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2005 Canadian MO Problems/Problem 2: Difference between revisions

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==Solution==
==Solution==


We have <P><center><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></center></P>.  <div align=left>By [[AM-GM]], we have </div><P><center><math>x + \frac 1x > 2,</math></center></P><div align=left> where <math>x</math> is a [[positive]] [[real number]] not equal to one.  If <math>a = b</math>, then <math>c \not\in \mathbb{Z}</math>. Thus <math>\displaystyle a \neq b</math> and <math>\frac ab \neq 1\implies \frac{a^2}{b^2}\neq 1</math>. Therefore, </div><P><center><math>\left(\frac ca + \frac cb\right)^2 > 2 + 2 + 2(2) = 8.</math></center></P>
We have  
 
:<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>
 
where <math>x</math> is a [[positive]] [[real number]] not equal to one.  If <math>a = b</math>, then <math>c \not\in \mathbb{Z}</math>. Thus <math>\displaystyle a \neq b</math> and <math>\frac ab \neq 1\implies \frac{a^2}{b^2}\neq 1</math>. Therefore,  
 
:<math>\left(\frac ca + \frac cb\right)^2 > 2 + 2 + 2(2) = 8.</math>


==See also==
==See also==

Revision as of 17:25, 7 February 2007

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

We have

$\left(\frac ca + \frac cb\right)^2 = \frac{c^2}{a^2} + 2\frac{c^2}{ab} + \frac{c^2}{b^2} = \frac{a^2 + b^2}{a^2} + 2\frac{a^2 + b^2}{ab} + \frac{a^2+b^2}{b^2} = 2 + \left(\frac{a^2}{b^2} + \frac{b^2}{a^2}\right) + 2\left(\frac ab + \frac ba\right)$

By AM-GM, we have

$x + \frac 1x > 2,$

where $x$ is a positive real number not equal to one. If $a = b$, then $c \not\in \mathbb{Z}$. Thus $\displaystyle a \neq b$ and $\frac ab \neq 1\implies \frac{a^2}{b^2}\neq 1$. Therefore,

$\left(\frac ca + \frac cb\right)^2 > 2 + 2 + 2(2) = 8.$

See also

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