Putnam 2009, Problem B-4

Say that a polynomial with real coefficients in two variables, $x, y$, is balanced if the average value of the polynomial on each circle centered at the origin is $0$. The balanced polynomials of degree at most $2009$ form a vector space $V$ over $\mathbb{R}$. Find the dimension of $V$.

Solution:
Any polynomial \( P(x,y) \) of degree at most 2009 can be written uniquely as a sum \( \sum_{i=0}^{2009} P_i(x,y) \) in which \( P_i(x,y) \) is a homogeneous polynomial of degree \( i \). For \( r > 0 \), let \( C_r \) be the path \( (r\cos\theta, r\sin\theta) \) for \( 0 \leq \theta \leq 2\pi \). Put \( \lambda(P_i) = \oint_{C_1} P_i \); then for \( r > 0 \),
\[
\oint_{C_r} P = \sum_{i=0}^{2009} r^i \lambda(P_i).
\]
For fixed \( P \), the right side is a polynomial in \( r \), which vanishes for all \( r > 0 \) if and only if its coefficients vanish. In other words, \( P \) is balanced if and only if \( \lambda(P_i) = 0 \) for \( i = 0, \ldots, 2009 \).

For \( i \) odd, we have \( P_i(-x,-y) = -P_i(x,y) \). Hence \( \lambda(P_i) = 0 \), e.g., because the contributions to the integral from \( \theta \) and \( \theta + \pi \) cancel.

For \( i \) even, \( \lambda(P_i) \) is a linear function of the coefficients of \( P_i \). This function is not identically zero, e.g., because for \( P_i = (x^2 + y^2)^{i/2} \), the integrand is always positive and so \( \lambda(P_i) > 0 \). The kernel of \( \lambda \) on the space of homogeneous polynomials of degree \( i \) is thus a subspace of codimension 1.

It follows that the dimension of \( V \) is
\[
(1 + \cdots + 2010) - 1005 = (2011 - 1) \times 1005 = 2020050.
\]