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

Difference between revisions of "2005 Canadian MO Problems/Problem 2"
Line 5: Line 5:
 
* Prove that there does not exist any integer <math>n</math> for which we can find a Pythagorean triple <math>(a,b,c)</math> satisfying <math>(c/a + c/b)^2 = n</math>.
 
* Prove that there does not exist any integer <math>n</math> for which we can find a Pythagorean triple <math>(a,b,c)</math> satisfying <math>(c/a + c/b)^2 = n</math>.
 
==Solution==
 
==Solution==
 +
First part:
 +
 +
<math>\left(\frac ca + \frac cb\right)^2 = \frac{c^2}{a^2} + 2\frac{c^2}{ab} + \frac{c^2}{b^2} = \frac{a^2 + b^2}{a^2} + 2\frac{a^2 + b^2}{ab} + \frac{a^2+b^2}{b^2} = 2 + \left(\frac{a^2}{b^2} + \frac{b^2}{a^2}\right) + 2\left(\frac ab + \frac ba\right)</math>.  By [[AM-GM]] we have <math>x + \frac 1x > 2</math> if <math>x</math> is a [[positive]] [[real number]] other than 1.  If <math>a = b</math> then <math>c \not\in \mathbb{Z}</math> so <math>\displaystyle a \neq b</math> and <math>\frac ab \neq 1</math> and <math>\frac{a^2}{b^2}\neq 1</math> and thus <math>\left(\frac ca + \frac cb\right)^2 > 2 + 2 + 2(2) = 8</math>.
 +
 +
 
==See also==
 
==See also==
 +
*[[2005 Canadian MO Problems/Problem 3|Next problem]]
 +
*[[2005 Canadian MO Problems/Problem 1|Previous problem]]
 
*[[2005 Canadian MO]]
 
*[[2005 Canadian MO]]

Revision as of 10:10, 16 August 2006

Problem

Let $(a,b,c)$ be a Pythagorean triple, i.e., a triplet of positive integers with ${a}^2+{b}^2={c}^2$.

  • Prove that $(c/a + c/b)^2 > 8$.
  • Prove that there does not exist any integer $n$ for which we can find a Pythagorean triple $(a,b,c)$ satisfying $(c/a + c/b)^2 = n$.

Solution

First part:

$\left(\frac ca + \frac cb\right)^2 = \frac{c^2}{a^2} + 2\frac{c^2}{ab} + \frac{c^2}{b^2} = \frac{a^2 + b^2}{a^2} + 2\frac{a^2 + b^2}{ab} + \frac{a^2+b^2}{b^2} = 2 + \left(\frac{a^2}{b^2} + \frac{b^2}{a^2}\right) + 2\left(\frac ab + \frac ba\right)$. By AM-GM we have $x + \frac 1x > 2$ if $x$ is a positive real number other than 1. If $a = b$ then $c \not\in \mathbb{Z}$ so $\displaystyle a \neq b$ and $\frac ab \neq 1$ and $\frac{a^2}{b^2}\neq 1$ and thus $\left(\frac ca + \frac cb\right)^2 > 2 + 2 + 2(2) = 8$.


See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_Canadian_MO_Problems/Problem_2&oldid=9585"