Difference between revisions of "1993 USAMO Problems/Problem 1"

(Solution)
m (See also)
 
(6 intermediate revisions by 5 users not shown)
Line 4: Line 4:
 
is larger.
 
is larger.
  
==Solution==
+
==Solutions==
 +
===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}
Line 19: Line 20:
 
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 </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$.
+
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 also ==
+
===Solution 2===
{{USAMO box|year=1993|before=First question|num-a=2}}
+
 
 +
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>.
 +
 
 +
 
 +
 
 +
== 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 $n\ge 2$, determine, with proof, which of the two positive real numbers $a$ and $b$ satisfying \[a^n=a+1,\qquad b^{2n}=b+3a\] is larger.

Solutions

Solution 1

Square and rearrange the first equation and also rearrange the second. \begin{align} a^{2n}-a&=a^2+a+1\\ b^{2n}-b&=3a \end{align} It is trivial that \begin{align*} (a-1)^2 > 0 \tag{3} \end{align*} since $a-1$ clearly cannot equal $0$ (Otherwise $a^n=1\neq 1+1$). Thus \begin{align*} a^2+a+1&>3a \tag{4}\\ a^{2n}-a&>b^{2n}-b \tag{5} \end{align*} where we substituted in equations (1) and (2) to achieve (5). Notice that from $a^{n}=a+1$ we have $a>1$. Thus, if $b>a$, then $b^{2n-1}-1>a^{2n-1}-1$. Since $a>1\Rightarrow a^{2n-1}-1>0$, multiplying the two inequalities yields $b^{2n}-b>a^{2n}-a$, a contradiction, so $a> b$. However, when $n$ equals $0$ or $1$, the first equation becomes meaningless, so we conclude that for each integer $n\ge 2$, we always have $a>b$.

Solution 2

Define $f(x)=x^n-x-1$ and $g(x)=x^{2n}-x-3a$. By Descarte's Rule of Signs, both polynomials' only positive roots are $a$ and $b$, respectively. With the Intermediate Value Theorem and the fact that $f(1)=-1$ and $f(2)=2^n-3>0$, we have $a\in(1,2)$. Thus, $-3a\in(-6,-3)$, which means that $g(1)=-3a<0$. Also, we find that $g(a)=a^{2n}-4a$. All that remains to prove is that $g(a)>0$, or $a^{2n}-4a>0$. We can then conclude that $b$ is between $1$ and $a$ from the Intermediate Value Theorem. From the first equation given, $a^{2n}=(a+1)^2=a^2+2a+1$. Subtracting $4a$ gives us $a^2-2a+1>0$, which is clearly true, as $a\neq1$. Therefore, we conclude that $1<b<a<2$.


See Also

1993 USAMO (ProblemsResources)
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. AMC logo.png

Invalid username
Login to AoPS