1959 IMO Problems/Problem 5

Problem

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

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

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

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

Solutions

Part A

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

Part B

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

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

Part C

Denote the midpoint of $\displaystyle PQ$ as $\displaystyle R$ . It is clear that $\displaystyle R$ 's distance from $\displaystyle AB$ is the average of the distances of $\displaystyle P$ and $\displaystyle Q$ from $\displaystyle AB$ , i.e., half the length of $\displaystyle AB$ , 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.

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