2006 Cyprus Seniors Provincial/2nd grade/Problems

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 $\frac{\alpha^2}{(1+\alpha^2)^2} + \frac{\beta^2}{(1+\beta^2)^2} = \frac{1}{4}$ .