Problem:
A quadrilateral is inscribed in a circle of radius 2002. Three of the sides of this quadrilateral have length 200. What is the length of its fourth side?
Answer Choices:
A. 200
B. 2002
C. 2003
D. 3002
E. 500 Solution:
Let ABCD be the given cyclic quadrilateral with AB=BC=CD=200, and let E and F be the feet of the perpendicular segments from B and C, respectively, to AD, as shown in the figure. Let the center of the circle be O, and let ∠AOB=∠BOC=∠COD=θ. Because inscribed ∠BAD is half the
size of central ∠BOD=2θ, it follows that ∠BAD=θ. Let M be the midpoint of AB. Then sin(2θ)=AOAM=2002100=221. Then cosθ=1−2sin2(2θ)=43. Hence AE=ABcosθ=200⋅43=150, and FD=150 as well. Because EF=BC=200, the remaining side AD=AE+EF+FD=150+200+150=500.
OR
Label the quadrilateral ABCD and the center of the circle as in the first solution. Because the chords AB,BC, and CD are shorter than the radius, each of ∠AOB,∠BOC, and ∠COD is less than 60∘, so O is outside the quadrilateral ABCD. Let G and H be the intersections of AD with OB and OC, respectively. Because AD and BC are parallel, and △OAB and △OBC are congruent and isosceles, it follows that ∠ABO=∠OBC=∠OGH=∠AGB. Thus △ABG, △OGH, and △OBC are similar and isosceles with BGAB=GHOG=BCOB=2002002=2. Then AG=AB=200,BG=2AB=2200=1002, and GH=2OG=2BO−BG=22002−1002=100. Therefore AD=AG+GH+HD=200+100+200=500.
OR
Let θ be the central angle that subtends the side of length 200 . Then by the Law of Cosines, (2002)2+(2002)2−2(2002)2cosθ=2002, which gives cosθ=43. The Law of Cosines also gives the square of the fourth side of the quadrilateral as
