Any solutions?

Nicest configuration I've seen in a while. Let $S\in (ABC)$ be such that $AS$ is the A-symmedian on $\triangle{ABC}$, $L$ be the midpoint of $\overline{AS}.$ Also let $B_1,C_1$ be the reflections of $B,C$ in $L$, $F=BD\cap CE$ and $M_1,M_2,N_1,N_2,N$ be the midpoints of $\overline{BD},\overline{CE},\overline{BF},\overline{CF},\overline{AF}.$ First note that $\angle{ABF}=\angle{(DE,BC)}=\angle{ACF}$, so $F\in (ABC).$ Step 1. $L\in (FM_1M_2)$ Proof: $\sqrt{bc}$ inversion in $A$ takes $L$ to the reflection of $A$ in $M$,so $\angle{ALB}=\angle{ALC}=180^0-\angle{BAC}.$ Therefore $$\angle{AB_1B}=\angle{LBS}=\angle{LBC}+\angle{LAC}=\angle{ABC}=\angle{ADB}$$meaning that $B_1\in (ABD)$. In the same fashion we prove that $C_1\in (ACE)$, and since $LM_1\parallel B_1D,LM_2\parallel C_1E$ and $\angle{ADB_1}=\angle{ABB_1}=\angle{BAM}$, and in the same way $\angle{AEC_1}=\angle{CAM}$, we obtain that $$\angle{M_1LM_1}=\angle{(DB_1,EC_1)}=\angle{BAC}=\angle{M_1FM_2}$$which proves the claim. $\blacksquare$ Step 2. $N\in (FM_1M_2)$ Proof: Notice that $\triangle{FDE}\sim\triangle{ABC}$, so $$\frac{NN_1}{NN_2}=\frac{\frac{AB}{2}}{\frac{AC}{2}}=\frac{\frac{FD}{2}}{\frac{FE}{2}}=\frac{M_1N_1}{M_2N_2}$$combined with the fact that $$\angle{NN_1M_1}=180^0-\angle{NN_1F}=180^0-\angle{ABF}=180^0-\angle{ACF}=180^0-\angle{NN_2F}=\angle{NN_2M_2}$$yields that $\triangle{NN_1M_1}\sim\triangle{NN_2M_2}$, i.e. $N$ is the Miquel point of $M_1N_1N_2M_2.$ This is enough to prove the claim. $\blacksquare$ Now, from Steps 1 and 2, it follows that $L\in (PNF)$, and since $\angle{ALO}=90^0=\angle{ANO}$, we also have that $L\in (AON).$ We conclude that there is a spiral similarity that sends $A$ to $O$ and $F$ to $P$. But $\triangle{LAO}\sim\triangle{LOK}$, so the spiral similarity also sends $O$ to $K$. Overall, we infer that $\triangle{OKP}\sim\triangle{AOF}$, which proves that $KO=KP.$

Official Solution: Consider the isosceles triangles $ASB$ and $CTA$ drawn to the interior of $\Delta ABC$, and similar to triangles $COA$ and $AOB$, respectively. From the similarities we obtain $\angle BAS+\angle TAC=\angle OAC+\angle BAO=\angle BAC$, so $A$, $S$, $T$ are collinear, as well as $\frac{AS}{AO}=\frac{AB}{AC}=\frac{AO}{AT}$, from which $AO^2=AS\cdot AT$, from which in turn it follows that $AT$ is tangent to the circumcircle of $\Delta SOT$, let's say of circumcenter $K'$. Since $S$ and $O$ are on the perpendicular bisector of $AB$, $SO$ is the internal angle bisector of $\angle AOB$, and we have $\angle ACB=\frac{1}{2}\angle AOB=\angle AOS=\angle OTS$. Similarly, $\angle CBA=\angle TSO$. We then infer that $OST$ and $ABC$ are similar triangles. We show now $K=K'$. Let $N$ be the midpoint of $ST$. Then $\angle K'NA=\angle AOK=90^\circ$, and so the quadrilateral $AOK'N$ is cyclic. Using this along with $\angle NOK'=\angle MAO$, which we obtain from $\Delta ABC$ and $\Delta OST$, we calculate $$\angle BAK'=\angle BAS+\angle NAK=\angle OAC+\angle NOK'=\angle OAC+\angle MAO=\angle MAC.$$ This means that $K'$ satisfies $\angle BAK'=\angle MAC$ and $AO\perp OK$, from which we conclude $K=K'$. To finish, it's enough to show that $P$ lies on the circumcircle of $\Delta OST$ From $\angle ASB=\angle COA=2\angle CBA=2\angle ADB$ and $AS=SB$, We get that $S$ is the circumcenter of $ADB$, and analogously $T$ is the circumcenter of $AEC$. Therefore, $S$ lies on the perpendicular bisector of $BD$ and $T$ on that of $CE$, so $P$ is no more than the intersection of the lines perpendicular to $BD$ and $CE$ passing through $S$ and $T$, respectively. Using the notation $\angle (l, m)$ referring to the least non-negative angle we need to rotate $l$ anti-clockwise so as to obtain a line parallel to $m$, we calculate modulo $180^\circ$ $$\measuredangle SPT=\angle (SP, PT)=\angle (BD, CE)\equiv \angle (BD, AD)+\angle (AD, CE)\equiv -\measuredangle ADB-\measuredangle CEA=$$$$-\measuredangle CBA-\measuredangle ACB\equiv \measuredangle BAC=\measuredangle SOT.$$ It follows that $P$ lies on the circumcircle of $SOT$, and finally $KO=KP$.

Interesting problem, hopefully my solution works plagueis wrote: Let $ABC$ be a scalene triangle with circumcenter $O$, and let $D$ and $E$ be points inside angle $\measuredangle BAC$ such that $A$ lies on line $DE$, and $\angle ADB=\angle CBA$ and $\angle AEC=\angle BCA$. Let $M$ be the midpoint of $BC$ and $K$ be a point such that $OK$ is perpendicular to $AO$ and $\angle BAK=\angle MAC$. Finally, let $P$ be the intersection of the perpendicular bisectors of $BD$ and $CE$. Show that $KO=KP$. Lemma: Let $ABC$ be a triangle, $\omega_1,\omega_2$ be circles through $A,B$ and $A,C$ respectively. $X$ is a variable point on $\omega_1$ and $AX$ meets $\omega_2$ at $Y$. Then the locus of the perpendicular bisectors of $BX,CY$ is a circle through the centers of $\omega_1,\omega_2$. (Proof): Let $O_1,O_2$ be the centers of $\omega_1,\omega_2$. Now, $\angle BO_1Z + \angle ZO_2C = \angle A$, this gives that $\angle O_1ZO_2$ is fixed so we are done. $\hfill{\square}$ Main Problem. Let $B',C'$ be the reflection of $B$ and $C$ in the $A-$ altitude. Note that $D,E$ varies on $\odot(ABB'),\odot(ACC')$ respectively. Let $O_1,O_2$ be the center of these circles respectively. Now from the Lemma, we just need to show the problem for 3 cases of $D$. For $D = E = A$, $P\equiv O$. When $D = B'$, the perpendicular bisector of $BD$ is $A-$ altitude and perpendicular bisector of $CE$ passes through $O_2$, so when $D = B'$, $P\equiv O_2$. Similarly when $D = C'$, $P \equiv O_1$, so we need to show that $K$ is the circumcenter of $\triangle O_1O_2O$. Now note that $O_1$ lies on perpendicular bisector of $AB$ and $O_2$ lies on perpendicular bisector of $AC$, $\angle OO_2O_1 = \angle OO_2A = C = \angle AOO_1$. This gives that $\triangle ABC\sim \triangle OO_1O_2$ and $AO$ is tangent to $\odot(OO_1O_2)$. Let $K'$ be the circumcenter of $\triangle OO_1O_2$ and $N$ be the midpoint of $O_1O_2$. Now, $\angle ANK' = 90^{\circ} = \angle AOK'$. As $\triangle OO_1O_2\sim \triangle ABC$, so $\triangle OK'N \sim \triangle AOM$. This gives that $MAO=\angle NOK' = \angle NAK'$ which gives that $K'$ lies on the $A-$ symmedian and we are done. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(6cm); 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 = -4.618468119884228, xmax = 9.404971407348569, ymin = -0.9275912378296551, ymax = 6.150069464002545; /* image dimensions */ pair A = (0.,5.2899431606514336), B = (-1.4876313893336057,0.), C = (4.292234737910577,0.), O = (1.402301674288486,2.0414430609646925), K = (0.541162529618346,1.6697093908679173), O_1 = (0.,2.435796634270331), O_2 = (0.,0.9036221474655032), B_1 = (2.1461173689552884,2.6449715803257168), C_1 = (-0.7438156946668029,2.6449715803257168), M = (1.4023016742884855,0.); /* draw figures */ draw(A--B, linewidth(1.)); draw(B--C, linewidth(1.)); draw(C--A, linewidth(1.)); draw(A--O, linewidth(1.)); draw(K--O, linewidth(1.)); draw(O_1--A, linewidth(1.)); draw(O_2--O_1, linewidth(1.)); draw(circle((0.5411625296183468,1.6697093908679173), 0.9379480518487351), linewidth(1.)); draw((0.,1.6697093908679173)--K, linewidth(1.)); draw(O--(0.,1.6697093908679173), linewidth(1.)); draw(C_1--O, linewidth(1.)); draw(O_2--B_1, linewidth(1.)); draw(M--O, linewidth(1.)); draw(M--A, linewidth(1.)); /* dots and labels */ dot(A,dotstyle); label("$A$", (-0.057553226622777565,5.391748674520524), NE * labelscalefactor); dot(B,dotstyle); label("$B$", (-1.9,-0.4), NE * labelscalefactor); dot(C,dotstyle); label("$C$", (4.349499477468407,-0.4), NE * labelscalefactor); dot(O,linewidth(4.pt) + dotstyle); label("$O$", (1.3,2.2), NE * labelscalefactor); dot(K,linewidth(4.pt) + dotstyle); label("$K$", (0.3,1.3), NE * labelscalefactor); dot(O_1,linewidth(4.pt) + dotstyle); label("$O_1$", (-0.45,2.5), NE * labelscalefactor); dot(O_2,linewidth(4.pt) + dotstyle); label("$O_2$", (-0.5,0.7), NE * labelscalefactor); dot((0.,1.6697093908679173),linewidth(4.pt) + dotstyle); label("$N$", (-0.35,1.611134883479721), NE * labelscalefactor); dot(B_1,linewidth(4.pt) + dotstyle); dot(C_1,linewidth(4.pt) + dotstyle); dot(M,linewidth(4.pt) + dotstyle); label("$M$", (1.448098195971966,-0.4), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Note. The last part was pretty much same as IMO SL 2017 G3.

Is there any solution with the method of moving points?

Let $X $, $Y $ be the intersection of the perpendicular bisector of $CA $, $AB $ with $A $-altitude of $\triangle ABC $ respectively. It is easy to show that $D $ lies on $ (Y, YB) $, and $E $ lies on $(X, XC)$. Hence, $XP\perp CE $ and $YP\perp BD $. By angle chasing, $OXPY $ is concyclic quadrilateral. Only show that $KO=KY $. $(\cot C>\cot B)OY=\frac {1}{2}AB (\cot C-\cot B) $ $\implies \frac {KO}{OA}=\frac {DM}{DA}=\frac {1}{2}\frac {(DC-DB)}{DA} $ $D $ is the foot of $A $-altitude of $\triangle ABC$. $M $ is the midpoint of $BC $. Now, $\frac {KO}{KY}=\frac {OA}{AB} $ Hence, $\triangle KOY $ is similar to $\triangle OAB $ $\blacksquare $