IMO 2014/4

Points PP and QQ lie on side BCBC of acute-angled triangle ABCABC so that PAB=BCA \angle PAB = \angle BCA and CAQ=ABC \angle CAQ = \angle ABC. Points MM and NN lie on lines APAP and AQAQ respectively, such that PP is the midpoint of AMAM, and QQ is the midpoint of ANAN. Prove that lines BMBM and CNCN intersect on the circumcircle of triangle ABCABC.

#Geometry #IMO

Note by Calvin Lin
6 years, 11 months ago

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Comments

I'll give a solution using complex numbers, hopefully no mistakes were made. The triangles QCA and ACB are similar, so (C-Q)/(A-Q)=[(C-A)/(B-A)]£, where £ refers to complex conjugation. Also, triangles ABC, PBA are similar, so (A-P)/(B-P)=[(C-A)/(B-A)]£ . But it is obvious that M-P=P-A and N-Q=Q-A, using these relations we get that (C-Q)/(N-Q)=(M-P)/(B-P), which means that the triangles CQN and MPB are similar. Now we observe that angle(ABM) +angle(ACN)=180 degrees, which means that ABCD is cyclic, so obviously D is on the circumcircle of the triangle ABC.

Adrian Stefan - 6 years, 11 months ago

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Alternatively, you could note that BPMCQN\triangle BPM \sim \triangle CQN and then perform a not too tedious trig bash.

Sreejato Bhattacharya - 6 years, 11 months ago

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My friend said that he solved it with Barycentric coordinates in 5 minutes :O

Zi Song Yeoh - 6 years, 11 months ago
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