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

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[[Image:IMO19595A.png]]
 
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=== Part a) ===
 
=== Part a) ===
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Notice that <math>\angle BAF = \arctan (\frac{MF}{AM}) = \frac{MB}{AM} </math> and <math>\angle ABC = </math>arctan (\frac{MC}{MB}) = \frac{AM}{MB} <math>.
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Notice that <math>\angle BAF = \arctan (\frac{MF}{AM}) = \frac{MB}{AM} </math> and <math>\angle ABC = \arctan (\frac{MC}{MB}) = \frac{AM}{MB} </math>.
  
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</math>\implies <math> </math>\angle BAF + \angle ABC = 90^{\circ} <math>.  
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<math>\implies </math> <math>\angle BAF + \angle ABC = 90^{\circ} </math>.  
  
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Now notice that </math>\angle FNB = 90^{\circ} <math>. Considering </math>\angle BAF <math> as </math>\theta <math>, this gives </math>\angle MBN = 90^{\circ} - \theta <math> and thus </math>\angle MFN = 90^{\circ} + \theta <math>. But notice that </math>\angle MFA = 90^{\circ} - \theta <math>, which means that </math>\angle AFN = 180^{\circ} <math>. Therefore points </math>A, F, N <math> are collinear. Now </math>\angle BNF = 90^{\circ} and \angle ANC = 90^{\circ} <math>. Therefore, </math>\angle BNC = 180^{\circ} <math> and thus points </math>B, N, C $ are collinear. Therefore, AF and BC intersect at N.
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Now notice that <math>\angle FNB = 90^{\circ} </math>. Considering <math>\angle BAF </math> as <math>\theta </math>, this gives <math>\angle MBN = 90^{\circ} - \theta </math> and thus <math>\angle MFN = 90^{\circ} + \theta </math>. But notice that <math>\angle MFA = 90^{\circ} - \theta </math>, which means that <math>\angle AFN = 180^{\circ} </math>. Therefore points <math>A, F, N </math> are collinear. Now <math>\angle BNF = 90^{\circ} </math> and <math>\angle ANC = 90^{\circ} </math>. Therefore, <math>\angle BNC = 180^{\circ} </math> and thus points <math>B, N, C </math> are collinear. Therefore, AF and BC intersect at N.
  
 
=== Part b) ===
 
=== Part b) ===
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=== Part c) ===
 
=== Part c) ===
 
Let the midpoint of <math>PQ </math> be <math>R </math>. Let <math>G </math> be the midpoint of <math>AM </math>. Let <math>F </math> be the midpoint of <math>MB </math>. Let <math>I </math> be the foot of the perpendicular from <math>R </math> onto <math>AB </math>. Therefore by Midpoint Theorem, <math>RI = \frac{PG + HQ}{2} = \frac{AG + HB}{2} = \frac{AB}{4} </math>. Therefore the distance <math>RI </math> is a constant and thus the locus is a straight line parallel to <math>AB </math> at a distance (to <math>AB </math>, of course) of <math>\frac{AB}{4} </math>.
 
Let the midpoint of <math>PQ </math> be <math>R </math>. Let <math>G </math> be the midpoint of <math>AM </math>. Let <math>F </math> be the midpoint of <math>MB </math>. Let <math>I </math> be the foot of the perpendicular from <math>R </math> onto <math>AB </math>. Therefore by Midpoint Theorem, <math>RI = \frac{PG + HQ}{2} = \frac{AG + HB}{2} = \frac{AB}{4} </math>. Therefore the distance <math>RI </math> is a constant and thus the locus is a straight line parallel to <math>AB </math> at a distance (to <math>AB </math>, of course) of <math>\frac{AB}{4} </math>.
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%alternate solutions%
 
  
 
== See Also ==
 
== See Also ==

Latest revision as of 07:23, 23 May 2024

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.

Solution 2

IMO19595A.png

Part a)

Notice that $\angle BAF = \arctan (\frac{MF}{AM}) = \frac{MB}{AM}$ and $\angle ABC = \arctan (\frac{MC}{MB}) = \frac{AM}{MB}$.

$\implies$ $\angle BAF + \angle ABC = 90^{\circ}$.

Now notice that $\angle FNB = 90^{\circ}$. Considering $\angle BAF$ as $\theta$, this gives $\angle MBN = 90^{\circ} - \theta$ and thus $\angle MFN = 90^{\circ} + \theta$. But notice that $\angle MFA = 90^{\circ} - \theta$, which means that $\angle AFN = 180^{\circ}$. Therefore points $A, F, N$ are collinear. Now $\angle BNF = 90^{\circ}$ and $\angle ANC = 90^{\circ}$. Therefore, $\angle BNC = 180^{\circ}$ and thus points $B, N, C$ are collinear. Therefore, AF and BC intersect at N.

Part b)

Construct the bisector of arc AB above AB. Call it X. $\angle ANM = \angle ACM = 45^{\circ}$. Now $\angle ANB = 90^{\circ}$ which means N lies on the circle with AB as diameter.

$\implies$ $\angle ANX = \angle ABX = 45^{\circ} = \angle ANM$. Therefore since M and X are on the same side of $N$, $NM$ passes through $X$ wherever we choose $M$ on $AB$.

Part c)

Let the midpoint of $PQ$ be $R$. Let $G$ be the midpoint of $AM$. Let $F$ be the midpoint of $MB$. Let $I$ be the foot of the perpendicular from $R$ onto $AB$. Therefore by Midpoint Theorem, $RI = \frac{PG + HQ}{2} = \frac{AG + HB}{2} = \frac{AB}{4}$. Therefore the distance $RI$ is a constant and thus the locus is a straight line parallel to $AB$ at a distance (to $AB$, of course) of $\frac{AB}{4}$.

See Also

Quadrados e Circulos circunscritos / IMO 1959-#5 Link do vídeo: https://youtu.be/UNcHD5JI6wU

1959 IMO (Problems) • Resources
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Problem 4 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 6
All IMO Problems and Solutions
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