Problem:
Consider all quadrilaterals ABCD such that AB=14,BC=9,CD=7, and DA=12. What is the radius of the largest possible circle that fits inside or on the boundary of such a quadrilateral?
Answer Choices:
A. 15
B. 21
C. 26
D. 5
E. 27 Solution:
Because AB+CD=21=BC+DA, it follows that ABCD always has an inscribed circle tangent to its four sides. Let r be the radius of the inscribed circle. Note that [ABCD]=21r(AB+BC+CD+DA)=21r. Thus the radius is maximum when the area is maximized. Note that [ABC]=21⋅14⋅9sinB=63sinB and [ACD]=21⋅12⋅7sinD=42sinD. On the one hand,
with equality if and only if B+D=π (that is ABCD is cyclic). Therefore [ABCD]2≤1052−212=212(52−1)=422⋅6, and the required maximum r=211[ABCD]=26.
\section*{OR}
Establish as in the first solution that r is maximized when the area is maximized. Bretschneider's formula, which generalizes Brahmagupta's formula, states that the area of an arbitrary quadrilateral with side lengths a,b,c, and d, is given by
(s−a)(s−b)(s−c)(s−d)−abcdcos2θ
where s=21(a+b+c+d) and θ is half the sum of either pair of opposite angles. For a,b,c, and d fixed, the area is maximized when cosθ=0. Thus the area is maximized when θ=21π, that is, when the quadrilateral is cyclic. In this case, the area equals 7⋅12⋅14⋅9=426, and the required maximum radius r=211⋅426=26.