2005 Canadian MO Problems/Problem 2: Difference between revisions

Revision as of 18:46, 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,

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

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 ==See also==
-*[[2005 Canadian MO Problems/Problem 3|Next problem]]
-*[[2005 Canadian MO Problems/Problem 1|Previous problem]]
 *[[2005 Canadian MO]]
+{{CanadaMO box|year=2005|num-b=1|num-a=3}}
 [[Category:Olympiad Algebra Problems]]
 [[Category:Olympiad Number Theory Problems]]

2005 Canadian MO (Problems)
Preceded by Problem 1	1 • 2 • 3 • 4 • 5	Followed by Problem 3

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2005 Canadian MO Problems/Problem 2: Difference between revisions

Revision as of 18:46, 7 February 2007

Problem

Solution

See also