## Friday, December 30, 2005

### Pick one, any one! Topic: Algebra. Level: Olympiad

Problem: (1989 USAMO - #5) 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: Consider the functions

$f(x) = 1+x+\cdots+x^8+10x^9$ and $g(x) = 1+x+\cdots+x^{10}+10x^{11}$,

which are monotonically increasing on the interval $(0,\infty)$.

We are given that $f(u) = g(v) = 9$.

We have $g(x)-f(x) = 10x^{11}+x^{10}-9x^9 = x^9(10x-9)(x+1)$.

$f\left(\frac{9}{10}\right) = 1+\frac{9}{10}+\cdots+\frac{9^8}{10^8}+10 \cdot \frac{9^9}{10^9}$.

By summing the geometric series,

$f\left(\frac{9}{10}\right) = \frac{1-\frac{9^9}{10^9}}{1-\frac{9}{10}} + 10 \cdot \frac{9^9}{10^9}= 10-10 \cdot \frac{9^9}{10^9}+10 \cdot \frac{9^9}{10^9} = 10 > 9$.

Since f is an increasing function, $f(u) < f\left(\frac{9}{10}\right) \Rightarrow u < \frac{9}{10}$.

Hence $g(u)-f(u) = u^9(10u-9)(u+1) < 0 \Rightarrow g(u) < f(u) = 9$.

But since $g$ is an increasing function as well, we know $g(u) < g(v) = 9 \Rightarrow u < v$. QED.

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Comment: The last problem of the 1989 USAMO, so supposedly it is difficult, but that wasn't really the case. A simple analysis of $f$ and $g$ gives us the answer quickly.

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Practice Problem: (2002 USAMO - #4) Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f: \mathbb{R} \to \mathbb{R}$ such that

$f(x^2 - y^2) = x f(x) - y f(y)$

for all pairs of real numbers $x$ and $y$.