This condition imply also that $AF+AE=BC$. Assume that bisector of angle $BAC$ intersect $BC$ and circumcircle of triangle $ABC$ at $M$ and $N$ respectively. Let circumcircle of triangle $AEF$ intersect line $AM$ at $L$. Assume that $R$ is point on $AB$ ($F$ lies between $A$ and $R$) such that $\angle ARL = \angle RAL$. Since $AELF$ is cyclic, and $FL=LE$, we have that triangles $AEL$ and $FLR$ are congruent, so $FR=AE \Longrightarrow AR=BC \Longrightarrow$ triangles $BNC$ and $ALR$ are congruent $\Longrightarrow AL=BN$. By the well known lemma $IN=BN=NC=AL$, so point $I$ is image of $M$ wrt midpoint of $AN \Longrightarrow OI=OL$. Miquel theorem imply that $P$ lies on circumcircle (call it $\omega$) of triangle $BFD$. If $AI \cap \omega = S$, then last argument show that $OS=OL=OI$. Wlog, $S$ lies beetwen $B, I$, and $P$ lies in the same side of $AI$ as $C$, we have (simple angle chasing with other cases): \[\angle SIL=\angle PDB=\angle PEC=\angle ILP,\] so quadrangle $ISLP$ is cyclic, but points $S, L, I$ lie on the circle with center $O$, so $OP=OI$.

The following result is also true: Let $D,E,F$ be arbitrary points on the sides of $\triangle ABC$ and $D',E',F'$ the reflections of $D,E,F$ on the midpoints of the respective sides. Then the Miquel points $M$ of $D,E,F$ and $M'$ of $D',E',F'$ are equidistant from the circumcenter of $ \triangle ABC.$

the main idea is to prove that thos 2 circumcircles pass through $2$ fixed points on lines $BI$ and $CI$($R,Q$) and that $IRQP$ is cyclic(circle $k$. Now just consider the case when $D$ is where the $A-excircle$ meets $BC$ to prove that the circumcenter $T$(where $T=P$ for previously described $D$) of the triangle formed by excenters of $ABC$ and is on line $OI$ where $O$ is the midpoint of $IT$. Now it's easy to get that $IT$ is the diameter of $k$ which finishes the problem.

Here is a very bad solution. I had way too much fun writing it.

Here are brief sketches of two barycentric solutions from MOP, the first by Bobby Shen and the second by me. Linearity approach. (This seems to be equivalent to v_enHance's circle equations.) Let $(Q,\Gamma)$ be the power of $Q$ with respect to $\Gamma$. If $P = (x,y,z)$ ($x+y+z=1$) with respect to $\triangle{ABC}$, $\omega_A = (AEF)$, etc., and $\omega = (ABC)$, then the function $f(Q) = (Q,\omega_A) - (Q,\omega)$ is linear in $Q$. Hence \begin{align*} 0 - (P,\omega) = f(P) &= xf(A) + yf(B) + zf(C) \\ &= x(0-0) + y(BF\cdot BA - 0) + z(CE\cdot CA - 0), \end{align*}and cyclic. But then \[ -(P,\omega)\sum{a} = \sum yca(BF) + zba(CE) = \sum xbc(AE+AF) = abc\sum x = abc \]gives $-(P,\omega) = 2Rr = R^2 - OI^2$, as desired. Comment. This also establishes proglote's claim, since if $P' = (x',y',z')$ is the Miquel point of $\triangle{D'E'F'}$, \[ - (P',\omega) = y'(AF\cdot AB) + z'(AE\cdot AC) \]yields \[ -(P',\omega)\sum{x} = \sum x'(yBF\cdot BA+zCE\cdot CA) = -(P,\omega)\sum{x'}, \]as desired. (In the original problem, we have $P'=I$, so $(a+b+c)P' = (a,b,c)$.) Spiral similarity approach. Given $P$, let $D_0,E_0,F_0$ be the feet from $P$ to $BC,CA,AB$. It is well-known and easy to prove that the cevian triangles $DEF$ with Miquel point $P$ are precisely those (directly) similar to $\triangle{D_0E_0F_0}$. If $P = (x,y,z)$ as in the previous solution, we can easily parameterize $BD+BF$ and $CD+CE$ as linear functions in $\cot\alpha$ (or $\tan\alpha$; I don't really remember), where $\alpha$ is the (directed) angle of rotation $\measuredangle{DPD_0} = \measuredangle{EPE_0} = \measuredangle{FPF_0}$. In general (i.e. if $P$ doesn't lie on any angle bisector), it's easy to solve this system, which eventually leads to $(xA+yB+zC)^2 = R^2-2Rr = OI^2$.

We can also use complex numbers; if the circumcenter is the unit circle, the Miquel point $M$ of $D,E,F$ satisfies \[ \overline{m} = \frac{ \sum_{\text{cyc}} d(ab - ac)}{\sum_{\text{cyc}} d (b-c)},\] so we have \[{|m| = \frac{ \sum_{\text{cyc}} d(ab - ac)}{\sum_{\text{cyc}} d (b-c)} \cdot \frac{\sum_{\text{cyc}} \overline{d} (b-c)}{ \sum_{\text{cyc}} \overline{d}(ab - ac)}}.\] Since $d' = bc\overline{d} = b+c-d$, the numerators in $|m|$ and $|m'|$ cancel out; it then remains to check that the denominators are equal, which is not hard using $\sum_{\text{cyc}} d'(b-c) = -\sum_{\text{cyc}} d(b-c).$

Lemma Given a $\triangle ABC$, let $D, E, F$ be any points on $BC, CA, AB$ and P an arbitrary point on the same plane. $AP$ meets the circle $(AEF)$ at $X$ and $Y, Z$ are defined similarly. If $M$ is the intersection of circles $(AEF), (BFD), (CDE)$, then $M, P, X, Y, Z$ are concyclic. This lemma can be proved by just angle chasing, and it kills this shortlisted problem. I have a different solution. Solution Let $A', B', C'$ be the intersection of $A$-excircle, $B$-excircle, $C$-excircle and $BC, CA, AB$, respectively. The condition implies that $D, E, F$ are points that are translated constant length from $A', B', C'$ all clockwise or all counter-clockwise (just call this same direction). Let $D', E, F'$ be the reflection of $D, E, F$ wrt midpoints of $BC, CA, AB$, respectively. If the incircle of $\triangle ABC$ meets $BC, CA, AB$ at $A'', B'', C''$, then $D', E', F$ are the points that are translated constant length from $A'', B'', C''$. If $l_1 , l_2 , l_3$ denote the lines $ID'$, $IE'$, $IF'$, then $\measuredangle (l_1 , BC) = \measuredangle (l_2 , CA) = \measuredangle (l_3 , AB)$. (They are directed angles, measured clockwise from the first line to the second line.) Call this equal angle $\theta$. Now rotate $l_1 , l_2 , l_3 $ clockwise by $2 \theta$ and get $l_1 ', l_2 ', l_3 '$. Easy to see that $l_1 ' $, $l_1$, $BC$ from an isosceles triangle, thus $l_1 '$ passes through $D$. Also, $l_1 ' , l_2 ' , l_3 ' $ meet at some point $P'$. But $\measuredangle ( l_1 ' , l_2 ' ) = \measuredangle (l_1 , l_2) = \measuredangle (BC, AC)$, so quadrilateral made by lines $l_1 ' , l_2 ' , BC, AC$ or quadrilateral $CDEP'$ is cyclic. Similarly, $BDFP'$ is cyclic, implying $P=P'$. However, $P'$ is just a rotation of $I$ centered at $O$, thus $OP=OI$.

Hehehehe, beautiful problem. For convenience, forget about the original definition of P, call the intersection of the two circles Q. First, a lemma: LEMMA: In triangle XYZ, the bisector of angle X intersects the circumcircle again at W. Then XW=(XY+XZ)/(2cos(<X/2)) This is well-known. Now, we can easily see AF+AE=BC. Draw the circumcircle of AFE ($w_a$, we define $w_b$ and $w_c$ analogously), say it intersect the angle bisector of A at X, and the circle of center O, radius OI (circle $w$) at P. Then by the lemma AX=BC/2(sinA/2)=GB=GC=GI, where G is the intersection of the circumcircle of ABC and the angle bisector of A. Then we can easily see OX=OI so $X \in w$. Analogously $Y, Z \in w$, if we define $Y$ and $Z$ analogously. Now, we have that BFYD is cyclic. But <BYP = < IYP = 180 - <IXP = <AXP = <AFP = 180-<AFP and therefore BFPYD is cyclic pentagon, so $P \in w_a, w_b$. Analogously $P \in w_c$ and so $P = w \cap w_a \cap w_b \cap w_c$. Therefore Q=P. So $Q \in w$ so OQ=OI, which is what we wanted.

Another proof after proving that $P$ lies on circumcurcle of $BFD$. Let $O\neq I$ and suppose that $\exists$ points $K_{A}, K_{B}, K_{C}$ s.t. $K_{A}=\{ OM_{A}\cap ID'\}, K_{B}=\{ OM_{B}\cap IE'\}, K_{C}=\{ OM_{C}\cap IF'\}.$ But by the lemma above $\angle (OM_{A}, D'I)=\angle (OM_{B}, E'I)=\angle (OM_{C}, F'I)\Rightarrow\angle (OK_{A}, K_{A}I)=\angle (OK_{B}, K_{B}I)=\angle (OK_{C}, K_{C}I).$ So $O, I, K_{A}, K_{B}, K_{C}$ are concyclic. Denote this circle as $\omega$. Let $P'\neq K_{A}$ to be point on $\omega$ such that $P'K_{A}$ is the reflection of line $IK_{A}$ in line $OK_{A}.$ It can be seen that $OP'=OI.$ So that $\angle (P'K_{A}, OK_{A})=\angle (OK_{A}, K_{A}I)=\angle (P'K_{B}, OK_{B})=\angle (OK_{B}, K_{B}I)=\angle (P'K_{C}, OK_{C})=\angle (OK_{C}, K_{C}I),$ since $P', O, I, K_{A}, K_{B}, K_{C}$ are concyclic. So $P'K_{B}, P'K_{C}$ are reflection of lines $IK_{B}, IK_{C}$ in lines $OK_{B}, OK_{C}$ respectively. Let $\{K_{A}P'\cap BC\}=D_{1},$ since $K_{A}M{_A}\perp BC$ and $K_{A}M_{A}$ bisects $\angle D'K_{A}D_{1},$ we know that $K_{A}D'=K_{A}D_{1}$ and so $K_{A}M_{A}$ is the perpendicular bisector of segment $D'D_{1}.$ So $D_{1}, D$ are both reflection of $D'$ in $M_{A},$ hence $D=D_{1}\Rightarrow D\in\{K_{A}P'\}.$ Similarly $E\in\{K_{B}P'\}$ and $F\in\{K_{C}P'\}$. Finally, since $\angle (K_{A}D', BC)=-\angle (K_{A}D, BC)=\angle (K_{B}E', AC)=-\angle (K_{B}E, AC)=\angle (K_{C}F', AB)=-\angle (K_{C}F, AB),$ it implies that $\angle (DP', BC)=\angle (EP', AC)=\angle (FP', AB)$ so $BDP'F, CDP'E, AEP'F$ are cyclic. Hence $P=P',$ and by the definition above $OP=OI.$ Case where $O=I$ and $OM_{A}\parallel ID', OM_{B}\parallel IE', OM_{C}\parallel IF'$ are limit cases.

I have a very nice and new solution! If you seen it, you know it is very easy. here my solution: We have that $AE+AF=BC$. Let $ A^', B^', C^' $ be the tangent points of excircles to $BC,CA,AB$ respectively of $A,B,C$ and $J$ is intersection point of perpendiculars to $BC,CA,AB$ at the points $A^', B^', C^'$. We have that $I,O,J$ are collinear and $O$ is midpoint of the segment $IJ$. Let $k_a,k_b,k_c$ be circumcircles of the triangles $AFE$, $BDF$, $CEF$ and $AA_1,BB_1,CC_1$ be diameters of the $k_a,k_b,k_c$ respectively. So, the triangles $B_1PC_1$, $BPC$ are similar and analoguosly ... $ \Rightarrow $ the triangles $ABC$ and $A_1B_1C_1$ are similar. We have that $DA^'=EB^'=FC^'$ and $J$ is incenter of the $A_1B_1C_1$. The triangle $A_1B_1C_1$ is change of the $ABC$ to $90^\circ$ with coeffitsiyent $ \alpha $, around of the point $P$. (center ir $P$) Hence $ \angle IPJ=90^\circ $ and $P$ lies on the circle with diameter $IJ$, center $O$ $ \Rightarrow $ $OP=OI=OJ$. End.

@JuanOrtiz, you have a small mistake! (see lemma? )

Clearly, $D,E,F$ are contact points of excircles with the sides. If the incircle touches the sides at $D',E',F'$ respectively, $D,E,F$ are their reflections about midpoints of sides; consequently, as per Carnot lemma, the perpendiculars at $D,E,F$ onto $BC, CA, AB$ respectively are concurrent, and the common point is $P$ (Miquel point of $\Delta ABC$ and $D,E,F$). Since the perpendicular bisectors of $DD', EE', FF'$ are the perpendicular bisectors of $BC, CA, AB$ respectively, they concur at $O$, but they concur at midpoint of $PI$, as well, and $O$ is midpoint of $IP$, thus we are done. Best regards, sunken rock

sunken rock wrote: Clearly, $D,E,F$ are contact points of excircles with the sides. This is false. Actually, if we take the contact points and move them a constant distance in the same direction along the sides, you get infinitely many $D, E, F$ that satisfy the condition.

Let $BI\cap (BDF), (O)=X,M, CI\cap (CED), (O)=Y,N$. We apply Ptolemy theorem to cyclic $BEXD$: $DF*BX=FX*BD+XD*BF=DX(BF*BD)=DX*AC\Rightarrow BX$ $=AC\frac {DX}{BX}$. Note that $\triangle FXD\sim \triangle AMC$, hence $BX=AC\frac {AM}{AC}=AM=MI$, since $OM=OB$, therefore $OX=OI$. Analogously we can obtain $OI=OX$, hence $O$ is the circumcenter of $\triangle IXY$. Note that $\angle (XI,XP)=\angle (BC,DP)=\angle (YI,YP)$, hence $P$ lies on $(IXY)\Rightarrow OP=OI$.

proglote wrote: The following result is also true: Let $D,E,F$ be arbitrary points on the sides of $\triangle ABC$ and $D',E',F'$ the reflections of $D,E,F$ on the midpoints of the respective sides. Then the Miquel points $M$ of $D,E,F$ and $M'$ of $D',E',F'$ are equidistant from the circumcenter of $ \triangle ABC.$ proglote wrote: We can also use complex numbers; if the circumcenter is the unit circle, the Miquel point $M$ of $D,E,F$ satisfies \[ \overline{m} = \frac{ \sum_{\text{cyc}} d(ab - ac)}{\sum_{\text{cyc}} d (b-c)},\]so we have \[{|m| = \frac{ \sum_{\text{cyc}} d(ab - ac)}{\sum_{\text{cyc}} d (b-c)} \cdot \frac{\sum_{\text{cyc}} \overline{d} (b-c)}{ \sum_{\text{cyc}} \overline{d}(ab - ac)}}.\]Since $d' = bc\overline{d} = b+c-d$, the numerators in $|m|$ and $|m'|$ cancel out; it then remains to check that the denominators are equal, which is not hard using $\sum_{\text{cyc}} d'(b-c) = -\sum_{\text{cyc}} d(b-c).$ sorry, do you have synthetic proof of that fact? really thankful

@^ I am pretty sure that that proof is about as synthetic as it gets

Here's a bary bash. We start with a lemma: Lemma: If $x, y, z$ satisfy $dy+ez=fz+gx=hx+iy$ then $(x, y, z)=w(if-ie-fd, eh-hf-ge, dg-dh-ig)=w((i-d)(f-e)-ed,...)$ for some $w$. Moreover, $dy+ez=-w\cdot (dfh+egi)$. Proof Just check that the ordered pair $(x, y, z)$ solves the system with a direct computation. $\blacksquare$ Let $BD=s-c+t$. Then it follows that $CD=s-b-t, CE=s-a+t, EA=s-c-t, AF=s-b+t, FB=s-a-t$. Then we have the barycentric coordinates $D=(0: s-b-t: s-c+t)$ $E=(s-a+t: 0: s-c-t)$ $F=(s-a-t: s-b+t: 0)$ Using the barycentric circle formula, one deduces that the formula for circle $AEF$ is given by $$a^2yz+b^2zx+c^2xy=(x+y+z)(c(s-a+t)y+b(s-a-t)z)$$and cyclic variations for circle $BFD, CDE$. Note that $P$ is the intersection of these 3 circles. In other words, if $P=(x:y:z)$ then $$c(s-a+t)y+b(s-a-t)z=a(s-b+t)z+c(s-b-t)x=b(s-c+t)x+a(s-c-t)y$$With the aid of our lemma we get $$P=(x:y:z)=\boxed{(bct^2-bc(s-a)^2-[s(c-a)+t(c+a)][s(a-b)+t(a+b)]:...)}$$with cyclic variations for the other variables. Set $O$ as the origin. Suppose $P=(x, y, z)$ such that $x+y+z=1$; i.e. the coordinates are normalized. Note that $\vec{A}\cdot\vec{A}=R^2$ where $R$ is the circumradius. Also note $\vec{A}\cdot\vec{B}=R^2-\frac{c^2}{2}$. Thus it follows that $$|\vec{P}|^2=(x\vec{A}+y\vec{B}+z\vec{C})\cdot (x\vec{A}+y\vec{B}+z\vec{C})=R^2(x+y+z)^2-\sum_{\text{cyc}} c^2xy=R^2-\sum_{\text{cyc}} c^2xy$$We wish to show that $OP^2=OI^2=R^2-2rR$. Thus, it remains to show that $\sum_{\text{cyc}} c^2xy=2rR=\frac{abc}{2s}$. But recall that $\sum_{\text{cyc}} c^2xy=(x+y+z)(c(s-a+t)y+b(s-a-t)z)$. After homogenizing, we want to show $c(s-a+t)y+b(s-a-t)z=\frac{abc}{2s}(x+y+z)$. By our lemma, the left hand side is just $abc[(s-a-t)(s-b-t)(s-c-t)+(s-a+t)(s-b+t)(s-c+t)]$. The right hand side is equivalent to $-\frac{abc}{2s}\sum_{\text{cyc}} bct^2-bc(s-a)^2-[s(c-a)+t(c+a)][s(a-b)+t(a+b)]$. It remains to check that $$2s[(s-a-t)(s-b-t)(s-c-t)+(s-a+t)(s-b+t)(s-c+t)]=\sum_{\text{cyc}} t^2[(a+b)(a+c)-bc]+t[s(c+a)(a-b)+s(a+b)(c-a)]+s^2(c-a)(a-b)-bc(s-a)^2$$A direct computation shows that the $t, t^3$ terms on both sides vanish. On the other hand, $\sum_{\text{cyc}} (a+b)(a+c)-bc=\sum_{\text{cyc}} 2as=4s^2=2s(2(s-a)+2(s-b)+2(s-c))$ which confirms the $t^2$ coefficients of both sides match. To confirm the constant terms match, we need $4s(s-a)(s-b)(s-c)=\sum_{\text{cyc}} s^2(c-a)(a-b)-bc(s-a)^2$ which ends up being true. Therefore, $\sum_{\text{cyc}} c^2xy=-2rR$ and hence $OP^2=R^2-2rR$ as desired. $\square$ Note that this nicely shows the power of a normalized point $(x, y, z)$ WRT the circumcircle of $\triangle ABC$ is $-\sum_{\text{cyc}} c^2xy$

proglote wrote: The following result is also true: Let $D,E,F$ be arbitrary points on the sides of $\triangle ABC$ and $D',E',F'$ the reflections of $D,E,F$ on the midpoints of the respective sides. Then the Miquel points $M$ of $D,E,F$ and $M'$ of $D',E',F'$ are equidistant from the circumcenter of $ \triangle ABC.$ A purely synthetic proof of this can be found on my blog. Also, in the proof, I show that the following assertion is also true: Let $P$ be a point in the plane of triangle $ABC$ with its pedal triangle $P_aP_bP_c$. Let $D,E,F$ be arbitrary points on the sides of $\triangle ABC$ and $D',E',F'$ the reflections of $D,E,F$ in $P_a, P_b, P_c$ respectively. Then the Miquel points $M$ of $D,E,F$ and $M'$ of $D',E',F'$ are equidistant from $P$.

Let $J$ be the Bevan's point of triangle $ABC$. It suffices to show that $P$ lies on the circle $(IJ)$ as $O$ is the midpoint of $IJ$. Let $X,Y$ denote the projections of $J$ on $BI,CI$, respectively. The crucial claim is that for all points $D,F$ on $BA,BC$ respectively such that $BD+BF=CA$, the circumcircle of triangle $BDF$ passes through $X$ and the same holds where $(CDE)$ passes through $Y$. Indeed, let $A_1,B_1,C_1$ be the touch points of the excircles on the sides $BC,CA,AB$ respectively. Then, we see that $A_1D=B_1E=C_1F$ with lengths oriented along $BC,CA,$ and $AB$. As $X$ is the midpoint of arc $A_1C_1$ in $(BA_1C_1)$, it follows that $\triangle XFC_1$ and $\triangle XDA_1$ are congruent, hence similar, yielding that $X$ lies on $(BDF)$. The claim is established. Now, by Miquel's Theorem, for the points $X,Y,$ and $D$, and the triangle $BIC$, it follows that $P=(BDF) \cap (CDE) \not=D$ lies on $(IXY)=(IJ)$ and the result holds.

Wait, what? Let $\omega$ be circle with center $O$ and radius $OI$. Let $S=BI\cap (BDF)$ and $T=CI\cap (CDE)$. I contend that $S,T$ lie on $\omega$. If $X=BI\cap (ABC)$, then by Ptolemy, \begin{align*} BS=\frac{(BD+BF)\cdot SD}{DF}=\frac{AC}{2\cos{\tfrac{\angle B}{2}}}=XA=XI, \end{align*}thus $O$ lies on the perpendicular bisector of $SI$. Similarly, $O$ lies on the perpendicular bisector of $TI$. Therefore, \begin{align*} \measuredangle ISP=\measuredangle BSP=\measuredangle BDP=\measuredangle CDP=\measuredangle CTP=\measuredangle ITP, \end{align*}hence $P$ lies on $\omega$, i.e. $OP=OI$, as desired.

Firstly we observe that $|AE|+|AF|=|AC|-|CE|+|AB|-|BF|=|BD|+|CD|=|BC|.$ Throughout the solution $\mathcal{S} _\ell$ denotes reflection over arbitrary line $\ell.$ Denote by $D'$ reflection of $D$ wrt midpoint of $BC,$ by $A'$ intersection of $D'I$ with perpendicular to $BC$ from $O.$ Define $E',F',B',C'$ analogously. Claim. $\odot (AF'E'),\odot (BD'F'),\odot (CE'D')$ concur at $I.$ Proof. It's suffice to show that $AF'IE'$ is cyclic. By length condition $\mathcal{S} _{BI} (F') =\mathcal{S} _{CI} (E') =G\in BC,$ moreover $I$ is the circumcenter of $E'F'G,$ so $\angle E'IF'=2\angle E'GF'=\angle ABC+\angle ACB=\pi-\angle E'AF'\text{ } \Box$ Now from $\measuredangle OA'I=90^\circ +\measuredangle BD'I=90^\circ +\measuredangle AF'I=90^\circ +\measuredangle AE'I=\measuredangle OB'I$ and all similar statements we deduce that $O,I,A',B',C'$ are concyclic, and clearly $\mathcal{S} _{A'O} (D'I) ,\mathcal{S} _{B'O} (E'I) ,\mathcal{S} _{C'O} (F'I)$ concur at $Q$ lying on same circle and with $|OQ|=|OI|.$ But $D\in \mathcal{S} _{A'O} (D'I)$ and cyclically, so $\measuredangle QDC=-\measuredangle ID'B=-\measuredangle IE'A=\measuredangle CEQ,$ and all analogous equalities hold. Thus $Q=P$ and we are done.

Here is a significantly different solution that doesn't use the fixed point on $\odot(AEF)$. Thanks to NN37252 who suggested the idea. Note that the condition implies $AE+AF=BC$. Let $D'$, $E'$, $F'$ be the reflection of $D$, $E$, $F$ across the midpoints of respective sides. Also, let the incircle of $\triangle ABC$ touches $BC,CA,AB$ at $T_A$, $T_B$, $T_C$. Then, the length condition implies that $BE'+CF'=BC\implies T_BE'=T_CF'$. Thus, $\triangle IT_AD'\cong\triangle IT_BE'\cong\triangle IT_CF'$, implying that $\triangle D'E'F'\sim\triangle T_AT_BT_C$. Now, motivated by ISL 2006 G9, we let $\odot(AEF)$, $\odot(BDF)$, $\odot(CDE)$ meet $\odot(ABC)$ again at $X,Y,Z$. By that problem, $\triangle XYZ\sim\triangle D'E'F'$. Moreover, since $$\frac{XB}{XC} = \frac{BF}{CE} = \frac{AF'}{AE'},$$we have $\triangle XBC\sim\triangle AF'E'$. Similarly, we have $\triangle YAC\sim\triangle YF'D'$, and $\triangle ZAB\sim\triangle ZE'D'$ (this is the key step of 2006 G9). Here is the main claim. Claim: $P$ is the orthocenter of $\triangle XYZ$. Proof: This is just angle chasing \begin{align*} \measuredangle YPZ &= \measuredangle YPD + \measuredangle DPZ \\ &= \measuredangle YBD + \measuredangle DCZ \\ &= \measuredangle YBC + \measuredangle BCZ \\ &= \measuredangle YAC + \measuredangle BAZ \\ &= \measuredangle BF'D' + \measuredangle CE'D' \\ &= \measuredangle AF'D' + \measuredangle AE'D' \\ &= \measuredangle F'AE' + \measuredangle E'D'F'. \end{align*}Thus, $\angle YPZ = \angle BAC + \angle E'D'F' = \angle A + 90^\circ - 0.5\angle A = 90^\circ + 0.5\angle A$, implying the result. $\blacksquare$ Finally, let $AI, BI, CI$ intersects $\odot(ABC)$ again at $M_A, M_B, M_C$. Notice that $\triangle M_AM_BM_C\cup O\cup I$ and $\triangle XYZ\cup O\cup P$ are congruent, done!

Let circle with radius $OI$ intersect $BI, CI$ at $S, T \ne I$ respectively. Let $H$ be the foot of $S$ to $AB$ and let $BI$ meet $(ABC)$ at $M \ne A.$ It's known that if any circle passing through $S,B$ intersects $BC,BA$ at $D',F'$ respectively then $BD'+BF' = 2BH.$ But $$BH = BS \cdot \cos \left(\frac{1}{2} \angle ABC \right) = IM \cdot \cos \left(\frac{1}{2} \angle ABC \right) = MA \cdot \cos \left(\frac{1}{2} \angle ABC \right) = \frac{1}{2}AC$$so in fact $BFSD$ cyclic, similarly $CDTE$ cyclic. Then $$\angle TPS = 360^\circ - \angle TPD - \angle SPD = \frac{1}{2} \angle ABC + \frac{1}{2} \angle ACB = \angle TIS$$so $P$ lies on $(TIS)$ centered at $O,$ done. $\blacksquare$

lyukhson wrote: Let $ABC$ be a triangle with circumcenter $O$ and incenter $I$. The points $D,E$ and $F$ on the sides $BC,CA$ and $AB$ respectively are such that $BD+BF=CA$ and $CD+CE=AB$. The circumcircles of the triangles $BFD$ and $CDE$ intersect at $P \neq D$. Prove that $OP=OI$. Flip99 wrote: proglote wrote: The following result is also true: Let $D,E,F$ be arbitrary points on the sides of $\triangle ABC$ and $D',E',F'$ the reflections of $D,E,F$ on the midpoints of the respective sides. Then the Miquel points $M$ of $D,E,F$ and $M'$ of $D',E',F'$ are equidistant from the circumcenter of $ \triangle ABC.$ How can we solve this problem using this lemma? Let $D',E',F'$ be isotonic be reflection of midpoints in the respective side. We will show $I$ is the the Miquel point of $D',E',F'$ wrt $\triangle ABC$. Note this completes the proof due to the Lemma. It suffices to show $I \in \odot(AE'F')$ and others follow analogously. Let $X,Y,Z$ be touch point of incircle with sides $BC,CA,AB$, respectively. We have that $$ AE' + AF' = AB + AC - (AE + AF) = AB + AC - BC = AY+ AZ $$It follows $YE' = ZF'$ with oriented lengths (i.e. exactly one of $D',E'$ lie on segments $AX,AY$, respectively). This means the rotation at $I$ sending $Y$ to $Z$ sends $E'$ to $F'$. So $$ \angle E'IF' = \angle YIZ ~~ \implies ~~ I \in \odot(AE'F') \qquad (\text{since } I \in \odot(AYZ))$$This completes the proof. $\blacksquare$ Remark: This Lemma was also 2014 ELMO Shortlist G11.

Let $\omega$ be circle with center $O$ radius $OI$. we want to show $AFE$ and $BFD$ and $CED$ are concurrent on $\omega$. Let $AEF$ meet $\omega$ at $X$ and $P_1$. $BDF$ meets it at $Y$ and $P_2$ and $CDE$ meets it at $Z$ and $P_3$. with simple angle chasing we conclude that for proving $P_1,P_2,P_3$ are same, we need to show $AX,BY,CZ$ are concurrent on $\omega$. Let $AI,BI,CI$ meet $\omega$ at $S_1,S_2,S_3$. simply $AS_1,BS_2,CS_3$ have that option so we may prove the following claim. Claim $: AFS_1E$ is cyclic. Proof $:$ $P_{\omega}(A) = AO^2 - OI^2 = AS_1.AI$ so $AS_1 = \frac{2Rr}{AI} = \frac{2Rr}{\frac{r}{\sin{A/2}}} = 2R\sin{\frac{A}{2}} = \frac{a}{2\cos{A/2}} = \frac{AF+AE}{2\cos{A/2}}$. Let $AFE$ meet $AI$ at $S_1'$, since $S_1'$ is midpoint of arc $FE$ then $AS_1' = \frac{AF+AE}{2\cos{A/2}}$ so $S_1'$ is $S_1$. Let $AFE$ meet $\omega$ at $P$ for second time. $PS_3I = \angle AS_1P = \angle PEC \implies PECS_3$ is cyclic so $P$ lies on $CES_3D$. with same approach $P$ lies on $BDS_2F$.

It will be automatically true that $AF+AE=BC$. Now let's define $D_1,E_1,F_1$ as the reflections of $D,E,F$ about perpendicular bisectors of $BC,AC,AB$.

Let $I_1,I_2,I_3$ be the reflections of $I$ about perpendicular bisectors of $BC,AC,AB$. (Note the $II_1I_2I_3$ lies on a circle with radius of $OI$. Let $M,N,K$ be the midpoints of $AB,BC,AC$.

Now that point is the $P$. Let us call the intersection of $FI_1,EI_2,DI_3$, a $P'$. If we look back at the solution for Claim 2, it is mentioned that $B,F,D$ and $P'$ are concyclic. Similarly, all the others are concyclic so it is true that $P'$ is in fact the miquel point of $D,E,F$ which also means $P=P'$ and $OP = OP' = OI$.

A simple exercise in believing. First, let $P, Q, R$ be the projections of $I$ onto $BC, CA, AB$, then the problem becomes equivalent to freely shifting $P, Q, R$ along their respective sides by a fixed directed length to yield $P', Q', R'$, then reflecting each of these over the perpendicular bisector of its reflective side to yield the desired problem in $D, E, F$. Now notice that by spiral similarity at $I$, it follows that $(BP'R'I)$ and $(CP'Q'I)$. Let $B', C'$ be the antipodes of $B, C$ on their respective circles, then it follows by taking a coordinate system along the sides of the triangle that the antipodes $B'', C''$ of $B, C$ in $(BFD), (CDE)$ are the reflections of $B', C'$ over $O$ respectively. Furthermore, the projection $I_B$ of $B''$ onto $BI$ must lie on $(BFD)$, and since $B''I_B \perp BI \perp B'I$, it follows that $OI = OI_B$. Now let $K_1$ be the reflection of $I_B$ over the line between the centers of $(BFD)$ and $(O, OI)$. Similarly, define everything for $C$ to yield $K_2$. I claim $K_1 = K_2$ is the desired intersection point. Notice that the final reflection is a reflection over the vector $\frac{B''+B}{2} - O = \frac{2O - B' + B}{2} - O = \frac{B- B'}{2}$. So the reflection direction is given by $BB'$. The first reflection vector is parallel to $B'I$, so it follows that the combined reflection from $O$ as an action on $I$ yielding $K_1$ is a rotation with a directed angle (this one over 360 degrees) of $2\angle(BB', B'I) = 2\angle BP'I$. Notice that the action for $K_2$ is a rotation with a directed angle (this one over 360 degrees) of $2\angle(CC', C'I) = 2\angle CP'I$. Since the two directed angles are equal, it follows that $K_1 = K_2$, and thus the resulting point lies on all three circles $(O, OI), (BFD), (CDE)$ as desired.

Let $A'$, $B'$, $C'$ be the feet of the altitude from $I$ to $BC$, $CA$, and $AB$. Let $D'$, $E'$, $F'$ be the isotomic conjugates of $D$, $E$, $F$. Let $D''$, $E''$, $F''$ be the reflection of $I$ across the perpendicular bisectors of $BC$, $CA$, and $AB$. Clearly, \[AF'+AE'=AB+AC-BF'-CE'=AB+AC-AF-AE=AB+AC-BC=AB'+AC'\]so $F'C'=B'E'$. Additionally, $IB'=IC'$ so $\angle IF'C' = \angle IE'B'$, which implies $AE'IF'$ is cyclic. Note that $I$ is the Miquel point of $D'$, $E'$, $F'$. Note that $OI=OD''$ since $O$ is on the perpendicular bisector of $BC$. Similarly for others. Note that \[\angle ID''D=\angle ID'B=\angle IE'C=\angle IE''E\]so $D''D$ and $E''E$ intersect on the circle $(D''E''F'')$. Similarly, $F''F$ also concurs with those. Clearly, $P$ is the intersection, so we are done.

This problem examines the nature of BEVAN points and it has fundamental significance for solving Pure Geometry Problem 3002 (Root axis IG),but I haven't solve Root axis IG yet

What a wonderful intuitive geometry problem! Here's a sketch of my solution: The side condition implies $AE+AF=BC$. Let $\angle A=2a$ and $Pow Q$ denote the power of point $Q$ wrt $(ABC)$. Main claim: Each of the three circles has a fixed point. Take the intersections $X,Y,Z$ of the respective bisectors of the three angles of $A,B,C$ with the circles. Then, it can be shown with Ptolomeu on $AEFX$ that $XA=BC/2cos a$, and thus $X$ is fixed, and it is analogous for $Y,Z$. Then, bash length implies that $Pow I=Pow X$ (using the midpoint $U$ of arc $BC$ we may calculate $XA.XU$ and see it equals $IA.IU$). Thus $OX=OI=OY=OZ$ (analogously). But now, to see that $PXYZ$ is cyclic we invert wrt $P$ and bash with Menelaus theorem on triangle $D'E'F'$, giving us that $X',Y',Z'$ are collinear using the facts that $XE=XF,YD=YF,ZE=ZD$. Actually, this last cyclicity is true for any point P inside ABC (it was a problem of Brazil Como Sur Test 2019). Thus, we're done! Idk what motivated me to take the fixed points except for training and seeing other geometry problems with random points on sides (I didn't use ggb :0)

We start by letting $\gamma$ be the circle with center $O$ and radius $OI$. The proof is based on one lemma : Lemma : Let $ABC$ be a triangle, and let $S$ be the midpoint of the arc $BC$ not containing $A$. If $\alpha=\angle BAC$, $b=CA$ and $c=AB$, then \[AS=\frac{b+c}{2\cdot \cos\frac{\alpha}{2}}\]Proof : Let $a=BC$, and let $I$ be the incenter of triangle $ABC$. Note that if $M$ and $N$ denote the touchpoints of the incircle with sides $AB$ and $AC$, then we have \[AM=AN=\frac{b+c-a}{2}\]It follows that $AI=\frac{b+c-a}{2\cdot \cos\frac{\alpha}{2}}$. Moreover, since $S$ is the circumcenter of triangle $BIC$, we have by applying the law of sines that \[R=SI=\frac{a}{2\cdot \sin\widehat{BIC}}=\frac{a}{2\cdot \sin\left(90+\frac{\alpha}{2}\right)}=\frac{a}{2\cdot \cos\frac{\alpha}{2}}\]The conclusion follows by summing the expressions for $AI$ and $SI$. $\square$ Back to the problem. Using this lemma, we get one big claim : Claim : Circle $\gamma$, line $(BI)$, and circle $(BDF)$ are concurrent at a point $X$. Similarly, circle $\gamma$, line $(CI)$, and circle $(CDE)$ are concurrent at a point $Y$. Proof : We prove the result for $B$, the other will then obviously hold by symmetry. Let $X\ne B$ be the point where line $(BI)$ meets circle $(BDF)$. Applying the lemma, we get \[BX=\frac{BD+BF}{2\cdot \cos \frac{\beta}{2}}=\frac{CA}{2\cdot \cos\frac{\beta}{2}}\]However, note that if we let $S$ be the midpoint of arc $CA$ not containing $B$, we get by law of sines that \[R_{(AIC)}=\frac{CA}{2\cdot \sin\widehat{AIC}}=\frac{CA}{2\sin\left(90+\frac{\beta}{2}\right)}=\frac{CA}{2\cdot \cos\frac{\beta}{2}}\]It follows that $BX=IS$, so $I$ and $X$ have the same power w.r.t $(ABC)$. We thus have $OI=OX$, which is equivalent to $X\in \gamma$. $\square$ It's quite straightforward to conclude from here ; We have, by angle chasing, \[\angle YPX=\angle DPX-\angle DPY=180-\frac{\angle B}{2}-\frac{\angle C}{2}=90+\frac{A}{2}=180-\angle YIX,\]hence $P\in (IYX)=\gamma$, as desired. $\blacksquare$

Here is a cute synthetic finish: As in above solutions, we get $IXYZP$ cyclic, where $X,Y,Z$ are fixed points on $A-$bisector et cetera. Call this circle $\Omega$ We calculate using Ptolomy and LoS: \[Pow(A,\Omega)=AI\cdot AX=2R\cdot \sin(\alpha/2)\cdot 4R \cdot \sin (\beta/2) \sin (\gamma/2)\], which is symmetric in $A,B,C$. This implies that $A,B,C$ have the same power wrt. $\Omega$, so circles $ABC$ and $\Omega$ are concentric and we conclude.

Beautiful problem. Apparently there does exist a clean linpop solution, presented below. First, note that $P$ lies on $(AEF)$ by Miquel point. Claim: The midpoint of arc $DF$ on $(BDF)$ is fixed. Symmetric variants hold. Proof. Let $L$ be the intersection of the interior angle bisector of $\angle B$ and $\overline{AC}$. It suffices to show that $\text{Pow}(L, (BDF))$ is constant. Note that \[ L = \frac{c}{a+c} \cdot A + \frac{a}{a+c} \cdot C, \]so we have by linearity on the function $\text{Pow}(\bullet, (BDF))-\text{Pow}(\bullet, (ABC))$ that \begin{align*} \text{Pow}(L, (BDF)) = \frac{c}{a+c} \cdot \text{Pow}(A, (BDF)) + \frac{a}{a+c} \cdot \text{Pow}(C, (BDF)) + \text{Pow}(L, (ABC)) \\ = \frac{ac(a-BD)+ac(c-BF)}{a+c} - \frac{acb^2}{(a+c)^2} \\ = \frac{ac(a+c-b)}{a+c} - \frac{acb^2}{(a+c)^2}, \end{align*}which is indeed constant. Let $M$ denote the midpoint of arc $DF$ on $(BDF)$ and $N$ denote the midpoint of arc $DE$ on $(CDE)$. Angle chasing gives that \begin{align*} \angle MPN = 360^{\circ} - \angle MPD = \angle NPD = 360^{\circ} - (180^{\circ}-\angle B/2) - (180^{\circ}-\angle C/2) \\ = \angle B/2 + \angle C/2 = 180^{\circ} - \angle BIC \\ = 180^{\circ} - \angle MIN, \end{align*}so $MPNI$ is cyclic. It remains to show that $O$ is the center of $(MIN)$. Since $M$ and $N$ are fixed by the claim, we can take a favorable case (specifically, choose $BD=b/2$) and compute $\text{Pow}(B, (MIN))$. Let $\omega$ denote the circle with center $O$ and radius $OI$. Write \[ \text{Pow}(B, (MIN)) = BM \cdot BI = \frac{b/2}{\cos(\angle B/2)} \cdot \frac{r}{\sin(\angle B/2)} = r \cdot \frac{b}{\sin(\angle B)} = 2Rr = \text{Pow}(B, \omega), \]so $O$ is the center of $(MIN)$, as desired.

Yet another instance of "dang it just draw the circle"... Note that $AF+AE=BC$ as well. Let $AI$ intersect $(AEF)$ again at $X$, and define $Y$ and $Z$ similarly. We will show two important claims in order to show that $I,X,Y,Z,P$ all lie on a circle centered at $O$. Claim 1: $OI=OX$ (and therefore $OY,OZ$ as well). Let $L$ be the arc midpoint of $BC$ on $(ABC)$. Then, it suffices to show that $AX=IL$. By Ptolemy's, $$AX=\frac{(AF+AE)\cdot XE}{FE}=\frac{BC\cdot XE}{FE}.$$As $IL=LB=LC$, it suffices to show $$\frac{BC}{LC}=\frac{FE}{XE},$$which is clear as $\triangle BCL$ is similar to $\triangle FEX$ since they are both isosceles triangles with vertex angle $180-\angle A$. Claim 2: $XYZP$ is cyclic. Invert around $P$, we have $X'$ on $E'F'$ and similarly for the others. We have $$\frac{X'E'}{X'F'}=\frac{\frac{XE}{PX\cdot PE}}{\frac{XF}{PX\cdot PF}}=\frac{XE}{XF}\cdot\frac{PF}{PE}.$$Since $X$ is a arc midpoint, $\frac{XE}{XF}$ is just $1$, and the cyclic product of $\frac{PF}{PE}$ is clearly also $1$. Thus, by Menalaus, $X',Y',Z'$ are collinear, so $XYZP$ are concylic and we are done as due to Claim 1 $(XYZ)$ is the circle centered at $O$ with radius $OI$. remark: this approach is heavily motivated if you actually draw the circle centered at $O$ through $I$ that the problem wants you to show goes through $P$. You'll notice that the circles seem to coincide on the angle bisector, and after you realize the existence of the points $X,Y,Z$ this is quite simple