## Sunday, April 15, 2007

### Sense Of Poise And Rationality. Topic: Algebra. Level: AIME/Olympiad.

Problem: (1993 Canada National Olympiad - #2) Show that $x$ is rational if and only if three distinct terms that form a geometric progression can be chosen from the sequence $x, x+1, x+2, x+3, \ldots$.

Solution: We will prove the "if" statement first, i.e. $x$ is rational implies we can find the geometric progression. Let $x = \frac{p}{q}$. The terms of the sequence are then

$\frac{p}{q}$, $\frac{p+q}{q}$, $\frac{p+2q}{q}$, $\frac{p+3q}{q}, \ldots$.

We simply choose the terms

$\frac{p}{q}$, $\frac{p+pq}{q}$, $\frac{p+(2p+pq)q}{q}$ which are the same as $\frac{p}{q}$, $\frac{p(1+q)}{q}$, $\frac{p(1+q)^2}{q}$,

clearly a geometric progression.

Now for the other direction, assume $x+a, x+b, x+c$ are a geometric progression with $a < b < c$ positive integers. Then

$\frac{x+a}{x+b} = \frac{x+b}{x+c} \Rightarrow x^2+(a+c)x+ac = x^2+2bx+b^2 \Rightarrow (a+c)x+ac = 2bx+b^2$.

Solving for $x$ yields $x = \frac{b^2-ac}{a+c-2b}$ which is clearly rational, as desired. QED.

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Comment: An interesting "definition" of a rational number, pretty neat in fact. Though it may not seem so at first, the "if" direction was actually considerably harder than the "only if" direction because the approach was not as straightforward. The "only if" proof required simple algebra, while the "if" proof needed a little creative picking and choosing. Another approach for the "if" direction is to set $x = \frac{p}{q}$ and try to find corresponding $a, b, c$ but that makes it a little more difficult than necessary.

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Practice Problem: (1993 Canada National Olympiad - #3) In triangle $ABC$, medians to the sides $\overline{AB}$ and $\overline{AC}$ are perpendicular. Show that $\cot{B}+\cot{C} \ge \frac{2}{3}$.