Just apply Pascal's Theorem in the cyclic hexagon $AFNPME$.

Yeah, that's what I meant by the title. What if the problem were rephrased as: For acute triangle $ABC$, ... , show that the intersection of lines $ME$ and $NF$ lies on side $BC$. Anyone who lost a mark on IMO2004#1 will understand what I mean.

billzhao wrote: Anyone who lost a mark on IMO2004#1 will understand what I mean. Ouch. I wonder how the students taking the selection test would have done if it had been phrased that way (remembering what happened to the China team last year because of that question)...

Thanks, Yufei, for this nice remainder about Athens 2004: billzhao wrote: What if the problem were rephrased as: For acute triangle $ABC$, ... , show that the intersection of lines $ME$ and $NF$ lies on side $BC$. Well, as a matter of principle, I find such chicanes really annoying, but in this particular case, the problem is just too trivial and this additional assertion makes it somewhat more interesting. So let's solve your problem: It was shown in the original problem that the lines ME, NF and BC are concurrent at some point X. What we have to prove is that this point X lies inside the segment BC, if the triangle ABC is acute-angled. At first, if we have M = E and N = F, then the problem becomes nonsense, since the lines ME and NF are not defined anymore. So we can assume that at least one of the relations M = E and N = F doesn't hold. WLOG, let's say that the relation N = F doesn't hold. Since the point P lies inside the triangle ABC, the point of intersection T of the line CP with the side AB of this triangle lies inside the segment AB. Now, we distinguish two cases: Case 1. The angle < CTA is acute. Case 2. The angle < CTA is obtuse. The case when the angle < CTA is right is impossible; in fact, in this case, we would have N = F, what we have assumed not to hold. So let's consider Case 1. In this case, the angle < CTA is acute. The angle < ACT is also acute, since it is smaller than < ACB, and < ACB is acute since the triangle ABC is acute-angled. Hence, the orthogonal projection N of the point A on the line CT lies inside the segment CT. On the other hand, since the angle < CTA is acute, its complementary angle < CTB is obtuse; thus, the orthogonal projection F of the point P on the line AB lies outside the segment BT. Now, consider the triangle BCT, and the collinear points N, F and X on its sidelines CT, BT and BC. By the Pasch axiom, either two of these points lie inside the respective sides and the third one lies outside, or all three of them lie outside the respective sides. Since the point N lies inside the side CT, the first of these cases must be in existence; but since the point F lies outside the side BT, it follows that the point X lies inside the side BC. This solves the problem in the Case 1. Remains to handle Case 2. In this case, the angle < CTA is obtuse. Its complementary angle < PTB must therefore be acute. The angle < TBP is also acute; in fact, it is smaller than the angle < ABC, which is acute since the triangle ABC is acute-angled. Hence, the orthogonal projection F of the point P on the line AB lies inside the segment BT. Also, since the angle < CTA is obtuse, the orthogonal projection N of the point A on the line CT lies outside the segment CT. Now, look at the triangle BCT, and the collinear points N, F and X on its sidelines CT, BT and BC. By the Pasch axiom, either two of these points lie inside the respective sides and the third one lies outside, or all three of them lie outside the respective sides. Since the point F lies inside the side BT, the first of these cases must be in existence; but since the point N lies outside the side CT, it follows that the point X lies inside the side BC. Hence, the problem is solved in the Case 2, too. Thus, the problem is completely solved. Darij

jaja, I had 6 points in this problem for the same reason. Any coordinators wanted substract another point because I didn't 'prove' that if AB<AC, then AHc>AHb (Hc in AB : <CHcA=90º). It's interesting the Darij's solution of the chinese problem. This prove require many willpower, je.

Pascal go brrrrrrrrrrrrrr By Pascal's Theorem, $FN\cap ME,$ $NP\cap EA,$ and $PM\cap AF$ are collinear. The second and third points are $B$ and $C.$

EMPNFA pascal theorem Very easy problem