Haha, you know, one of the contestants got a special prize that year for an outstanding solution. His name is Frits van Zyl. All other solutions given at that contest were analytical, I believe Try to find the "beautiful" solution, it's really cool.

Did you went to that year's contest Arne?

No, but I have been trainer of the South African IMO and PAMO (Pan African) teams this year, that's why I know the story. I went to IMO 2003 and 2004 myself.

What medal did you get at the IMO, Arne? And this problem. I never know how to solve the " fixed point " type of problems. They really confuses me...so can anyone please tell me some general ways of proving "line pass a fixed point" and "circle passes a fixed point" kinds of prolems? For this problem, I constructed the perpendicular bisector of AB, since when C and D are very close to A and B, we will include AB as one of the lines. And then I found out that it seemed $\angle{APB}=90^{o}$. The geometer's sketchpad helped me to confirm it. Thus this problem becomes much easies. Construct as I show in the figure: The perpendicular line meet the circumcircle of $\triangle{OAB}$ at $P$. Then it is well-known that $PC=PD$. Also $\angle{PAC}=\angle{PBD}$ is trivial. Since we are gived $AC=BD$, so $\triangle{PAC} \cong \triangle{PBD}$, which gives $PA=PB$. Since $\angle{APB}=\angle{AOB}=90^{o}$, so $P$ is the intersection of the perpendicular bisector of $AB$ and the circumcircle of $\triangle{AOB}$, which is obviously "fixed".

Attachments:

Oh I made a little mistake in my proof: My three conditions of congruence only holds if $\angle{ACP}, \angle{BDP} > 90^{o}$. Maybe this one is better: $P, C, D, O$ are cyclic because of $\angle{CPD}=\angle{COD}=90^{o}$. So $\angle{PCO}=\angle{PDO}$. Hence: $\angle{PCA}=\angle{PDB}$. This time it should be perfect.

A nice solution. I got silver

Take $C,D$ on $AO$ and $BO$ such that $AC=BD$,now let the perpendicular bisector of $AB$ and $CD$ intersect at $P$.now take two points $X,Y$ at $AO$ and $BO$ such that $AX=BY$,from triangles $PAX$ and $PBY$ we get $AP=BP$,$AX=BY$ and ${\angle}PAX={\angle}PBY$,hence we get $XP=YP$,so $P$ is a point on perpendicular bisector of $XY$,hence we get the result.

is that necessary for AOB to be a right triangle?!! i think the bisector of CD passes through a fixed point anyway...(which is the midpoint of the arc AOB in circumcircle of the triangle)

I believe even Cartesian coordinates can do it !!

There is a fixed center of rotation mapping $A\mapsto B$ and mapping line $AO$ to line $BO$. as this rotation also maps $C\mapsto D$, it lies on perpendicular bisector of $CD$. $\square$