Let $M$ be midpoint of $BC$; $P$ be $A$ - Humpty point of $\triangle ABC$; $D$ $\equiv$ $(ABE)$ $\cap$ $BC$; $I$ be Miquel point of $E$, $D$, $F$ with respect to $\triangle ABC$ We have: $\overline{BD} . \overline{BC} = BC^2 - \overline{CD} . \overline{CB} = BC^2 - \overline{CE} . \overline{CA} = \overline{BF} . \overline{BA}$ Then: $A$, $F$, $D$, $C$ lie on a circle So: $(IB; IC) \equiv (IB; ID) + (ID; IC) \equiv (FB; FD) + (ED; EC) \equiv (CA; CB)$ $+ (BC; BA) \equiv (PB; PC)$ (mod $\pi$) or $B$, $C$, $P$, $I$ lie on a circle Hence: $(ID; IP) \equiv (ID; IB) + (IB; IP) \equiv (FD; FB) + (CB; CP) \equiv (CB; CA)$ $+ (AC; AM) \equiv (MB; MA)$ (mod $\pi$) or $D$, $M$, $P$, $I$ lie on a circle Then: $(PA; PI) \equiv (PM; PI) \equiv (DM; DI) \equiv (DB; DI) \equiv (FB; FI) \equiv (FA; FI)$ (mod $\pi$) or $P$ $\in$ $(AEF)$ So: $(AEF)$ passes through fixed point $P$

K= intersection of Wacf and BC so KEAB is cyclic so we can easily prove /S is intersection of cf and be/ so SFAE is cyclic am i right?

Let $BE$ intersect $(AEF)$ at $X\neq E$, and let $(BXC)$ intersect $(AEF)$ again at $P$. Suppose $(CEX)$ interescts $BC$ again at $D$. It's well-known that $(AFXD)$ is cyclic. Then by repeated applications of PoP, \begin{align*}BC^2&= BF\cdot BA+CE\cdot CA \\ BC^2 &= BD\cdot BC +CE\cdot CA\\ \implies CD\cdot CB&=CE\cdot CA \end{align*}hence $AEDB$ is cyclic. If $CF$ interescts $(AEF)$ at $Y\neq F$, then $CY\cdot CF=CE\cdot CA=CD\cdot CB$ so $BFYD$ would be cyclic too, implying $Y=X$. Hence $F,X,C$ is collinear. Now $\angle BPC=\angle BXC=\angle FXE=180^\circ - \angle BAC$, meaning that $P$ lies on a fixed circle through $B$ and $C$, (in fact, this circle passes through the orthocentre of $ABC$) and $$\angle PAC=\angle BAC-\angle BAP=180^\circ -\angle BXC-\angle FXP=\angle PXB=\angle PCB$$meaning that $(APC)$ is tangent to $BC$. Hence $P$ is the interesction of a fixed circle through $B$ and $C$ with the circle tangent to $BC$ at $C$ passing through $A$. Since $P$ is independent of $E$ and $F$, and always lies on $(AEF)$, $P$ is our desired `fixed point'.

khanhnx wrote: Let $M$ be midpoint of $BC$; $P$ be $A$ - Humpty point of $\triangle ABC$; $D$ $\equiv$ $(ABF)$ $\cap$ $BC$; $I$ be Miquel point of $E$, $D$, $F$ with respect to $\triangle ABC$ We have: $\overline{BD} . \overline{BC} = BC^2 - \overline{CD} . \overline{CB} = BC^2 - \overline{CF} . \overline{CA} = \overline{BE} . \overline{BA}$ Then: $A$, $E$, $D$, $C$ lie on a circle So: $(IB; IC) \equiv (IB; ID) + (ID; IC) \equiv (EB; ED) + (FD; FC) \equiv (CA; CB)$ $+ (BC; BA) \equiv (PB; PC)$ (mod $\pi$) or $B$, $C$, $P$, $I$ lie on a circle Hence: $(ID; IP) \equiv (ID; IB) + (IB; IP) \equiv (ED; EB) + (CB; CP) \equiv (CB; CA)$ $+ (AC; AM) \equiv (MB; MA)$ (mod $\pi$) or $D$, $M$, $P$, $I$ lie on a circle Then: $(PA; PI) \equiv (PM; PI) \equiv (DM; DI) \equiv (DB; DI) \equiv (EB; EI) \equiv (EA; EI)$ (mod $\pi$) or $P$ $\in$ $(AEF)$ So: $(AEF)$ passes through fixed point $P$ You have switched E and F

Solved together with @blastoor Let $K$ be a point on $BC$ such that quadrilateral $BFKC$ is cyclic. From PoP follows that $BF \cdot BA = BK \cdot BC$. Note that: \begin{align*} BA \cdot BF = BC^2 - CE \cdot CA \\ BK \cdot BC = BC^2 - CE \cdot CA \\ CE \cdot CA = CK \cdot BC \end{align*}This implies that quadrilateral $ABKE$ is cyclic. It is easy to see that: $$ \angle BKF = \angle CKE = \angle BAC $$This implies that $K$ is center of spiral similarity sending $\overline{BF}$ to $\overline{EC}$. This implies that $X= BE \cap CF$ lies on $\odot(BFK)$ and $\odot(CKE)$. From Miquel's theorem we have that $X$ also lies on $\odot(AFE)$. Now let $\odot(BXC)$ and $\odot(AFE)$ intersects at point $P \neq X$. We claim that $P$ is $A$ Humpty point, which is clearly fixed. Note that: $$ PBC = \angle PXA = \angle PAB $$$$ PCB = \angle PXB = \angle PAC $$From Humpty 's point definition it follows that $P$ is $A$ Humpty point as desired.

Let the two circles passing through $A$ tangent to $BC$ at $B$ and $C$ respectively intersect again at $P$. It suffices to prove that: For any circle $\omega$ passing through $A$ and $P$, intersecting $AC,AB$ at $E,F$ respectively, then $BA\cdot BF + CA\cdot CE=BC^2$. It is well-known that $P$ lies on the median $AM$, where $M\in BC$. Since the geometric function $f:\mathbb{R}^2 \rightarrow \mathbb{R}$: $$f(T)=\text{pow}_{\omega}(T)-\text{pow}_{(ABC)}(T)$$is linear, $$BA\cdot BF + CA\cdot CE = f(B)+f(C) = 2f(M) = 2(MA\cdot MP - (- BM\cdot MC)) =BC^2.$$QED

Vaijan_Mama wrote: khanhnx wrote: Let $M$ be midpoint of $BC$; $P$ be $A$ - Humpty point of $\triangle ABC$; $D$ $\equiv$ $(ABF)$ $\cap$ $BC$; $I$ be Miquel point of $E$, $D$, $F$ with respect to $\triangle ABC$ We have: $\overline{BD} . \overline{BC} = BC^2 - \overline{CD} . \overline{CB} = BC^2 - \overline{CF} . \overline{CA} = \overline{BE} . \overline{BA}$ Then: $A$, $E$, $D$, $C$ lie on a circle So: $(IB; IC) \equiv (IB; ID) + (ID; IC) \equiv (EB; ED) + (FD; FC) \equiv (CA; CB)$ $+ (BC; BA) \equiv (PB; PC)$ (mod $\pi$) or $B$, $C$, $P$, $I$ lie on a circle Hence: $(ID; IP) \equiv (ID; IB) + (IB; IP) \equiv (ED; EB) + (CB; CP) \equiv (CB; CA)$ $+ (AC; AM) \equiv (MB; MA)$ (mod $\pi$) or $D$, $M$, $P$, $I$ lie on a circle Then: $(PA; PI) \equiv (PM; PI) \equiv (DM; DI) \equiv (DB; DI) \equiv (EB; EI) \equiv (EA; EI)$ (mod $\pi$) or $P$ $\in$ $(AEF)$ So: $(AEF)$ passes through fixed point $P$ You have switched E and F I have edited the solution. Thank you, dear Vaijan_Mama

we will use barycentric coordinates. Let $ABC$ be the reference triangle hence $A(1,0,0),B(0,1,),C(0,0,1)$. Since $E \in AC$, parametrize $E(e,0,1-e)$. We will calculate the length of $CE$. The displacement vector has coordinates $\overrightarrow{EC}(e,0,-e)$.Putting the coordinates to the length formula we get: $$ |EC|^2=b^2e^2\rightarrow |EC|= be$$Similiarly, $|FB|=fc$. Then the condition gives $a^2=c^2f+b^2e$. Then $f= \frac{a^2-b^2e}{c^2}$ WLOG $AB=1$. THen $f=a^2-b^2e$. meaning $F(a^2-b^2e,1+b^2e-a^2,0)$. Now it is trivial to calculate formula for circle $(AEF)$. It is well known that the circle equation is $-a^2yz-b^2xz-c^2xy+(ux+vy+wz)(x+y+z)=0$ Putting $A$ gives $u=0$. Putting $E$ gives $b^2e^2-we=0$ or $w=b^2e$.Puttinf $F$ gives: \begin{align*} -&c^2(a^2-b^2e)(1+b^2e-a^2)+((1-b^2e-a^2)(w))=0 \\ -&c^2(a^2-b^2e)+v=0 \\ &b^2e-a^2+v=0 \\ &v= a^2-b^2e \end{align*}Therefore $(AEF):-a^2yz-b^2xz-xy+((a^2-b^2e)y+b^2ez)(x+y+z)=0$ Now we have to prove that there is some $x,y,z$ such that this circle equation is right for all $e \in (0,1)$.Obviously, this meants that for different values of $e$, say $e_i,e_j$, the Below is true: $$(a^2-b^2e_i)y+b^2e_iz=(a^2-b^2e_j)y+b^2e_jz$$Let $e_i=\frac{1}{2}$,$e_j=\frac{1}{3}$. This gives $y=z$ which means that if fixed point exists,fixed point is on line $AM$ where $M$ is midpoint of $BC$.Let this fixed point be $T$. From the last fact, one may guess $T$ is $A-HM$ point, and use synthetic methods to finish it, or using the fact $y=z$, we can keep barybash. Parametrize $T(1-2t,t,t)$.we can show $t$ in terms of $a,b,1$ (note that $c=$) by putting it to the circle equation.Since it is shown by $a,b,c$ only, it is fixed point and we are done. \begin{align*} &-a^2t^2-(1-2t)(t)(b^2+1)+(ta^2)=0 \\ &a^2t-(1-2t)(b^2+1)+a^2=0 \\ &t=\frac{1+b^2-a^2}{2+a^2+2b^2} \end{align*}which is need $A-HM$ point and is fixed. BMOSL 2017 G2 is the same as this one?

Let $M$ be the midpoint of $BC$ and $X=AM\cap (AEF)$. We claim $X$ is the desired fixed point. To prove our claim it is sufficient to prove the power of $M$ wrt $(AEF)$ is fixed. Let $f:R^2\to R$ such that $f(\bullet)=P(\bullet, (AEF))-P(\bullet, (ABC))$. It is known that $f$ is linear over the plane. Now equivalently it is sufficient to prove $f(M)$ is fixed. $M=(0, 1/2, 1/2)$. So \[ f(M)=\frac {1}{2}[f(C)+f(B)]=\frac{1}{2}(CA\cdot CE+BA\cdot BF)=\frac{1}{2}BC^2 \]which is fixed. Remark. Obviously the problem is true if $BA\cdot BF+CA\cdot CE$ is any constant instead of $BC^2$.

Let $(ABE)$ meet $BC$ again at $D$. Then, the length condition and POP yields $D \in (ACF)$. Now, the problem is equivalent to ELMO SL 2013/G3.

Proving that $(AEB)$ and $(AFC)$ intersect at $BC$ (at some point $D$) is similar to the other solutions. Now, inversion about $A$ with an arbitrary radius leads to a cyclic quad $(A'B'D'C')$, where points $A', B', C'$ are fixed and $D'$ is moving (we are moving point $D$ since it uniquely defines $E$ and $F$). $E' = A'B' \cap C'D', F'=A'C' \cap B'D'$. we wanna show that line $E'F'$ passes through some fixed point. Let $P$ be a pole of $B'C'$ wrt $(A'B'C')$, clearly $P$ is fixed and it is well-known that $P$ lies on $E'F'$. $\blacksquare$