IMO 1994 · Problem 2

\(ABC\) is an isosceles triangle with \(AB = AC\). Suppose that

  1. \(M\) is the midpoint of \(BC\) and \(O\) is the point on the line \(AM\) such that \(OB\) is perpendicular to \(AB\);
  2. \(Q\) is an arbitrary point on the segment \(BC\) different from \(B\) and \(C\);
  3. \(E\) lies on the line \(AB\) and \(F\) lies on the line \(AC\) such that \(E, Q, F\) are distinct and collinear.

Prove that \(OQ\) is perpendicular to \(EF\) if and only if \(QE = QF\).

Official text of the problem. Official problem paper (PDF)

Solution

Choose Cartesian coordinates, scaling lengths so that \[ B=(-1,0),\qquad C=(1,0),\qquad A=(0,a),\qquad a>0. \] Then \(M=(0,0)\). Write \(O=(0,o)\). Since \(OB\perp AB\), \[ (1,o)\cdot(1,a)=0, \] so \(o=-1/a\). Thus \[ O=\left(0,-\frac1a\right),\qquad Q=(q,0),\qquad -1<q<1. \]

Take any nonzero direction vector \((u,v)\) of \(EF\). Because \(E,Q,F\) are distinct and collinear, there are distinct nonzero real numbers \(s,t\) such that \[ E=(q+su,sv),\qquad F=(q+tu,tv). \] The equations of \(AB\) and \(AC\) are respectively \[ x-\frac ya=-1,\qquad x+\frac ya=1. \] Consequently, \[ s\left(u-\frac va\right)=-(1+q), \qquad t\left(u+\frac va\right)=1-q. \] Both right-hand sides are nonzero, so both factors \(u-v/a\) and \(u+v/a\) are nonzero. We may therefore calculate \[ \begin{aligned} s+t &=-\frac{1+q}{u-v/a}+\frac{1-q}{u+v/a}\\ &=\frac{-2(qu+v/a)}{u^2-v^2/a^2}. \end{aligned} \] It follows that \[ s+t=0\quad\Longleftrightarrow\quad qu+\frac va=0. \]

On the one hand, \[ QE=|s|\sqrt{u^2+v^2},\qquad QF=|t|\sqrt{u^2+v^2}. \] Since \(s\ne t\), equality of these lengths is equivalent to \(s=-t\), or \(s+t=0\).

On the other hand, \(\overrightarrow{OQ}=(q,1/a)\), so \[ OQ\perp EF \quad\Longleftrightarrow\quad (q,1/a)\cdot(u,v)=0 \quad\Longleftrightarrow\quad qu+\frac va=0. \] Combining these equivalences proves the assertion. The parametrization imposes no restriction on the direction of \(EF\) or on whether \(E,F\) lie on the sides or their extensions. \(\square\)

The solution was checked against John Scholes' IMO solution archive (Kalva).