Squares in Circles

The applet below illustrates Problem 5 from the 1959 International Mathematics Olympiad:

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 circumscribed about these squares, with 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.

Prove that the points N and N' coincide.
Prove that the straight lines MN pass through a fixed point S independent of the choice of M:
Find the locus of the midpoints of the segments PQ as M varies between A and B.

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Solution

Prove that the points N and N' coincide.
Prove that the straight lines MN pass through a fixed point S independent of the choice of M:
Find the locus of the midpoints of the segments PQ as M varies between A and B.

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Solution

Let P and Q be the centers of squares AMCD and MBEF and (P) and (Q) the circumscribed circles.

Triangles ABN' and CBM are right and similar. Their corresponding legs are at right angles; therefore, so are the hypotenuses: AN' ⊥BC. Since ∠AN'C is right and AC is the diameter of (P), N' lies on that circle. Similarly, N' lies on (Q), implying N' = N.

In (P), ∠ANM subtends a quarter circle. Thus ∠ANM = 45°. Similarly, in (Q) ∠BNM = 45°. As we just saw, ∠ANB = 90°, implying that MN is the bisector of the angle at N. MN, therefore passes through the lowest point of the circle with AB as a diameter, proving #2.

Let T be the midpoint of PQ, P₀ and Q₀ projections on AB of P and Q, respectively. P₀ is the midpoint of AM, Q₀ the midpoint of BM. P lies on the diagonal AC of square AMCD, Q on the diagonal BF of square MBEF. It follows that AP₀ = PP₀ and BQ₀ = QQ₀ from which

PP₀ + QQ₀ = AP₀ + BQ₀ = AB/2.

The height of the midpoint T of PQ is the average of PP₀ and QQ₀ which is AB/4, independent of M. Thus if P' and Q' are at the distance AB/4 from AB and lie on the diagonals of the two squares, T ranges over P'Q'. However, note that if M moves outside AB and one of the length AM or BM becomes negative, with the corresponding square is drawn on the other side of AB, T still belongs to the line P'Q' but is no longer located inside the segment P'Q'.

The problem's configuration exhibits additional properties. For example, if U denotes the point of intersection of the two diagonals, then the circle with PQ as diameter passes through M and U.

Note: Part 1 of the problem has been included as #367 in the collection by Barbeau et al.

References

E. J. Barbeau, M. S. Klamkin, W. O. J. Moser, Five Hundred Mathematical Challenges, MAA, 1995.

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