Problem:
Cyclic quadrilateral ABCD has lengths BC=CD=3 and DA=5 with β CDA=120β. What is the length of the shorter diagonal of ABCD ?
Answer Choices:
A. 731β
B. 733β
C. 5
D. 739β
E. 741β
Solution:
Place the figure in the coordinate plane with D=(0,0) and A=(5,0). Because β CDA=120β and CD=3, it follows that C=(β23β,23β3β). The perpendicular bisectors of CD and AD intersect at the center of the circumscribing circle. The midpoint of CD is (β43β,43β3β). The line through CD has slope β3β, so a line perpendicular to that has slope 33ββ. The perpendicular bisector of CD therefore has equation
yβ43β3β=33ββ(x+43β)
The perpendicular bisector of AD has equation x=25β. Solving this system of equations locates the center of the circumscribing circle at O(25β,611β3β).
By the Distance Formula, the radius of the circle is
r=OD=(25β)2+(611β3β)2β=373ββ.
Let ΞΈ be the measure of β BCO. By the Law of Cosines applied to β³BCO,
which gives cosΞΈ=143β3β. A Double Angle Formula then gives cos2ΞΈ=2β 19627ββ1=β9871β. Triangles β³BOC and β³COD are congruent isosceles triangles, so β BCD has measure 2ΞΈ. By the Law of Cosines applied to β³BCD,
so BD=739β. The Law of Cosines applied to β³ADC gives AC=7, so the shorter diagonal has length (D)739ββ.
OR
The Law of Cosines applied to β³ADC gives
CA2=32+52β2β 3β 5β cos120β.
Because cos120β=β21β, diagonal AC has length 9+25+15β=7. Because ABCD is cyclic, β ABC is supplementary to β ADC, so β ABC=60β. The Law of Cosines applied to β³ABC gives
72=32+BA2β2β 3β BAβ cos60β
Simplifying yields BA2β3β BAβ40=0, so BA=8. By Ptolemy's Theorem
CAβ BD=BAβ CD+BCβ AD
Substituting gives 7β BD=8β 3+3β 5, so BD=(D)739ββ, which is less than the length of diagonal AC.
