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

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== Problem 1 ==
 
== Problem 1 ==
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Let <math>\Alpha\Beta\Gamma\Delta</math> be a parallelogram. Let <math>(\epsilon)</math> be a straight line passing through <math>\Alpha</math> without cutting <math>\Alpha\Beta\Gamma\Delta</math>. If <math>\Beta ', \Gamma ', \Delta ' </math> are the projections of <math>\Beta, \Gamma, \Delta</math> on <math>(\epsilon)</math> respectively, show that
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Let <math>\text{A}\text{B}\Gamma\Delta</math> be a parallelogram. Let <math>(\epsilon)</math> be a straight line passing through <math>\text{A}</math> without cutting <math>\text{A}\text{B}\Gamma\Delta</math>. If <math>\text{B} ', \Gamma ', \Delta ' </math> are the projections of <math>\text{B}, \Gamma, \Delta</math> on <math>(\epsilon)</math> respectively, show that
  
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a) the distance of <math>\Gamma</math> from <math>(\epsilon)</math> is equal to the sum of the distances <math>\Beta, \Delta</math> from <math>(\epsilon)</math>.
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a) the distance of <math>\Gamma</math> from <math>(\epsilon)</math> is equal to the sum of the distances <math>\text{B}</math> , <math>\Delta</math> from <math>(\epsilon)</math>.
  
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b)Area<math>(\Beta\Gamma\Delta)</math>=Area<math>(\Beta '\Gamma '\Delta ')</math>.
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b) <math>\text{Area}(\text{B}\Gamma\Delta)=\text{Area}(\text{B} '\Gamma '\Delta ')</math>.
  
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[[2006 Specimen Seniors Provincial/2nd grade/Problem 1|Solution]]
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[[Specimen Cyprus Seniors Provincial/2nd grade/Problem 1|Solution]]
  
 
== Problem 2 ==
 
== Problem 2 ==
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== Problem ==
 
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If <math>\alpha=sinx_{1}</math>,<math>\beta=cosx_{1}</math><math>sinx_{2}</math>, <math>\gamma=cosx_{1}cosx_{2} sinx_{3}</math> and <math>\delta=cosx_{1}cosx_{2}cosx_{3}</math> prove that <math>\alpha^2+\beta^2+\gamma^2+\delta^2=1</math>
 
  
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[[2006 Specimen Seniors Provincial/2nd grade/Problem 2|Solution]]
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If <math>\alpha=\sin x_{1}</math>,<math>\beta=\cos x_{1}\sin x_{2}</math>, <math>\gamma=\cos x_{1}\cos x_{2} \sin x_{3}</math> and <math>\delta=\cos x_{1}\cos x_{2}\cos x_{3}</math> prove that <math>\alpha^2+\beta^2+\gamma^2+\delta^2=1</math>
 +
 
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[[Specimen Cyprus Seniors Provincial/2nd grade/Problem 2|Solution]]
  
 
== Problem 3 ==
 
== Problem 3 ==
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Prove that if <math>\kappa, \lambda, \nu</math> are positive integers, then the equation <math>x^2-(\nu +2)\kappa\lambda x+\kappa^2\lambda^2 = 0</math> has irratioanl roots.
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Prove that if <math>\kappa, \lambda, \nu</math> are positive integers, then the equation <math>x^2-(\nu +2)\kappa\lambda x+\kappa^2\lambda^2 = 0</math> has irrational roots.
  
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[[2006 Specimen Seniors Provincial/2nd grade/Problem 3|Solution]]
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[[Specimen Cyprus Seniors Provincial/2nd grade/Problem 3|Solution]]
  
 
== Problem 4 ==
 
== Problem 4 ==
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b) Calculate the value of: <math>\rho_{1}^{2006} + \rho_{2}^{2006}</math>.
 
b) Calculate the value of: <math>\rho_{1}^{2006} + \rho_{2}^{2006}</math>.
  
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[[2006 Specimen Seniors Provincial/2nd grade/Problem 4|Solution]]
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[[Specimen Cyprus Seniors Provincial/2nd grade/Problem 4|Solution]]
  
 
== See also ==
 
== See also ==

Latest revision as of 19:19, 18 January 2021

Problem 1

Let $\text{A}\text{B}\Gamma\Delta$ be a parallelogram. Let $(\epsilon)$ be a straight line passing through $\text{A}$ without cutting $\text{A}\text{B}\Gamma\Delta$. If $\text{B} ', \Gamma ', \Delta '$ are the projections of $\text{B}, \Gamma, \Delta$ on $(\epsilon)$ respectively, show that

a) the distance of $\Gamma$ from $(\epsilon)$ is equal to the sum of the distances $\text{B}$ , $\Delta$ from $(\epsilon)$.

b) $\text{Area}(\text{B}\Gamma\Delta)=\text{Area}(\text{B} '\Gamma '\Delta ')$.

Solution

Problem 2

If $\alpha=\sin x_{1}$,$\beta=\cos x_{1}\sin x_{2}$, $\gamma=\cos x_{1}\cos x_{2} \sin x_{3}$ and $\delta=\cos x_{1}\cos x_{2}\cos x_{3}$ prove that $\alpha^2+\beta^2+\gamma^2+\delta^2=1$

Solution

Problem 3

Prove that if $\kappa, \lambda, \nu$ are positive integers, then the equation $x^2-(\nu +2)\kappa\lambda x+\kappa^2\lambda^2 = 0$ has irrational roots.

Solution

Problem 4

If $\rho_{1}, \rho_{2}$ are the roots of equation $x^2-x+1=0$ then:

a) Prove that $\rho_{1}^3=\rho_{2}^3 = -1$ and

b) Calculate the value of: $\rho_{1}^{2006} + \rho_{2}^{2006}$.

Solution

See also

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