The problem states that if $AA_1, BB_1$ and $CC_1$ are the altitudes of the triangle $\bigtriangleup ABC$ and $\vec{AA_1} + \vec{BB_1} + \vec{CC_1} = 0$ then the triangle is equilateral.
My solution:
WLOG we can assume that $AB \geq BC \geq CA$. This implies that $CC_1 \leq AA_1 \leq BB_1$ (i).From $\vec{AA_1} + \vec{BB_1} + \vec{CC_1} = 0$ we can create a triangle whose edges are the segments $AA_1, BB_1$ and $CC_1$. Furthermore since $AA_1 \perp BC$, $BB_1 \perp AC$ and $CC_1 \perp AB$ this triangle is similar to the triangle $\bigtriangleup ABC$. Hence: $$\frac{CC_1}{AB}=\frac{AA_1}{BC}=\frac{BB_1}{AC} \Longrightarrow CC_1\geq AA_1 \geq BB_1$$ But with (i) this implies that $CC_1=AA_1=BB_1$ and it follows that $AB=BC=CA$.
qed
My problem is that I am looking for something more elegant and easier to understand for people with a basic understanding of vectors (basic properties, addition, scalar multiplication etc).
Possible way: let vectors $a,\ b,\ a-b$ be the sides of the triangle.
Then the two altitudes, corresponding to $a,b$ are $-a+\frac{(a,b)}{b^2}b,\ -b+\frac{(a,b)}{a^2}a$.
Let's find the third as $ta+(1-t)b$: $$(ta+(1-t)b)(a-b)=0$$ $$ta^2-t(a,b)+(1-t)(a,b)-(1-t)b^2=0$$ $$t(a^2-(a,b)-(a,b)+b^2)+(a,b)-b^2=0$$ $$t=\frac{-b(a-b)}{(a-b)^2}$$ And the altitude: $$\frac{-b(a-b)}{(a-b)^2}a+\frac{-a(a-b)}{(a-b)^2}b$$ Let's combine: $$-a+\frac{(a,b)}{b^2}b-b+\frac{(a,b)}{a^2}a+\frac{-b(a-b)}{(a-b)^2}a+\frac{-a(a-b)}{(a-b)^2}b=0$$ We see it's a linear combination of basis vectors $a,b$, so it's coefficients must be $=0$: $$\begin{cases} -1+\frac{(a,b)}{a^2}+\frac{-b(a-b)}{(a-b)^2}=0\\ -1+\frac{(a,b)}{b^2}+\frac{-a(a-b)}{(a-b)^2}=0 \end{cases}$$ Hence we derive $a^2=b^2=(a,b)$
P.S. all the products are dot products.