An algebra problem by Asher Joy

Expanding the longer equation: $2a^2+ 2b^2 + 2c^2 +2ab + 2bc + 2ac = 2013$ We know that: $2a^2+ 2b^2 + 2c^2 = 2*(a^2+ b^2 + c^2) = 989*2 = 1978$ Therefore $2ab + 2bc + 2ac = 2013-1978 = 35$

With some random thinking, we stumble upon this: $(a+b+c)^2 = a^2 + 2ab + b^2 + 2ac + 2bc + c^2$ ; $(a+b+c)^2 = (a^2 + b^2 + c^2) + (2ab + 2ac + 2bc) = 989 + 35 = 1024.$ Taking the square root we get the desired answer:

$\boxed{a+b+c = 32}$

(oops...forgot a title for the problem!)

This is my solution : ( $\sum_{a,b,c}{f(a,b,c)}$ is the sum of the 3 expressions obtained by permuting circularly a, b and c hence $f(a,b,c)+f(b,c,a)+f(c,a,b)$ ) $(a+b+c)^2 = \sum_{a,b,c} {(a^2+2ab)} = \sum_{a,b,c}{(2a^2+2ab)}-\sum_{a,b,c}{a^2} = 2013-989 = 1024$ . Then $a+b+c = 32$ .

(a+b+c)^2=a^2+b^2+c^2 +2ab +2ac +2bc...=>(a+b+c)^2=989+[a+b]^2+[b+c]^2+[a+c]^2-2[a^2+b^2+c^2]...=>[a+b=c]^2=2013-989=1024....=>[a+b+c]=32

$(a+b)^2+(a+c)^2+(b+c)^2=2013$ $2a^2+2b^2+2c^2+2ab+2ac+2bc=2013$ if we subtact $a^2+b^2+c^2=989$ from this equation we get: $a^2+b^2+c^2+2ab+2ac+2bc=2013-989$ $989+2ab+2ac+2bc=1024$ $2ab+2ac+2bc=1024-989=35$ Recall that: $(a+b+c)^2=(a^2+b^2+c^2)+(2ac+2bc+2ab)$ $a+b+c=\sqrt{989+35}$ $a+b+c=\sqrt{1024}=\boxed{32}$