8
$\begingroup$

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.

Improve this question
$\endgroup$
3

1 Answer 1

Reset to default
6
$\begingroup$

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.

Improve this answer
$\endgroup$
10
  • 10
    $\begingroup$ The same W. Kuperberg who asked the question? $\endgroup$
    – Robert Israel
    Commented Jul 11, 2013 at 4:14
  • 8
    $\begingroup$ I would hope not. I would hope that anyone posting a question here who had previously published on the question would remember what was published and include it in the statement of the question. In any event, even if my information is not news to OP, it may be news to others reading MO. $\endgroup$
    – Gerry Myerson
    Commented Jul 11, 2013 at 5:55
  • 14
    $\begingroup$ Wlodzimierz, please don't do anything like that again. $\endgroup$
    – Kim Morrison
    Commented Jul 11, 2013 at 14:15
  • 4
    $\begingroup$ @Scott: Why not? Is it against the rules? Have I offended you or anyone else? Besides, even with Gerry's comments, the question remains valid. Any relevant remarks will be appreciated. $\endgroup$
    – Wlodek Kuperberg
    Commented Jul 11, 2013 at 16:13
  • 7
    $\begingroup$ Prof. Kuperberg, the issue is not whether your question is good or bad; the issue is whether this is the place for it. MathOverflow has a purpose different from that of pursuing open problems: it is to get existing answers quickly to definite questions in a certain realm. Instead of presenting a chllenge, you might ask if there is recent progress, or if someone has tried a specific approach on a small part of the problem. $\endgroup$
    – The Masked Avenger
    Commented Jul 11, 2013 at 16:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .