IMC 2012 · Day 2 · P6

We show that Homer has a winning strategy in both part (a) and part (b).

(a) Notice that the last move is Homer's, and only the last move matters. Homer wins if and only if $f(2012) = 0$, i.e. $$\tag{1} 2012^{2012} + a_{2011} 2012^{2011} + \dots + a_k 2012^k + \dots + a_1 2012 + a_0 = 0.$$ Suppose that all of the coefficients except for $a_k$ have been assigned values. Then Homer's goal is to establish (1) which is a linear equation on $a_k$. Clearly, it has a solution and hence Homer can win.

(b) Define the polynomials $$g(y) = a_0 + a_2 y + a_4 y^2 + \dots + a_{2010} y^{1005} + y^{1006} \quad \text{and} \quad h(y) = a_1 + a_3 y + a_5 y^2 + \dots + a_{2011} y^{1005},$$ so $f(x) = g(x^2) + h(x^2) \cdot x$. Homer wins if he can achieve that $g(y)$ and $h(y)$ are divisible by $y + 1$, i.e.\ $g(-1) = h(-1) = 0$.

Notice that both $g(y)$ and $h(y)$ have an even number of undetermined coefficients in the beginning of the game. A possible strategy for Homer is to follow Albert: whenever Albert assigns a value to a coefficient in $g$ or $h$, in the next move Homer chooses the value for a coefficient in the same polynomial. This way Homer defines the last coefficient in $g$ and he also chooses the last coefficient in $h$. Similarly to part (a), Homer can choose these two last coefficients in such a way that both $g(-1) = 0$ and $h(-1) = 0$ hold.