**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} $.

Does median AB mean the median to side AB? Because then I think you mean cot A + cot C.

ReplyDeleteSo anyway, the medians divide each other in the ratio 2 : 1, so let median AB be divided 2a : a and let median BC be divided 2b : b. After some easy trig we find that cot C = (2a^2 - b^2) / 3ab and cot A = (2b^2 - a^2) / 3ab, so

cot C + cot A = a^2 + b^2 / 3ab ≥ 2ab / 3ab = 2/3

Wow, I totally typed that problem wrong on so many levels. In any case, your proof should apply to the new problem.

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