1999 Putnam A-5 as a direct result of Archimedean Property I was working on question A-5 from the 1999 Putnam Exam:
Prove that there is a constant C such that, if $p(x)$ is a polynomial of degree 1999, then
$|p(0)|≤C\int_{-1}^1|p(x)|dx$
This seemed like a direct consequence of the Archimedean Property of the reals to me, since the value of the integral must be a finite, positive real number, and the lefthand side is a finite, nonnegative real number. However, all of the solutions (they gave 3: see this page http://kskedlaya.org/putnam-archive/) were much more complicated; is there a case I am missing by merely viewing this as a result of the Archimedean property? Or is there some missing condition that makes it not applicable for this problem?

Posting this because I can't comment:
The norm trick would work for a lot of similar Putnam questions. That's why in recent years they ask you to find C explicitly, not just prove it exists.
