There are $8$ different quadratic trinomials written on the board, among them there are no two that add up to a zero polynomial. It turns out that if we choose any two trinomials $g_1(x), g_2(X)$ from the board, then the remaining $6$ trinomials can be denoted as $g_3(x),g_4(x),\dots,g_8(x)$ so that all four polynomials $g_1(x)+g_2(x),g_3(x)+g_4(x),g_5(x)+g_6(x)$ and $g_7(x)+g_8(x)$ have a common root. Do all trinomials on the board necessarily have a common root? Proposed by S. Berlov
Problem
Source: All-Russian MO 2024 9.7
Tags: algebra, algebra proposed
R8kt
23.04.2024 20:48
Must the root be real?
NO_SQUARES
07.06.2024 17:42
R8kt wrote: Must the root be real? In fact, it is not necessary, because the answer in the problem is "no".
sami1618
24.06.2024 05:41
I think this should work Let $\mathcal{L}(a,b,c)$ denote the quadratic trinomial $p(x)$ such that $p(0)=a$, $p(1)=b$, and $p(2)=c$. We can simply choose \begin{align*} p_1(x) &=\mathcal{L}(1,1,3) \\ p_2(x)&=\mathcal{L}(1,1,-3) \\ p_3(x)&=\mathcal{L}(1,-1,4) \\ p_4(x)&=\mathcal{L}(1,-1,-4) \\ p_5(x)&=\mathcal{L}(-1,1,5) \\ p_6(x)&=\mathcal{L}(-1,1,-5) \\ p_7(x)&=\mathcal{L}(-1,-1,6) \\ p_8(x)&=\mathcal{L}(-1,-1,-6) \end{align*}Notice that clearly no two of these sum to $0$ or are identical and the first two do not share a root. If the question wanted trinomials to have all non-zero terms then we could shift all polynomials $p(x)\mapsto p(x+c)$ for a suitable choice of $c$. Then if the first two terms of the generating values of $g_1(x)$ and $g_2(x)$ are the same then just pair up the polynomials from top to bottom. Otherwise in one value one is $1$ and the other is $-1$. The remaining polynomials will have three values $1$ and three values $-1$ so it is easy to pair them up.