Two intersecting circles of radius \(R\) are given, and the distance between their centers is greater than \(R\). Prove that \(\beta = 3\alpha\) (Fig.).
On the sides \(AB\), \(BC\) and \(AC\) of the triangle \(ABC\) points \(P\), \(M\) and \(K\) are chosen so that the segments \(AM\), \(BK\) and \(CP\) intersect at one point and \[\vec{AM} + \vec{BK}+\vec{CP} = 0\] Prove that \(P\), \(M\) and \(K\) are the midpoints of the sides of the triangle \(ABC\).