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

Revision as of 11:17, 20 July 2018

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 $N$ .

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.

@@ Line 19: / Line 19: @@
 === 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, 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.

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