In triangle ABC point M lays on the BC such that CM/MB=3/2 and point N lays on the AB. AM and CN intersect at point O. AO/OM=5:1. what is the area of the triangle ABC if NBMO quadrilateral's area is equal to 6?
2026-03-31 08:51:57.1774947117
In triangle ABC point M lays on the BC such that CM/MB=3/2
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Let $S_{OMB}=2x$, then $S_{OMC}=3/2 \cdot S_{OMB}=3x$, then $S_{AOC}=5/1\cdot S_{OMC}=15x$, then $S_{AOB}=2/3 \cdot 15x = 10x$. $S_{BOC}=2x+3x=5x$. Then $AN:NB=S_{AOC}/S_{BOC}=15x / (5x)=3$. Then $S_{AON} / S_{BON} = 3$, then $S_{AON} / S_{AOB} = 3 / (3+1) = 3/4$, then $S_{AON}=3/4 \cdot 10x = 15x/2$, then $S_{BON}=10x-15x/2=5x/2$. Then $S_{BMON}=2x+5x/2=9x/2 = 6$, then $x=12/9=4/3$. $S_{ABC}=S_{BOC}+S_{AOC}+S_{AOB}=5x+15x+10x= 30x=40$