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. 15β
B. 21β
C. 26β
D. 5
E. 27β 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]=21βr(AB+BC+CD+DA)=21r. Thus the radius is maximum when the area is maximized. Note that [ABC]=21ββ 14β 9sinB=63sinB and [ACD]=21ββ 12β 7sinD=42sinD. On the one hand,
with equality if and only if B+D=Ο (that is ABCD is cyclic). Therefore [ABCD]2β€1052β212=212(52β1)=422β 6, and the required maximum r=211β[ABCD]=26β.
\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
(sβa)(sβb)(sβc)(sβd)βabcdcos2ΞΈβ
where s=21β(a+b+c+d) and ΞΈ is half the sum of either pair of opposite angles. For a,b,c, and d fixed, the area is maximized when cosΞΈ=0. Thus the area is maximized when ΞΈ=21βΟ, that is, when the quadrilateral is cyclic. In this case, the area equals 7β 12β 14β 9β=426β, and the required maximum radius r=211ββ 426β=26ββ.