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Difference between revisions of "1969 Canadian MO Problems/Problem 1"

 
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== Problem ==
 
== Problem ==
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Show that if <math>\displaystyle a_1/b_1=a_2/b_2=a_3/b_3</math> and <math>\displaystyle p_1,p_2,p_3</math> are not all zero, then <math>\displaystyle\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>\displaystyle n.</math>
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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 ==
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Subtracting the LHS from the RHS,
 
Subtracting the LHS from the RHS,
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<math>0=\displaystyle \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>
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<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
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<math>\displaystyle 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>
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<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)
  
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From <math>\displaystyle a_1/b_1=a_2b_2,</math> <math>\displaystyle a_1^nb_2^n=a_2^nb_1^n.</math> Similarly, <math>\displaystyle a_1^nb_3^n=a_3^nb_1^n</math> from <math>\displaystyle a_1/b_1=a_3/b_3.</math>  
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From <math>a_1/b_1=a_2/b_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>  
  
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Hence, <math>\displaystyle 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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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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{{Old CanadaMO box|before=First question|num-a=2|year=1969}}
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* [[1969 Canadian MO Problems/Problem 2|Next Problem]]
 
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* [[1969 Canadian MO Problems|Back to Exam]]
 

Latest revision as of 14:06, 18 October 2015

Problem

Show that if $a_1/b_1=a_2/b_2=a_3/b_3$ and $p_1,p_2,p_3$ are not all zero, then $\left(\frac{a_1}{b_1} \right)^n=\frac{p_1a_1^n+p_2a_2^n+p_3a_3^n}{p_1b_1^n+p_2b_2^n+p_3b_3^n}$ for every positive integer $n.$

Solution

Instead of proving the two expressions equal, we prove that their difference equals zero.

Subtracting the LHS from the RHS, $0=\frac{p_1a_1^n+p_2a_2^n+p_3a_3^n}{p_1b_1^n+p_2b_2^n+p_3b_3^n}-\frac{a_1^n}{b_1^n}.$

Finding a common denominator, the numerator becomes $b_1^n(p_1a_1^n+p_2a_2^n+p_3a_3^n)-a_1^n(p_1b_1^n+p_2b_2^n+p_3b_3^n)=p_2(a_2^nb_1^n-a_1^nb_2^n)+p_3(a_3^nb_1^n-a_1^nb_3^n)=0.$ (The denominator is irrelevant since it never equals zero)

From $a_1/b_1=a_2/b_2,$ $a_1^nb_2^n=a_2^nb_1^n.$ Similarly, $a_1^nb_3^n=a_3^nb_1^n$ from $a_1/b_1=a_3/b_3.$

Hence, $a_2^nb_1^n-a_1^nb_2^n=a_3^nb_1^n-a_1^nb_3^n=0$ and our proof is complete.

1969 Canadian MO (Problems)
Preceded by
First question 		1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • Followed by
Problem 2


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