Inequalities (Unit 1 – Unit 3) Solutions to Exercises Inequalities (Unit ...

5. Let AB = BC = CD = a and DE = EF =FA = b. As shown in the figure, constructequilateral triangles ABX and DEY. Since∠AXB = ∠DYE = 60°, AX = XB = BD = aand DY = YE = EA = b, the two hexagonsABCDEF and AXBDYE are congruentand so CF = XY.AXBSince∠AXB + ∠AGB = 180° = ∠DHE + ∠DYE,FEHGDCAXBG and DYEH are cyclic quadrilaterals.Hence by the Ptolemy’s theorem,AB ⋅ XG = AX ⋅ BG + XB ⋅ AG ,Ywhich is equivalent toaXG = aBG + aAG , orXG = BG + AG .Similarly, we have YH = DH + EH and henceAG + GB + GH + DH + HE = XG + GH + YH ≥ XY = EF .6. As shown in the figure, extend BC and EF todraw a rectangle PQRS enclosing the hexagon.Since opposite sides of the hexagon are parallel,opposite angles are equal (i.e. ∠A = ∠D, ∠B =∠E and ∠C = ∠F). Let a, b, c, d, e, f denote thelengths of AB, BC, CD, DE, EF and FArespectively. We havePASBFCEQDR2BF ≥ PA + AS + QD + DR= asin B+ f sin F + csin C+dsinE= asin B+ f sin C+ csin C+dsinBHenceRABF 1 ⎛asin B f sin C csin C dsinB⎞= = ⎜ + + + ⎟2sin A 4 ⎝ sin A sin A sin A sin A ⎠ .Similarly, we have14