Difference between revisions of "1989 USAMO Problems/Problem 5"

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Let <math>u</math> and <math>v</math> be real numbers such that
 
Let <math>u</math> and <math>v</math> be real numbers such that
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<cmath> (u + u^2 + u^3 + \cdots + u^8) + 10u^9 = (v + v^2 + v^3 + \cdots + v^{10}) + 10v^{11} = 8. </cmath>
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Determine, with proof, which of the two numbers, <math>u</math> or <math>v</math>, is larger.
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==Solution==
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The answer is <math>v</math>.
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We define real functions <math>U</math> and <math>V</math> as follows:
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<cmath>\begin{align*}
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U(x) &= (x+x^2 + \dotsb + x^8) + 10x^9 = \frac{x^{10}-x}{x-1} + 9x^9 \\
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V(x) &= (x+x^2 + \dotsb + x^{10}) + 10x^{11} = \frac{x^{12}-x}{x-1} + 9x^{11} .
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\end{align*} </cmath>
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We wish to show that if <math>U(u)=V(v)=8</math>, then <math>u >v</math>.
  
<math>
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We first note that when <math>x \le 0</math>, <math>x^{12}-x \ge 0</math>, <math>x-1 < 0</math>, and <math>9x^9 \le 0</math>, so
(u + u^2 + u^3 + \cdots + u^8) + 10u^9 = (v + v^2 + v^3 + \cdots + v^{10}) + 10v^{11} = 8.
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<cmath> U(x) = \frac{x^{10}-x}{x-1} + 9x^9 \le 0 < 8 .</cmath>
</math>
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Similarly, <math>V(x) \le 0 < 8</math>.
  
Determine, with proof, which of the two numbers, <math>u</math> or <math>v</math>, is larger.
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We also note that if <math>x \ge 9/10 </math>, then
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<cmath> \begin{align*}
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U(x) &= \frac{x-x^{10}}{1-x} + 9x^9 \ge \frac{9/10 - 9^9/10^9}{1/10} + 9 \cdot \frac{9^{9}}{10^9} \\
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&= 9 - 10 \cdot \frac{9^9}{10^9} + 9 \cdot \frac{9^9}{10^9} = 9 - \frac{9^9}{10^9} > 8.
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\end{align*} </cmath>
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Similarly <math>V(x) > 8</math>.  It then follows that <math>u, v \in (0,9/10)</math>.
  
==Solution==
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Now, for all <math>x \in (0,9/10)</math>,
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<cmath> \begin{align*}
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V(x) &= U(x) + V(x)-U(x) = U(x) + 10x^{11}+x^{10} -9x^9 \\
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&= U(x) + x^9 (10x -9) (x+1) < U(x) .
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\end{align*} </cmath>
 +
Since <math>V</math> and <math>U</math> are both strictly increasing functions over the nonnegative reals, it then follows that
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<cmath> V(u) < U(u) = 8 = V(v), </cmath>
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so <math>u<v</math>, as desired.  <math>\blacksquare</math>
  
{{solution}}
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== Resources ==
  
==See Also==
 
 
{{USAMO box|year=1989|num-b=4|after=Final Question}}
 
{{USAMO box|year=1989|num-b=4|after=Final Question}}
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* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356639#356639 Discussion on AoPS/MathLinks]
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[[Category:Olympiad Algebra Problems]]

Revision as of 19:37, 10 January 2008

Problem

Let $u$ and $v$ be real numbers such that \[(u + u^2 + u^3 + \cdots + u^8) + 10u^9 = (v + v^2 + v^3 + \cdots + v^{10}) + 10v^{11} = 8.\] Determine, with proof, which of the two numbers, $u$ or $v$, is larger.

Solution

The answer is $v$.

We define real functions $U$ and $V$ as follows: \begin{align*} U(x) &= (x+x^2 + \dotsb + x^8) + 10x^9 = \frac{x^{10}-x}{x-1} + 9x^9 \\ V(x) &= (x+x^2 + \dotsb + x^{10}) + 10x^{11} = \frac{x^{12}-x}{x-1} + 9x^{11} . \end{align*} We wish to show that if $U(u)=V(v)=8$, then $u >v$.

We first note that when $x \le 0$, $x^{12}-x \ge 0$, $x-1 < 0$, and $9x^9 \le 0$, so \[U(x) = \frac{x^{10}-x}{x-1} + 9x^9 \le 0 < 8 .\] Similarly, $V(x) \le 0 < 8$.

We also note that if $x \ge 9/10$, then \begin{align*} U(x) &= \frac{x-x^{10}}{1-x} + 9x^9 \ge \frac{9/10 - 9^9/10^9}{1/10} + 9 \cdot \frac{9^{9}}{10^9} \\ &= 9 - 10 \cdot \frac{9^9}{10^9} + 9 \cdot \frac{9^9}{10^9} = 9 - \frac{9^9}{10^9} > 8. \end{align*} Similarly $V(x) > 8$. It then follows that $u, v \in (0,9/10)$.

Now, for all $x \in (0,9/10)$, \begin{align*} V(x) &= U(x) + V(x)-U(x) = U(x) + 10x^{11}+x^{10} -9x^9 \\ &= U(x) + x^9 (10x -9) (x+1) < U(x) . \end{align*} Since $V$ and $U$ are both strictly increasing functions over the nonnegative reals, it then follows that \[V(u) < U(u) = 8 = V(v),\] so $u<v$, as desired. $\blacksquare$

Resources

1989 USAMO (ProblemsResources)
Preceded by
Problem 4
Followed by
Final Question
1 2 3 4 5
All USAMO Problems and Solutions
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