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Problems

2010

Croatian Mathematical Olympiad 2010 Problem M-3

Unutar trokuta ABCABC dana je točka PP takva da je

ABP=PCA=13(ABC+BCA).\measuredangle ABP = \measuredangle PCA = \frac{1}{3} (\measuredangle ABC + \measuredangle BCA).

Dokaži da je ABAC+PB=ACAB+PC\frac{|AB|}{|AC| + |PB|} = \frac{|AC|}{|AB| + |PC|}.