From - Wed Aug 19 14:15:28 1998 Path: ar4dec01!ar4news.dlh.de!newsfeed2.de.ibm.net!newsm2.ibm.net!ibm.net!news-peer.gip.net!news-raspail.gip.net!news.gsl.net!gip.net!oleane!rain.fr!wanadoo.fr!not-for-mail From: "Laurent Morel" Newsgroups: rec.puzzles Subject: Re: A triangle in a convex polygon Date: 19 Aug 1998 07:27:55 GMT Organization: Wanadoo - (Client of French Internet Provider) Lines: 46 Message-ID: <01bdcaf0$2c2313a0$LocalHost@dell> References: <01bdc4a2$e5207ea0$LocalHost@dell> <6rcffj$7bf@fridge.shore.net> NNTP-Posting-Host: gre1-146.abo.wanadoo.fr X-Newsreader: Microsoft Internet News 4.70.1155 David A Karr a écrit : > Laurent Morel wrote: > >Prove that a convex polygon of area 1 contains a triangle of area >= 1/4 > > ObPuzzle 2: What is the maximum value you can substitute for "1/4" in > the first puzzle while the statement is still true? (That is, there > is some k for which a convex polygon of area 1 exists that contains no > triangle of area > k, but for which every convex polygon of area 1 > contains a triangle of area < k. Find k.) > > > For Laurent's puzzle I offer the following : langlois bruno found another solution : If P is a convex polygone of >4 vertices, let ABC be a triangle of P with maximal area (A,B,C are vertices of P). Draw three straightlines : D1 = parallel to (AB) through C. D2 = parallel to (AC) through B. D3 = parallel to (CB) through A. A' = intersection of D1 and D2 B' = intersection of D1 and D3 C' = intersection of D2 and D3 Now, all other vertices of P are in the triangle A'B'C' because ABC has been choosen with a maximal area (proof by contradiction : if a vertex S of P were outside A'B'C', one could construct another triangle ABS, BCS or ACS of greater area) Then conclude that area(ABC) = area(A'B'C')/4 >= area(P)/4. Concerning obPuzzle 2, I am afraid it is a hard problem. Nobody in news:fr.sci.maths found an answer (the same problem had been asked). We may consider the almost reverse problem : Given a triangle ABC, find a convex polygone of maximal area that contains ABC without containing any triangle of greater area. We may also suppose WLOG that ABC is equilateral if it can help. We know that k is between 1/4 and 1/2. 9/(4.pi.sqrt(3)) ~ 0.41 was conjectured... -- Laurent