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Difference between revisions of "2006 Cyprus Seniors Provincial/2nd grade/Problem 1"

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== Problem ==
 
== Problem ==
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If <math>\alpha, \beta, \gamma \in \Re- \{0\}</math> with <math>\alpha + \beta + \gamma = 0</math>, prove that
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If <math>\alpha, \beta, \gamma \in \mathbb{R}- \{0\}</math> with <math>\alpha + \beta + \gamma = 0</math>, prove that
  
 
i) <math>\alpha^2 + \beta^2 - \gamma^2 = -2(\beta + \gamma)(\alpha + \gamma) </math>
 
i) <math>\alpha^2 + \beta^2 - \gamma^2 = -2(\beta + \gamma)(\alpha + \gamma) </math>
  
−
ii) <math>\frac{1}{\beta^2 + \gamma^2 - \alpha^2} + \frac{1}{\gamma^2 + \alpha^2 - \beta^2} + \frac{1}{\alpha^2 + \beta^2 - \gamma^2} = 0</math>
+
ii) <math>\frac{1}{\beta^2 + \gamma^2 - \alpha^2} + \frac{1}{\gamma^2 + \alpha^2 - \beta^2} + \frac{1}{\alpha^2 + \beta^2 - \gamma^2} = 0</math>.
  
 
== Solution ==
 
== Solution ==
  
−
i)<math>(\alpha + \beta + \gamma)^2 = 0</math>
+
i) <math>(\alpha + \beta + \gamma)^2 = 0</math>
  
 
<math>\alpha^2 + \beta^2 - \gamma^2 +2\gamma^2 -2\alpha\beta -2\beta\gamma -2\alpha\gamma = 0</math>
 
<math>\alpha^2 + \beta^2 - \gamma^2 +2\gamma^2 -2\alpha\beta -2\beta\gamma -2\alpha\gamma = 0</math>
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<math>\alpha^2 + \beta^2 - \gamma^2 = -2(\beta + \gamma)(\alpha + \gamma) </math>
 
<math>\alpha^2 + \beta^2 - \gamma^2 = -2(\beta + \gamma)(\alpha + \gamma) </math>
  
−
ii)<math>2(\alpha + \beta + \gamma) = 0</math>
+
ii) <math>2(\alpha + \beta + \gamma) = 0</math>
  
 
<math>(\alpha + \beta) + (\alpha + \gamma) + (\beta + \gamma)= 0</math>
 
<math>(\alpha + \beta) + (\alpha + \gamma) + (\beta + \gamma)= 0</math>
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<math>\frac{1}{-2(\beta + \gamma)(\alpha + \gamma)} + \frac{1}{-2(\alpha + \beta)(\beta + \gamma)} + \frac{1}{-2(\alpha + \beta)(\alpha + \gamma)} = 0</math>
 
<math>\frac{1}{-2(\beta + \gamma)(\alpha + \gamma)} + \frac{1}{-2(\alpha + \beta)(\beta + \gamma)} + \frac{1}{-2(\alpha + \beta)(\alpha + \gamma)} = 0</math>
  
−
Form part i)
+
Form part i) it becomes
  
 
<math>\frac{1}{\beta^2 + \gamma^2 - \alpha^2} + \frac{1}{\gamma^2 + \alpha^2 - \beta^2} + \frac{1}{\alpha^2 + \beta^2 - \gamma^2} = 0</math>
 
<math>\frac{1}{\beta^2 + \gamma^2 - \alpha^2} + \frac{1}{\gamma^2 + \alpha^2 - \beta^2} + \frac{1}{\alpha^2 + \beta^2 - \gamma^2} = 0</math>

Latest revision as of 13:41, 24 November 2006

Problem

If $\alpha, \beta, \gamma \in \mathbb{R}- \{0\}$ with $\alpha + \beta + \gamma = 0$, prove that

i) $\alpha^2 + \beta^2 - \gamma^2 = -2(\beta + \gamma)(\alpha + \gamma)$

ii) $\frac{1}{\beta^2 + \gamma^2 - \alpha^2} + \frac{1}{\gamma^2 + \alpha^2 - \beta^2} + \frac{1}{\alpha^2 + \beta^2 - \gamma^2} = 0$.

Solution

i) $(\alpha + \beta + \gamma)^2 = 0$

$\alpha^2 + \beta^2 - \gamma^2 +2\gamma^2 -2\alpha\beta -2\beta\gamma -2\alpha\gamma = 0$

$\alpha^2 + \beta^2 - \gamma^2 = 2(\gamma^2 -\alpha\beta -\beta\gamma -\alpha\gamma)$

$\alpha^2 + \beta^2 - \gamma^2 = -2(\beta + \gamma)(\alpha + \gamma)$

ii) $2(\alpha + \beta + \gamma) = 0$

$(\alpha + \beta) + (\alpha + \gamma) + (\beta + \gamma)= 0$

$\frac{\alpha + \beta}{-2(\alpha + \beta)(\beta + \gamma)(\alpha + \gamma)} + \frac{\alpha + \gamma}{-2(\alpha + \beta)(\beta + \gamma)(\alpha + \gamma)} + \frac{\beta + \gamma}{-2(\alpha + \beta)(\beta + \gamma)(\alpha + \gamma)} = 0$

$\frac{1}{-2(\beta + \gamma)(\alpha + \gamma)} + \frac{1}{-2(\alpha + \beta)(\beta + \gamma)} + \frac{1}{-2(\alpha + \beta)(\alpha + \gamma)} = 0$

Form part i) it becomes

$\frac{1}{\beta^2 + \gamma^2 - \alpha^2} + \frac{1}{\gamma^2 + \alpha^2 - \beta^2} + \frac{1}{\alpha^2 + \beta^2 - \gamma^2} = 0$


See also

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