Problem:
Cyclic quadrilateral ABCD has lengths BC=CD=3 and DA=5 with ∠CDA=120∘. What is the length of the shorter diagonal of ABCD ?
Answer Choices:
A. 731
B. 733
C. 5
D. 739
E. 741
Solution:
Place the figure in the coordinate plane with D=(0,0) and A=(5,0). Because ∠CDA=120∘ and CD=3, it follows that C=(−23,233). The perpendicular bisectors of CD and AD intersect at the center of the circumscribing circle. The midpoint of CD is (−43,433). The line through CD has slope −3, so a line perpendicular to that has slope 33. The perpendicular bisector of CD therefore has equation
y−433=33(x+43)
The perpendicular bisector of AD has equation x=25. Solving this system of equations locates the center of the circumscribing circle at O(25,6113).
By the Distance Formula, the radius of the circle is
r=OD=(25)2+(6113)2=373.
Let θ be the measure of ∠BCO. By the Law of Cosines applied to △BCO,
349=BO2=BC2+CO2−2BC⋅COcosθ=9+349−2⋅3⋅373cosθ
which gives cosθ=1433. A Double Angle Formula then gives cos2θ=2⋅19627−1=−9871. Triangles △BOC and △COD are congruent isosceles triangles, so ∠BCD has measure 2θ. By the Law of Cosines applied to △BCD,
so BD=739. The Law of Cosines applied to △ADC gives AC=7, so the shorter diagonal has length (D)739.
OR
The Law of Cosines applied to △ADC gives
CA2=32+52−2⋅3⋅5⋅cos120∘.
Because cos120∘=−21, diagonal AC has length 9+25+15=7. Because ABCD is cyclic, ∠ABC is supplementary to ∠ADC, so ∠ABC=60∘. The Law of Cosines applied to △ABC gives
72=32+BA2−2⋅3⋅BA⋅cos60∘
Simplifying yields BA2−3⋅BA−40=0, so BA=8. By Ptolemy's Theorem
CA⋅BD=BA⋅CD+BC⋅AD
Substituting gives 7⋅BD=8⋅3+3⋅5, so BD=(D)739, which is less than the length of diagonal AC.
