The Young Scientist and the Old Scientist play the following game, taking turns in an alternating fashion, with the Young Scientist starting first. The player on turn fills in one of the stars in the equation \[ x^4 + *x^3 + *x^2 + *x + * = 0 \] with a positive real number. Who has a winning strategy if the goals of the players are: a) the Young Scientist - to make the resulting equation have no real roots, and the Old Scientist -- to make it have real roots? b) the Young Scientist - to make the resulting equation have real roots, and the Old Scientist -- to make it have none?
Problem
Source: XIII International Festival of Young Mathematicians Sozopol 2024, Theme for 10-12 grade
Tags: algebra, polynomial
10.09.2024 22:27
Probably one of my favourite problems from Sozopol (if not my favourite). Sadly we didn't get the opportunity to present this cool solution (by @GeoGuesrr and me). Here I will show a generalized solution for degree $n$ (where $n$ is even). Denote the polynomial to be $P(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_{0}$, the polynomial after the $n$-th move to be $P_n(x)$ and the players to be $A$ and $B$. a) Answer: Player B wins. Proof: Case 1: At the last move player B is left to choose $a_m$, where $m$ is odd. Then we want to find $a_m$ s.t. $f(x)=\frac{P_{n-1}(x)}{x^m}=-a_m$. Taking $x\rightarrow -\infty$ we know that $f(x)\rightarrow -\infty$, meaning there $\exists x_1$ s.t. $f(x_1)<0$, so choosing $a_m=-f(x_1)$ we get our desired root. Case 2: Player $B$ is left to choose $a_0$ at the end. We want to find $a_0$ s.t. $P_{n-1}(x)=-a_0$, which is just asking can we find a negative value of $P_{n-1}$. But this is obviously true, because $0$ is a single root of this polynomial. Thus the strategy for Player $B$ is just to delay taking any of the above mentioned coefficients and as they are $\frac{n}{2}+1$, it is possible to insure that he's left with choosing such a value. b) Answer: Player B wins. Proof: We consider $x<0$ as they are no non-negative roots. If Player $B$ is left to choose $a_m$, where $m$ is even. He then wants to insure that for no $x$, $f(x)=\frac{P_{n-1}(x)}{x^m}=-a_m$, which is $\Leftrightarrow$ to proving that $P$ is not surjective. If FTSOC we assume that the last is impossible, then this means that $\sup(f)=-\infty$ in $(-\infty, 0)$. As $f(x)\rightarrow +\infty, x\rightarrow -\infty$ and $f(x)\rightarrow +\infty ,x\rightarrow 0^-$ and $f$ is continuous in our interval it is possible to show that $\sup(f)\neq-\infty$ (Doing this rigorously was quite difficult) from which it follows that we can find a desired value for $a_m$. This means that if $B$ only chooses odd degrees, $A$ must only choose even ones. But now we will show that after an odd number of moves $k$, it is possible for $B$ to choose an odd coefficient $m$ such that if $P_k(x)$ has no roots, then $P_{k+1}(x)$ also has no roots. We want to show that $f(x)=\frac{P_k(x)}{x^m}$ is not surjective for $(0,-\infty)$. We know that $f$ has no roots so by IVT we have either have $\inf(f)<0\Rightarrow f$ is not surjective on our interval or $\inf(f)=0$. Using the fact that $f$ is continuous in $(-\infty,0)$ one can again show that $\inf(f)=0 \Rightarrow f$ has real roots (I'll again omit the details). Thus one can conclude that as $P_0$ has no real root (in our interval), then if $B$ always chooses odd degrees he can insure that no polynomial $P_i$ has a real root (When $A$ adds an even degree he obviously can't make a new real root), from which we are done.