GeometryDifficulty 5.6AIME, harderProve itSouth Africa
Let P be a point inside triangle ABC. Construct the points P1, P2, P3 such that PP1⊥BC, PP2⊥CA, PP3⊥AB and BP3=BP1, CP2=CP1. Prove that AP3=AP2.
Solution
Place the points A, B and C in the complex plane such that P is at the origin. Then we have the following equations: PP1⊥BC⟹P1B−C=Pˉ1Cˉ−Bˉ(3) PP2⊥AC⟹P2A−C=Pˉ2Cˉ−Aˉ(4) PP3⊥AB⟹P3B−A=Pˉ3Aˉ−Bˉ(5) BP1=BP3⟹(B−P1)(Bˉ−Pˉ1)=(B−P3)(Bˉ−Pˉ3)⟹BPˉ3+BˉP3−P3Pˉ3=BPˉ1+BˉP1−P1Pˉ1(6) CP1=CP2⟹(C−P1)(Cˉ−Pˉ1)=(C−P2)(Cˉ−Pˉ2)⟹CPˉ2+CˉP2−P2Pˉ2=CPˉ1+CˉP1−P1Pˉ1(7) Then, AP22−AP32=(A−P2)(Aˉ−Pˉ2)−(A−P3)(Aˉ−Pˉ3)=APˉ3+AˉP3−APˉ2−AˉP2+P2Pˉ2−P3Pˉ3=BPˉ3+BˉP3−CPˉ2−CˉP2+P2P2−2P3Pˉ3(using (4) and (5))=(BPˉ3+BˉP3−P3Pˉ3)−(CPˉ2+CˉP2−P2Pˉ2) =(BPˉ1+BˉP1−P1Pˉ1)−(CPˉ1+CˉP1−P2Pˉ1(using (6) and (7))=(B−C)Pˉ1−(Cˉ−Bˉ)P1=0,(using (3)) as required.
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 reproduced verbatim; metadata (topic, difficulty) added by this project.