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

(Solution)
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a^{2n}-a&>b^{2n}-b \tag{5}
 
a^{2n}-a&>b^{2n}-b \tag{5}
 
\end{align*}</cmath>
 
\end{align*}</cmath>
where we substituted in equations (1) and (2) to achieve (5).  If <math>b>a</math>, then <math>b^{2n}>a^{2n}</math> since <math>a</math>, <math>b</math>, and <math>n</math> are all positive.  Adding the two would mean <math>b^{2n}-b>a^{2n}-a</math>, a contradiction, so <math>a>b</math>.  However, when <math>n</math> equals <math>0</math> or <math>1</math>, the first equation becomes meaningless, so we conclude that for each integer <math>n\ge 2</math>, we always have <math>a>b</math>.
+
where we substituted in equations (1) and (2) to achieve (5).  Notice that from <math>a^{2n}=a+1</math> we have <math>a>1</math>.  Thus, if <math>b>a</math>, then <math>b^{2n-1}-1>a^{2n-1}-1.  Since </math>a>1\Rightarrow a^{2n-1}-1>0<math>, multiplying the two inequalities yields </math>b^{2n}-b>a^{2n}-a<math>, a contradiction, so </math>a> b<math>.  However, when </math>n<math> equals </math>0<math> or </math>1<math>, the first equation becomes meaningless, so we conclude that for each integer </math>n\ge 2<math>, we always have </math>a>b$.
  
 
== See also ==
 
== See also ==

Revision as of 17:24, 10 November 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. \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 $a^{2n}=a+1$ 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$a>b$.

See also

1993 USAMO (ProblemsResources)
Preceded by
First question
Followed by
Problem 2
1 2 3 4 5
All USAMO Problems and Solutions
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