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2006 Seniors Pancyprian/2nd grade/Problems

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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) into product.

ii)Prove that: If in a triangle \Alpha\Beta\Gamma is true that \alpha sin \Beta + \beta sin \Gamma + \gamma sin \Alpha= \frac {\alpha+\beta+\gamma}{2}, 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

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