TITLE: Small quadrilaterals containing a given convex region
QUESTION [9 upvotes]: It is easy to prove that
(*) Every convex planar set of area 1 is contained in a quadrilateral of area 2.
It is also easy to see that statement (*) remains true if the constant 2 is replaced with a somewhat smaller one. Contest: Find such a constant, the smaller the better.
Update:
Reaching $\sqrt 2$ and even a strictly smaller value was proved by Chakerian (1973) and Kuperberg (1983) and the research challenge offered is to improve it even further, and perhaps even to verify the conjecture that the minimum is attained by a regular pentagon. But any nice arguments for bounds below 2 are welcome.

REPLY [5 votes]: G. D. Chakerian, Minimum area of circumscribed polygons, Elem. Math. 28 (1973), 108–111, MR0322682 (48 #1044) proved that if $K$ is a convex body of area 1 in the plane then $K$ is contained in a quadrilateral of area at most $\sqrt2$. 
W. Kuperberg, On minimum area quadrilaterals and triangles circumscribed about convex plane regions, Elem. Math. 38 (1983), no. 3, 57–61, MR0703939 (85a:52009), proved that the infimum of the area ratio is strictly less than $\sqrt2$, and suggested the infimum might be attained when $K$ is a regular pentagon.