BMO 2017

Let $H$ be the ortocenter of $ABC$ and $K,L,M$ be the feet of perpendiculars respectively from $A,B,C$ to their opposite sides of $ABC$. Also let $D$ be the intersection point of lines $BE$ and $CF$. From power of point we have $$BA\cdot BM=BC\cdot BK(1)$$ and $$CA\cdot CL=CB\cdot CK(2)$$ Adding (1) and (2) we have: $$CA\cdot CL+BA\cdot BM=BC\cdot BK+CB\cdot CK=BC(BK+CK)=B{C}^2(3)$$ Combining (3) with the problem statement $B{C}^2=BA\cdot BF+CE\cdot CA$ we have: $$BA\cdot BF-BA\cdot BM=CA\cdot CL-CE\cdot CA$$ $$BA(BF-BM)=CA(CL-CE)$$ $$BA\cdot FM=CA\cdot LE$$ $$\frac{LE}{FM}=\frac{AB}{AC}=\frac{BL}{CM}(4)$$



Where the last equality follows from $△AMC\sim △ALB$. Now since $\frac{LE}{FM}=\frac{BL}{CM}$ and $\angle FMC=\angle ELB=90^{\circ }$ we get that triangles $△FMC\sim △ELB$. From this similarity we get $\angle AED=\angle AEB=\angle LEB=\angle MFC=180^{\circ }-\angle AFC=180^{\circ }-\angle AFD$, meaning points $A,D,E,F$ are concylic. Since both pairs $\{E,F\}$ and $\{M,L\}$ satisfy the problem condition, we must have this fixed point we are looking for is the second intersection of the circumcircles around $AFDE$ and $AMHL$. Let this point be $X$. We now prove that $X$ is fixed on the circumcircle of $AMHL$ (which would imply $X$ is fixed). From the concylicity we have $\angle XLE=180^{\circ }-\angle XLA=\angle XMA=\angle XMF$ and $\angle XEL=\angle XEA=180^{\circ }-\angle XFA=\angle XFM$ and from here we get $△XLE\sim △XMF$. This similarity gives us $$\frac{XL}{XM}=\frac{LE}{MF}(5)$$ Now combining (4) and (5) we get $\frac{XL}{XM}=\frac{AB}{AC}$ which is a fixed quantity. Since points $M,L$, the circumcircle of $AML$, and ratio $\frac{XL}{XM}$ are fixed, this implies that point $X$ is fixed.

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Alternative solution.

Let the $D$ be the intersection of $BE$ and $CF$ and let circumcircle of triangle $CFA$ intersect $BC$ at point $G$. From power of point we have $$BG\cdot BC=BF\cdot BA(6)$$

Combining (6) with the problem statement we get $$B{C}^2=BA\cdot BF+CE\cdot CA=BG\cdot BC+CE\cdot CA$$ and from here we get $$CE\cdot CA=BC(BC-BG)=BC\cdot CG$$ (7) implies that E, A, B, G are concyclic as well. This gives us $$\angle GAC=\angle GAE=\angle GBE=\angle CBD$$ and $$\angle BAC-\angle GAC=\angle GAB=\angle GAF=\angle GCF=\angle BCD$$ Adding these two equalities gives us $$\angle BAC=\angle CBD+\angle BCD=180^{\circ }-\angle BDC$$ This implies that A, E, D, F are concyclic. Now let the second intersection of the circumcircles of BDC and AFDE be X. We have $$\angle XAB=\angle XAF=\angle XDF=180^{\circ }-\angle XDC=\angle XBC$$ and $$\angle XAC=\angle XAE=180^{\circ }-\angle XDE=\angle XDB=\angle XCB$$ (8) and (9) imply that BC is tangent to the circumcircles of $△XAB$ and $△XAC$ respectively. Let AX, the radical axis of the two circumcircles, intersect BC at Q. Now we have by power of point $$Q{B}^2=QX\cdot QA=Q{C}^2$$ giving up that AX bisects BC. So X is the point on the median from A to side BC such that $\angle BXC=180^{\circ }-\angle BAC$. This point is unique and we have proven that it is always on the circumcircle of AEDF.