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

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[[ 2006 Cyprus Seniors Provincial/2nd grade/Problem 4|Solution]]
 
[[ 2006 Cyprus Seniors Provincial/2nd grade/Problem 4|Solution]]
 
  
 
== See also ==
 
== See also ==

Revision as of 04:33, 12 November 2006

Problem 1

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

Problem 2

Let $\Alpha, \Beta, \Gamma$ (Error compiling LaTeX. Unknown error_msg) be consecutive points on a straight line $(\epsilon)$. We construct equilateral triangles $\Alpha\Beta\Delta$ (Error compiling LaTeX. Unknown error_msg) and $\Beta\Gamma\Epsilon$ (Error compiling LaTeX. Unknown error_msg) to the same side of $(\epsilon)$.

a) Prove that $\angle\Alpha\Epsilon\Beta = \angle\Delta\Gamma\Beta$ (Error compiling LaTeX. Unknown error_msg)

b) If $x_{1}$ is the distance of $A$ form $\Gamma\Delta$ and $x_{2}$ is the distance of $\Gamma$ form $\Alpha\Gamma$ (Error compiling LaTeX. Unknown error_msg) prove that

$\frac{x_{1}}{x_{2}} = \frac{Area(\Alpha\Gamma\Delta)}{Area(\Alpha\Gamma\Epsilon)} = \frac{\Alpha\Beta}{\Beta\Gamma}$ (Error compiling LaTeX. Unknown error_msg).

Solution

Problem 3

If $\Alpha=\frac{1-cos\theta}{sin\theta}$ (Error compiling LaTeX. Unknown error_msg) and $\Beta=\frac{1-sin\theta}{cos\theta}$ (Error compiling LaTeX. Unknown error_msg), prove that $\frac{\Alpha^2}{(1+\Alpha^2)^2} + \frac{\Beta^2}{(1+\Beta^2)^2} = \frac{1}{4}$ (Error compiling LaTeX. Unknown error_msg).

Solution

Problem 4

Find all integers pairs (x,y) that verify at the same time the inequalities $x^2\leq\frac{y^2+2x-1}{2}$ and $y^2\leq\frac{x^2-2y-1}{2}$.

Solution

See also