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Difference between revisions of "1965 IMO Problems/Problem 5"
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The equation for <math>MP</math> is an equation of degree <math>1</math> in <math>x, y</math>
 
The equation for <math>MP</math> is an equation of degree <math>1</math> in <math>x, y</math>
−
whose constant term is an expression of degree <math>1</math> in <math>\lambda</math>.
+
with constant coefficients for <math>x, y</math>, and whose constant term
 +
is an expression of degree <math>1</math> in <math>\lambda</math>.
 +
 
 +
The coordinates of <math>P</math> (the intersection of <math>MP</math> and <math>OA</math>) are
 +
expressions of degree <math>1</math> in <math>\lambda</math>.
 +
 
 +
The equation of the perpendicular from <math>P</math> to <math>OB</math> is of degree
 +
<math>1</math> in <math>x, y</math>, with constant coefficients for <math>x, y</math>, and whose
 +
constant term is an expression of degree <math>1</math> in <math>\lambda</math>.  This
 +
is equation (2) in the above solution.
 +
 
 +
Similarly, the equation of the perpendicular from <math>Q</math> to <math>OA</math>
 +
is of degree <math>1</math> in <math>x, y</math>, with constant coefficients for
 +
<math>x, y</math>, and whose constant term is an expression of degree
 +
<math>1</math> in <math>\lambda</math>.  This is equation (1) in the above solution.
 +
 
 +
 
  
  

Revision as of 16:35, 30 October 2024

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, 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 is 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 is equation (1) in the above solution.




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


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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