IMO 1967 LL SWE53

Origin: SWE

Problem

In making Euclidean constructions in geometry it is permit- ted to use a straightedge and compass. In the constructions considered in this question, no compasses are permitted, but the straightedge is as- sumed to have two parallel edges, which can be used for constructing two parallel lines through two given points whose distance is at least equal to the breadth of the ruler. Then the distance between the parallel lines is equal to the breadth of the straightedge. Carry through the following constructions with such a straightedge. Construct: (a) The bisector of a given angle. (b) The midpoint of a given rectilinear segment. (c) The center of a circle through three given noncollinear points. (d) A line through a given point parallel to a given line.

Solution

(a) We can construct two lines parallel to the rays of the angle, at equal distances from the rays. The intersection of these two lines lies on the bisector of the angle. (b) If the length of a segment AB exceeds the breadth of the ruler, we can construct parallel lines through A and B in two different ways. The diagonal in the resulting rhombus is the perpendicular bisector of the segment AB.

If the segment AB is too short, we can construct a line l parallel to AB and centrally project AB onto l from a point C chosen sufficiently close to the segment, thus obtaining an arbitrarily long segment A′B′ \parallel AB. Then we construct the midpoint D′ of A′B′ as above. The line D′C intersects the segment AB at its midpoint D. By means of lines parallel to DC the segment AB can be prolonged symmetrically, and then the perpendicular bisector can be found as above. (c) follows immediately from part (b). (d) Let there be given a point P and a line l. We draw an arbitrary line through P that intersects l at A, and two lines l1 and l2 parallel to AP, at equal distances from AP and on either side of AP. Line l1 intersects l at B. We can construct the midpoint C of AP. If BC intersects l2 at D, then PD is parallel to l.