AoPSWiki
Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!
Personal tools

1989 USAMO Problems/Problem 5

From AoPSWiki

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}... 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} \\&= ... 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{... 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 (Problems)
Preceded by
Problem 4
1 2 3 4 5 Followed by
Final Question
All USAMO Problems and Solutions
Add a glimpse of the Art of Problem Solving Forum to your own site!
Click here for details!
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us