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Difference between revisions of "1993 USAMO Problems/Problem 1"
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==Solution==
 
==Solution==
 
Square and rearrange the first equation and also rearrange the second.
 
Square and rearrange the first equation and also rearrange the second.
−
<math>\begin{gather}
+
<cmath>\begin{align}
−
a^{2n}-a=a^2+a+1\\
+
a^{2n}-a&=a^2+a+1\\
−
b^{2n}-b=3a
+
b^{2n}-b&=3a
−
\end{gather}</math>
+
\end{align}</cmath>
 
It is trivial that
 
It is trivial that
−
\begin{equation}
+
<cmath>\begin{align}
−
(a-1)^2>0
+
(a-1)^2&>0
−
\end{equation}
+
\end{align}</cmath>
 
since <math>a</math> clearly cannot equal 0 (Otherwise <math>a^n=0\ne0+1</math>).  Thus
 
since <math>a</math> clearly cannot equal 0 (Otherwise <math>a^n=0\ne0+1</math>).  Thus
−
\begin{gather}
+
<cmath>\begin{align}
−
a^2+a+1>3a\\
+
a^2+a+1&>3a\\
−
a^{2n}-a>b^{2n}-b
+
a^{2n}-a&>b^{2n}-b
−
\end{gather}
+
\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 0 or 1, 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).  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>.
 +
 
 +
== See also ==
 +
{{USAMO box|year=1993|before=First question|num-a=2}}
 +
 
 +
[[Category:Olympiad Algebra Problems]]

Revision as of 15:03, 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. \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 \end{align} since $a$ clearly cannot equal 0 (Otherwise $a^n=0\ne0+1$). Thus \begin{align} a^2+a+1&>3a\\ a^{2n}-a&>b^{2n}-b \end{align} 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 $a>b$.

See also

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