Putnam problem of the day, roots of a polynomial I've been lately having fun with some Putnam problems (http://www.math.harvard.edu/putnam/) and I would like to see how todays problem can be solved and for somebody more experienced to check my attempted solution.

Find all real polynomials $p(x)$ of degree $n \geq 2$ for which there exists real numbers $ r_1 < r_2 < ... < r_n$ such that 
  
  
*
  
*$p(r_i) = 0$,  $i = 1,2,...,n,$ and
  
*$p'(\frac{r_i + r_{i+1}}{2}) = 0$,  $i = 1,2,...,n-1$,
  
  
  where $p'(x)$ denotes the derivative of $p(x)$

(messy) attempt involving checking coefficients;
Any polynomial of degree $n$ with roots in $r_1, r_2, ...,r_n$ will have the following form:
$p(x) = c(x-r_1)(x-r_2)...(x-r_n)$
Expanding this we must get that the coefficient of $x^{n-1}$ is:
$-c(r_1 + r_2 + ... + r_n)$. 
This would imply that the coefficient of $x^{n-2}$ of the (monic equivalent) derivative $p'(x)$ is:
$\frac{(1-n)}{n}(r_1 + r_2 + ... + r_n)$ (1)
So the derivative function must have same coefficient in front of $x^{n-2}$, and from description we get that it is supposed to have following form:
$p'(x) = (x-\frac{r_1 + r_2}{2}) (x-\frac{r_2 + r_3}{2})... (x-\frac{r_n-1 + r_n}{2})$
Expanding this will yield $x^{n-2}$ term with following coefficient:
$-(\frac{r_1}{2} + r_2 + r_3 + ... + r_{n-1} + \frac{r_n}{2})$ (2)
By inspecting equations (1) and (2) we see that they are equal in case of $n = 2$ and never else. So the answer would be the set of polynomials of degree 2?
(btw, how do I make the numeration of equation look nice? :)) 
 A: I don't think you can conclude what you think you can conclude.  You are given
$$
p(x) \propto \prod_{i=1}^{n} (x-r_i)
$$
and
$$
p^{\prime}(x) \propto \prod_{i=1}^{n-1} \left(x - \frac{r_i + r_{i+1}}{2}\right),
$$
since all the roots of each polynomial are accounted for.  By differentiating the first equation you have
$$
p^{\prime}(x) \propto nx^{n-1} - (n-1)\left(\sum_{i}r_i\right) x^{n-2} + O(x^{n-3}),
$$
and by inspection of the second equation you have
$$
p^{\prime}(x) \propto x^{n-1} - \left(\frac{r_1}{2} + r_2 + \dots + r_{n-1} + \frac{r_{n}}{2}\right)x^{n-2} + O(x^{n-3}).
$$
I think that all you can conclude by comparing the coefficients of the $x^{n-2}$ terms is that
$$
\frac{n-1}{n}\sum_{i}r_i = \sum_{i}r_i - \frac{1}{2}(r_1 + r_n),
$$
or that
$$
r_1 + r_n = \frac{2}{n}\sum_{i}r_i = \frac{2}{n-2}\sum_{1<i<n}r_i
$$
for $n>2$ (and the equation is automatically satisfied for $n=2$, regardless of $r_1$ and $r_2$).
A: Here's another attempt :
Because $p$ is of degree $n$ and has $n$ distinct roots, we can write
$$p(x) = \lambda \prod_{i = 1}^n (x-r_i)$$
By differentiating, we have
$$\frac{p'(x)}{p(x)} = \sum_{i = 1}^n \frac{1}{x-r_i}$$
And setting $x = \frac{r_1+r_2}{2}$, we get
$$0 = \sum_{i = 1}^n \frac{1}{\frac{r_1+r_2}{2}-r_i}$$
Notice the first two terms cancel, which leaves
$$0 = \sum_{i = 3}^n \frac{1}{\frac{r_1+r_2}{2}-r_i}$$
But if you assume $n > 2$, since for $i > 2$, we have $r_i > r_2 > \frac{r_1+r_2}{2}$ we would get
$$\sum_{i = 3}^n \frac{1}{\frac{r_1+r_2}{2}-r_i} < 0$$
So the only polynomials that can possibly satisfy your conditions are degree $2$ polynomial with positive discriminant (and they all work).
