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Difference between revisions of "1994 IMO Problems/Problem 2"

(Created page with "Let <math> ABC</math> be an isosceles triangle with <math> AB = AC</math>. <math> M</math> is the midpoint of <math> BC</math> and <math> O</math> is the point on the line <ma...")
 
 
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==Solution==
 
==Solution==
 
Let <math> E'</math> and <math> F'</math> be on <math> AB</math> and <math> AC</math> respectively such that <math> E'F'\perp OQ</math>. Then, by the first part of the problem, <math> QE'=QF'</math>. Hence, <math> Q</math> is the midpoint of <math> EF</math> and <math> E'F'</math>, which means that <math> EE'FF'</math> is a parallelogram. Unless <math> E=E'</math> and <math> F=F'</math>, this is a contradiction since <math> EE'</math> and <math> F'F</math> meet at <math> A</math>. Therefore, <math> E=E'</math> and <math> F=F'</math>, so <math> OQ\perp EF</math>, as desired.
 
Let <math> E'</math> and <math> F'</math> be on <math> AB</math> and <math> AC</math> respectively such that <math> E'F'\perp OQ</math>. Then, by the first part of the problem, <math> QE'=QF'</math>. Hence, <math> Q</math> is the midpoint of <math> EF</math> and <math> E'F'</math>, which means that <math> EE'FF'</math> is a parallelogram. Unless <math> E=E'</math> and <math> F=F'</math>, this is a contradiction since <math> EE'</math> and <math> F'F</math> meet at <math> A</math>. Therefore, <math> E=E'</math> and <math> F=F'</math>, so <math> OQ\perp EF</math>, as desired.
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==See Also==
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{{IMO box|year=1994|num-b=1|num-a=3}}

Latest revision as of 01:28, 22 November 2023

Let $ABC$ be an isosceles triangle with $AB = AC$. $M$ is the midpoint of $BC$ and $O$ is the point on the line $AM$ such that $OB$ is perpendicular to $AB$. $Q$ is an arbitrary point on $BC$ different from $B$ and $C$. $E$ lies on the line $AB$ and $F$ lies on the line $AC$ such that $E, Q, F$ are distinct and collinear. Prove that $OQ$ is perpendicular to $EF$ if and only if $QE = QF$.

Solution

Let $E'$ and $F'$ be on $AB$ and $AC$ respectively such that $E'F'\perp OQ$. Then, by the first part of the problem, $QE'=QF'$. Hence, $Q$ is the midpoint of $EF$ and $E'F'$, which means that $EE'FF'$ is a parallelogram. Unless $E=E'$ and $F=F'$, this is a contradiction since $EE'$ and $F'F$ meet at $A$. Therefore, $E=E'$ and $F=F'$, so $OQ\perp EF$, as desired.

See Also

1994 IMO (Problems) • Resources
Preceded by
Problem 1 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 3
All IMO Problems and Solutions
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