1993 USAMO Problems/Problem 1: Difference between revisions

Revision as of 14:29, 15 April 2012

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. $\begin{align} a^{2n}-a&=a^2+a+1\\ b^{2n}-b&=3a \end{align}$ It is trivial that $\begin{align*} (a-1)^2 > 0 \tag{3} \end{align*}$ since $a-1$ clearly cannot equal $0$ (Otherwise $a^n=1\neq 1+1$ ). Thus $\begin{align*} a^2+a+1&>3a \tag{4}\\ a^{2n}-a&>b^{2n}-b \tag{5} \end{align*}$ where we substituted in equations (1) and (2) to achieve (5). Notice that from $a^{n}=a+1$ we have $a>1$ . Thus, if $b>a$ , then $b^{2n-1}-1>a^{2n-1}-1$ . Since $a>1\Rightarrow a^{2n-1}-1>0$ , multiplying the two inequalities yields $b^{2n}-b>a^{2n}-a$ , 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 $a>b$ .

@@ Line 22: / Line 22: @@
 == See also ==
-{{USAMO box|year=1993|before=First Question|num-a=2}}
+{{USAMO box|year=1993|before=First Problem|num-a=2}}
 [[Category:Olympiad Algebra Problems]]

1993 USAMO (Problems • Resources)
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1993 USAMO Problems/Problem 1: Difference between revisions

Revision as of 14:29, 15 April 2012

Problem

Solution

See also