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Difference between revisions of "1993 USAMO Problems/Problem 1"

(New page: ==Problem== For each integer <math>n\ge 2</math>, determine, with proof, which of the two positive real numbers <math>a</math> and <math>b</math> satisfying <cmath>a^n=a+1,\qquad b^{2n}=b+...)
(No difference)

Revision as of 13:57, 22 March 2008

Problem

For each integer $n\ge 2$, determine, with proof, which of the two positive real numbers $a$ and $b$ satisfying \[a^n=a+1,\qquad b^{2n}=b+3a\] is larger.

Solution

Square and rearrange the first equation and also rearrange the second. a2n−a=a2+a+1b2n−b=3a It is trivial that (a−1)2>0 since $a$ clearly cannot equal 0 (Otherwise $a^n=0\ne0+1$). Thus a2+a+1>3aa2n−a>b2n−b where we substituted in equations (1) and (2) to achieve (5). If $b>a$, then $b^{2n}>a^{2n}$ since $a$, $b$, and $n$ are all positive. Adding the two would mean $b^{2n}-b>a^{2n}-a$, a contradiction, so $a>b$. However, when $n$ equals 0 or 1, the first equation becomes meaningless, so we conclude that for each integer $n\ge 2$, we always have $a>b$.

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