on $(AB_2C_2)$ by Fact 5.

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Ywgh1

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Ywgh1

#24 h Jan 20, 2025, 2:44 PM

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$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(11.281882046676591cm); 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 = -7.810625290444857, xmax = 6.471256756231734, ymin = -5.601370414301284, ymax = 0.7419622161542905; /* image dimensions */ pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen ffwwqq = rgb(1.,0.4,0.); pen qqffff = rgb(0.,1.,1.); pen qqccqq = rgb(0.,0.8,0.); pen ffzztt = rgb(1.,0.6,0.2); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen ffcctt = rgb(1.,0.8,0.2); pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); /* draw figures */ draw(circle((-1.1333183935657343,-1.9230902496389544), 2.246924449764216), linewidth(1.) + qqwuqq); draw((0.7993241017788306,-3.069197778685229)--(-2.1418963035901135,0.08475438637227817), linewidth(1.) + ffwwqq); draw((-3.0426529553799324,-3.1076201200981206)--(-2.1418963035901135,0.08475438637227817), linewidth(1.) + ffwwqq); draw((0.7993241017788306,-3.069197778685229)--(-3.0426529553799324,-3.1076201200981206), linewidth(1.) + ffwwqq); draw(circle((-0.6616599985701797,-1.4832448673615106), 2.1563212141072086), linewidth(1.) + qqffff); draw(circle((-1.7540068272761298,-1.7479575281426314), 1.873310227228806), linewidth(1.) + qqffff); draw((-3.0426529553799324,-3.1076201200981206)--(0.047202520399816095,-2.2626769597603187), linewidth(1.) + qqccqq); draw((-2.7431815848701495,-2.046262686063586)--(1.4167121369496614,-0.908709886091022), linewidth(1.) + qqccqq); draw((0.7993241017788306,-3.069197778685229)--(-2.7431815848701495,-2.046262686063586), linewidth(1.) + qqccqq); draw((0.047202520399816095,-2.2626769597603187)--(-3.5523384591232348,-1.223272337444277), linewidth(1.) + qqccqq); draw(circle((-0.9712071016649549,-2.5920486405323233), 2.9216070324993746), linewidth(1.) + dotted + ffzztt); draw((-2.1418963035901135,0.08475438637227817)--(-1.259878807867243,-3.55492424279874), linewidth(1.) + wrwrwr); draw(circle((-1.155788017979065,0.3237218472945994), 2.852480503800896), linewidth(1.) + ffcctt); draw((-1.259878807867243,-3.55492424279874)--(1.4167121369496614,-0.908709886091022), linewidth(1.)); draw((-1.259878807867243,-3.55492424279874)--(-3.5523384591232348,-1.223272337444277), linewidth(1.)); draw(circle((-1.282348432280575,-1.3081121458651874), 1.636734468293493), linewidth(1.) + cqcqcq); draw((0.047202520399816095,-2.2626769597603187)--(-1.1557880179790645,0.32372184729459846), linewidth(1.) + wrwrwr); draw((-2.7431815848701495,-2.046262686063586)--(-1.1557880179790645,0.32372184729459846), linewidth(1.) + wrwrwr); draw((1.4167121369496614,-0.908709886091022)--(-2.1418963035901135,0.08475438637227817), linewidth(1.) + wrwrwr); draw((-2.1418963035901135,0.08475438637227817)--(-3.5523384591232348,-1.223272337444277), linewidth(1.) + wrwrwr); /* dots and labels */ dot((-3.0426529553799324,-3.1076201200981206),dotstyle); label("$B$", (-3.006207676845374,-3.0114890444703097), NE * labelscalefactor); dot((0.7993241017788306,-3.069197778685229),dotstyle); label("$C$", (0.841079865294838,-2.973954531864064), NE * labelscalefactor); dot((-2.1418963035901135,0.08475438637227817),dotstyle); label("$A$", (-2.1053793742954707,0.17894452706060046), NE * labelscalefactor); dot((-2.7431815848701495,-2.046262686063586),dotstyle); label("$B_2$", (-2.705931575995406,-1.9511390633438603), NE * labelscalefactor); dot((0.047202520399816095,-2.2626769597603187),linewidth(4.pt) + dotstyle); label("$C_1$", (0.0810059850183572,-2.185729767132898), NE * labelscalefactor); dot((1.4167121369496614,-0.908709886091022),linewidth(4.pt) + dotstyle); label("$B_2$", (1.451015695146335,-0.8344873133080416), NE * labelscalefactor); dot((-3.5523384591232348,-1.223272337444277),linewidth(4.pt) + dotstyle); label("$D$", (-1.120098418381514,0.39476797454651497), NE * labelscalefactor); dot((-1.259878807867243,-3.55492424279874),linewidth(4.pt) + dotstyle); label("$Y$", (-1.2233183280486903,-3.480670452048385), NE * labelscalefactor); dot((-1.694943236930499,-1.7596143850546555),linewidth(4.pt) + dotstyle); label("$J$", (-1.654965223020519,-1.6883974751001383), NE * labelscalefactor); dot((-1.001035827539835,-2.549325260769249),linewidth(4.pt) + dotstyle); label("$C_2$", (-3.5129235970296944,-1.1441470423095712), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Here is a Sketch

Let $J$ be the intersection of $B_2B_1$ and $C_2C_1$ , and let $(AB_1C)$ intersect $(ABC_1)$ again at $Y$ .

We claim that the fixed point is the mid point of arc $BAC$ . And let it be $D$ .

Firstly we show that $D$ lies on $AB_1C_1$ and then we show that $AY$ is the angle bisector of $\angle BAC$ , then we show $Y-C_1-B_2$ and $Y-B_1-C_2$ are collinear.

After that we get that $A$ is the Miquel point of $YB_1C_1D$ and we get that $J$ lies on $AY$ . Finally we show that $D$ is the Center of $(B_1B_2C_1C_2)$ . After that it’s easy to see that $D$ lies on circle $(AB_2C_2)$ hence we are done.

This post has been edited 4 times. Last edited by Ywgh1, Jan 20, 2025, 2:58 PM

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Ilikeminecraft

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Ilikeminecraft

#25 h Mar 15, 2025, 8:45 PM

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Let $P$ be the second intersection of $(ABC_1), (AB_1C).$
Note that $\angle AC_2C_1 = \angle ABC_1 = \angle AB_1B_2,$ with implies $AB_1B_2\sim AC_2C_1.$
By Spiral similarly, we get that $P = B_1C_2\cap B_2C_1.$
Furthermore, there is spiral similarity centered at $P$ mapping $BB_1$ to $CC_1$ , and so $P$ lies on the angle bisector of $\angle BAC.$
Take $\angle C_1C_2P=\angle PAC_1 = \angle BAP = \angle B_11B_2P,$ so $B_1C_2B_2C_1$ is cyclic.

Let $M$ be Miquel point of quad $BB_1C_1C.$
By the given length condition, we have $M$ is the midpoint of arc $BAC.$

Take a $\sqrt{bc}$ -inversion and we get:
Let $ABC$ be a triangle
Let $M$ be the intersection of the $A$ -external bisector and $BC$
Let $B_1$ be on $AB,$ and $C_1$ be the intersection of $MB_1, AC.$
Let $Q$ be a point on the $A$ angle bisector
Let $B_2$ be intersections of $(AB_1Q)$ with $CB_1.$
Define $C_2$ similarly.
Show that $B_2C_2$ passes through $M$ .

Note that from inversion, we also have that $B_1C_2B_2C_2$ are cyclic. Recall that $P$ is the second intersection $(AB_1C_2), (AB_2C_1).$
Observe that the Miquel point of $B_1B_2C_1C_2$ is $A.$ If we define $T$ to be the intersection of $B_1C_1, B_2C_2,$ by an application of Miquel’s master theorem, we get $A, T_0$ are inverses with respect to $(B_1C_2B_2C_1).$
By Ceva-Menelaus, note that $(M, B_1C_1\cap AP; B_1C_1) = -1$ , so $M$ lies on the polar of $AP\cap B_1C_1$ . By Brokard's on $B_1C_2B_2C_1,$ we get $M$ lies on the polar of $P.$ By La Hire's, $AP$ is the polar of $M.$ However, $AP\perp AM,$ which implies that $M$ and $A$ are inverses.
Thus, $M = T_0.$

This post has been edited 1 time. Last edited by Ilikeminecraft, Mar 15, 2025, 9:04 PM

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ohiorizzler1434

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ohiorizzler1434

#26 h Mar 30, 2025, 5:41 AM

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Solved with mcmp and Scilyse in 36 minutes.

Define $M_a$ , $N_a$ as the midpoints of arcs $(BC)$ and $(BAC)$ .

Claim: $N_aAB_1C_1$ cyclic.
Consider that $\measuredangle N_aBB_1 = \measuredangle N_aBA = \measuredangle N_aCA =\measuredangle N_aCC_1$ , and that as $N_aB=N_aC$ , $BB_1 = CC_1$ , there is a spiral similarity (congruence) at $N_a$ and it is thus the miquel point. This gives $N_aAB_1C_1$ concyclic.

Analogously define $A_1$ to be the spiral center of $BB_1$ to $C_1C$ . Therefore, as $BB_1=CC_1$ , the perpendicular distances of $A_1$ to $AB$ and $AC$ are equal, and thus $A_1$ lies on the angle bisector at $A$ , or $A A_1 M_a$ collinear.

Claim: $A_1B_1C_2$ collinear, $A_1C_1B_2$ collinear.
Note that $\measuredangle AC_1C_2 = \measuredangle ACB_1 = \measuredangle AB_2B_1$ , and similarly $\measuredangle AC_2C_1 = \measuredangle AB_1B_2$ , which gives $AC_2C_1 \sim AB_1B_2$ , or that $A$ is a miquel point. Consequently $(AC_2C_1) \cap (AB_1B_2)$ gives the intersection of $B_1C_2$ and $C_1B_2$ , which is $A_1$ .

Claim: $C_1C_2B_1B_2$ concyclic.
Recollect that by spiral congruence, $N_aB_1 = N_aC_1$ . Observe that $2\measuredangle B_1C_2C_1 = 2\measuredangle A_1C_2C_1 = 2\measuredangle A_1AC_1 = \measuredangle B_1AC_1 = \measuredangle B_1N_aC_1$ , giving $N_a$ the center of $(C_2B_1C_1)$ , or $N_aC_2=N_aB_1=N_aC_1$ . Analogously $N_a$ is the center of $(B_1C_1B_2)$ , giving $B_1B_2C_1C_2$ concyclic.

Claim: $C_1C_2$ , $B_1B_2$ , and $AA_1$ concur.
This is by considering cyclics $(AC_1A_1C_2)$ , $(AB_1A_1B_2)$ , and $(B_1B_2C_1C_2)$ , and applying radical axis theorem.

Claim: $AN_a$ , $B_2C_2$ , $B_1C_1$ concur.
Let $C_2B_2$ and $C_1B_1$ meet at $Q$ . Observe that in the complete quadrilateral, $Q$ is the pole of $AM_a$ as $AM_a$ passes through $C_1C_2 \cap B_1B_2$ . Then because $N_a$ is the center, $KN_a$ is perpendicular to $AM_a$ . However, as $AM_a$ is perpendicular to $AN_a$ by diameter, we know $K$ , $A$ , $N_a$ collinear.

This is enough to finish as now applying radical axis theorem gives that $AB_2C_2N_a$ is concyclic, and that is the fixed point.

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This post has been edited 1 time. Last edited by ohiorizzler1434, Mar 30, 2025, 5:41 AM

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