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

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 $1<b<a<2$ .

@@ Line 4: / Line 4: @@
 is larger.
-==Solution==
+==Solutions==
+===Solution 1===
 Square and rearrange the first equation and also rearrange the second.
 <cmath>\begin{align}
@@ Line 19: / Line 20: @@
 a^{2n}-a&>b^{2n}-b \tag{5}
 \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).  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]]
+{{MAA Notice}}

1993 USAMO (Problems • Resources)
Preceded by First Problem	Followed by Problem 2
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