Difference between revisions of "1993 USAMO Problems/Problem 1"
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It is [[Trivial Inequality|trivial]] that | It is [[Trivial Inequality|trivial]] that | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
| − | (a-1)^2 | + | (a-1)^2 > 0 \tag{3} |
\end{align*}</cmath> | \end{align*}</cmath> | ||
since <math>a-1</math> clearly cannot equal <math>0</math> (Otherwise <math>a^n=1\neq 1+1</math>). Thus | since <math>a-1</math> clearly cannot equal <math>0</math> (Otherwise <math>a^n=1\neq 1+1</math>). Thus | ||
Revision as of 14:07, 22 March 2008
Problem
For each integer 
, determine, with proof, which of the two positive real numbers 
and 
satisfying
![\[a^n=a+1,\qquad b^{2n}=b+3a\]](http://latex.artofproblemsolving.com/9/a/6/9a63b2ade70110e8161ea351e1b94044db3fc522.png)
is larger.
Solution
Square and rearrange the first equation and also rearrange the second.
It is trivial that
since 
clearly cannot equal 
(Otherwise 
). Thus
where we substituted in equations (1) and (2) to achieve (5). If 
, then 
since , , and 
are all positive. Adding the two would mean 
, a contradiction, so 
. However, when equals or 
, the first equation becomes meaningless, so we conclude that for each integer , we always have .
See also
| 1993 USAMO (Problems • Resources) | ||
| Preceded by First question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
