Same problem as: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3338548&sid=18791f5c99978f09fcf6cef32e379ef1#p3338548

This is just a generalisation of 2012 G6! Couldn't be any more like a G6!

I see someone posted this generalization in the 2012 G6 thread. Also, as XmL pointed out, see Proofathon December 2013 geometry contest.

Yeah,its like ISL 2012 G6. I drew two diagrams for this one,one for $P$ as Miquel point and the other for $Q$ as miquel point.Let $BI$ meet the circle $BDF$ at $Y$,circle $BTR$ at $Y'$ and circle $ABC$ at $X$,where $I$ is the incenter of $\triangle{ABC}$.Then applying Ptolemy's theorem in $BDYF$ we get $BY=\frac{(BD+BF) \cdot DY}{FD}$.Also note that $\triangle{DYF} \sim \triangle{CXA}$ so $\frac{DY}{FD}=\frac{CX}{AC}$.Combining we get $BY=\frac{(BD+BF) \cdot CX}{AC}$.Analogously we get $BY'=\frac{(BR+BT) \cdot CX}{AC}$.Adding the two relations we get $BY+BY'=\frac{(AB+BC) \cdot CX}{AC}$.(Here we have used the reflection property)Also by applying Ptolemy's theorem in $ABCX$ we get $BX=\frac{(AB+BC) \cdot CX}{AC}$.So $BY+BY'=BX \implies BY=Y'X$.This means that $\triangle{OBY} \cong \triangle{OXY'}$ so $OY=OY'$.Analogously if $CI \cap CDE=Z,CI \cap CRS=Z'$ we get $OZ=OZ'$ and if $AI \cap AEF=X,AI \cap ATS=X'$ we get $OX=OX'$. Next note that $\angle{ZPY}=\angle{ZCB}+\angle{YBC}=\angle{IBC}+\angle{ICB}=\angle{YIZ}$ so $Y,Z,P,I$ lie on a circle.By combination of the analogous facts we get $Y,Z,X,P,I$ lie on a circle.Similarly $Y',Z',X',Q,I$ lie on a circle. Its now clear that $OP=OQ$.

Generalization: Let $ D, E, F $ be the point on $ BC, CA, AB $ , respectively . Let $ K $ be a point and $ X, Y, Z $ be the projection of $ K $ on $ BC, CA, AB $, respectively . Let $ R, S, T $ be the reflection of $ D, E, F $ in $ X, Y, Z $ , respectively . Let $ P $ be the Miquel point of $ \{ D, E, F \} $ and $ Q $ be the Miquel point of $ \{ R, S, T \} $ . Then $ KP=KQ $ . Proof: Let $ P_x, P_y, P_z $ be the reflection of $ P $ in $ KX, KY, KZ $ , respectively . Let $ Q_x=P_xR \cap \odot (P_xP_yP_z), Q_y=P_yS \cap \odot (P_xP_yP_z), Q_z=P_zT \cap \odot (P_xP_yP_z) $ . From $ \angle Q_xP_xP=\angle RDP=\angle SEP=\angle Q_yP_yP \Longrightarrow Q_x \equiv Q_y $ . Similarly, we can prove $ Q_x \equiv Q_y \Longrightarrow P_xR, P_yS, P_zT $ are concurrent at $ Q' \in \odot (K, KP) $ . From $ \angle TAS=\angle TQ'S \Longrightarrow A, T, S, Q' $ are concyclic . Similarly, we can prove $ Q' \in \odot (BRT) $ and $ Q' \in \odot (CRS) $ , so $ Q' \equiv Q $ which is the Miquel point of $ \{ R, S, T \} $ and lie on $ \odot (K, KP) $ . i.e. $ KP=KQ $ Q.E.D

The referred generalization was posted before: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=298400 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=363074 (Lemma)

Here's a funny solution with a few interesting well-known results. Change $R, S, T$ in the problem to $D', E', F'$ (to avoid confusion between $R$ and the circumradius of $\triangle ABC$). WLOG assume $BD < BR.$ Recall that the signed area of the pedal triangle of point $P$ w.r.t. $\triangle ABC$ is $[\triangle ABC] \cdot \frac{R^2 - OP^2}{4R^2},$ where $R$ is the circumradius of $\triangle ABC.$ Hence, it suffices to show that the areas of the pedal triangles of $P, Q$ in $\triangle ABC$ have the same area. Lemma 1. $\triangle DEF, \triangle RST$ have the same area. Proof. This is pretty straightforward to verify with barycentric coordinates. $\blacksquare$ Let $\theta = \angle PDR.$ Note that $PE, PF$ make the same angles with $AC, AB.$ Let $D_1, E_1, F_1$ be the feet from $P$ to $BC, CA, AB.$ Notice that $(BD_1^2 - CD_1^2) + (CE_1^2 - E_1A^2) + (AF_1^2 - F_1B^2) = 0$. Hence, since we have $BD^2 - CD^2 = BC \cdot (BD - DC) = (BD_1^2 - CD_1^2) + 2 \cdot BC \cdot DD_1$ and similar relations, we've: $$BD^2 - CD^2 + CE^2 - EA^2 + AF^2 - FB^2 = 2 (BC \cdot DD_1 + CA \cdot EE_1 + AB \cdot FF_1) = 4 \delta \cot \theta.$$ Doing a similar computation with $D, E, F$ replaced by $R, S, T$, we easily find that line $RQ$ also meets $BC$ at the angle $\theta$, but reflected. In other words, $RQ$ is the reflection of line $DP$ over the perpendicular bisector of $BC$, and similar results hold for $AC$ and $AB.$ Finally, it remains to notice that $[\triangle D_1 E_1 F_1] = (\sin \theta)^2 [\triangle DEF].$ A similar result for $Q$, combined with Lemma $1$, implies that the pedal triangles of $P$ and $Q$ have equal areas. Hence, we're done. $\square$

TelvCohl wrote: Generalization: Let $ D, E, F $ be the point on $ BC, CA, AB $ , respectively . Let $ K $ be a point and $ X, Y, Z $ be the projection of $ K $ on $ BC, CA, AB $, respectively . Let $ R, S, T $ be the reflection of $ D, E, F $ in $ X, Y, Z $ , respectively . Let $ P $ be the Miquel point of $ \{ D, E, F \} $ and $ Q $ be the Miquel point of $ \{ R, S, T \} $ . Then $ KP=KQ $ . Proof: Let $ P_x, P_y, P_z $ be the reflection of $ P $ in $ KX, KY, KZ $ , respectively . Let $ Q_x=P_xR \cap \odot (P_xP_yP_z), Q_y=P_yS \cap \odot (P_xP_yP_z), Q_z=P_zT \cap \odot (P_xP_yP_z) $ . From $ \angle Q_xP_xP=\angle RDP=\angle SEP=\angle Q_yP_yP \Longrightarrow Q_x \equiv Q_y $ . Similarly, we can prove $ Q_x \equiv Q_y \Longrightarrow P_xR, P_yS, P_zT $ are concurrent at $ Q' \in \odot (K, KP) $ . From $ \angle TAS=\angle TQ'S \Longrightarrow A, T, S, Q' $ are concyclic . Similarly, we can prove $ Q' \in \odot (BRT) $ and $ Q' \in \odot (CRS) $ , so $ Q' \equiv Q $ which is the Miquel point of $ \{ R, S, T \} $ and lie on $ \odot (K, KP) $ . i.e. $ KP=KQ $ Q.E.D [asy][asy] size(300); pair A=dir(110),B=dir(-150),C=dir(-30),D=0.65*B+0.35*C,E=0.26*C+0.74*A,F=0.33*A+0.67*B,K=(-0.08,-0.03),X=foot(K,B,C),Y=foot(K,C,A),Z=foot(K,A,B),R=2*X-D,S=2*Y-E,T=2*Z-F,P=2*foot(F,circumcenter(B,F,D),circumcenter(A,F,E))-F,Q=2*foot(R,circumcenter(C,R,S),circumcenter(B,R,T))-R; fill(A--B--C--cycle,cyan); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(-90)); dot("$E$",E,dir(60)); dot("$F$",F,dir(-70)); dot("$R$",R,dir(-50)); dot("$S$",S,dir(-100)); dot("$T$",T,dir(T)); dot("$P$",P,dir(-125)); dot("$Q$",Q,dir(Q)); draw(circumcircle(B,F,D)^^circumcircle(C,D,E)^^circumcircle(A,F,E),red); draw(circumcircle(B,R,T)^^circumcircle(C,R,S)^^circumcircle(A,T,S),royalblue); draw(A--B--C--A,magenta); dot("$K$",K,dir(-135)); pair Px=2*foot(P,K,X)-P,Py=2*foot(P,K,Y)-P,Pz=2*foot(P,K,Z)-P; dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$Z$",Z,dir(Z)); dot("$P_x$",Px,dir(Px)); dot("$P_y$",Py,dir(90)); dot("$P_z$",Pz,dir(-80)); draw(circle(K,abs(K-P)),brown+linewidth(0.8)); draw(Px--R,dashed+brown+linewidth(1)); draw(T--Q,dashed+brown+linewidth(1)); draw(Py--S,dashed+brown+linewidth(1)); draw(K--X,magenta); draw(K--Y,magenta); draw(K--Z,magenta); markscalefactor=0.007; draw(rightanglemark(K,X,C)^^rightanglemark(K,Y,A)^^rightanglemark(K,Z,B)); [/asy][/asy] Can someone please tell motivation behind the above solution. Like, how to think all these constructions and those nice results.

TelvCohl is just a geo god.