# Difference between revisions of "1987 USAMO Problems/Problem 3"

## Problem $X$ is the smallest set of polynomials $p(x)$ such that:

1. $p(x) = x$ belongs to $X$.
2. If $r(x)$ belongs to $X$, then $x\cdot r(x)$ and $(x + (1 - x) \cdot r(x) )$ both belong to $X$.

Show that if $r(x)$ and $s(x)$ are distinct elements of $X$, then $r(x) \neq s(x)$ for any $0 < x < 1$.

## Solution

Let $s(x)$ be an arbitrary polynomial in $X.$ Then $0 when $0 Define $X_1=\{s(x)\in X:s(x)=x\cdot s_1(x)$ for some $s_1(x)\in X\},$ and $X_2=\{t(x)\in X: t(x)=x+(1-x)t_1(x)$ for some $t_1(x)\in X\}.$

If $s(x) \in X_1$ and $t(x)\in X_2,$ we have $s(x) for all $x$ with $0 Therefore $s(x)\ne t(x)$ for any $0

For any $s(x), t(x) \in X_1$, Let $s(x)=x\cdot s_1(x)$ and $t(x)=x\cdot t_1(x)$ for $s_1(x), t_1(x) \in X.$ If $s_1(x) \ne t_1(x)$ for $0 then $s(x)-t(x)=x(s_1(x)-t_1(x))\ne 0$ for $0

Similarly, for any $s(x), t(x) \in X_2$, Let $s(x)=x+(1-x) s_1(x)$ and $t(x)=x+(1-x) t_1(x)$ for $s_1(x), t_1(x) \in X.$ If $s_1(x) \ne t_1(x)$ for $0 then $s(x)-t(x)=(1-x)(s_1(x)-t_1(x))\ne 0$ for $0

The proof is done by an induction.

J.Z.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 