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

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== Problem 1 ==
 
== Problem 1 ==
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Let <math>\Alpha\Beta\Gamma</math> be a given triangle and <math>\Mu</math> the midpoint of the side <math>\Beta\Gamma</math>. The circle with diameter <math>\Alpha\Beta</math> cuts <math>\Alpha\Gamma</math> at <math>\Delta</math> and form <math>\Delta</math> we bring <math>\Delta\Zeta=//\Mu\Gamma</math> (<math>\Delta</math> is out of the triangle). Prove that the area of the quadrilateral <math>\Alpha\Mu\Gamma\Zeta</math> is equal to the area of the triangle <math>\Alpha\Beta\Gamma</math>.
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Let <math>\alpha\beta\gamma</math> be a given triangle and <math>\mu</math> the midpoint of the side <math>\beta\gamma</math>. The circle with diameter <math>\alpha\beta</math> cuts <math>\alpha\gamma</math> at <math>\delta</math> and form <math>\delta</math> we bring <math>\delta\zeta=\mu\gamma</math> (<math>\delta</math> is out of the triangle). Prove that the area of the quadrilateral <math>\alpha\mu\gamma\zeta</math> is equal to the area of the triangle <math>\alpha\beta\gamma</math>.
  
 
[[2006 Seniors Pancyprian/2nd grade/Problem 1|Solution]]
 
[[2006 Seniors Pancyprian/2nd grade/Problem 1|Solution]]

Revision as of 20:16, 9 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 $\Alpha\Beta\Gamma\Delta$ (Error compiling LaTeX. Unknown error_msg), that has no parallel sides, is inscribed in a circle, its sides $\Delta\Alpha$ (Error compiling LaTeX. Unknown error_msg), $\Gamma\Beta$ (Error compiling LaTeX. Unknown error_msg) meet at $\Epsilon$ (Error compiling LaTeX. Unknown error_msg) and its sides $\Beta\Alpha$ (Error compiling LaTeX. Unknown error_msg), $\Gamma\Delta$ meet at $\Zeta$ (Error compiling LaTeX. Unknown error_msg). If the bisectors of of $\angle\Delta\Epsilon\Gamma$ (Error compiling LaTeX. Unknown error_msg) and $\angle\Gamma\Zeta\Beta$ (Error compiling LaTeX. Unknown error_msg) intersect the sides of the quadrilateral at th points $\Kappa, \Lambda, \Mu, \Nu$ (Error compiling LaTeX. Unknown error_msg) prove that

i)the bisectors intersect normally

ii)the points $\Kappa, \Lambda, \Mu, \Nu$ (Error compiling LaTeX. Unknown error_msg) 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 out of 50, who is sitting between two girls.

Solution

See also

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