Baltic Way 1998 · Problem 6

Algebra

Let $P$ be a polynomial of degree 6 and let $a, b$ be real numbers such that $0<a<b$ . Suppose that $P(a)=P(-a), P(b)=P(-b)$ and $P^{\prime}(0)=0$ . Prove that $P(x)=P(-x)$ for all real $x$ .

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Review

Topics

Polynomials

Solutions

Solution

Solution: The polynomial $Q(x) = P(x) - P(-x)$ , of degree at most $5$ , has roots at $-b$ , $-a$ , $0$ , $a$ and $b$ ; these are five distinct numbers. Moreover, $Q'(0) = 0$ , showing that $Q$ has a multiple root at $0$ . Thus $Q$ must be the constant $0$ , i.e. $P(x) = P(-x)$ for all $x$ .

Contest context

Results from Baltic Way 1998

11 teams

Mean score: 4.9 / 5
Scores of 4 or 5: 11 / 11
Estonia: 5 / 5

Score distribution

510

All team scores

Team	Score
Latvia	5 / 5
Estonia	5 / 5
Poland	4 / 5
Finland	5 / 5
St. Petersburg	5 / 5
Sweden	5 / 5
Denmark	5 / 5
Iceland	5 / 5
Norway	5 / 5
Germany	5 / 5
Lithuania	5 / 5

Elsewhere

AoPS · Baltic Way 1998
Official results