Problem proposed by Armenia/Australia for the 35th international mathematical olympiad (held in Hong Kong, July 12–19, 1994). $ABC$ is an isosceles triangle with $AB = AC$. Suppose that (i) $M$ is the midpoint of $BC$ and $O$ is the point on the line $AM$ such that $OB$ is perpendicular to $AB$; (ii) $Q$ is an arbitrary point on the segment $BC$ different from $B$ and $C$; and (iii) $E$ lies on the line $AB$ and $F$ lies on the line $AC$ such that $E, Q,$ and $F$ are distinct and collinear. Prove, with Vi`te’s method, that $OQ$ is perpendicular to $EF$ if and only if $QE = QF$ .
Proof:Let $M=(0,0),Q=(r,0),B=(-a,0),C=(a,0),A=(0,b),O=(0,k)$.Then
\begin{equation}\label{eq:28.15.36}(a,k)\cdot (a,b)=0\end{equation}
Let the equation of the line EF be $y=t(x-r)$.The intersection point of the line EQ and AB is
$$E=(\frac{ab+tra}{at-b},\frac{tb(r+a)}{at-b})$$
Similary,the intersection point of the line EF and AC is
$$F=(\frac{tra+ab}{at+b},\frac{tb(a-r)}{at+b})$$
\begin{align*}
EQ^2=f(a)&=(\frac{tra+ab}{at-b}-r)^2+\frac{t^2b^2(r+a)^2}{(at-b)^2}\\&=(\frac{ab+br}{at-b})^2+\frac{t^2b^2(r+a)^2}{(at-b)^2}\\&=\frac{a^2b^2+b^2r^2+2ab^2r+t^2b^2r^2+t^2b^2a^2+2t^2b^2ar}{a^2t^2+b^2-2abt}
\end{align*}
\begin{equation}FQ^2=f(-a)\end{equation}
It is easy to verify that
\begin{align*}
\frac{a^2b^2+b^2r^2+2ab^2r+t^2b^2r^2+t^2b^2a^2+2t^2b^2ar}{a^2t^2+b^2-2abt}=\frac{a^2b^2+b^2r^2-2ab^2r+t^2b^2r^2+t^2b^2a^2-2t^2b^2ar}{a^2t^2+b^2+2abt}
\end{align*}
if and only if
\begin{equation}\label{eq:28.23.31}
t(a^2b+br^2+t^2br^2+t^2ba^2)+(r+t^2r)(a^2t^2+b^2)=0
\end{equation}
\ref{eq:28.23.31} holds if and only if \begin{equation}\label{eq:28.23.32}br^2t+ba^2t+a^2t^2r+b^2r=0\end{equation}
\ref{eq:28.23.32} holds if and only if \begin{equation}\label{eq:28.23.36}(b+rt)(a^2t+br)=0\end{equation}It is easy to verify that $b+rt\neq 0$,so \ref{eq:28.23.36} holds if and only if \begin{equation}a^2t+br=0\end{equation}
$OQ$ is perpendicular to $EF$ if and only if $tk=r$.Because $a^2+kb=0$,and $tk=r$,so $a^2t+rb=0$.