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.

Page

Toolbox

Search

2000 APMO Problems/Problem 3

Statement

Solution