MathDB
a trig approach... (long)

Source: flanders '92

September 27, 2005
trigonometrygeometryparameterizationinequalitiescyclic quadrilateral

Problem Statement

Let A,B,PA,B,P positive reals with PA+BP\le A+B. (a) Choose reals θ1,θ2\theta_1,\theta_2 with Acosθ1+Bcosθ2=PA\cos\theta_1 + B\cos\theta_2=P and prove that Asinθ1+Bsinθ2(A+BP)(A+B+P) A\sin\theta_1 + B\sin\theta_2 \le \sqrt{(A+B-P)(A+B+P)} (b) Prove equality is attained when θ1=θ2=arccos(PA+B)\theta_1=\theta_2=\arccos\left(\dfrac{P}{A+B}\right). (c) Take A=12xy,B=12wzA=\dfrac{1}{2}xy, B=\dfrac{1}{2}wz and P=14(x2+y2z2w2)P=\dfrac14 \left(x^2+y^2-z^2-w^2\right) with 0<xyx+z+w0<x\le y\le x+z+w, z,w>0z,w>0 and z2+w2<x2+y2z^2+w^2<x^2+y^2. Show that we can translate (a) and (b) into the following theorem: from all quadrilaterals with (ordered) sidelenghts (x,y,z,w)(x,y,z,w), the cyclical one has the greatest area.