Problem:
A quadrilateral is inscribed in a circle of radius 2002β. Three of the sides of this quadrilateral have length 200. What is the length of its fourth side?
Answer Choices:
A. 200
B. 2002β
C. 2003β
D. 3002β
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 β AOB=β BOC=β COD=ΞΈ. Because inscribed β BAD is half the
size of central β BOD=2ΞΈ, it follows that β BAD=ΞΈ. Let M be the midpoint of AB. Then sin(2ΞΈβ)=AOAMβ=2002β100β=22β1β. Then cosΞΈ=1β2sin2(2ΞΈβ)=43β. Hence AE=ABcosΞΈ=200β 43β=150, and FD=150 as well. Because EF=BC=200, the remaining side AD=AE+EF+FD=150+200+150=500β.
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 β AOB,β BOC, and β COD is less than 60β, 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 β³OAB and β³OBC are congruent and isosceles, it follows that β ABO=β OBC=β OGH=β AGB. Thus β³ABG, β³OGH, and β³OBC are similar and isosceles with BGABβ=GHOGβ=BCOBβ=2002002ββ=2β. Then AG=AB=200,BG=2βABβ=2β200β=1002β, and GH=2βOGβ=2βBOβBGβ=2β2002ββ1002ββ=100. Therefore AD=AG+GH+HD=200+100+200=500β.
OR
Let ΞΈ be the central angle that subtends the side of length 200 . Then by the Law of Cosines, (2002β)2+(2002β)2β2(2002β)2cosΞΈ=2002, which gives cosΞΈ=43β. The Law of Cosines also gives the square of the fourth side of the quadrilateral as
(2002β)2+(2002β)2β2(2002β)2cos(3ΞΈ)
=160,000β160,000(4cos3ΞΈβ3cosΞΈ)=250,000.
Thus the fourth side has length 250,000β=500β.
