Difference between revisions of "1993 USAMO Problems/Problem 1"
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is larger. | is larger. | ||
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+ | ===Solution 1=== | ||
Square and rearrange the first equation and also rearrange the second. | Square and rearrange the first equation and also rearrange the second. | ||
<cmath>\begin{align} | <cmath>\begin{align} | ||
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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^{ | + | where we substituted in equations (1) and (2) to achieve (5). Notice that from <math>a^{n}=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 | + | ===Solution 2=== |
− | {{USAMO box|year=1993|before=First | + | |
+ | Define <math>f(x)=x^n-x-1</math> and <math>g(x)=x^{2n}-x-3a</math>. By Descarte's Rule of Signs, both polynomials' only positive roots are <math>a</math> and <math>b</math>, respectively. With the Intermediate Value Theorem and the fact that <math>f(1)=-1</math> and <math>f(2)=2^n-3>0</math>, we have <math>a\in(1,2)</math>. | ||
+ | Thus, <math>-3a\in(-6,-3)</math>, which means that <math>g(1)=-3a<0</math>. Also, we find that <math>g(a)=a^{2n}-4a</math>. All that remains to prove is that <math>g(a)>0</math>, or <math>a^{2n}-4a>0</math>. We can then conclude that <math>b</math> is between <math>1</math> and <math>a</math> from the Intermediate Value Theorem. From the first equation given, <math>a^{2n}=(a+1)^2=a^2+2a+1</math>. Subtracting <math>4a</math> gives us <math>a^2-2a+1>0</math>, which is clearly true, as <math>a\neq1</math>. Therefore, we conclude that <math>1<b<a<2</math>. | ||
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+ | == See Also == | ||
+ | {{USAMO box|year=1993|before=First Problem|num-a=2}} | ||
[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 06:54, 19 July 2016
Problem
For each integer , determine, with proof, which of the two positive real numbers and satisfying is larger.
Solutions
Solution 1
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 .
Solution 2
Define and . By Descarte's Rule of Signs, both polynomials' only positive roots are and , respectively. With the Intermediate Value Theorem and the fact that and , we have . Thus, , which means that . Also, we find that . All that remains to prove is that , or . We can then conclude that is between and from the Intermediate Value Theorem. From the first equation given, . Subtracting gives us , which is clearly true, as . Therefore, we conclude that .
See Also
1993 USAMO (Problems • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.