Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1969 Canadian MO Problems/Problem 1: Difference between revisions

← Older editNewer edit →
1=2 (talk | contribs)
4,129 edits
No edit summary
1=2 (talk | contribs)
4,129 edits
No edit summary
Line 17: Line 17:


----
----
{{CanadaMO box|year=1969|before=First problem|num-a=2}}
[[1969 Canadian MO Problems/Problem 2 | Next problem]]

Revision as of 12:05, 8 October 2007

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 $a_1/b_1=a_2b_2,$ $a_1^nb_2^n=a_2^nb_1^n.$ 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.

Next problem

Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=1969_Canadian_MO_Problems/Problem_1&oldid=17601"