Let $\{Y,Z\}$ be the midpoints of minor arcs $AC, AB$ and $\{V,W\}$ be the midpoints of major arcs $AC, AB$. It's well-known that $BB_1\cap CC_1 \in OI$, so Pascal on $BB_1C_2CC_1B_2$ and $WC_1B_2BYZ$ implies $\{B_1C_2, C_1B_2, OI\}$ are concurrent on $YZ$, which is the perpendicular bisector of $\overline{AI}$.

What are $B_2, C_2$?

tastymath75025 wrote: If $B_2, C_2$ are the points opposite to $B,C$

EulerMacaroni wrote: tastymath75025 wrote: If $B_2, C_2$ are the points opposite to $B,C$ You should read the problem carefully next time, the statement doesn't say $ B_2,C_2\in\odot(ABC).$

DSD wrote: EulerMacaroni wrote: tastymath75025 wrote: If $B_2, C_2$ are the points opposite to $B,C$ You should read the problem carefully next time, the statement doesn't say $ B_2,C_2\in\odot(ABC).$ Ok congrats on ur reading abilities

Everyone's upvoting the previous post, but no one is saying anything. Lol.

Well, the thing is there is a certain "someone" who "trivialises" every problem by just writing two lines about it and wastes half of his/her posts criticising other comments. So these upvotes are just a reflection of the "repercussions" of his immature conduct.

babu2001 wrote: Well, the thing is there is a certain "someone" who "trivialises" every problem by just writing two lines about it and wastes half of his/her posts criticising other comments. So these upvotes are just a reflection of the "repercussions" of his immature conduct. The first post is precisely an entire , complete , total , perfect and beautiful solution with Pascal’s theorem!

Basically the same as post #02, but this is too beautiful! Denote the incircle as $\Omega$. $\textbf{Claim 01.}$ $O, I, CC_1 \cap BB_1$ are collinear. $\textit{Proof.}$ Monge Theorem on $(\omega_B, \omega, \Omega), (\omega_C, \omega, \Omega)$ respectively gives us that $BB_1 \cap CC_1$, $O$, and $I$ are collinear. $\textbf{Claim 02.}$ $X,I,O$ are collinear. $\textit{Proof.}$ Pascal Theorem on $B_1 C_2 C C_1 B_2 B$ gives us $B_1 C_2 \cap C_1 B_2 = X$, $CC_2 \cap BB_2 = O$ and $CC_1 \cap BB_1$ are collinear. Therefore, by $\textbf{Claim 01}$, we have $X,I,O$ are collinear. Now, denote $Y$ and $Z$ as the midpoints of minor arc $AB$ and $AC$ respectively. $\textbf{Claim 03.}$ $X,Y,Z$ are collinear. $\textit{Proof.}$ Notice that by Pascal on $WC_1 B_2 BYZ$ on $WC_1 \cap BY = I$, $C_1 B_2 \cap YZ$ and $B_2 B \cap WZ = O$ are collinear. This gives us $C_1 B_2, YZ$ and $OI$ are collinear. But notice that $C_1 B_2 \cap OI = X$, which gives us $X,Y,Z$ are collinear. Now, notice that $YZ$ is the perpendicular bisector of $AI$, and therefore $XA = XI$.

Overcomplicated it whoops. tastymath75025 wrote: Let $ABC$ be an acute triangle with incenter $I$ and circumcircle $\omega$. Suppose a circle $\omega_B$ is tangent to $BA,BC$, and internally tangent to $\omega$ at $B_1$, while a circle $\omega_C$ is tangent to $CA, CB$, and internally tangent to $\omega$ at $C_1$. If $B_2, C_2$ are the points opposite to $B,C$ on $\omega$, respectively, and $X$ denotes the intersection of $B_1C_2, B_2C_1$, prove that $XA=XI$. Proposed by Vincent Huang and Nathan Weckwerth Let $B_0, C_0$ be the midpoint of the arcs $ABC, BCA$ respectively, and let $B_0^\ast, C_0^\ast$ be the midpoints of the opposite arcs. Then, it is well known that $B_1=B_0I \cap \omega$ and $C_1=C_0I \cap \omega.$ We will show that $B_1C_2 \cap B_2C_1$ is the same as $B_0^\ast C_0^\ast \cap OI.$ This is enough since $B_0^\ast C_0^\ast$ is the perpendicular bisector of $AI.$ To do this, define $X=B_0^\ast C_0^\ast \cap B_2C_1.$ We show $X$ lies on $OI,$ which is sufficient. Apply Pascal: \begin{align*} B_0^\ast C_0^\ast CB_2C_1C_0 &\implies \{X,I,CB_2 \cap C_0B_0^\ast\} \text{ collinear}\\ CB_2BC_0B_0^\ast B_0 &\implies \{CB_2 \cap C_0B_0^\ast, O, BC_0 \cap B_0C \} \text{ collinear} \\ BC_0C_0^\ast B_0CC_2 &\implies \{BC_0 \cap B_0C, O, C_0^\ast B_0 \cap BC_2\} \text{ collinear} \\ C_0^\ast B_0B_0^\ast BC_2C &\implies \{C_0^\ast B_0 \cap BC_2,O,I \} \text{ collinear} \end{align*}Hence, $X,I,O$ are collinear. 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We define some points first . Let $E$ and $F$ denote the midpoint of arcs $\widehat{ABC}$ and $\widehat {ACB}$ respectively. Let $E',F'$ denote their respective antipodes on $\odot(ABC)$ . Set $D \stackrel{\text{def}}{:=} \overline{BB_1}\cap \overline{CC_1}.$ Note that $ \overline{E'F'}$ is the perpendicular bisector of $ \overline{AI}$ . So it suffices to prove that $X \in \overline{E'F'}$ . Set $:=$$$X_1 \stackrel{\text{def}}{:=} \overline{B_1C_2}\cap \overline{E'F'} \quad X_2 \stackrel{\text{def}}{:=} \overline{B_2C_1}\cap \overline{E'F'}.$$ Claim $:=$ $X \in \overline{OI}$ Proof $:=$ We will use the well known fact that $D \in \overline{OI}$ . Note that by Pascal on $B_1C_2CC_1B_2B$, we have $:=$ $$ \underbrace {\overline{B_1C_2}\cap \overline{B_2C_1}}_{X} -\underbrace {\overline{C_2C}\cap \overline{B_2B}}_{O} -\underbrace {\overline{CC_1}\cap \overline{BB_1}}_{D} \quad \text{collinear} \quad \implies X\in \overline {OD} \equiv \overline {OI}$$ Claim $:=$ $X_1,X_2 \in \overline{OI}$ Proof $:=$ We will use the well known fact that $EB_1 \cap FC_1 \equiv I$ By Pascal on $E'F'CC_2B_1E$ , we have $:=$ $$ \underbrace {\overline{B_1C_2}\cap \overline{E'F'}}_{X_1} -\underbrace {\overline{F'C}\cap \overline{B_1E}}_{I} -\underbrace {\overline{CC_2}\cap \overline{EE'}}_{O} \quad \text{collinear} \quad \implies X_1 \in \overline {OI}$$ Similarly , $X_2 \in \overline {OI}$ . Hence $X \equiv X_1 \equiv X_2$ and we conclude $\blacksquare$

Let $M$ and $D$ be the midpoints of arc $ABC$ and arc $AC$, and define $N$ and $E$ similarly for $C$. It is known that $M,I,B_1$ are collinear and so are $N,I,C_1$. Since $DE$ is the perpendicular bisector of $AI$, it suffices to show that $X$ lies on $DE$. Let $O$ be the circumcenter of $\triangle ABC$. Then by Pascal's theorem on hexagon $CC_2B_1MDE$, it follows that $CC_2\cap MD=O$, $C_2B_1\cap DE$, and $B_1M\cap EC=I$ are collinear so $OI$, $DE$ and $C_2B_1$ are concurrent. Similarly, $OI,DE$ and $C_1B_2$ are also concurrent and we're done.

What do opposite here means ?? anyone ??

@above, it means the antipode. I will use Wizard 32's diagram. From,Pascal on $C_{2}B_{1}C_{0}^{*}B_{2}C_{1}B_{0}^{*}$ $CC_{0}^{*}C_{0}BB_{0}^{*}B_{0}$ and $B_{1}BC_{0}C_{1}CB_{0}$, We get, $X--O--I--BC_{0} \cap CB_{0}--BB_{1} \cap CC_{1}$ Now since we know $C_{0}^{*}B_{0}^{*}$ is the perpendicular bisector of $AI$,it remains to show that $X$ lies on $C_{0}^{*}B_{0}^{*}$ Let $X'=C_{2}B_{1} \cap C_{0}^{*}B_{0}^{*}$. Now, combining the above result with Pascal on $CC_{0}^{*}B_{0}^{*}CC_{2}B_{1}$ and $BB_{0}^{*}CC_{1}C_{0}B_{1}$ gives $X'=X$ and hence we are done. So much Pascal....

Here's another way of interpreting the Pascals, lol. Let $B_3=B_2I\cap\omega,C_3=C_2I\cap\omega$ and $M,M'$ be the midpoints of small and large arc $AC$ respectively and $N,N'$ be the same for arc $AB$. We first prove the following claim: Claim: $B_3N',BC$ and $C_3M'$ are concurrent at $T$. Proof: We use complex numbers with $\omega$ as the unit circle. With $A=a^2,B=b^2,C=c^2$, then $I=-ab-bc-ca,N'=ab$. One can see that $b_3=\frac{a^2bc+abc^2+a^2c^2-ab^2c}{ab+b^2+bc-ac}$ from $\angle BB_3I=90^\circ$ and then check that $B_3N'\cap BC$ is symmetric in $b$ and $c$ (it's $\frac{a(b^2+c^2)(ab+bc+ca)}{a^2(b+c)+2abc+bc(b+c)}$). Let $\varphi$ be the composition of involutions on $\omega$ with centers $I,T$ and again $I$. They're collinear so by Ping Pong lemma, $\varphi$ is also an involution on $\omega$. Also, note $\varphi(M)=N,\varphi(B_1)=C_2$ and $\varphi(C_1)=B_2$. Thus, $MN,B_1C_2$ and $B_2C_1$ all go through the center of $\varphi$ (which is $X$) which is equivalent to $XA=XI$. Remark: Note we also proved that $T,I$ and $X$ are collinear (by using Ping Pong lemma).

Let $S_B$ and $S_C$ denote the midpoints of arcs $AC$ and $BC$ of $\omega$ not containing the third point, and $M_B, M_C$ be their antipodes. Also, let $O$ be the center of $(ABC)$. It's well known that $C_1, I$, and $M_C$ are collinear, as are $B_1, I$, and $M_C$. Additionally, by fact $5$, we find that $S_BS_C$ is the perpendicular bisector of $\overline{AI}$, so it suffices to show that $X$ lies on this line. Now, let $X_1 = \overline{S_BS_C} \cap \overline{B_1C_2}$ and $X_2 = \overline{S_BS_C} \cap \overline{B_2C_1}$. By Pascal's on $S_CS_BM_BB_1C_2C$ and $S_BS_CM_CC_1B_2B$, we find that $X_1 = X_2 = \overline{OI} \cap \overline{B_1C_2} \cap \overline{B_2C_1} = X$, as desired.

oops Let $M_A,M_B,N_B,M_C,N_C$ be the midpoints of arcs $\widehat{BC},\widehat{AC},\widehat{ABC},\widehat{AB},\widehat{ACB}$ respectively, let $O$ be the circumcenter, let $T = OI\cap (BIC)$, and let $S = B_2I\cap (ABC)$. By Pascal's on $N_BB_1C_2N_CC_1B_2$, $X$ lies on $OI$ (since $C_2N_C\cap B_2N_B$ is the Bevan point). By Pascal's on $N_CSB_2BCM_C$, $P = N_CS\cap BC$ lies on $OI$. Then by PoP at $P$, $SITN_C$ is cyclic, so the negative inversion at $I$ fixing $(ABC)$ swaps $T$ and $X$. Since $\triangle M_AIT$ is isosceles, it follows that $\triangle XAI$ is isosceles, as desired.

Define the midpoints of arcs $AC$, $ABC$, $AB$, $ACB$ be $M$, $N$, $P$, and $Q$. Also let $O$ be the circumcenter, and $IO$ be line $\ell$. Then $BB_1 \cap CC_1$ is known to be the exsimilicenter of the incircle and circumcircle, so it lies on $\ell$. Pascal's on $B_1BB_2C_1CC_2$ tells us $X \in \ell$. Pascal's on $BB_2C_1QPM$ tells us $X \in MP$, the perpendicular bisector of $AI$. $\blacksquare$

Let $M_B$ be the midpoint of arc $AC$ and $M_C$ be the midpoint of arc $AB$. Let $N_C$ and $N_B$ be the antipodes of $M_C$ and $M_B$. It is well known that $N_C$, $I$, and $C_1$ are collinear and similarly $N_B$, $I$, and $B_1$ are collinear. It is sufficient to show that $X$ lies on the perpendicular bisector of $AI$ that is $M_BM_C$(This is because $M_BA=M_BI$ and $M_CA=M_CI$). It is equivalent to show that $M_BM_C\cap B_1C_2$ and $M_BM_C\cap C_1B_2$ both lie on $IO$, where $O$ is the circumcenter of $ABC$. But this is just two applications of pascal; pascal on $C_2B_1N_BM_BM_CC$ and on $B_2C_1N_CM_CM_BB$.

Let $Y = \overline{BB_1} \cap \overline{CC_1}$. It is well-known (say, see EGMO 7.42) that $Y$ lies on $\overline{OI}$. Also, let $M_B$ and $M_C$ be the minor arc midpoints. Now, we apply Pascal twice: Pascal on $B_1C_2CC_1B_2B$ implies $X$, $O$ (the circumcenter), $Y$ collinear, hence $X$ lies on $\overline{OI}$, and Pascal on $B_2C_1QM_CM_BB$ implies that $I$, $O$, and $\overline{M_BM_C} \cap \overline{B_2C_1}$ are collinear. The second condition implies that $X$ lies on $\overline{M_BM_C}$, i.e. $XA=XI$.

Let $D$ be the second intersection of $AI$ with $\omega$. Define $E,F$ similarly for $B,C$, respectively. Let $M_A,M_B,M_C$ be the mid-points of arcs $BAC,CBA,ACB$, respectively. It is well-known that $M_BB_1$ and $M_CC_1$ intersect at $I$. Let $I'=\overline{B_2M_B}\cap\overline{C_2M_C}$. Since $I=\overline{BE}\cap\overline{CF}$, reflecting about $O$, $B$ goes to $B_2$, $E$ goes to $M_B$, $C$ goes to $C_2$ and $F$ goes to $M_C$. Hence, $I$ goes to $I'$. So, $I,O,I'$ are collinear. Claim: $X\in IO$. Proof. Applying Pascal's Theorem on $(C_2B_1M_BB_2C_1M_C)$, we get that $C_2B_1\cap B_2C_1=X$, $B_1M_B\cap C_1M_C=I$, $M_BB_2\cap M_CC_2=I'$ are collinear. Hence, $X,I,I'$ are collinear. Since $I,O,I'$ are collinear as well, the claim follows. Claim: $X\in EF$. Proof. Applying Pascal's Theorem on $(B_1C_2CFEM_B)$, we get that $B_1C_2\cap FE$, $C_2C\cap EM_B=O$, $CF\cap M_BB_1=I$ are collinear. Hence, $B_1C_2\cap FE\in IO$. However, $X=B_1C_2\cap IO$, hence $X\in EF$, as claimed. Note that $E$ is the center of circle $(AIC)$ and $F$ is the center of $(AIB)$. Hence, $EF$ is the perpendicular bisector of $\overline{AI}$. Since $X\in EF$, we have that $XA=XI$, as desired. $\blacksquare$

solved with Kanad, AN1729, Om245