Angles in a convex pentagon?

[Tom Shermer]

You can get an angle approaching pi + arccos(1/4) > pi + pi/3
by the following method:
	1) Construct an equilateral not-strictly-convex pentagon
	    with EA and AB colinear, and BC and CD colinear.
	2) Wiggle it. (shrink the appropriate edges slightly,
	    causing slight changes in the angles).
	    
	    Tom

	
> Date: Fri, 18 May 2001 14:21:41 +0200 (MEST)
> From: Jobst Heitzig <heitzig at mbox.math.uni-hannover.de>
> X-Sender: nhabheit at unics
> To: compgeom-discuss at research.bell-labs.com
> Subject: Angles in a convex pentagon?
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> Does anyone know whether the following is true in a convex pentagon ABCDE:
> Given that EA>AB>BC and CD<DE<EA, the angle sum DEA+EAB is at most 4*pi/3.
> 
> I tried hard but could neither prove nor disprove it :-)
> 
> Jobst Heitzig
> 
> 
> 
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