Belarusian Mathematical Olympiad

Answer: $ab+bc+ac$. Let $f(x)=\alpha x^2+\beta x+\gamma$. Then $$\alpha a^2+\beta a+\gamma =bc,\alpha b^2+\beta b+\gamma =ac,\alpha c^2+\beta c+\gamma =ab.(1)$$ Subtracting the second and the third equations from the first one, we have $$\alpha (a^2-b^2)+\beta (a-b)=c(b-a),\alpha (a^2-c^2)+\beta (a-c)=b(c-a).$$ Since $a,b,c$ are pairwise distinct, we have $$\alpha (a+b)+\beta =-c,\alpha (a+c)+\beta =-b.(2)$$ Subtracting the second equation from the first one, we have $\alpha (b-c)=b-c$. Since $b-c\ne 0$, we have $\alpha =1$. Then it follows from (2) that $\beta =-(a+b+c)$, and it follows from (1) that $\gamma =ab+bc+ac$. Therefore, $f(x)=x^2-(a+b+c)x+(ab+bc+ac)$, so $f(a+b+c)=ab+bc+ac$.