IMO 1968 SL 18

Origin: ITA

Problem

If an acute-angled triangle ABC is given, construct an equilat- eral triangle A′B′C′ in space such that lines AA′, BB′, CC′ pass through a given point.

Solution

The required construction is not feasible. In fact, let us consider the special case \angleBOC = 135◦, \angleAOC = 120◦, \angleAOB = 90◦, where AA′ \capBB′ \cap CC′ = {O}. Denoting OA′, OB′, OC′ by a, b, c respectively we obtain the system of equations a2+b2 = a2+c2+ac = b2+c2+ \sqrt 2bc. Assuming w.l.o.g. c = 1 we easily obtain a3 −a2 −a−1 = 0, which is an irreducible equation of third degree. By a known theorem, its solution a is not constructible by ruler and compass.