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Difference between revisions of "1965 IMO Problems/Problem 5"
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== Problem ==
 
== Problem ==
 +
 
Consider <math>\triangle OAB</math> with acute angle <math>AOB</math>. Through a point <math>M \neq O</math> perpendiculars are drawn to <math>OA</math> and <math>OB</math>, the feet of which are <math>P</math> and <math>Q</math> respectively. The point of intersection of the altitudes of <math>\triangle OPQ</math> is <math>H</math>. What is the locus of <math>H</math> if <math>M</math> is permitted to range over (a) the side <math>AB</math>, (b) the interior of <math>\triangle OAB</math>?
 
Consider <math>\triangle OAB</math> with acute angle <math>AOB</math>. Through a point <math>M \neq O</math> perpendiculars are drawn to <math>OA</math> and <math>OB</math>, the feet of which are <math>P</math> and <math>Q</math> respectively. The point of intersection of the altitudes of <math>\triangle OPQ</math> is <math>H</math>. What is the locus of <math>H</math> if <math>M</math> is permitted to range over (a) the side <math>AB</math>, (b) the interior of <math>\triangle OAB</math>?
 +
  
 
== Solution ==
 
== Solution ==
Line 16: Line 18:
 
a line segment <math>MN , M \in OA , N \in OB</math>.
 
a line segment <math>MN , M \in OA , N \in OB</math>.
 
Second question: the locus consists in the <math>\triangle OMN</math>.
 
Second question: the locus consists in the <math>\triangle OMN</math>.
 +
 +
 +
== 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 <math>\lambda, x, y</math>
 +
involved.  If we manage to conclude that the equation for <math>H</math>
 +
is an equation of degree <math>1</math>, 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 <math>M</math> are expressions of degree <math>1</math> in <math>\lambda</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>.
 +
 +
 +
 +
 +
 +
TO BE CONTINUED.  SAVING MID WAY, SO I DON'T LOSE WORK DONE SO FAR.
 +
  
 
== See Also ==  
 
== See Also ==  
 
{{IMO box|year=1965|num-b=4|num-a=6}}
 
{{IMO box|year=1965|num-b=4|num-a=6}}

Revision as of 15:58, 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 $x, y$ whose constant term is an expression of degree $1$ in $\lambda$.



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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