Let ABCA′B′C′ be a triangular regular prism, with lateral edges AA′, BB′, CC′. Consider the midpoint D of the edge BC and the parallelogram ADB′E. Let F be the orthogonal projection of the point A′ on the line AE, d be the intersection of the planes (ADE) and (A′CF) and P be the intersection of the line d with the plane (ABC). Prove that P is the baricentre of the triangle ABC if and only if AB=AA′2. Valeriu Bărbieru
Solution
AD is a median in △ABC, so the point P is the baricenter of the triangle ABC if and only if AP=2PD. Since PF∥DE, this is equivalent to AF=2FE. Because AA′∥BB′, BD∥EA′ and BB′⊥BD, the triangle AEA′ has a right angle at A′. Then A′E and A′A are legs of this triangle, therefore AF=2FE⟺AF⋅AE=2EF⋅AE⟺A′A2=2A′E2⟺A′A2=21BC2⟺BC=A′A2.
Want a route through all this instead of an archive? The track
puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.
Source: MathNet,
licensed CC-BY-4.0.
Statement and solution reproduced as published; topic and difficulty added by this site.