We can prove that these 6 points are on a circle whose center is the midpoint of the orthocenters of ABC and DEF.

CZRorz wrote: We can prove that these 6 points are on a circle whose center is the midpoint of the orthocenters of ABC and DEF. And then, complex numbers makes it a number 1 problem.

It is enough to prove that $B_{EF}, A_{EF}, E_{AB}E_{AC}\cap F_{AB}F_{AC}, A_{EF}A_{ED}\cap B_{EF}B_{FD}$ are cyclic (because from similar facts we get that $E_{AB}E_{AC}\cap F_{AB}F_{AC}$ is Miquel point of $EF, A_{FD}A_{DE}, B_{DE}B_{EF}, C_{EF}C_{FD}$ and so $E_{AB}E_{AC}\cap F_{AB}F_{AC}$ lies on the circumircle of a triangle formed by lines $A_{EF}A_{DE}, B_{EF}B_{DE}, C_{EF}C_{DE}$). In fact we will prove that $E_{AB}, F_{AB}, B_{EF}, A_{EF}, E_{AB}E_{AC}\cap F_{AB}F_{AC}, A_{EF}A_{ED}\cap B_{EF}B_{FD}$ are cyclic. It's well-known that $E_{AB}, F_{AB}, B_{EF}, A_{EF}$ cyclic, so it's enough to prove that $E_{AB}E_{AC}\cap F_{AB}F_{AC}\in (E_{AB}F_{AB}B_{EF}A_{EF})$. And it is some angle chasing : we know that $E_{AB}E_{AC}, F_{AB}F_{AC}$ goes through $E_{AB}, F_{AB}$ and angle between them can be computed. Done P.S. I think that this problem has been already posted on mathlinks.

toto1234567890 wrote: And then, complex numbers makes it a number 1 problem. This is weird point here: According to marking scheme of this exam, you cannot receive over 2 points if you don't prove some lemmae for using complex numbers here. There are some compactification issues, I think, and also some policies to recommend synthetic proofs..

IsoLyS wrote: This is weird point here: According to marking scheme of this exam, you cannot receive over 2 points if you don't prove some lemmae for using complex numbers here. There are some compactification issues, I think, and also some policies to recommend synthetic proofs.. Well... I agree. That's why I use complex thingies as the last thing I could do.