Advanced Geometric Problem & Solution (Twice a week update) Now 4691 problems posted.
<Tri-2025-2269>, hint
IA' cuts BC at P, IP=PA'=r (inradius).
Since IA'=IB'=IC'=2r , circumcircle of A'B'C', centered at I, pass through B.
IB=2r, IP=r. Thus Sin <IBP=r/2r=1/2. <IBP=30 ,<ABC=60.
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