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

Latest revision as of 18:16, 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. $\blacksquare$

@@ Line 38: / Line 38: @@
 == See Also ==
-{{USAMO box|year=1989|num-b=4|after=Final Question}}
+{{USAMO box|year=1989|num-b=4|after=Last Question}}
 * [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356639#356639 Discussion on AoPS/MathLinks]

1989 USAMO (Problems • Resources)
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Difference between revisions of "1989 USAMO Problems/Problem 5"

Latest revision as of 18:16, 18 July 2016

Problem

Solution

See Also