jmc

First, we can move everything to one side, to get $$a^2+b^2+c^2-ab-ac-bc=0.$$Then $$2a^2+2b^2+2c^2-2ab-2ac-2bc=0.$$We can write this as $$(a-b{)}^2+(a-c{)}^2+(b-c{)}^2=0.$$This forces $a=b=c.$ Thus, the triples that work are of the form $(a,b,c)=(k,k,k),$ where $1\le k\le 100,$ and there are $\boxed{100}$ such triples.