On phone right now so can't type out all the details but the mysterious fixed point is actually the HM point corresponding to vertex $A$ of $\triangle ABC$.

babu2001 wrote: On phone right now so can't type out all the details but the mysterious fixed point is actually the HM point corresponding to vertex $A$ of $\triangle ABC$. You are definitely true,my friend!Solution is just applying spiral similarity!

Yeah, the fixed point is HM-point. There is an easy proof with bary bash: We easily get that $P=(-a^2,b^2+c^2-a^2,b^2+c^2-a^2) $ which we are done since it doesnot depend on $E,F$. Btw the spiral similarity solution is similar to Allen Liu's ELMO SL problem

Actually the question's been stated pretty dumbly, for the circles to pass through a fixed point you just need $AB\times FB+AC\times EC$ to be constant. Moreover this fixed point always lies on the $A$-median. Heres a power bash. Label $\odot (AEF)$ as $\Gamma$ and midpoint of $BC$ as $M$. Then $AB\times FB+AC\times EC=\text{pow}(B,\Gamma)+\text{pow}(C,\Gamma)$ so $\text{pow}(M,\Gamma)$ is constant so all circles intersect at a point on $AM$.

tenplusten wrote: Yeah, the fixed point is HM-point. There is an easy proof with bary bash: We easily get that $P=(-a^2,b^2+c^2-a^2,b^2+c^2-a^2) $ which we are done since it doesnot depend on $E,F$. Btw the spiral similarity solution is similar to Allen Liu's ELMO SL problem @above:It is also possible to draw the same circles as ELMO's

Note that the given condition is equivalent to $AE.AC+AF.AB=AB^2+AC^2-BC^2$. Let $H$ be the orthocentre of $\triangle ABC$ and $D=AH\cap BC$. Then perform $\sqrt{AH.AD}$ inversion. Let point $P$ get mapped to $P'$ under this inversion. Then we want to show that $E'F'$ passes through a fixed point. We'll show that the fixed point is midpoint of $BC$. Note that $2AH.AD=AB^2+AC^2-BC^2$. So the given condition becomes equivalent to $\frac{AB}{AF'}+\frac{AC}{AE'}=2$. Then it can be verified easily using Menelaus that $E'F'$ passes through midpoint of $BC$. Q.E.D.

Who can explain, în detail, the elementary solution of this problem? Thanks

Let the circle tangent to $BC$ at $C$ cuts $AB$ at $C'$, the circle tangent to $BC$ at $B$ cuts $AC$ at $B'$. so $B',B$ and $C,C'$ fulfil the condition .if $BF=k BC'$ hence $CE=(1-k) CB'$ thus $B'E=k B'C$ then $\frac{p_{(BB'A)}(E)}{P_{(CC'A)}(E)}=\frac{EB'.EA}{EC.EA}=\frac{EB'}{EC}=-\frac{k}{1-k};$ $\frac{P_{(BB'A)}(F)}{P_{(CC'A)}(F)}=\frac{FB.FA}{FC'.FA}=\frac{FB}{FC'}=-\frac{k}{1-k}$ therefore $E,F$ are on a circle coaxal with $(ABB'),(ACC')$

Absolutely,k=1/2.Otherwise,I do not understand...

The fixed point required is on the median to A, at the intersection of the circle through A and the foots of altitudes from Bând C with the circle through B, H,, C.(H, orthocenter of the triangle ABC).

Solution using Linearity of PoP. Let $\omega=(ABC)$ and $\omega_1=(AEF)$. Let $M$ be midpoint of $BC$ . $M=\frac{1}{2} (B+C)$ motivates us to use Linearity of PoP. So, $\mathbb{P}(M,\omega,\omega_1)=\frac{1}{2}((\mathbb{P}(B,\omega,\omega_1)+\mathbb{P}(C,\omega,\omega_1))$. $\mathbb{P}(B,\omega,\omega_1)=\mathbb{P}(B,\omega)-\mathbb{P}(B,\omega_1)=0-BF.FA$. $\mathbb{P}(C,\omega,\omega_1)=\mathbb{P}(C,\omega)-\mathbb{P}(C,\omega_1)=0-CE.CA$. Thus, $\mathbb{P}(M,\omega,\omega_1)=-\frac{1}{2}BC^2$ which is equal to $\mathbb{P}(M,\omega)-\mathbb{P}(M,\omega_1)=-\frac{1}{4}BC^2-\mathbb{P}(M,\omega_1)$. Thus, $\mathbb{P}(M,\omega_1)$ is constant.So, $\omega_1$ pass through fixed point on $AM$.