Sedrakyan $H$.

Two parabolas with different vertices are the graphs of quadratic trinomials with leading coefficients $p$ and $q$. It is known that the vertex of each parabola lies on the other parabola. What can $p+q$ be?
We can assume that the vertex of the first parabola is the point $(0,0)$. Let the vertex of the second be $(a, b)$. Then the equations of the parabolas are: $y=p x^{2}$ and $y=q(x-a)^{2}+b$, with $b=p a^{2}$ and $0=q a^{2}+b$. From this, $(p+q) a^{2}=0$. If $a=0$, then $b=$ 0, but the vertices are different. Therefore, $p+q=0$.

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