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

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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>.
  
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[[ 2006 Cyprus Seniors Provincial/2nd grade/Problem 1|Solution]]
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[[2006 Cyprus Seniors Provincial/2nd grade/Problem 1|Solution]]
  
 
== Problem 2 ==
 
== Problem 2 ==
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Let <math>\Alpha, \Beta, \Gamma</math> be consecutive points on a straight line <math>(\epsilon)</math>. We construct equilateral triangles <math>\Alpha\Beta\Delta</math> and <math>\Beta\Gamma\Epsilon</math> to the same side of <math>(\epsilon)</math>.
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Let <math>\text{A}, \text{B}, \Gamma</math> be consecutive points on a straight line <math>(\epsilon)</math>. We construct equilateral triangles <math>\text{AB}\Delta</math> and <math>\text{B}\Gamma\text{E}</math> to the same side of <math>(\epsilon)</math>.
  
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a) Prove that <math>\angle\Alpha\Epsilon\Beta = \angle\Delta\Gamma\Beta</math>
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a) Prove that <math>\angle \text{AEB} = \angle\Delta\Gamma\text{B}</math>
  
 
b) If <math>x_{1}</math> is the distance of <math>A</math> form <math>\Gamma\Delta</math> and <math>x_{2}</math> is the distance of <math>\Gamma</math> form <math>\Alpha\Gamma</math> prove that
 
b) If <math>x_{1}</math> is the distance of <math>A</math> form <math>\Gamma\Delta</math> and <math>x_{2}</math> is the distance of <math>\Gamma</math> form <math>\Alpha\Gamma</math> prove that
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<math>\frac{x_{1}}{x_{2}} = \frac{Area(\Alpha\Gamma\Delta)}{Area(\Alpha\Gamma\Epsilon)} = \frac{\Alpha\Beta}{\Beta\Gamma}</math>.
 
<math>\frac{x_{1}}{x_{2}} = \frac{Area(\Alpha\Gamma\Delta)}{Area(\Alpha\Gamma\Epsilon)} = \frac{\Alpha\Beta}{\Beta\Gamma}</math>.
  
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[[ 2006 Cyprus Seniors Provincial/2nd grade/Problem 2|Solution]]
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[[2006 Cyprus Seniors Provincial/2nd grade/Problem 2|Solution]]
  
 
== Problem 3 ==
 
== Problem 3 ==
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If <math>\Alpha=\frac{1-cos\theta}{sin\theta}</math> and <math>\Beta=\frac{1-sin\theta}{cos\theta}</math>, prove that  
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If <math>\alpha=\frac{1-\cos \theta}{\sin \theta}</math> and <math>\beta=\frac{1-sin\theta}{cos\theta}</math>, prove that  
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<math>\frac{\Alpha^2}{(1+\Alpha^2)^2} + \frac{\Beta^2}{(1+\Beta^2)^2} = \frac{1}{4}</math>.
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<math>\frac{\alpha^2}{(1+\alpha^2)^2} + \frac{\beta^2}{(1+\beta^2)^2} = \frac{1}{4}</math>.
  
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[[ 2006 Cyprus Seniors Provincial/2nd grade/Problem 3|Solution]]
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[[2006 Cyprus Seniors Provincial/2nd grade/Problem 3|Solution]]
  
 
== Problem 4 ==
 
== Problem 4 ==
 
Find all integers pairs (x,y) that verify at the same time the inequalities <math>x^2\leq\frac{y^2+2x-1}{2}</math> and <math>y^2\leq\frac{x^2-2y-1}{2}</math>.
 
Find all integers pairs (x,y) that verify at the same time the inequalities <math>x^2\leq\frac{y^2+2x-1}{2}</math> and <math>y^2\leq\frac{x^2-2y-1}{2}</math>.
  
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[[ 2006 Cyprus Seniors Provincial/2nd grade/Problem 4|Solution]]
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[[2006 Cyprus Seniors Provincial/2nd grade/Problem 4|Solution]]
  
 
== See also ==
 
== See also ==

Latest revision as of 21:07, 10 April 2013

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 $\text{A}, \text{B}, \Gamma$ be consecutive points on a straight line $(\epsilon)$. We construct equilateral triangles $\text{AB}\Delta$ and $\text{B}\Gamma\text{E}$ to the same side of $(\epsilon)$.

a) Prove that $\angle \text{AEB} = \angle\Delta\Gamma\text{B}$

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 $\alpha=\frac{1-\cos \theta}{\sin \theta}$ and $\beta=\frac{1-sin\theta}{cos\theta}$, prove that $\frac{\alpha^2}{(1+\alpha^2)^2} + \frac{\beta^2}{(1+\beta^2)^2} = \frac{1}{4}$.

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

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