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Difference between revisions of "1965 IMO Problems/Problem 5"
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== Solution ==
 
== Solution ==
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{{solution}}
 
 
Let <math>O(0,0),A(a,0),B(b,c)</math>.
 
Let <math>O(0,0),A(a,0),B(b,c)</math>.
 
Equation of the line <math>AB: y=\frac{c}{b-a}(x-a)</math>.
 
Equation of the line <math>AB: y=\frac{c}{b-a}(x-a)</math>.
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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>.
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 +
== See Also ==
 +
{{IMO box|year=1965|num-b=4|num-a=6}}

Revision as of 23:23, 8 December 2022

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 $\triangle OMN$.

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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