If $a+b+c=15$ and $a^2+b^2+c^2=83$, then $ab+bc+ca$ equals:

From $(a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)$ we get $225=83+2(ab+bc+ca)$, so $ab+bc+ca=142/2=71$. (142 forgets to halve, 154 adds the squares instead of subtracting them before halving, and 35.5 halves a second time.) The values a, $b,c=3$, 5, 7 fit both conditions, and 15 + 35 + 21 = 71.