Difference between revisions of "1993 USAMO Problems/Problem 1"
(→Solution) |
(→Solution) |
||
Line 19: | Line 19: | ||
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). 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 < | + | 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</math>. 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</math>. |
== See also == | == See also == |
Revision as of 18:25, 10 November 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). Notice that from we have . Thus, if , then . Since , multiplying the two inequalities yields , 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 |