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 quadrilateral inscribed in the circle centered at O, with AB=BC=CD=200, as shown in the figure. 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=(E)500.
