Difference between revisions of "1993 USAMO Problems/Problem 1"
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b^{2n}-b&=3a | b^{2n}-b&=3a | ||
\end{align}</cmath> | \end{align}</cmath> | ||
| − | It is trivial that | + | It is [[Trivial Inequality|trivial]] that |
| − | <cmath>\begin{align} | + | <cmath>\begin{align*} |
| − | (a-1)^2&>0 | + | (a-1)^2&>0 \tag{3} |
| − | \end{align}</cmath> | + | \end{align*}</cmath> |
| − | since <math>a</math> clearly cannot equal 0 (Otherwise <math>a^n= | + | since <math>a-1</math> clearly cannot equal <math>0</math> (Otherwise <math>a^n=1\neq 1+1</math>). Thus |
| − | <cmath>\begin{align} | + | <cmath>\begin{align*} |
| − | a^2+a+1&>3a\\ | + | a^2+a+1&>3a \tag{4}\\ |
| − | a^{2n}-a&>b^{2n}-b | + | 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). 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>. | ||
Revision as of 15:06, 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 | ||
