# 2000 APMO Problems/Problem 3

## Statement

Let $ABC$ be a triangle. Let $M$ and $N$ be the points in which the median and the angle bisector, respectively, at $A$ meet the side $BC$. Let $Q$ and $P$ be the points in which the perpendicular at $N$ to $NA$ meets $MA$ and $BA$, respectively, and $O$ the point in which the perpendicular at $P$ to $BA$ meets $AN$ produced. Prove that $QO$ is perpendicular to $BC$.

## Solution

The problem can be solved by using coordinate geometry.

Let $(0,0)$ be the coordinate of $A$, $y=-mx$ and $y=mx$ be the equation of straight line of $AC$ and $AB$. Let $(b,mb)$, $(c,-mc)$, $(n,0)$ be the coordinate of $B$, $C$ and $N$.

Now the coordinates of $P$ and $M$ are $(n,mn)$ and $(\frac{b+c}{2}, \frac{m(b-c)}{2})$ respectively. If the coordinate of $Q$ is $(n,q)$, then $\frac{q}{n}=\frac{m(b-c)}{b+c}$ so $q=\frac{mn(b-c)}{b+c}$.

Let $(\delta ,0)$ be the coordinate of $O$. $m_{OP}=\frac{mn-0}{n-\delta}$. $OP \perp AB \Rightarrow \frac{mn}{n-\delta}=-\frac{1}{m} \Rightarrow (n-\delta)=-m^2n$ $m_{BC}=\frac{m(b+c)}{b-c}$ and $m_{OQ}=\frac{mn(b-c)}{(b+c)(n-\delta)}=\frac{-(b-c)}{(b+c)m}$.

Clearly, $m_{BC} \times m_{OQ}=-1$. Hence, $QO$ is perpendicular to $BC$. The proof is done.