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

Revision as of 10:56, 11 November 2006

Problem

If $\alpha, \beta, \gamma \in \Re- \{0\}$ 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

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

@@ Line 24: / Line 24: @@
 <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>

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

Revision as of 10:56, 11 November 2006

Problem

Solution

See also