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

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== Problem 4 ==
 
== Problem 4 ==
A quadrilateral <math>\Alpha\Beta\Gamma\Delta</math>, that has no parallel sides, is inscribed in a circle, its sides <math>\Delta\Alpha</math>, <math>\Gamma\Beta</math> meet at <math>\Epsilon</math> and its sides <math>\Beta\Alpha</math>, <math>\Gamma\Delta</math> meet at <math>\Zeta</math>.
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A quadrilateral <math>\alpha\beta\gamma\delta</math>, that has no parallel sides, is inscribed in a circle, its sides <math>\delta\alpha</math>, <math>\gamma\beta</math> meet at <math>\epsilon</math> and its sides <math>\beta\alpha</math>, <math>\gamma\delta</math> meet at <math>\zeta</math>.
If the bisectors of of <math>\angle\Delta\Epsilon\Gamma</math> and <math>\angle\Gamma\Zeta\Beta</math> intersect the sides of the quadrilateral at th points <math>\Kappa, \Lambda, \Mu, \Nu</math> prove that
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If the bisectors of <math>\angle\delta\epsilon\gamma</math> and <math>\angle\gamma\zeta\beta</math> intersect the sides of the quadrilateral at th points <math>\kappa, \lambda, \mu, \nu</math> prove that
  
 
i)the bisectors intersect normally
 
i)the bisectors intersect normally
  
ii)the points <math>\Kappa, \Lambda, \Mu, \Nu</math> are vertices of a rhombus.
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ii)the points <math>\kappa, \lambda, \mu, \nu</math> are vertices of a rhombus.
  
 
[[2006 Seniors Pancyprian/2nd grade/Problem 4|Solution]]
 
[[2006 Seniors Pancyprian/2nd grade/Problem 4|Solution]]
  
 
== Problem 5 ==
 
== Problem 5 ==
Fifty persons, twenty five boys and twenty five girls are sitting around a table. Prove that there is a person out out of 50, who is sitting between two girls.
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Fifty persons, twenty five boys and twenty five girls are sitting around a table. Prove that there is a person out of 50 who is sitting between two girls.
  
 
[[2006 Seniors Pancyprian/2nd grade/Problem 5|Solution]]
 
[[2006 Seniors Pancyprian/2nd grade/Problem 5|Solution]]

Latest revision as of 11:19, 10 October 2007

Problem 1

Let $\alpha\beta\gamma$ be a given triangle and $\mu$ the midpoint of the side $\beta\gamma$. The circle with diameter $\alpha\beta$ cuts $\alpha\gamma$ at $\delta$ and form $\delta$ we bring $\delta\zeta=\mu\gamma$ ($\delta$ is out of the triangle). Prove that the area of the quadrilateral $\alpha\mu\gamma\zeta$ is equal to the area of the triangle $\alpha\beta\gamma$.

Solution

Problem 2

Find all three digit numbers $\overline{xyz}$(=100x+10y+z) for which $\frac {7}{4}(\overline{xyz})=\overline{zyx}$.

Solution

Problem 3

i)Convert $\Alpha=sin(x-y)+sin(y-z)+sin(z-x)$ (Error compiling LaTeX. Unknown error_msg) into product.

ii)Prove that: If in a triangle $\Alpha\Beta\Gamma$ (Error compiling LaTeX. Unknown error_msg) is true that $\alpha sin \Beta + \beta sin \Gamma + \gamma sin \Alpha= \frac {\alpha+\beta+\gamma}{2}$ (Error compiling LaTeX. Unknown error_msg), then the triangle is isosceles.

Solution

Problem 4

A quadrilateral $\alpha\beta\gamma\delta$, that has no parallel sides, is inscribed in a circle, its sides $\delta\alpha$, $\gamma\beta$ meet at $\epsilon$ and its sides $\beta\alpha$, $\gamma\delta$ meet at $\zeta$. If the bisectors of $\angle\delta\epsilon\gamma$ and $\angle\gamma\zeta\beta$ intersect the sides of the quadrilateral at th points $\kappa, \lambda, \mu, \nu$ prove that

i)the bisectors intersect normally

ii)the points $\kappa, \lambda, \mu, \nu$ are vertices of a rhombus.

Solution

Problem 5

Fifty persons, twenty five boys and twenty five girls are sitting around a table. Prove that there is a person out of 50 who is sitting between two girls.

Solution

See also