# 1959 IMO, Problem 5

Note: This is the exact same problem #5 of Romania IMO 1959.

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 N1: Let N denote the point of intersection of the straight lines AF and BC: (a) Prove that the points N1 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 by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to former schoolmate Trương Viết Hiệp) (a) Two right triangles AFM and CBM are congruent because their sides are equal AM = CM and MF = MB. Therefore, ∠AFM = ∠CBM but since FM is perpendicular with BM, AF must perpendicular with CB or ∠ANB = 90° as always. So the locus of N is half the circle on top of diameter AB.

AC and FB are the two diagonal lines of the two squares AMCD and MBEF, respectively, and thus they are also diameters of the two circumcircles of those two squares, respective.

Moreover, ∠ANC = 90° and ∠FNB = 90°, therefore, N is on both the circumcircles of those two squares or N1 coincides with N.

(b) As mentioned earlier, the locus of N is the top half circle with diameter AB. ∠ANM = ∠ACM = ∠MNB = ∠MFB = 45°, or ∠ANS = 45° and since N is on the circle, arc AS = ¼ the circumference of the circle with diameter AB and since A is fixed, S is also a fixed point independent of location of M.

(c) Let T be the midpoint of AB. Draw the three perpendicular lines PU, TR and QV to AB. We have: TR = ½( PU + QV) = ¼ (CM + EB) = ¼ AB

Therefore, when M moves between A and B the distance from the midpoint of PQ to AB is constant and is equal to ¼ AB. The locus is a straight segment parallel to AB, ¼ AB away from AB, and its length is equal to ½ AB the leftmost point of the locus is when M->A; the rightmost is when M->B. The segment that connects the midpoint of locus and midpoint of AB is perpendicular to AB.

(There is a dynamic version of the proof.)