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1965 IMO Problems/Problem 5

Revision as of 18:31, 31 October 2024 by Pf02 (talk | contribs)

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.

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

Let $A, B, O$ have coordinates $(a_1, a_2), (b_1, b_2), (c_1, c_2)$, and let $M = (\lambda a_1 + (1 - \lambda b_1, \lambda a_2 + (1 - \lambda b_2)$ with $\lambda \in [0, 1]$. The idea is to just follow the degrees of the expressions and equations in $\lambda, x, y$ involved as we make the computations for obtaining the coordinates of $H$, and the equation of the curve $H$ is on. We will see that the equation for $H$ is an equation of degree $1$, so we will know that it is a line. We don't need to write out the equation explicitly.

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

The equation for $MP$ (the perpendicular from $M$ to $OA$) 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 foot of the perpendicular from $P$ to $OA$) 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 (1) and (2) to obtain the coordinates of $H$ as expressions of $\lambda$, and then eliminate $\lambda$ to obtain the equation in $x, y$ for $H$. As a shortcut, we can eliminate $\lambda$ directly from the two equations (1) and (2). Either way, the result is an equation of degree $1$ in $x, y$.

This tells us that the locus is on a line. We just need to specify which set of points on this line is the locus. And, we want to make the line explicit.

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).

For part (b) of the problem, 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

This solution is elementary, it does not use analytic geometry.

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 $OA$. Construct $H$ as described in the statement of the problem.

Prob 1965 5.png

We will prove that $H \in A_1B_1$.

First, as shown in the second picture, take $MQ \perp OB$, and let $H_1$ be the intersection of $A_1B_1$ with the perpendicular from $Q$ to $OA$. From the triangle $\triangle BAA_1$ we have $\frac{AM}{MB} = \frac{A_1Q}{QB}$ because $AA_1 \parallel MQ$. From the triangle $\triangle A_1BB_1$ we have $\frac{A_1Q}{QB} = \frac{A_1H_1}{H_1B_1}$ because $QH_1 \parallel BB_1$. So, $\frac{AM}{MB} = \frac{A_1H_1}{H_1B_1}$.

Second, as shown in the third picture, take $MP \perp OA$, and let $H_2$ be the intersection of $A_1B_1$ with the perpendicular from $P$ to $OB$. By a similar argument, we have $\frac{AM}{MB} = \frac{A_1H_2}{H_2B_1}$.

It follows that $H_1 = H_2$ since the two points divide $A_1B_1$ in the same ratio. This is the point $H$ from the statement of the problem.

The solution to the problem is now completed by repeating the last two paragraphs from Solution 2.

(Solution by pf02, October 2024)


See Also

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