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
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 |