Prove that the interval $\lbrack 0,1 \rbrack$ can be split into black and white intervals for any quadratic polynomial $P(x)$, such that the sum of weights of the black intervals is equal to the sum of weights of the white intervals. (Define the weight of the interval $\lbrack a,b \rbrack$ as $P(b) - P(a)$.) Does the same result hold with a degree 3 or degree 5 polynomial?
Problem
Source: China TST 1995, problem 6
Tags: quadratics, algebra, polynomial, algebra unsolved
fleeting_guest
20.05.2005 00:21
Beautiful problem. I think that points $(1+cos(\frac{\pi i}{2n}))/2$ define such a partition for polynomials of degree less than $n$.
Singular
26.05.2005 23:53
Can't we just split [0,1] into two intervals? Consider $A = \frac{P(0) + P(1)}{2}$. Consider some horizontal line $y = A$. The arc from $(0, P(0))$ to $(1, P(1))$ must cross this line somewhere, say at ordinate x. It is clear $0 \leq x \leq 1$ and thus [0,x] and [x,1] are intervals with equal weight. ?? how am I reading the question wrong??
fleeting_guest
27.05.2005 07:50
Apparently the order of quantifiers can be different in Chinese and English, this kind of confusion has appeared in several of the China TST problems that were translated here. I think what is required is to find one partition that works for all polynomials of the given degree; "there Exists a splitting for All $p(x)$"). The other interpretation, "for All $p(x)$ there Exists a splitting", trivializes the problem.