This is a very nice problem! Here is a generalization of the original one. Problem. Let $P, Q$ be two isogonal conjugate points WRT $\triangle ABC$. $BP, CP$ intersect $CA, AB$ at points $B_1, C_1$, respectively. Let $B_2 \neq B$ be a point on $BP$ such that $AB = AB_2$. Let $C_2 \neq C$ be a point on $CP$ such that $AC = AC_2$. $O_b, O_c$ are the centers of $\odot (BC_1C_2), \odot (CB_1B_2)$, respectively. Prove that $\angle O_bAO_c = \angle BQC$.

Ya_pank wrote: In triangle $ABC$ angle bisectors $AA_{1}$ and $CC_{1}$ intersect at $I$. Line through $B$ parallel to $AC$ intersects rays $AA_{1}$ and $CC_{1}$ at points $A_{2}$ and $C_{2}$ respectively. Let $O_{a}$ and $O_{c}$ be the circumcenters of triangles $AC_{1}C_{2}$ and $CA_{1}A_{2}$ respectively. Prove that $\angle{O_{a}BO_{c}} = \angle{AIC} $ It is really nice problem. I know a synthetic proof but it seems that the calculations with barycentric coordinate are not too hard Rewrite this problem to vertex $A$. In triangle $ABC$ angle bisectors $BB_{1}$ and $CC_{1}$ intersect at $I$. Line through $A$ parallel to $BC$ intersects rays $BB_{1}$ and $CC_{1}$ at points $B_{2}$ and $C_{2}$ respectively. Let $O_{b}$ and $O_{c}$ be the circumcenters of triangles $BB_{1}B_{2}$ and $BC_{1}C_{2}$ respectively. Prove that $\angle{O_{b}AO_{c}} = \angle{BIC} $. Take $A(1,0,0),\ B(0,1,0),\ C(0,0,1),\ I = \left( \frac{a}{a + b + c}, \frac{b}{a + b + c}, \frac{c}{a + b + c} \right) $ We easily seen the homogenous barycentric coordinates of points $B_1 = \left(\frac{a}{a + c}, 0, \frac{c}{a + c} \right), C_1 = \left(\frac{a}{a + b}, \frac{b}{a + b}, 0 \right)$ and $B_2 = \left(1,\frac{-c}{a}, \frac{c}{a} \right) ,C_2 = \left(1, \frac{b}{a}, \frac{-b}{a} \right).$ The square of distances $d_c^2 =B_1B_2^2= \frac{c^{5} + 2ac^{4} + a^{2}c^{3} - b^{2}c^{3}}{a^{3} + ac^{2} + 2a^{2}c}, d_{b1}^2 =B_2C^2= \frac{c^{3} + ab^{2} + ac^{2} - a^{2}c - b^{2}c}{a}, d_{b2} =CB_1^2=\frac{a^{2}b^{2}}{a^{2} + c^{2} + 2ac}.$ Therefore, $O_b=\frac{d_c^2(d_{b1}^2+d_{b2}^2-d_c^2)C+d_{b1}^2(-d_{b1}^2+d_{b2}^2+d_c^2)B_1+d_{b2}^2(d_{b1}^2-d_{b2}^2+d_c^2)B_2}{d_c^2(d_{b1}^2+d_{b2}^2-d_c^2)+d_{b1}^2(-d_{b1}^2+d_{b2}^2+d_c^2)+d_{b2}^2(d_{b1}^2-d_{b2}^2+d_c^2)}=$ $=\left( \frac{-b^{2}c + c^{3} - 2ca^{2} + a^{3}}{\left(-c - a \right)\left(b - c + a \right)\left(b + c - a \right)},\frac{b^2(2c - a)}{\left(-a - c \right)\left(b - a + c \right)\left(b + a - c \right)}, \frac{c(a^{2} - ac - 2b^{2})}{\left(-c - a \right)\left(b - c + a \right)\left(b + c - a \right)} \right).$ Similarly, $d_b^2 =C_1C_2^2 \frac{b^{5} + 2ab^{4} + a^{2}b^{3} - b^{3}c^{2}}{a^{3} + ab^{2} + 2a^{2}b},d_{c1}^2 = \frac{b^{3} + ab^{2} + ac^{2} - bc^{2} - a^{2}b}{a},d_{c2}^2 = \frac{a^{2}c^{2}}{a^{2} + b^{2} + 2ab}.$ Therefore, $O_c=\frac{(d_b^2(d_{c1}^2+d_{c2}^2-d_b^2)B+d_{c1}^2(-d_{c1}^2+d_{c2}^2+d_b^2)C_1+d_{c2}^2(d_{c1}^2-d_{c2}^2+d_b^2)C_2}{d_b^2(d_{c1}^2+d_{c2}^2-d_b^2)+d_{c1}^2(-d_{c1}^2+d_{c2}^2+d_b^2)+d_{c2}^2(d_{c1}^2-d_{c2}^2+d_b^2)}$ $=\left( \frac{-c^{2}b + b^{3} - 2ba^{2} + a^{3}}{\left(-b - a \right)\left(c - b + a \right)\left(c + b - a \right)}, \frac{b(a^{2} - ab - 2c^{2})}{\left(-b - a \right)\left(c - b + a \right)\left(c + b - a \right)}, \frac{c^{2}(2b - a)}{\left(-a - b \right)\left(c - a + b \right)\left(c + a - b \right)} \right).$ From these, $AO_b^2 = \frac{ab^2c(2b^{2}+2c^{2}-a^2)}{\left(a + c \right)^{2}\left(b - a + c \right)\left(b + a - c \right)},$ $AO_c^2 = \frac{abc^2(2b^{2} + 2c^{2}-a^{2})}{\left(b + a \right)^{2}\left(c - b + a \right)\left(c + b - a \right)}$ and $O_bO_c^2 = \frac{abc(a + b + c)(2b^{2} + 2c^{2}-a^{2})(a^{3} + a^{2}b + a^{2}c - ab^{2} + 3abc - ac^{2} - b^{3} + b^{2}c + bc^{2} - c^{3})}{\left(a + c \right)^{2}\left(b + a \right)^{2}\left(b - a - c \right)\left(b - a + c \right)\left(b + a - c \right)}.$ Thus $$\cos\angle O_bAO_c=\frac{AO_b^2+AO_c^2-O_bO_c^2}{2AO_b\cdot AO_c}=\sin\frac{A}{2}=\cos\angle BIC.$$Done! P/s: I always admire Russian geometry problems .

Attachments:

Inspiration Let $K$ be circumcenter of triangle $AO_bO_c$. Let $L$ be circumcenter of triangle $KO_bO_c$. Then $AL$ is the symmedian line of $ABC$.

Click to reveal hidden text

Attachments:

Solution using complex numbers. It seems to be much shorter and requires far less calculations than that bary bash above Denote $A_{0}, B_{0}, C_{0}$ the midpoints of minor arcs $BC, CA, AB$ and $B'$ the second intersection of $A_{2}C_{2}$ with circle $(ABC)$. Then we have $A = a^2;\; B = b^2;\; C = c^2;\; A_{0} = -bc;\; B_{0} = -ca;\; C_{0} = -ab;\; B' = \dfrac{a^2c^2}{b^2}$ $A_{1} \in AA_{0} \iff a_{1} - a^2bc \overline{a_{1}} = a^2 - bc \quad | \cdot bc$ $A_{1} \in BC \iff a_{1} + b^2c^2 \overline{a_{1}} = b^2 + c^2 \quad | \cdot a^2$ Adding these we get $(a^2 + bc) \cdot a_{1} = a^2b^2 + a^2c^2 +a^2bc - b^2c^2$ $A_{2} \in AA_{0} \iff a_{2} - a^2bc \overline{a_{2}} = a^2 - bc \quad | \cdot bc^2$ $A_{2} \in BB' \iff a_{2} - a^2c^2 \overline{a_{2}} = b^2 - \dfrac{a^2c^2}{b^2} \quad | \cdot b^2c$ Adding these we get $bc(b+c) \cdot a_{2} = b^4c + a^2c^3 + a^2bc^2 - b^2c^3 = (b+c)(a^2c^2 + b^3c - b^2c^2) \Rightarrow bc \cdot a_{2} = a^2c^2 + b^3c - b^2c^2$ $O_{c} = x;\; O_{a} = y$. Projection of $O_{c}$ on $AA_{0}$ is the midpoint of $A_{1}A_{2}$ $(x + a^2 - bc + a^2bc \overline{x})/2 = (a_{1} + a_{2})/2 \iff x + a^2bc \overline{x} = a_{1} + a_{2} + bc - a^2 \quad | \cdot bc$ Projection of $O_{c}$ on $BC$ is the midpoint of $A_{1}C$ $(x + b^2 + c^2 - b^2c^2 \overline{x})/2 = (a_{1} + c^2)/2 \iff x - b^2c^2 \overline{x} = a_{1} - b^2 \quad | \cdot a^2$ Adding these we get $(a^2+bc) \cdot x = (a^2 + bc) \cdot a_{1} + bc \cdot a_{2} - a^2b^2 - a^2bc + b^2c^2 = b^3c - b^2c^2 + 2a^2c^2$ Clearly if $x = f(a,b,c)$ then $y = f(c,b,a) \Rightarrow (c^2+ba) \cdot y = b^3a - b^2a^2 + 2a^2c^2$ Consider $r = \dfrac{\overrightarrow{O_{c}B}}{\overrightarrow{O_{a}B}} = \dfrac{b^2 - x}{b^2 - y} = \left(\dfrac{2a^2c^2 - a^2b^2 - c^2b^2}{a^2+bc}\right) / \left(\dfrac{2a^2c^2 - a^2b^2 - c^2b^2}{c^2+ab}\right) = \dfrac{c^2+ab}{a^2+bc}$ Therefore $\dfrac{\overrightarrow{O_{c}B}}{\overrightarrow{O_{a}B}} : \dfrac{\overrightarrow{CC_{0}}}{\overrightarrow{AA_{0}}} = 1 \Rightarrow \dfrac{\overrightarrow{O_{c}B}}{\overrightarrow{O_{a}B}} : \dfrac{\overrightarrow{CI}}{\overrightarrow{AI}} \in \mathbb{R} \Rightarrow \measuredangle{O_{a}BO_{c}} \equiv \measuredangle{AIC}$ Actually this is my solution that I came up with suring the contest. It took me something like 10 minutes to destroy the problem with complex bash

Here is a solution without coordinates. We re-write the problem to vertex $A$. Ya_pank wrote: In triangle $ABC$, angle bisectors $BB_{1}$ and $CC_{1}$ intersect at $I$. The line through $A$ parallel to $BC$ intersects rays $BB_{1}$ and $CC_{1}$ at points $B_{2}$ and $C_{2}$ respectively. Let $O_{b}$ and $O_{c}$ be the circumcenters of triangles $BC_{1}C_{2}$ and $CB_{1}B_{2}$ respectively. Prove that $\angle{O_{b}AO_{c}} = \angle{BIC} $ [asy][asy] size(10cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); draw(A--B--C--A--cycle); draw(unitcircle); pair I = incenter(A,B,C); pair C_1 = extension(A,B,C,I); pair B_1 = extension(A,C,B,I); pair B_2 = extension(A,A+B-C,B,I); pair C_2 = extension(A,A+B-C,C,I); pair CC = dir(160); pair T = intersectionpoints(A--extension(B,B+B*dir(90),C,C+C*dir(90)),unitcircle)[0]; pair S = extension(T,CC,A,B); draw(C--C_2); draw(B--B_2); draw(B_2--C_2); draw(circumcircle(B,C_1,C_2)); pair O_b = circumcenter(B,C_1,C_2); draw(T--CC); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$B_1$",B_1,dir(B_1)); dot("$C_1$",C_1,dir(C_1)); dot("$B_2$",B_2,dir(B_2)); dot("$C_2$",C_2,dir(C_2)); dot("$T$",T,dir(T)); dot("$S$",S,dir(S)); dot("$C'$",CC,dir(CC)); pair P = extension(B,CC,A,C_2); draw(B--P,dashed); dot("$P$",P,dir(P)); dot("$I$",I,dir(I)); dot("$O_b$",O_b,dir(O_b)); [/asy][/asy] Let $C'$ be the midpoint of $\widehat{AB}$. Claim: $BC'$ and $AC_2$ intersect at a point $P$ on $(BC_1C_2)$. Proof: Since we have $$\frac{AC_2}{BC}=\frac{AC_1}{BC_1}=\frac{AC}{BC}$$this means that $AC_2=AC$, and that $$\angle AC_2C=\angle C_2 CA=\angle C'BA$$and the conclusion follows. $\square$ Now let $T$ be the point on $(ABC)$ such that $ABTC$ is harmonic, and let $TC'$ intersect $AB$ at $S$. We have $$-1=(A,T;B,C)\stackrel{C'}{=}(A,S;B,C_1)$$ This means that the polar of $A$ with respect to $(BC_1C_2)$ passes through $S$. And applying Brokard's theorem on $C_2PC_1B$, the polar of $C'$ passes through $A$, which means that the polar of $A$ passes through $C'$. This means that the polar of $A$ is in fact line $C'T$, which implies that $C'T\perp AO_b$. Similarly, if we define the point $B'$, we have $B'T\perp AO_c$. Now we just angle chase. $$\angle O_bAO_c=180^{\circ}-\angle B'TC'=\angle BIC$$

As I said I knew a synthetic proof, it was from here, page 18-19. Based on this nice proof, I give here a general for this nice problem with the same proof's idea. General problem. Let $ABC$ be a triangle. Points $B'$ and $C'$ lie on line $AB$ and $AC$, respectively, such that $BC'\parallel CB'$. Bisector of $\angle C'BA$ meets $AC$ at $P$. Bisector of $\angle B'CA$ meets $AB$ at $Q$. $BP$, $CQ$ meet parallel line from $A$ to $BC'$ ,$CB'$ at $M$, $N$, respectively. Let $K$ and $L$ be circumcenters of triangles $MPC$ and $NQB$, respectively. $BP$ meets $CQ$ at $I$. Prove that $\angle KAL=\angle BIC$. Proof. Let $(O)$ be circumcircle of $(ABC)$. Circle $(L)$ and $(K)$ meet $MN$ again at $S$ and $T$, respectively. Since $MN\parallel CB'$, $\angle SBA=\angle ANQ=\angle B'CN=\angle ACN$, this deduces that $BS$ and $CN$ meet at $U$ on $(O)$. Similarly, $CT$ and $BM$ meet at $V$ on $(O)$. Let $Y$, $Z$ lie on $AB$, $AC$, respectively, such that $$(AY,BQ)=(AZ,CP)=-1.$$From this construction, $U(AY,BQ)=V(AZ,CP)=-1$, this means $UY$ and $VZ$ meet at $X$ on $(O)$ such that $ABXC$ is a harmonic quadrilateral. Also by Brocard's theorem, $UY\perp AL$ and $VZ\perp AK$. Thus, $$\angle KAL=180^\circ-\angle UXV=180^\circ-\angle ACU-\angle ABV=180^\circ-\angle PBC'-\angle QCB'=\angle BIC.$$This completes the proof.

Attachments:

Amazing problem!!! I was told main claim by p_square. We rename everything as expected to a $A$ centric figure as expected. Magic Claim: Let the reflection of $C$ over $A$ symmedian be $C'$. Now, $C\in (BC_1C_2)$. Proof: The proof for this is infact easy angle chasing after a small inversion. Let us perform $\sqrt{bc}$ inverion and reflection to make the claim slightly simpler. Now, the rephrased claim is as follows- Let $T$ be a point on the $A$ tangent such that $AT=AB$ and $T$ is on the same side as $C$ of the angle bisector of $BAC$. Let $X$ be a point on the extension of $AC$ beyond $C$ such that $CX=BC$, $M$ midpoint of $BC$ and $B'$ be the reflection of $B$ in $AM$. Now, $TB'CX$ are cyclic. Proof: Remember that $\angle AXT=\angle AC_2C_1=\frac{\angle BCA}{2}$. So, we just want that $\angle AB'T=180-\frac{\angle BCA}{2}$. For convenience, let $\angle BAC=A$ and similarly for $B,C$. Now, observe that $\angle AB'C=180-\angle MAB'=180-\angle BAM$ . Also, $AB'=AY$, thus $\angle AB'Y=\angle AYB'\implies \angle ABY'=\frac{180-\angle BAY'}{2}=\frac{180-(A+B-2\angle BAM)}{2}=\frac{C}{2}+\angle BAM$. Now, by angle sum on point $B'$, we get that $\angle CB'Y=180-\frac{C}{2}$. Thus, the claim follows. Now, we have that the reflection of $C$ in the $A-$ symmedian lies on $BC_1C_2$. Let this point be $C'$. Now, $\angle C_2AO_B=\frac{A+B-\angle CAC'}{2}$. Thus, $\angle C_2AO_B+\angle B_2AO_C=\frac{A+B+C+A-\angle CAC'-\angle BAB'}{2}=90-\frac{A}{2}\implies \angle O_BAO_C=90+\frac{A}{2}$ and the problem statement follows.

Here is another general problem. Let $ABC$ be a triangle. Points $E$, $F$ lie on lines $CA$, $AB$, respectively, such that $BE\parallel CF$. Bisectors of angles $\angle AEB$, $\angle AFC$ meet at $J$. $JF$, $JE$ meet lines $CA$, $AB$ at $M$, $N$, and meet parallel line from $A$ to $BE$, $CF$ at $P$, $Q$, respectively. Let $K$, $L$ be circumcenters of triangles $BQN$, $CPM$, respectively. Prove that $\angle KAL=\angle EJF$.

Attachments:

Found fact that describes triangle AO_bO_c completely and it solves the problem. Let I_a, O be the excenter of ABC which is tangent to BC and the circumcenter of ABC. Circle through IBCI_a meet I_aO at I_a and J. Then JBC is similar to AO_cO_b.

buratinogigle wrote: Inspiration Let K be circumcenter of triangle AO_bO_c. Let L be circumcenter of triangle KO_bO_c. Then AL is the symmedian line of ABC.

Click to reveal hidden text

From buratinogigles proof we have that AO_b^2/AO_c^2 = b(b+a)^2(c-b+a)/c(a+c)^2(a+b-c) = IB_1/IC_1. But from problem angles A and I of AO_bO_c, IB_1C_1 are equal so AO_bO_c, IB_1C_1 are similar. From ACGNmaths proof we got that symmedian of ABC forms angles 90-C/2, 90-B/2 with lines AO_b, AO_c. So similarity sending IB_1C_1 to AO_bO_c sends line AI to line which is isogonal to the A symmedian line of ABC with respect to O_bAO_c. It is known fact that AI goes through nine-point center of IB_1C_1, so isogonal to nine-point center of AO_bO_c ie point X54 lies on the A symmedian line of ABC. And it well known that line AX54 goes through the circumcenter of BOC for any ABC.

Let the line through $B$ parallel to $AC$ intersect $(AC_1C_2)$, $(CA_1A_2)$ again at $D, E$. Let $M_c, M_a$ be the arc midpoints of minor arcs $BA, BC$. Because $\angle DAC_1 = \angle DC_2C = \frac12 \angle BCA = \angle M_cAC_1$, point $D$ lies on line $AM_c$. Similarly, $M_a = CE\cap A_1A_2$. Let $DC_1\cap AC_2 = F$, $EA_1\cap CA_2 = G$. We claim that lines $FM_c$ and $GM_a$ concur with the $A$-symmedian line and $(ABC)$. This is because $-1 = M_c(B, F; A, C_2) = (B, M_cF\cap (ABC); A, C)$ and similar for the other side. Suppose they concur at $K$. Then by Brokard, $\angle O_aBO_c = \pi - \angle M_cKM_a = \angle AIC$ as desired. In fact, we have $\triangle O_aBO_c \sim \triangle A_1IC_1$ with opposite orientation.

THVSH wrote: This is a very nice problem! Here is a generalization of the original one. Problem. Let $P, Q$ be two isogonal conjugate points WRT $\triangle ABC$. $BP, CP$ intersect $CA, AB$ at points $B_1, C_1$, respectively. Let $B_2 \neq B$ be a point on $BP$ such that $AB = AB_2$. Let $C_2 \neq C$ be a point on $CP$ such that $AC = AC_2$. $O_b, O_c$ are the centers of $\odot (BC_1C_2), \odot (CB_1B_2)$, respectively. Prove that $\angle O_bAO_c = \angle BQC$. Composition of rotations $\mathcal {R} (O_C,B_2\mapsto C) \circ \mathcal {R} (A,B\mapsto B_2)$ has summary rotation angle $$\measuredangle BAB_2+\measuredangle B_2O_CC=\pi -2\measuredangle B_2BA+2\measuredangle B_2B_1C=\pi -2\measuredangle BAC.$$Analogously it works for composition $\mathcal {R} (A,C_2\mapsto C) \circ \mathcal {R} (O_B,B\mapsto C_2),$ so these two mappings are coinciding rotations. If $O$ is the center of this rotation, then $$\measuredangle O_BAO_C=\measuredangle O_BAO+\measuredangle OAO_C=\frac{1}{2}(\measuredangle C_2AC+\measuredangle BAB_2)=$$$$=\pi -\measuredangle ACP-\measuredangle PBA=\pi-\measuredangle QCB-\measuredangle CBQ=\measuredangle BQC.$$

One of the greatest geometry olympiad problems I've ever seen. First, rewrite to vertex $A.$ Consider two following rotation compositions. Rotation around $A$ taking $B$ to $B_2,$ followed by rotation around $O_C$ taking $B_2$ to $C.$ Rotation around $O_B$ taking $B$ to $C_2,$ followed by rotation around $A$ taking $C_2$ to $C.$ It is not hard to verify that each of them is equivalent to the rotation $\mathcal {R}$ taking $B$ to $C$ around a point $O$ such that $OB = OC$ and $\measuredangle BOC = 2\measuredangle BAC.$ Then consider $\measuredangle O_BAO_C=\measuredangle O_BAO+\measuredangle OAO_C.$ Notice that $\mathcal {R}$ takes $O_B$ to a point $O_B'$ such that $O_BO = O_B'O.$ But also notice the rotation around $O_B$ taking $B$ to $C_2,$ followed by rotation around $A$ taking $C_2$ to $C,$ takes $O_B$ to $O_B'.$ So this means $O_BA = O_B'A,$ and $\measuredangle O_BAO = \frac{1}{2} \measuredangle O_BAO_B' = \frac{1}{2} \measuredangle C_2AC.$ Notice that the reverse of $\mathcal {R}$ takes $O_C$ to a point $O_C'$ such that $O_CO = O_C'O.$ But also notice the rotation around $O_C$ taking $C$ to $B_2,$ followed by rotation around $A$ taking $B_2$ to $B,$ takes $O_C$ to $O_C'.$ So this means $O_CA = O_C'A,$ and $\measuredangle OAO_C = \frac{1}{2} \measuredangle O_C'AO_C = \frac{1}{2} \measuredangle BAB_2.$ Thus $$ \measuredangle O_BAO+\measuredangle OAO_C=\frac{1}{2}\measuredangle C_2AC+ \frac{1}{2}\measuredangle BAB_2= -\frac{1}{2} \measuredangle ICB - \frac{1}{2} \measuredangle CBI = \measuredangle BIC$$ as desired. $\square$ Remark: The mystical rotation compositions were motivated by constructing $O$ first, then guessing several angle equalities.

Brilliant problem! I thought of this solution independently but unfortunately a lot of people had come across it before. Anyway I’m going to post it again.

Attachments:

squareman wrote: One of the greatest geometry olympiad problems I've ever seen. First, rewrite to vertex $A.$ Consider two following rotation compositions. Rotation around $A$ taking $B$ to $B_2,$ followed by rotation around $O_C$ taking $B_2$ to $C.$ Rotation around $O_B$ taking $B$ to $C_2,$ followed by rotation around $A$ taking $C_2$ to $C.$ It is not hard to verify that each of them is equivalent to the rotation $\mathcal {R}$ taking $B$ to $C$ around a point $O$ such that $OB = OC$ and $\measuredangle BOC = 2\measuredangle BAC.$ Then consider $\measuredangle O_BAO_C=\measuredangle O_BAO+\measuredangle OAO_C.$ Notice that $\mathcal {R}$ takes $O_B$ to a point $O_B'$ such that $O_BO = O_B'O.$ But also notice the rotation around $O_B$ taking $B$ to $C_2,$ followed by rotation around $A$ taking $C_2$ to $C,$ takes $O_B$ to $O_B'.$ So this means $O_BA = O_B'A,$ and $\measuredangle O_BAO = \frac{1}{2} \measuredangle O_BAO_B' = \frac{1}{2} \measuredangle C_2AC.$ Notice that the reverse of $\mathcal {R}$ takes $O_C$ to a point $O_C'$ such that $O_CO = O_C'O.$ But also notice the rotation around $O_C$ taking $C$ to $B_2,$ followed by rotation around $A$ taking $B_2$ to $B,$ takes $O_C$ to $O_C'.$ So this means $O_CA = O_C'A,$ and $\measuredangle OAO_C = \frac{1}{2} \measuredangle O_C'AO_C = \frac{1}{2} \measuredangle BAB_2.$ Thus $$ \measuredangle O_BAO+\measuredangle OAO_C=\frac{1}{2}\measuredangle C_2AC+ \frac{1}{2}\measuredangle BAB_2= -\frac{1}{2} \measuredangle ICB - \frac{1}{2} \measuredangle CBI = \measuredangle BIC$$ as desired. $\square$ Remark: The mystical rotation compositions were motivated by constructing $O$ first, then guessing several angle equalities. This solution, in hindsight, is really convoluted and kind of betrays the beauty of this problem. I revisited this problem yesterday, and found a much more direct, just angle chasing approach (which hasn't been found here it seems). This time, no rewriting. First, get $BC_2=BC$ and $BA_2=BA$ from the parallel condition. Let $X,Y$ be the intersections of the circles centered at $B$ through $C,A$ with $(O_A)$ and $(O_C)$ respectively. Note \begin{align*} \measuredangle CXA &= \measuredangle C_2XA + \measuredangle CXC_2 \\ &= \measuredangle C_2C_1A + \frac{1}{2} \measuredangle CBC_2\\ &= \frac{1}{2} \measuredangle ACB+ \measuredangle BAC + 90^\circ + \frac{1}{2} \measuredangle A_2BA + \frac{1}{2} ABC \\ &= \measuredangle A_2YA + \measuredangle CYA_2 \\ &= \measuredangle CYA \\ \end{align*}So $AXCY$ is cyclic. But the perpendicular bisectors of $AY,CX$ meet at $B,$ so $AXCY$ is an isosceles trapezoid and $\measuredangle ABC = \measuredangle XBY.$ Since $BO_A$ bisects $\angle C_2BX$ and $BO_C$ bisects $\angle A_2BY$ we're finished. $\blacksquare$

Attachments:

Nice problem! Rough sketch because I'm out of time: Invert at $B$ to get the above diagram. By angle chasing, $BC=BP$ and similar. Reflect $R$ and $O_B$ over $\overline{I_AI_C}$. By angle chasing, $AR'BO_B'$ is cyclic and similar. Then by angle chasing, $\triangle AO_B'O_C' \sim \triangle ABI_B \sim \triangle AI_CC$. Then, there is a spiral similarity centered at $A$ taking $BO_BI_CO_B'$ to $I_BO_CCO_C'$. Then, done by angle chasing.

Solved with an orange (or whatever he is) First, let's A-index the problem, because we're Americans. Define $M_B,M_C$ as the arc midpoints of $\overarc{AC}, \overarc{AB}$. Define $S_B = AC_2\cap BM_C, S_C = AB_2\cap CM_B, R_B = S_BC_1\cap BC_2$, and $R_C = S_CB_1\cap CB_2$. First, observe that \[\angle S_BC_2C_1 = \angle AC_2C_1 = \frac12 \angle C = \angle M_CBC_1 = \angle S_BBC_1\]so $(C_2S_BC_1B)$ and $(B_2S_CB_1C)$ are cyclic. By brocard, we have $R_BM_c\perp AO_B$, and $R_CM_B\perp AO_C$. Let $T$ be the intersection of $M_CR_B$ and $M_BR_C$. If $\angle M_CTM_B = 90- \frac12 \angle A$, then $\angle O_BAO_C = 180 - \angle M_CTM_B = 90 + \frac12A$, so we're done. Since $\angle M_CAM_B = 90 + \frac12A$, we need to show $T\in (ABC)$. Define $Q_B = AC_2\cap M_CR_B$, and define $T_B = M_CR_B\cap (ABC), T_C = M_BR_C\cap (ABC)$ . Then, we have \[-1 = (C_2S_B;Q_BA) \overset{M_C}{=} (CB;T_BA)\]Similarly, $-1 = (CB;T_CA)$, so $T_C = T_B$ and as such $T$ lies on $(ABC)$. We conclude that $\angle O_BAO_C = 90 + \frac12 A$.

Dear Mathlinkers, here [/ur] Problem 2 Sincerely Jean-Louis

Ya_pank wrote: In triangle $ABC$ angle bisectors $AA_{1}$ and $CC_{1}$ intersect at $I$. Line through $B$ parallel to $AC$ intersects rays $AA_{1}$ and $CC_{1}$ at points $A_{2}$ and $C_{2}$ respectively. Let $O_{a}$ and $O_{c}$ be the circumcenters of triangles $AC_{1}C_{2}$ and $CA_{1}A_{2}$ respectively. Prove that $\angle{O_{a}BO_{c}} = \angle{AIC} $ Let $A'$, $C'$ be the midpoints of arc $BC$, $AB$ not containing $A$, $C$ respectively $\hspace{0.5cm}$$A_3$ be the second intersection of line $BA_2$ and circle $CA_1A_2$; $C_3$ be the second intersection of line $BC_2$ and circle $AC_1C_2$ $\hspace{0.5cm}$$U$ be the intersection of lines $A_1A_3$ anf $CA_2$; $V$ be the intersection of lines $C_1C_3$ and $AC_2$ $\hspace{0.5cm}$$S$ be the point satisfying $ABCS$ is a harmonic quadrilateral Cuz $\angle A_1CA_3 = \angle A_1A_2A_3 = \angle A_1AC = \angle BAA_1 = \angle A_1CA'$ so $C$, $A'$, $A_3$ are colinear. Similar, $A$, $C'$, $C_3$ are colinear Since lines $CA_3$, $A_1A_2$, $UA'$ concur at $A'$ and $CA_1$ insect $A_2A_3$ at $B$, we have $A'(BU, A_1C) = -1 = A'(BS, AC)$ Therefore, $U$, $A'$, $S$ are colinear. Similar, $V$, $S$, $C'$ are colinear In the order hand, follow Brocard theorem we have : $BO_c \perp UA'$, $BO_a \perp VC'$ Thus, $\angle O_aBO_c = 180^\circ - \angle C'SA' = \angle AIC$, done

Attachments:
russian-2021.pdf (137kb)