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Difference between revisions of "1959 IMO Problems/Problem 5"

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(IMO box)
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== Problem ==
 
== Problem ==
  
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An arbitrary point <math>\displaystyle M </math> is selected in the interior of the segment <math>\displaystyle AB </math>.  The squares <math>\displaystyle AMCD </math> and <math>\displaystyle MBEF </math> are constructed on the same side of <math>\displaystyle AB </math>, with the segments <math>\displaystyle AM </math> and <math>\displaystyle MB </math> as their respective bases.  The circles about these squares, with respective centers <math>\displaystyle P </math> and <math>\displaystyle Q </math>, intersect at <math>\displaystyle M </math> and also at another point <math>\displaystyle N </math>.  Let <math>\displaystyle N' </math> denote the point of intersection of the straight lines <math>\displaystyle AF </math> and <math>\displaystyle BC </math>.
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An arbitrary point <math>M </math> is selected in the interior of the segment <math>AB </math>.  The squares <math>AMCD </math> and <math>MBEF </math> are constructed on the same side of <math>AB </math>, with the segments <math>AM </math> and <math>MB </math> as their respective bases.  The circles about these squares, with respective centers <math>P </math> and <math>Q </math>, intersect at <math>M </math> and also at another point <math>N </math>.  Let <math>N' </math> denote the point of intersection of the straight lines <math>AF </math> and <math>BC </math>.
  
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(a) Prove that the points <math>\displaystyle N </math> and <math>\displaystyle N' </math> coincide.
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(a) Prove that the points <math>N </math> and <math>N' </math> coincide.
  
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(b) Prove that the straight lines <math>\displaystyle MN </math> pass through a fixed point <math>\displaystyle S </math> independent of the choice of <math>\displaystyle M </math>.
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(b) Prove that the straight lines <math>MN </math> pass through a fixed point <math>S </math> independent of the choice of <math>M </math>.
  
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(c) Find the locus of the midpoints of the segments <math>\displaystyle PQ </math> as <math>\displaystyle M </math> varies between <math>\displaystyle A </math> and <math>\displaystyle B </math>.
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(c) Find the locus of the midpoints of the segments <math>PQ </math> as <math>M </math> varies between <math>A </math> and <math>B </math>.
  
 
== Solutions ==
 
== Solutions ==
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=== Part A ===
 
=== Part A ===
  
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Since the triangles <math>\displaystyle AFM, CBM</math> are congruent, the angles <math>\displaystyle AFM, CBM</math> are congruent; hence <math>\displaystyle AN'B </math> is a right angle.  Therefore <math>\displaystyle N' </math> must lie on the circumcircles of both quadrilaterals; hence it is the same point as <math>\displaystyle N </math>.
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Since the triangles <math>AFM, CBM</math> are congruent, the angles <math>AFM, CBM</math> are congruent; hence <math>AN'B </math> is a right angle.  Therefore <math>N' </math> must lie on the circumcircles of both quadrilaterals; hence it is the same point as <math>N </math>.
  
 
[[Image:1IMO5A.JPG]]
 
[[Image:1IMO5A.JPG]]
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=== Part B ===
 
=== Part B ===
  
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We observe that <math> \frac{AM}{MB} = \frac{CM}{MB} = \frac{AN}{NB} </math> since the triangles <math>\displaystyle ABN, BCN </math> are similar.  Then <math>\displaystyle NM </math> bisects <math>\displaystyle ANB </math>.
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We observe that <math> \frac{AM}{MB} = \frac{CM}{MB} = \frac{AN}{NB} </math> since the triangles <math>ABN, BCN </math> are similar.  Then <math>NM </math> bisects <math>ANB </math>.
  
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We now consider the circle with diameter <math>\displaystyle AB </math>.  Since <math>\displaystyle ANB </math> is a right angle, <math>\displaystyle N </math> lies on the circle, and since <math>\displaystyle MN </math> bisects <math>\displaystyle ANB </math>, the arcs it intercepts are congruent, i.e., it passes through the bisector of arc <math>\displaystyle AB </math> (going counterclockwise), which is a constant point.
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We now consider the circle with diameter <math>AB </math>.  Since <math>ANB </math> is a right angle, <math>N </math> lies on the circle, and since <math>MN </math> bisects <math>ANB </math>, the arcs it intercepts are congruent, i.e., it passes through the bisector of arc <math>AB </math> (going counterclockwise), which is a constant point.
  
 
=== Part C ===
 
=== Part C ===
  
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Denote the midpoint of <math>\displaystyle PQ </math> as <math>\displaystyle R </math>.  It is clear that <math>\displaystyle R </math>'s distance from <math>\displaystyle AB </math> is the average of the distances of <math>\displaystyle P </math> and <math>\displaystyle Q </math> from <math>\displaystyle AB </math>, i.e., half the length of <math>\displaystyle AB</math>, which is a constant.  Therefore the locus in question is a line segment.
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Denote the midpoint of <math>PQ </math> as <math>R </math>.  It is clear that <math>R </math>'s distance from <math>AB </math> is the average of the distances of <math>P </math> and <math>Q </math> from <math>AB </math>, i.e., half the length of <math>AB</math>, which is a constant.  Therefore the locus in question is a line segment.
  
  
 
{{alternate solutions}}
 
{{alternate solutions}}
  
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== Resources ==
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{{IMO box|year=1959|num-b=4|num-a=6}}
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* [[1959 IMO Problems]]
 
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* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=341530#p341530 Discussion on AoPS/MathLinks]
 
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[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]

Revision as of 20:23, 25 October 2007

Problem

An arbitrary point $M$ is selected in the interior of the segment $AB$. The squares $AMCD$ and $MBEF$ are constructed on the same side of $AB$, with the segments $AM$ and $MB$ as their respective bases. The circles about these squares, with respective centers $P$ and $Q$, intersect at $M$ and also at another point $N$. Let $N'$ denote the point of intersection of the straight lines $AF$ and $BC$.

(a) Prove that the points $N$ and $N'$ coincide.

(b) Prove that the straight lines $MN$ pass through a fixed point $S$ independent of the choice of $M$.

(c) Find the locus of the midpoints of the segments $PQ$ as $M$ varies between $A$ and $B$.

Solutions

Part A

Since the triangles $AFM, CBM$ are congruent, the angles $AFM, CBM$ are congruent; hence $AN'B$ is a right angle. Therefore $N'$ must lie on the circumcircles of both quadrilaterals; hence it is the same point as $N$.

1IMO5A.JPG

Part B

We observe that $\frac{AM}{MB} = \frac{CM}{MB} = \frac{AN}{NB}$ since the triangles $ABN, BCN$ are similar. Then $NM$ bisects $ANB$.

We now consider the circle with diameter $AB$. Since $ANB$ is a right angle, $N$ lies on the circle, and since $MN$ bisects $ANB$, the arcs it intercepts are congruent, i.e., it passes through the bisector of arc $AB$ (going counterclockwise), which is a constant point.

Part C

Denote the midpoint of $PQ$ as $R$. It is clear that $R$'s distance from $AB$ is the average of the distances of $P$ and $Q$ from $AB$, i.e., half the length of $AB$, which is a constant. Therefore the locus in question is a line segment.


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

1959 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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