I am preparing for BMO1 this year and found an old BMO1 paper from 1990 with this geometry question.
The diagonals of a convex quadrilateral ABCD intersect at O. The centroids of triangles AOD and BOC are P and Q; the orthocentres of triangles AOB and COD are R and S. Prove that PQ is perpendicular to RS.
After drawing out the diagram I can see a lot of cyclic quadrilaterals, and it looks like in the process of finding the orthocentres R and S, we have found the orthocentres of AOD and BOC too. I can't really find anywhere to start though, please could someone give me a hint on how to start solving this question?
According to the figure below, we just need to show that $RS$ is perpendicular to $KL$, where $K$ and $L$ are the midpoints of $AD$ and $BC$, respectively (notice that $PQ$ is parallel to $KL$). Now, we place the quadrilateral on the coordinate system such that $O=(0,0)$, and the slopes of the lines $BD$ and $AC$ are $a$ and $b$, respectively. In this manner, WLOG, we assume $A=(m,bm)$, $B=(t,at)$, $D=(s,as)$, and $C=(n,bn)$.
Now we can easily compute $R$, which is the intersection of the two lines (in fact, two of the altitudes of $\triangle AOB$) as below: $$y-bm=\frac{-1}{a}(x-m), \\ y-at=\frac{-1}{b}(x-t).$$
Hence; $$R=(\frac{(ab+1)(at-bm)}{a-b},\frac{(ab+1)(m-t)}{a-b}).$$
This immediately gives the fact that we must have (because the corresponding equations are exactly the same):
$$S=(\frac{(ab+1)(as-bn)}{a-b},\frac{(ab+1)(n-s)}{a-b}).$$
Therefore, the slope of the line passing through $RS$ is:
$$\frac{(n-s)-(m-t)}{(as-bn)-(at-bm)}.$$
Now, note that:
$$K=(\frac{m+s}{2}, \frac{as+bm}{2}), \\ L=(\frac{t+n}{2}, \frac{at+bn}{2}).$$
So, the slope of the line passing through $KL$ is: $$\frac{(at+bn)-(as+bm)}{(t+n)-(m+s)}.$$
Now that we have the slopes, it is almost obvious that $RS$ is perpendicular to $KL.$
We are done.