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December 24

quadrilateral question

If every 3 sides of a quadrilateral obey the triangle inequality, does that have a name, or does that perhaps imply some other named property, or is it implied by or equivalent to some other named property? Thanks76.218.104.120 (talk) 03:23, 24 December 2012 (UTC)[reply]

By "every 3 sides of a quadrilateral obey the triangle inequality", do you mean that for any three sides, lengths a, b, and c, we have

a+b>c

?—msh210℠ 05:30, 24 December 2012 (UTC)[reply]

yes.76.218.104.120 (talk) 05:33, 24 December 2012 (UTC)[reply]

There are lots of possible quadrilaterals for which the inequality does not hold, of course, including long thin parallelograms and kites, but I can't think of any property that corresponds to all sets of three sides obeying the inequality, other than the vague "not too irregular". I wonder if something could be proved about the distance of the centroid from the sides? Dbfirs 09:31, 24 December 2012 (UTC)[reply]

Plus on the other hand there are lengths for the sides which satisfy that but you can't form a quadrilateral, e.g. a square of size 2 with diagonals of length 3. You can always make a triangle from figures satisfying the triangle inequality. Dmcq (talk) 10:03, 24 December 2012 (UTC)[reply]

If every 3 sides fit the triangle inequality, then you always have a quadrilateral. The only way that four sides won't make a quadrilateral is if for some side d that d> a+b+c. However by the "every 3 sides obey the triangle inequality", you have d<a+b, so the "won't make a quadrilateral" condition won't apply. An example of an actual quadrilaterla where the "every three sides obey" is a rectangle of sides: 1,3,1,3. We need a property that is true for a rectangle of sides 1, 1.9, 1, 1.9 and not true for a rectangle of sides 1, 2.1, 1, 2.1.Naraht (talk) 20:11, 27 December 2012 (UTC)[reply]

And so the property would have to be "borderline true" for the rectangle 1, 2, 1, 2. The difficulty is that the property would also have to be borderline true for every parallelogram of sides 1, 2, 1, 2, no matter how far you tip the parallelogram over by swiveling on the vertices, no matter how close the area comes to zero. Such a property would be marvelous, but I doubt that one exists. Duoduoduo (talk) 20:44, 27 December 2012 (UTC)[reply]

Perhaps the property, if a property exists, is that as you swivel the vertices, the four #centers# of the 4 triangles formed by each st of 3 sides move with respect to each other in a restricted manner. By "#centers#" i mean incenter or circumcenter or some other of the many triangular centers that have been discovered.Thanks again.RPeterson199.33.32.40 (talk) 21:30, 27 December 2012 (UTC)[reply]

Oops that's a la what Dbfirs suggested above, sorry.199.33.32.40 (talk) 21:32, 27 December 2012 (UTC)[reply]