MathDB

5

Part of 2006 India National Olympiad

Problems(1)

In a cyclic quadrilateral ABCD, <ABC=120 degrees

Source: Indian National Maths Olympiad, Problem 5

2/5/2006
In a cyclic quadrilateral ABCDABCD, AB=aAB=a, BC=bBC=b, CD=cCD=c, ∠ABC=120∘\angle ABC = 120^\circ and ∠ABD=30∘\angle ABD = 30^\circ. Prove that (1) c≥a+bc \ge a + b; (2) ∣c+a−c+b∣=c−a−b|\sqrt{c + a} - \sqrt{c + b} | = \sqrt{c - a - b}.
trigonometryinequalitiesfunctiongeometrycircumcirclecyclic quadrilateralgeometry proposed