Difference between revisions of "1969 Canadian MO Problems/Problem 1"
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== Problem == | == Problem == | ||
| − | Show that if <math> | + | Show that if <math>a_1/b_1=a_2/b_2=a_3/b_3</math> and <math>p_1,p_2,p_3</math> are not all zero, then <math>\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}</math> for every positive integer <math>n.</math> |
== Solution == | == Solution == | ||
| Line 6: | Line 6: | ||
Subtracting the LHS from the RHS, | Subtracting the LHS from the RHS, | ||
| − | <math>0= | + | <math>0=\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}-\frac{a_1^n}{b_1^n}.</math> |
Finding a common denominator, the numerator becomes | Finding a common denominator, the numerator becomes | ||
| − | <math> | + | <math>b_1^n(p_1a_1^n+p_2a_2^n+p_3a_3^n)-a_1^n(p_1b_1^n+p_2b_2^n+p_3b_3^n)=p_2(a_2^nb_1^n-a_1^nb_2^n)+p_3(a_3^nb_1^n-a_1^nb_3^n)=0.</math> |
(The denominator is irrelevant since it never equals zero) | (The denominator is irrelevant since it never equals zero) | ||
| − | From <math> | + | From <math>a_1/b_1=a_2b_2,</math> <math>a_1^nb_2^n=a_2^nb_1^n.</math> Similarly, <math>a_1^nb_3^n=a_3^nb_1^n</math> from <math>a_1/b_1=a_3/b_3.</math> |
| − | Hence, <math> | + | Hence, <math>a_2^nb_1^n-a_1^nb_2^n=a_3^nb_1^n-a_1^nb_3^n=0</math> and our proof is complete. |
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| − | + | {{CanadaMO box|year=1969|before=First problem|num-a=2}} | |
| − | |||
Revision as of 12:03, 8 October 2007
Problem
Show that if 
and 
are not all zero, then 
for every positive integer 
Solution
Instead of proving the two expressions equal, we prove that their difference equals zero.
Subtracting the LHS from the RHS,
Finding a common denominator, the numerator becomes
(The denominator is irrelevant since it never equals zero)
From 
Similarly, 
from 
Hence, 
and our proof is complete.
| 1969 Canadian MO (Problems) | ||
| Preceded by First problem |
1 • 2 • 3 • 4 • 5 | Followed by Problem 2 |
