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

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 $\triangle OMN$ .

Solution 2

This solution is a simplified version of the previous solution, it fills in some gaps. and it provides more information. The idea is to just follow the degrees of the expressions and equations in $\lambda, x, y$ involved. If we manage to conclude that the equation for $H$ is an equation of degree $1$ , then we will know that it is a line. We don't need to know the equation explicitly.

Just like in the previous solution, we use analytic (coordinate) geometry, but we don't care how the axes are chosen.

The coordinates of $M$ are expressions of degree $1$ in $\lambda$ .

The equation for $MP$ is an equation of degree $1$ in $x, y$ with constant coefficients for $x, y$ , and whose constant term is an expression of degree $1$ in $\lambda$ .

The coordinates of $P$ (the intersection of $MP$ and $OA$ ) are expressions of degree $1$ in $\lambda$ .

The equation of the perpendicular from $P$ to $OB$ is of degree $1$ in $x, y$ , with constant coefficients for $x, y$ , and whose constant term is an expression of degree $1$ in $\lambda$ . This corresponds to equation (2) in the above solution.

Similarly, the equation of the perpendicular from $Q$ to $OA$ is of degree $1$ in $x, y$ , with constant coefficients for $x, y$ , and whose constant term is an expression of degree $1$ in $\lambda$ . This corresponds to equation (1) in the above solution.

Now, in principle, we would have to solve the system of two equations to obtain the coordinates of $H$ as expressions of $\lambda$ , and then eliminate $\lambda$ to obtain an equation in $x, y$ . Or, as a shortcut, we can eliminate $\lambda$ directly from the two equations. Either way, the result is an equation of degree $1$ in $x, y$ .

This tells us that the locus is on this line. We just need to specify which set of points on this line is the locus.

The previous solution, with a good amount of hand waving, tells us that the solution is "a line segment $B_1A_1, B_1 \in OA, A_1 \in OB$ ". (On top of the hand waving the solution uses the unhappy notation $M$ for $B_1$ and $N$ for $A_1$ , which is bad because $M$ has already been used!) We will do better than that.

Let $A_1$ be the foot of the perpendicular from $A$ to $OB$ , and $B_1$ be the foot of the perpendicular from $B$ to $OA$ . (For this paragraph see the picture shown in Solution 3.) Consider the limit situation when $M = A$ . Then $Q = A_1$ , and $P = A$ . It follows that the intersection $H$ of the perpendiculars from $P$ to $OB$ and $Q$ to $OA$ is $A_1$ . Similarly, the limit situation when $M = B$ yields $H = B_1$ . Now it is reasonable to say that when $M$ moves from $A$ to $B$ , $H$ moves from $A_1$ to $B_1$ . So, the locus is the line segment joining the feet $A_1, B_1$ of the perpendiculars in $\triangle OAB$ from $A, B$ . This answers question (a).

Again with a good amount of hand waving, the previous solution says "the locus consists in the $\triangle OB_1A_1$ ". We justify this by pointing out that if $M$ is inside $\triangle OAB$ , then we can take the triangle $\triangle OA'B'$ , such that $A' \in OA$ , $B' \in OB$ , $A'B'$ going through $M$ and parallel to $AB$ . Then $H$ will be on the corresponding segment $A_1'B_1'$ determined by the feet of the perpendiculars in $\triangle OA'B'$ . Conversely, it is easy to see that any point $H \in \triangle OA_1B_1$ is on a segment $A_1'B_1'$ obtained from a triangle $\triangle OA'B'$ , and $H$ is obtained from a point $M \in A'B'$ . This answers question (b).

(Solution by pf02, October 2024)

Solution 3

TO BE CONTINUED. SAVING MID WAY, SO I DON'T LOSE WORK DONE SO FAR.

1965 IMO (Problems) • Resources
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 6
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1965 IMO Problems/Problem 5

Contents

Problem

Solution

Solution 2

Solution 3

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