olymmath

JBMO-SL 2011 G4

  • Answer

Point DD lies on the side [BC][B C] of △ABC\triangle A B C. The circumcenters of △ADC\triangle A D C and △BAD\triangle B A D are O1O_{1} and O2O_{2}, respectively and O1O2∥ABO_{1} O_{2} \| A B. The orthocenter of △ADC\triangle A D C is HH and AH=O1O2A H=O_{1} O_{2}. Find the angles of △ABC\triangle A B C if 2m(<C)=3m(<B)2 m(<C)=3 m(<B).

Dataset text NuminaMath olympiads. Quoted with source.

Show the answer

\widehat{A B C}=30^{\circ}, \widehat{A C B}=45^{\circ}, \widehat{B A C}=105^{\circ}

Solutions

No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.

Add a solution

Locator jbmo-sl/2011/g4 · Tag 5GZ0MT2GP3 · Edit this page on GitHub · Keys: j next, k previous