Really nice one

Pretty bad solution imo [asy][asy] import olympiad; size(9cm); defaultpen(fontsize(10pt)); pair A = dir(120), B = dir(210), C = dir(330), D = foot(A,B,C), D1 = dir(300), P = dir(230), Q = dir(310), X = foot(B,A,P), Y = foot(B,A,Q), N = foot(C,A,P), M = foot(C,A,Q), X1 = extension(A,Q,C,D1), Y1 = extension(A,P,C,D1), N1 = extension(A,Q,B,D1), M1 = extension(A,P,B,D1), K = (B+C)/2, K1 = intersectionpoint(unitcircle,K--2D-A); dot("$A$", A, dir(120)); dot("$B$", B, dir(210)); dot("$C$", C, dir(330)); dot("$D'$", D1, dir(280)); dot("$K'$", K1, dir(255)); dot("$X'$", X1, dir(180)); dot("$Y'$", Y1, dir(225)); dot("$N'$", N1, dir(30)); dot("$M'$", M1, dir(30)); dot("$P$", extension(A,P,B,C), dir(135)); dot("$Q$", extension(A,Q,B,C), dir(45)); draw(A--B--C--A^^A--Y1^^A--N1); draw(C--Y1^^B--N1); draw(circumcircle(A,B,C)); draw(B--K1--A^^K1--C); draw(circumcircle(X1,Y1,N1)^^circumcircle(X1,D1,K1), dotted); draw(X1--K1--M1, dashed); [/asy][/asy] Let $D$ be the projection of $A$ to $\overline{BC}$, $K$ be the midpoint of $\overline{BC}$, and $\angle BAP = \angle CAQ = \theta$. Note that $ABXDY$ and $ANDMC$ are both cyclic with diameters $\overline{AB}$ and $\overline{AC}$, respectively. Let $\Psi$ be the inversion centered at $A$ with radius $\sqrt{AB \cdot AC}$ followed by a reflection across the angle bisector of $\angle BAC$, with $\Psi: J \leftrightarrow J'$ for any object $J$. Note that $\Psi: \overline{AP} \leftrightarrow \overline{AQ}$ since they are isogonal, $D'$ is the antipode of $A$ on $(ABC)$, and $K'$ is the point of intersection of the $A$-symmedian and $(ABC)$. Since $X$ lies on $\overline{AP}$ and $(ABD)$, we can redefine $X'=\overline{AQ} \cap \overline{CD}$. Similarly, $Y' = \overline{AP} \cap \overline{CD}$, $N' = \overline{AQ} \cap \overline{BD}$, and $M' = \overline{AP} \cap \overline{BD}.$ Claim 1: Points $N'$, $M'$, $X'$, $Y'$ are concyclic. Proof. Let $\measuredangle$ denote a directed angle. We have $$\measuredangle N'X'Y' = \measuredangle AX'C = 90^\circ - \theta = \measuredangle BM'A = \measuredangle N'M'Y'.$$The result follows. $\square$ Claim 2: $X'D'K'M'$ and $N'D'K'Y'$ are both cyclic. Proof. By property of symmedians, we have $\dfrac{BK'}{CK'} = \dfrac{AB}{AC}$. In addition, we know that $\triangle ABM' \sim \triangle ACX'$. Combining these information yields $$\dfrac{BK'}{CK'} = \dfrac{AB}{AC} = \dfrac{BM'}{CX'}.$$Since $\angle K'BD' = \angle K'CD'$, we must have $\triangle M'BK' \sim \triangle X'CK'$ by SAS, so $\angle BK'M' = \angle CK'X'$. Thus $$\measuredangle M'D'X' = \measuredangle BD'C = \measuredangle BK'C = \measuredangle M'K'X',$$which implies $X'D'K'M'$ is cyclic. Through an analogous argument, $N'D'K'Y'$ is cyclic as well. $\square$ Inverting back, we are done by the Radical Axis Theorem on $(NMXY)$, $(XDKM)$, and $(NDKY)$. $\blacksquare$

Inversion, nice!

Child's play for desergaues. 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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Proof: Note that by desergaues theorem we just need to prove that $\Delta NBX$ and $\Delta YCM$ are perspective with respect to a line or that if $H=BN \cap CY,I=BX\cap CM$ then $A-H-I$ are collinear.Let $K=BY \cap CN$ Firstly observe that since $\angle ANK+\angle AYK\equiv \angle ANC +\angle AYB=90+90=180$ we have that $ANKY$ is cyclic. Similiarly since $\angle AXI+\angle AMI=360-(\angle AXB+\angle AMC)=180$ we have $AXMI$ is cyclic .Note that by isogonal line lemma we have that $AH,AK$ are isogonal. So we just need to prove $AI,AK$ as isogonal. Now note that $\angle BAX=\angle MAC$ since $AP,AQ$ are isogonal and also $\angle BXA=\angle CMA=90$ so by $AA$ similiarity we have $$\Delta ABX\sim \Delta ACM\implies \frac{AB}{AC}=\frac{AX}{AM}.......(1)$$. Also note that as $\angle NAC\equiv \angle PAC=\angle QAB \equiv \angle YAB$ since $AP,AQ$ are isogonal and also as $\angle AYB=\angle ANB=90$ so by $AA$ similiarity we have $$\Delta AYB \sim \Delta ANC \implies \frac{AY}{AN}=\frac{AB}{AC}....(2)$$Now (1), (2) imply that $\Delta AYN \sim \Delta AXM$ by $SAS$ similiarity $\implies \angle ANY=\angle AMX$ and since $AKNY$ and $AXIM$ are cyclic we have $\angle AKY=\angle ANY=\angle AMX=\angle AIX \implies \angle YAK=90-\angle AKY=90-\angle AIX=\angle IAX \implies AI,AK$ are isogonal w.r.t $\angle PAQ \implies AI,AK$. Now note that by the isogonality we obtained we have $\angle BAK=\angle BAP+\angle XAK=\angle CAM+\angle MAO = \angle CAM\implies AK,AI $ are isogonal w.r.t $\angle BAC$ . We are done.

How many isogonal line lemma man this is isogonal year

Seriously?

Sharygin 2020 CR P16 wrote: Cevians $AP$ and $AQ$ of a triangle $ABC$ are symmetric with respect to its bisector. Let $X$, $Y$ be the projections of $B$ to $AP$ and $AQ$ respectively, and $N$, $M$ be the projections of $C$ to $AP$ and $AQ$ respectively. Prove that $XM$ and $NY$ meet on $BC$. First of all notice that $AYXB$ and $ANMC$ are cyclic quadrilaterals. Claim 1:- $X,Y,N,M$ are concyclic.

Drop a perpendicular from $A$ to $BC$, let this point be $H_A$. So, $H_A\in\odot(AXB),\odot(AMC)$. So, $AH_A$ is precisely the Radical Axis of $\odot(AXB),\odot(AMC)$. Now by Radical Axis Theorem on $\odot(AYXB),\odot(ANMC),\odot(YXNM)$ we get that $AH_A,XY,NM$ concurs at a point $R$ say and let $YN\cap XM=S$ so we need to prove that $S\in BC$. Claim 2:- Circumcenter of $\odot(YXNM)$ lies on $BC$, precisely the midpoint of $BC$ is its Circumcenter.

Now as $NX\cap YM=A, XY\cap NM=R$ and $NY\cap XM=S$. Hence, by Brocard's Theorem on $\odot(XYMN)$ we get that $OS\perp AR$ but $OH_A\perp AR$. This forces that $S\in BC$. Hence, $YN,XM,BC$ concurs. $\blacksquare$

My solution is attached. Nice problem! I actually solved this thrice, each time thinking I hadn't done this yet...

What was the morality of this act?

I have a very simple solution to this one. Let $BY$ meet $AC$ at $D$ and $CN$ meet $AB$ at $E$. If we can show $(A,P;N,X)=(A,Q;Y,M)$, that would solve the problem, because that would mean $XM$, $NY$, and $PQ$ concur (and line $PQ$ is line $BC$). Projecting through point $B$ onto line $CE$, $(A,P;N,X)=(E,C;N,CE_{\infty})=\frac{NE}{NC}$. Similarly, $(A,Q;Y,M)=\frac{YD}{YB}$. But $\triangle ABD \sim \triangle ACE$ (since $\angle ABD = 90 - \angle BAY = 90 - \angle CAN = \angle ACE$ and both share $\angle BAC$). Since $AN$ and $AY$ are respective feet of the altitudes from $A$ in both triangles, $\frac{NE}{NC}=\frac{YD}{YB}$. Thus, $(A,P;N,X)=(A,Q;Y,M)$, so $XM$, $NY$, and $BC$ concur.

NJOY wrote: Cevians $AP$ and $AQ$ of a triangle $ABC$ are symmetric with respect to its bisector. Let $X$, $Y$ be the projections of $B$ to $AP$ and $AQ$ respectively, and $N$, $M$ be the projections of $C$ to $AP$ and $AQ$ respectively. Prove that $XM$ and $NY$ meet on $BC$. BMO 2019

It suffices to show that $(X,N;P,A)=(M,Y;Q,A)$. Let $l_P$ and $l_Q$ be the lines through $A$ perpendicular to $AP$ and $AQ$. Since isogonal conjugation preserves the cross ratio (It can be seen as a reflection over the $A$- angle bisector and a projection from $A$ ), we have \[(X,N;P,A)=(B,C;P,l_P\cap BC)=(C,B,Q,l_Q\cap BC)=(M,Y;Q,A)\]as desired.

Note that $XYMN$ is cyclic quadrilateral as $$\frac{AM}{AX}=\frac{AC}{AB}=\frac{AN}{AY}$$by similar triangles $\triangle AMC\sim\triangle AXB$ and $\triangle ANC\sim\triangle AYB$. Note let $M',N',X',Y'$ be the antipodes of $M,N,X,Y$ wrt $(MNXY)$, respectively. Note that $M',N',X',Y'$ lie on $\overline{BY},\overline{BX},\overline{CY},\overline{CX}$, respectively. By Pascal's thereom on $XMM'YNN'$, we get that $\overline{MM'}\cap\overline{NN'}$, $B$ and $\overline{XM}\cap\overline{YN}$ are collinear. By Pascal's theorem on $MXX'NYY'$, we get that $\overline{XX'}\cap\overline{YY'}$, $C$ and $\overline{XM}\cap\overline{YN}$ are collinear. As $\overline{MM'},\overline{NN'},\overline{XX'},\overline{YY'}$ are concurrent at the center of $(XYMN)$, we conclude that $\overline{XM}\cap\overline{YN}$ lies on $\overline{BC}$. $\blacksquare$

Deleted (wrong) solution

sanyalarnab wrote: I haven't seen any solution using homothety here so.. It's obvious that $ABXY$ and $ANMC$ are cyclic using right angle arguments. $$\angle XBY = \angle XAY = \angle NAM=\angle NCM$$$$\angle BYX=\angle BAX=\angle MAC = \angle MNC$$Theefore $$BXY \sim CMN$$by AA test. So the lines formed by the corresponding vertices in the above similarity must concur. Thus $BC,XM,YN$ concur. Bhai similar triangles zruri nhi hai ki perspective ho