## Sunday, August 13, 2006

### Fraction, Fraction. Topic: Algebra. Level: AMC/AIME.

Problem: (1969 Canada - #1) If $\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3}$ and $p_1, p_2, p_3$ are not all zero, show that for all $n \in \mathbb{N}$,

$\left(\frac{a_1}{b_1}\right)^n = \frac{p_1a_1^n+p_2a_2^n+p_3a_3^n}{p_1b_1^n+p_2b_2^n+p_3b_3^n}$.

Solution: Well, a good first step seems to be expanding it out. We want to show

$p_1(a_1b_1)^n+p_2(a_1b_2)^n+p_3(a_1b_3)^n = p_1(a_1b_1)^n+p_2(a_2b_1)^n+p_3(a_3b_1)^n$,

which reduces to

$p_2(a_1b_2)^n+p_3(a_1b_3)^n = p_2(a_2b_1)^n+p_3(a_3b_1)^n$.

However, noting that $\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} \Rightarrow a_1b_2 = a_2b_1$ and $a_1b_3 = a_3b_1$, we simply substitute and the problem is solved. QED.

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Comment: Wow. Back in the old days, this qualified for an olympiad problem. It looks slightly complicated at first, but after expanding you're done before you know it.

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Practice Problem: Show that if $\frac{a_1}{b_1} = \frac{a_2}{b_2} = \cdots = \frac{a_n}{b_n}$, then

$\frac{a_i}{b_i} = \frac{a_1+a_2+\cdots+a_n}{b_1+b_2+\cdots+b_n}$

for $i = 1, 2, \ldots, n$.

1. Just express every a_i, b_i as (c_i a_1) / (c_i b_1) and the solution is immediate.

2. so ehh.. what does the rounded-E mean in the problem? and N.. where did that come from??

I get what QC's saying!!! that rarely happens...

3. the rounded E is for "in"

4. The bold-faced N means the natural numbers, aka {1,2,3,...}.

5. A note to commenters: Using the less-than symbol will cut short your reply.

6. Unless you're really cool and use & l t ; instead =D

7. <

Haha nice.