Indonesia RMO 2022/Essay 4. wrote: Suppose $ABC$ is a triangle with circumcenter $O$. Point $D$ is the reflection of $A$ with respect to $BC$. Suppose $\ell$ is the line which is parallel to $BC$ and passes through $O$. The line through $B$ and parallel to $CD$ meets $\ell$ at $B_1$. Lines $CB_1$ and $BD$ intersect at point $B_2$. The line through $C$ parallel to $BD$ and $\ell$ meet at $C_1$. Finally, $BC_1$ and $CD$ intersects at point $C_2$. Prove that points $A, B_2, C_2, D$ lie on a circle. [asy][asy] import graph; size(10cm); 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 = -8.92, xmax = 8.92, ymin = -5.16, ymax = 5.16; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen ccqqqq = rgb(0.8,0,0); pen zzttff = rgb(0.6,0.2,1); pen ffvvqq = rgb(1,0.3333333333333333,0); /* draw figures */ draw(circle((-2.535601848120987,1.0776965643215946), 2.48189193012609)); draw((-3.68,3.28)--(-4.6,-0.3)); draw((-4.6,-0.3)--(-0.42,-0.22)); draw((-0.42,-0.22)--(-3.68,3.28)); draw(circle((-1.8470782313493612,-0.24731250203539504), 3.97511455143227), linetype("2 2") + qqwuqq); draw((-4.254408748772609,1.044800738471803)--(-0.8167949474693652,1.1105923901713863), ccqqqq); draw((-3.543690413529533,-3.842175893081905)--(1.4593586984275286,1.95927094900745) ); draw((-5.235525403211309,1.831139195866038)--(-3.543690413529533,-3.842175893081905)); draw((-0.42,-0.22)--(-5.235525403211309,1.831139195866038)); draw((-4.6,-0.3)--(1.4593586984275286,1.95927094900745)); draw((-1.4763095864704665,3.3221758930819045)--(-4.6,-0.3)); draw((-1.4763095864704665,3.3221758930819045)--(-0.42,-0.22)); draw((-1.8470782313493583,-0.24731250203539445)--(-3.68,3.28), dotted + blue); draw((-3.543690413529533,-3.842175893081905)--(-1.4763095864704665,3.3221758930819045), dotted + zzttff); draw((-5.235525403211309,1.831139195866038)--(1.4593586984275286,1.95927094900745), dotted + ffvvqq); /* dots and labels */ dot((-3.68,3.28),dotstyle); label("$A$", (-3.8,3.4), NE * labelscalefactor); dot((-4.6,-0.3),dotstyle); label("$B$", (-4.98,-0.38), NE * labelscalefactor); dot((-0.42,-0.22),dotstyle); label("$C$", (-0.34,-0.32), NE * labelscalefactor); dot((-3.543690413529533,-3.842175893081905),dotstyle); label("$D$", (-3.6,-3.6), NE * labelscalefactor); dot((-2.535601848120987,1.0776965643215946),dotstyle); label("$O$", (-2.5,1.2), NE * labelscalefactor); dot((-4.254408748772609,1.044800738471803), dotstyle); label("$X$", (-4.5,1.15), NE * labelscalefactor); dot((-1.6445950858709713,1.094749325321595), dotstyle); label("$Y$", (-1.66,1.2), NE * labelscalefactor); dot((-3.426608610371003,1.0606438033215944), dotstyle); label("$B_1$", (-3.66,1.15), NE * labelscalefactor); dot((-0.8167949474693652,1.1105923901713863), dotstyle); label("$C_1$", (-0.8,1.28), NE * labelscalefactor); dot((-5.235525403211309,1.831139195866038), dotstyle); label("$B_2$", (-5.5,2), NE * labelscalefactor); dot((1.4593586984275286,1.95927094900745),dotstyle); label("$C_2$", (1.45,2.05), NE * labelscalefactor); dot((-1.4763095864704665,3.3221758930819045), dotstyle); label("$E$", (-1.5,3.43), NE * labelscalefactor); label("$\ell$", (-3.46,0.73), NE * labelscalefactor,ccqqqq); dot((-2.270952708764027,0.568400295450319),dotstyle); label("$F$", (-2.2,0.62), NE * labelscalefactor); dot((-1.8470782313493583,-0.24731250203539445), dotstyle); label("$G$", (-1.9,-0.08), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] This problem holds as long as $\ell$ passes through line $O$, but there's a surprisingly cute solution utilizing the fact that $\ell$ is parallel to $BC$, as follow. We will first define a set of points: $Y = \ell \cap AC, X = \ell \cap AB, E = BB_1 \cap CC_1, F = B_1 C\cap BC_1$ and $G = AO \cap BC$. Note that $B_1 C_1 CB$ is a trapezoid. Therefore, $EF$ passes through midpoint of $BC$ and therefore since $EBDC$ is a parallelogram, then $EF$ passes through $D$. Now, note that by reflection wrt perpendicular bisector of $BC$, we have $A \leftrightarrow E$ and $B \leftrightarrow C$ and therefore \[ \frac{B_1 O}{OC_1} = \frac{YO}{OX} = \frac{CG}{GB} \]where $G = AO \cap BC$. Therefore, this implies that $O,F,G$ are collinear. Now, $\triangle B_2 D F \sim \triangle CEF$ implies $ \frac{EF}{DF} = \frac{FC}{FB_2} $ and analogously, $\frac{EF}{DF} = \frac{FB}{FC_2}$. Furthermore, \[ \frac{EF}{DF} \stackrel{\triangle FBD \sim \triangle FC_1 E}{=} \frac{FC_1}{FB} = \frac{FO}{FG} \]Therefore, homothety centered at $F$ of ratio $\frac{FO}{FG}$ will send $\triangle ABC$ to $\triangle DC_2 B_2$. Furthermore, this homothety sends circumcenter of $ABC$, which is $O$ to $G$, which has to be circumcenter of $DB_2 C_2$. Therefore, $GB_2 = GC_2 = GD \stackrel{G \in BC}{=} GA$, and therefore $G$ is the circumcenter of $AB_2D C_2$, and hence $AB_2 D C_2$ is cyclic.

This is my proposal. The original problem is to prove that $AC_2 \perp AB$ (and similarly $AB_2 \perp AC$), and the solution I submitted for the problem relies on that fact. IndoMathXdZ wrote: This problem holds as long as $\ell$ passes through line $O$ Noo, how did I miss this . Here's the solution for the generalized version. Define $E = BB_1 \cap CC_1$ and $M$ as the midpoint of $BC$. Note that \[ \frac{DB}{BB_2} = \frac{CE}{BB_2} = \frac{EB_1}{BB_1} \implies \frac{DB}{DB_2} = \frac{EB_1}{EB}. \]Similarly, we have $DC/DC_2 = EC_1/EC$. Now let $O', B_3, C_3$ be the reflection of $O, B_1, C_1$ with respect to $M$ respectively. Obviously, $O'$ is the circumcenter of $\triangle DBC$. We have: \[ \frac{DB_3}{DC} = \frac{EB_1}{EB} = \frac{DB}{DB_2} \implies DB_3 \times DB_2 = DB \times DC. \]Similarly, $DC_3 \times DC_2 = DB \times DC$. Therefore, the composition of inversion with center $D$ and radius $\sqrt{DB \times DC}$ followed by reflection w.r.t. $\angle BDC$ will map $B \mapsto C$, $B_3 \mapsto B_2$, and $C_3 \mapsto C_2$. It's known by the property of $\sqrt{bc}$ inversion that $O' \mapsto A$. Since $B_3, O', C_3$ are collinear, therefore $D, B_2, A, C_2$ are concyclic, as desired. $\square$