IX OBM

Given any triangle $ABC$, we have to find the point $P$ inside the triangle, such that any line through $P$ divides the triangle into two pieces which are as equal as possible. We take $P$ to be the centroid $G$. We show first that in that case the smallest piece is at least $\frac{4}{9}$ of the total area. If we take a line through $G$ parallel to one of the sides, then the triangular piece has area $\frac{4}{9}$ of the total area.



Take any other line through $G$, say $RS$ as shown above. We claim that $RG>GS$. Take $S^{\prime }$ to be the point obtained by rotating $S$ through $180^{\circ }$ about $G$. Then $GSE$ and $G{S}^{\prime }D$ are congruent and $SE$ is parallel to $S^{\prime }D$. So $R$ cannot coincide with $S^{\prime }$ (because the lines $SE$ and $RD$ are not parallel, they meet at $A$). If $GR<G{S}^{\prime }$, then $RD$ and $SE$ will meet on the wrong side of $DE$. So we must have $GR>G{S}^{\prime }$ and hence $GR>GS$. The triangles $GRD$ and $GSE$ have the same height ($GD\sin \angle DGR$), so area $GRD>$ area $GSE$. Hence area $ARS>$ area $ADE=\frac{4}{9}$ area $ABC$.

Let $N$ be the midpoint of $AC$. As we move $R$ towards $B$, $S$ moves towards $A$. When $R$ reaches $B$, $S$ reaches $N$. So $S$ lies between $E$ and $N$. Now consider the triangles $BGR$, $NGS$. $BGR$ has the larger base, because $BG=2GN$, and the larger height because $RG>SG$, so $RG\sin \angle RGB>SG\sin \angle SGN$. So area $BGR>$ area $NGS$ and hence area $ABN>$ area $ARS$. So area $ARS<\frac{1}{2}$ area $ABC$. Thus $ARS$ is the smaller piece but its area is bigger than $\frac{4}{9}$ area $ABC$. It remains to show that no other choice of $P$ is better than $G$.



Take lines through $G$ parallel to the sides. That gives three overlapping triangles which cover $ABC$. If $P$ is not at $G$, then it must lie inside at least one of these triangles, say $ADE$. Now take a line $D^{\prime }{E}^{\prime }$ through $P$ parallel to $DE$. Then area $A{D}^{\prime }{E}^{\prime }<$ area $ADE=\frac{4}{9}$ area $ABC$, so $P$ is a worse choice than $G$ (from $A$'s point of view).