Let $I$ be the incenter of $ABC$. The central claim is that $(BDE)$ meets $BI$ at a fixed point $X$, and that $(CDF)$ meets $CI$ at a fixed point $Y$. Indeed, let $(BDE) \cap BI = X$; then, by Ptolemy on $BDXE$, we have that \[ \frac{BX}{BE+BD} = \frac{DX+EX}{DE} = \frac{2\sin \frac B2}{\sin B}, \]so $BX = BC\cdot \sec \frac B2$ is constant. So, $X$ is fixed; similarly, $Y = (CDF)\cap CI$ is fixed. Then, we see that \[ \measuredangle IXP = \measuredangle BXP = \measuredangle BDP = \measuredangle CDP = \measuredangle CYP = \measuredangle IYP, \]so $P$ is on $(IXY)$. So, set $Q$ to be the center of $(IXY)$, which clearly works. $\blacksquare$

This is a lemma used in IMOSL 2012 G6

See here

Full angle chasing solution: Lemma: Given $\Delta ABC$, $M$: midpoint of $BC$. $D \in BC$ and $F \in CA$ such that $BD=CF$. Let $X$ be the intersection of angle bisector at $\angle C$ with perpendicular bisector of $BC$. Then $X \in (CDF)$ Since the lemma is quite easy to prove (just notice some isosceles triangle and do some angle chasing) Back to the problem Let $M$ be the midpoint of $BC$, let $X,Y$ be the intersection of angle bisector at $\angle C,B$ with perpendicular bisector of $BC$. Let $Z=(BMY) \cap (CMX)$. We will prove $P \in (XYZ)$. Using the lemma we have $(B,D,P,Y,E)$ concyclic; $(C,D,P,X,F)$ concyclic. We have: $\angle XPY= (180^o-\angle DPY)+ (180^o-\angle DPX)=\angle DBY+ \angle DCX$ $\angle XZY= \angle YZM - \angle XZM=180^o - \angle DBY - DCX$. Thus,$\angle XPY+\angle XZY=180^o$, i.e. $P \in XYZ$ and this is a fixed cirlce. So the distance from center of $(XYZ)$ to $P$ is a constant.

My solution: Let $D'$ be a point on $BC$ such that $BD=CD'$, $M$ be the midpoint of $BC$, and $U,V$ be the midpoints of $FD'$ and $ED'$. Also, let $K$ be the antipode of $D'$ in $\odot D'MU$, and $L$ be the antipode of $D'$ in $\odot (D'MV)$. Then, as $BD'=BF$ and $CD'=CE$, we get that $K,U$ lie on the internal angle bisector of $\angle ABC$ and $L,V$ lie on the internal angle bisector of $\angle ACB$. Also, $M,K,L$ are collinear and lie on the perpendicular bisector of $BC$ (as well as $DD'$). This gives $KD=KD'=KF$, and so, by Fact 5, we get that $K$ lies on $\odot (BFD)$. Similarly, $L$ lies on $\odot (CED)$. Now, Let $BK \cap CL=I$ be the incenter of $\triangle ABC$. Then $$\angle KPL=360^{\circ}-\angle DPK-\angle DPL=\angle DBK+\angle DCL=180^{\circ}-\angle BIC=\angle KIL$$which shows that $P$ lies on $\odot (KIL)$, which is fixed, as $K$ and $L$ are also fixed (as proved before). Thus, the distance of $P$ from the center of $\odot (KIL)$ remains constant as $D$ moves on $BC$. $\blacksquare$

Let $I$ be the incenter of $\triangle ABC$, and let $X$ and $Y$ be the Miquel points of quadrilaterals $BEDC$ and $CFDB$, respectively. We first locate $X$ and $Y$. By definition $B,D,E,X$ are cyclic, and a spiral similarity at $X$ sends $\overline{DC}$ to $\overline{BE}$. But $DC = BE$, so this is just a rotation. Therefore, by symmetry $B,I,X$ collinear. Similarly, $C,I,Y$ collinear and $C,D,F,Y$ cyclic. Moreover, $X$ and $Y$ are fixed as they both lie on the perpendicular bisector of $\overline{BC}$. Finally, note that $$\measuredangle PXI = \measuredangle PXB = \measuredangle PDB = \measuredangle PDC = \measuredangle PYC = \measuredangle PYI,$$so $I,P,X,Y$ cyclic, as desired. ($Q$ is the center of $(IXY)$.) [asy][asy] size(10cm); defaultpen(fontsize(10pt)); pair A = dir(120); pair B = dir(200); pair C = dir(340); pair D = 0.47*B + 0.53*C; pair E = intersectionpoints(circle(B,abs(C-D)),A--B)[0]; pair F = intersectionpoints(circle(C,abs(B-D)),A--C)[0]; pair P = intersectionpoints(circumcircle(B,E,D),circumcircle(C,F,D))[0]; pair I = incenter(A,B,C); pair X = intersectionpoints(B--2I-B,circumcircle(B,E,D))[1]; pair Y = intersectionpoints(C--2I-C,circumcircle(C,F,D))[1]; filldraw(A--B--C--cycle,pink+white+white,pink); draw(circumcircle(B,E,D),orange); draw(circumcircle(C,F,D),orange); draw(circumcircle(A,E,F),red); draw(circumcircle(X,Y,I),dashed+blue+white); draw(B--X,gray(0.5)); draw(C--I,gray(0.5)); dot("$A$",A,dir(90)); dot("$B$",B,dir(225)); dot("$C$",C,dir(315)); dot("$D$",D,dir(270)); dot("$E$",E,dir(135)); dot("$F$",F,dir(0)); dot("$P$",P,dir(0)); dot("$I$",I,dir(250)); dot("$X$",X,dir(90)); dot("$Y$",Y,dir(110)); [/asy][/asy] Remark. Mostly inspired by the lemma that if two points $E$ and $F$ move linearly along $\overline{AB}$ and $\overline{AC}$, then $(AEF)$ passes through a fixed point.

This problem have some good points : First Lemma : If $D$ , $E$ , $F$ are some points on $BC$ , $AC$ , $AB$ and $T$ be tucker point of $D$ , $E$ , $F$ and $P$ be a arbitrary point on arc $EF$ . If lines $BP$ , $CP$ intersect arc $ED$ , $FD$ at $Q$ and $R$ , Then $TPQR$ is cyclic . Second Lemma : We have a point $A$ and two lines $x$ and $y$ passes through $A$ , Get points $X$ and $Y$ on lines $x$ , $y$ such that we can find fix $a$ and $b$ and $c$ such that $a.{AX}+b.{AY}=c$ then circumcircle of $AXY$ passes through a fix point .

Let $I$ be the incenter of $\triangle ABC$ and $\ell$ be the perpendicular bisector of $BC$. I claim that $(BDF) \cap \overrightarrow{BI}$ and $(CDE) \cap \overrightarrow{CI}$, which we denote $X$ and $Y$ respectively, are fixed under motion from $D, E, F$. We first show that $X = \ell \cap BI$, and then $Y = \ell \cap CI$ will follow similarly. Doing so will indeed prove that $X$ and $Y$ are fixed. Note that $XD = XF$ by angle bisection. Furthermore, $BF = CD$ by definition and $\angle XFB = \angle XDC$ by cyclic quadrilaterals. Therefore, $\triangle XFB \cong \triangle XDC$ and hence $X$ is the center of spiral similarity bringing $FB \to DC$ and therefore also $FD \to BC$. Since $XF = XD$, it must follow that $XB = XC$, hence $X \in \ell$. Similarly, $Y \in \ell$. We finish by showing $P \in (IXY)$. This does finish since $I, X, Y$ are all fixed under $D, E, F$ motion. This turns out to just be an angle chase:\[\angle IXP = \angle BXP = \angle PDC = \angle PYC = 180^{\circ} - \angle PYI\]as desired. $\blacksquare$

Generalization

Interesting problem. Let $D'$ be the reflection of $D$ over the midpoint of $\overline{BC}$, and set $O_B, O_C$ to be the circumcenters of $(FDD')$ and $(EDD')$ respectively. Claim. $O_B$ lies on $(BDF)$. Proof. Note that $BD'=BF$, so $$\angle FO_BD = 2\angle FD'B=180^\circ - B. \ \blacksquare$$ Similarly, $O_C$ lies on $(CDE)$. Note furthermore that these two points are fixed because the perpendicular bisectors of $\overline{D'F}, \overline{D'E}$, and $\overline{DD'}$ are all fixed. Finally, $$\measuredangle O_BPO_C = \measuredangle O_BPD+\measuredangle DPO_C = \measuredangle O_BBD + \measuredangle O_CCD$$is clearly fixed because $O_B, O_C$ lie on the $B$ and $C$ bisectors, respectively. Thus, $P$ lies on a fixed circle.

this is confusing Let $I$ be the incenter; let $(BDF)$ and $(CDE)$ intersect the perpendicular bisector of $BC$ at $M_B$ and $M_C$. By spiral similarity arguments $M_B,M_C$ are fixed. Now angle chase gives $P\in (IM_BM_C)$ done.

Let $D'$ be the reflection of $D$ in the midpoint $BC$ and let $X, Y$ be the circumcenters of $(FDD')$ and $(EDD')$, respectively. Claim: $X$ is the midpoint of arc $FD$ not containing $B$. Similarly $Y$ is the midpoint of arc $ED$ not containing $C$. Proof. Note that $\measuredangle FXD = 2\measuredangle FD'D = 2\measuredangle FD'B = \measuredangle BFD' + \measuredangle FD'B = \measuredangle FBD$, so $X$ lies on $(BFD)$. Since $X$ lies on the perpendicular bisector $FD$, thus $X$ is the midpoint of arc $FD$. Similarly $Y$ is the midpoint of arc $ED$. $\blacksquare$ Now we'll prove that $\measuredangle XPY$ is independent from $D$, completing the proof. Note that $\measuredangle XPY = \measuredangle XPD + \measuredangle DPY = \measuredangle XBD + \measuredangle DCY = \frac{\measuredangle ABC}{2} + \frac{\measuredangle BCA}{2}$ is independent from $D$, as needed. $\blacksquare$

This was pretty hard. I had to take a hint to find the key breakthrough but I guess it makes sense since it makes the weird length condition easier to handle. I think its hard for a Problem 1 even for China TST standards. We denote by $I$ the incenter of $\triangle ABC$, and by $\ell$ to perpendicular bisector of $BC$. Let $D'$ be the reflection of $D$ across the midpoint of segment $BC$. Now, let $O_B = \overline{BI} \cap \ell$ and $O_C = \overline{CI} \cap \ell$. We start off by proving the following key claims. Claim : Points $O_B$ and $O_C$ lie on circles $(BDF)$ and $(CDE)$, respectively. Proof : First, note that $\triangle O_BFB \cong \triangle O_BD'B$ since $BD' = CD = FB$ and $\measuredangle FBO_B = \measuredangle O_BBD'$. Thus, \[O_BF=O_BD' = O_BD\]which implies that $O_B$ is in fact the center of $(DD'F)$. Then, \[\measuredangle DO_BF = 2\measuredangle BD'F = \measuredangle D'BF = \measuredangle DBF\]which implies that $O_B$ indeed lies on $(BDF)$ as desired. Similarly, it follows that $O_C$ lies on $(CDE)$, which finishes the proof of the claim. Now, we can further note the following observations related to the incenter $I$. Claim : Point $I$ lies on $(AEF)$ and $\overline{CI} \cap \overline{D'F}$ lies on $(AEF)$. Proof : Let $T_A,T_B$ and $T_C$ denote the feet of the perpendiculars from $I$ to the sides of $\triangle ABC$. Now, we know that $IT_B=IT_C$. Further, $\measuredangle FT_CI = 90^\circ= \measuredangle ET_BI$. Finally, \[FT_C = FA - T_CB = CD - BT_A = CT_A - BD = CT_B - CE = ET_B\]and thus, $\triangle FT_BI \cong \triangle FT_BI$. Thus, $\measuredangle T_CIF = \measuredangle T_BIE$. This implies that $\measuredangle FIE = \measuredangle T_CIT_B = \measuredangle BAC = \measuredangle FAE$ which implies that indeed $I$ lies on $(AEF)$, as desired. Further, let $X = \overline{D'F} \cap \overline{CI}$. \[2\measuredangle IXF = 2\measuredangle CXD' = 2(\measuredangle BD'F + \measuredangle ICB) = \measuredangle CBA + \measuredangle ACB = \measuredangle CAB = 2\measuredangle IAF\]which implies that $\measuredangle IXF = \measuredangle IAF$, implying that $\overline{CI} \cap \overline{D'F}$ lies on $(AEF)$, as desired. Now, we claim that $P$ lies on $(IO_BO_C)$ irrespective of the position of $D$ on segment $BC$. This is simply an angle chase. First note that, $\measuredangle D'FB = \measuredangle O_CO_BI$. Now, \[\measuredangle XIP = \measuredangle XFP = \measuredangle D'FP = \measuredangle BFP + \measuredangle D'FB = \measuredangle IO_BP + \measuredangle O_CO_BI = \measuredangle O_CO_BP\]which implies the desired result.

We will prove that $P$ lies on a fixed circle. Let $I$ be incenter of $\triangle ABC,$ perpendicular bisector of $BC$ intersects $BI, CI$ at $U, V$. We have $(BU, BE) \equiv (BC, BU) \equiv (CV, CD) \pmod \pi$. But $BE = CD$ then $\triangle BEU \cong \triangle CDU$. So $UE = UD,$ which leads to $U \in (BDE)$. Similarly, we have $V \in (CDF)$. From this, we have $(PU, PV) \equiv (PU, PD) + (PD, PV) \equiv (BU, BD) + (CD, CV) \equiv (BU, CV) \equiv (IU, IV) \pmod \pi$ or $P \in (IUV)$