1965 IMO Problems/Problem 5

Problem

Consider $\triangle OAB$ with acute angle $AOB$ . Through a point $M \neq O$ perpendiculars are drawn to $OA$ and $OB$ , the feet of which are $P$ and $Q$ respectively. The point of intersection of the altitudes of $\triangle OPQ$ is $H$ . What is the locus of $H$ if $M$ is permitted to range over (a) the side $AB$ , (b) the interior of $\triangle OAB$ ?

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it. Let $O(0,0),A(a,0),B(b,c)$ . Equation of the line $AB: y=\frac{c}{b-a}(x-a)$ . Point $M \in AB : M(\lambda,\frac{c}{b-a}(\lambda-a))$ . Easy, point $P(\lambda,0)$ . Point $Q = OB \cap MQ$ , $MQ \bot OB$ . Equation of $OB : y=\frac{c}{b}x$ , equation of $MQ : y=-\frac{b}{c}(x-\lambda)+\frac{c}{b-a}(\lambda-a)$ . Solving: $x_{Q}=\frac{1}{b^{2}+c^{2}}\left[b^{2}\lambda+\frac{c^{2}(\lambda-a)b}{b-a}\right]$ . Equation of the first altitude: $x=\frac{1}{b^{2}+c^{2}}\left[b^{2}\lambda+\frac{c^{2}(\lambda-a)b}{b-a}\right] \quad (1)$ . Equation of the second altitude: $y=-\frac{b}{c}(x-\lambda)\quad\quad (2)$ . Eliminating $\lambda$ from (1) and (2): $ac \cdot x + (b^{2}+c^{2}-ab)y=abc$ a line segment $MN , M \in OA , N \in OB$ . Second question: the locus consists in the $\triangle OMN$ .

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1965 IMO Problems/Problem 5

Problem

Solution