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

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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>.
 
(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>.
  
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== Solutions ==
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== Solution ==
  
 
=== Part A ===
 
=== Part A ===
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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>ABN, BCN </math> are similar.  Then <math>NM </math> bisects <math>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, BCM</math> are similar.  Then <math>NM </math> bisects <math>ANB </math>.
  
 
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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{{alternate solutions}}
 
{{alternate solutions}}
  
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{{IMO box|year=1959|num-b=4|num-a=6}}
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== See Also ==
  
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[[Category:Olympiad Geometry Problems]]
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Quadrados e Circulos circunscritos / IMO 1959-#5
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Link do vídeo: https://youtu.be/UNcHD5JI6wU

Revision as of 19:18, 30 March 2021

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 $ABN, BCM$ 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

Quadrados e Circulos circunscritos / IMO 1959-#5

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Link do vídeo: https://youtu.be/UNcHD5JI6wU

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