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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
Coloring
demmy   6
N 5 minutes ago by Kaimiaku
Source: Thailand TST 2015
What is the maximum number of squares in an $8 \times 8$ board that can be colored so that for each square in the board, at most one square adjacent to it is colored.
6 replies
demmy
Dec 2, 2023
Kaimiaku
5 minutes ago
Nice and easy FE on R+
sttsmet   21
N 10 minutes ago by bo18
Source: EMC 2024 Problem 4, Seniors
Find all functions $ f: \mathbb{R}^{+} \to \mathbb{R}^{+}$ such that $f(x+yf(x)) = xf(1+y)$
for all x, y positive reals.
21 replies
sttsmet
Dec 23, 2024
bo18
10 minutes ago
max power of 2 that divides \lceil(1+\sqrt{3})^{2n}\rceil for pos. integer n
parmenides51   2
N 19 minutes ago by Inspector_Maygray
Source: Gulf Mathematical Olympiad GMO 2017 p4
1 - Prove that $55 < (1+\sqrt{3})^4 < 56$ .

2 - Find the largest power of $2$ that divides $\lceil(1+\sqrt{3})^{2n}\rceil$ for the positive integer $n$
2 replies
parmenides51
Aug 23, 2019
Inspector_Maygray
19 minutes ago
Points in general position
AshAuktober   1
N 21 minutes ago by Rdgm
Source: 2025 Nepal ptst p1 of 4
Shining tells Prajit a positive integer $n \ge 2025$. Prajit then tries to place n points such that no four points are concyclic and no $3$ points are collinear in Euclidean plane, such that Shining cannot find a group of three points such that their circumcircle contains none of the other remaining points. Is he always able to do so?

(Prajit Adhikari, Nepal and Shining Sun, USA)
1 reply
AshAuktober
Yesterday at 2:15 PM
Rdgm
21 minutes ago
No more topics!
GOTEEM #5: Circumcircle passes through fixed point
tworigami   21
N Yesterday at 8:45 PM by Ilikeminecraft
Source: GOTEEM: Mock Geometry Contest
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
21 replies
tworigami
Jan 2, 2020
Ilikeminecraft
Yesterday at 8:45 PM
GOTEEM #5: Circumcircle passes through fixed point
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G H BBookmark kLocked kLocked NReply
Source: GOTEEM: Mock Geometry Contest
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tworigami
844 posts
#1 • 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
#2 • 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
#3 • 2 Y
Y by Adventure10, Mango247
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
#4 • 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.

[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]

For this new problem, let $X = \overline{BC_1} \cap \overline{CB_1}$. Then by definition $A,B_1,C_2,X$ are concyclic, and similarly $A,C_1,B_2,X$ are concyclic. Moreover, by Ceva-Menelaus $X$ is on the $A$-internal bisector. So
$$\measuredangle B_1B_2C_1 = \measuredangle XAC_1 = \measuredangle B_1AX = \measuredangle B_1C_2C_1,$$and thus the self-intersecting quadrilateral $B_1B_2C_1C_2$ is cyclic and evidently has Miquel point $A$. Let $T_0 = \overline{B_1C_1} \cap \overline{B_2C_2}$; then it is well known that $\angle XAT_0 = 90^\circ$. Thus, $T=T_0$, as desired.
This post has been edited 1 time. Last edited by Th3Numb3rThr33, Jan 6, 2020, 1:06 AM
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pinetree1
1206 posts
#5 • 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
#6 • 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
#7 • 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
1739 posts
#8 • 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
6857 posts
#9 • 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.
  1. 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 $\angle A$.
  2. 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.
  3. 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 $(B_2AB_1)$.
  4. 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 $A$.)
  5. 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.
  6. By power of a point from $Y$, we find $B_1B_2C_1C_2$ is cyclic (with Miquel point $A$).
  7. Then it follows by Miquel theory, and from $XB_1 = XC_1$, that $X$ is the circumcenter of $B_1B_2C_1C_2$.
  8. By Miquel theory, $XAB_2C_2$ concyclic as needed.
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NJOY
495 posts
#11 • 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
#13 • 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. :D
tworigami wrote:
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

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
924 posts
#14 • 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
1051 posts
#16 • 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
888 posts
#17
Y by
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
238 posts
#18
Y by
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
480 posts
#19
Y by
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
548 posts
#20 • 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
87 posts
#21
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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
1294 posts
#22
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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
8844 posts
#23
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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.
This post has been edited 1 time. Last edited by HamstPan38825, Dec 22, 2024, 3:50 PM
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Ywgh1
136 posts
#24
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[asy]

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[/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
277 posts
#25
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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, Yesterday at 9:04 PM
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