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

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so <math>u<v</math>, as desired.  <math>\blacksquare</math>
 
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== Resources ==
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== See Also ==
  
{{USAMO box|year=1989|num-b=4|after=Final Question}}
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{{USAMO box|year=1989|num-b=4|after=Last Question}}
  
 
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356639#356639 Discussion on AoPS/MathLinks]
 
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356639#356639 Discussion on AoPS/MathLinks]

Latest revision as of 19:16, 18 July 2016

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. $\blacksquare$

See Also

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

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