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

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== Solution ==
 
== Solution ==
  
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=== Part a ===
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=== Part A ===
  
 
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>.
 
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>.
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[[Image:1IMO5A.JPG]]
 
[[Image:1IMO5A.JPG]]
  
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=== Part b ===
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=== Part B ===
  
 
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>.
 
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>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.
 
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.
  
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=== Part c ===
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=== Part C ===
  
 
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.
 
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.

Revision as of 10:01, 30 May 2012

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 $B$.

Solution

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.

See Also

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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