In following, ($x$) denotes condition $x$, and ($x$') denotes condition $\neg x$.

For a polygon of $2n$ sides, let it be given that (a) opposite sides are parallel. It is not difficult to find examples where (b') opposite interior angles differ and (c') opposite sides are of different lengths.

Example: Octagon (1,0), (2,1), (0,1), (0,2), (3,2), (4,3), (5,3), (5,0), (1,0).

I am tempted to conjecture that if the polygon is convex and opposite sides are parallel, then (b) opposite interior angles match and (c) opposite sides have equal length. But I don't see how to prove it and don't have a counterexample.

  1. Does anyone know of a counterexample? Update 1: Following the first remark by André Nicolas, here's a construction of convex polygon Q where (a) and (b) hold but not (c): Let P be a regular hexagon, with one edge extending from (-5,0) to (+5,0). Intersect P with all points where -6 =< x =< 7 to form an octagon Q.

  2. If the conjecture is false, what conditions would be sufficient to enforce (b) or (c) ? (It seems obvious that (a) and (b) imply (c), while (a) and (c) imply (b))

  • $\begingroup$ Opposite sides don't need to have equal length. $\endgroup$ – André Nicolas Sep 10 '11 at 2:32
  • $\begingroup$ @André: For $n>2$. :) $\endgroup$ – Blue Sep 10 '11 at 3:04
  • $\begingroup$ A countercounterexample. $\endgroup$ – André Nicolas Sep 10 '11 at 3:15
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    $\begingroup$ Did you mean to write \neg x ($\neg x$) instead of \not x ($\not x$)? $\endgroup$ – Rahul Sep 10 '11 at 4:36
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    $\begingroup$ ( Convex + (a) ) $\implies$ (b): draw the diagonal and consider the "Z"s (or are they "N"s?) made with the sides. $\endgroup$ – Blue Sep 10 '11 at 6:19

For (c), consider a non-square rectangle. For (b) if you define opposite sides as half the sides around the polygon, you should have them equal by the theorem that corresponding angles with a transversal of parallel lines are equal.

  • $\begingroup$ A rectangle satisfies (a), (b), and (c), no? $\endgroup$ – Gerry Myerson Sep 10 '11 at 5:03
  • $\begingroup$ @Gerry Myerson: Not (c), that's why I said a non-square rectangle. Maybe you missed the prime in opposite sides are different lengths. $\endgroup$ – Ross Millikan Sep 10 '11 at 5:12
  • $\begingroup$ (c) is "opposite sides have equal length". This is true of rectangles, square or not. (c') is "opposite sides have different lengths". Well, that's how I read it. But you are quite right that if opposite sides are parallel then opposite angles are equal, whether (c) holds or not. $\endgroup$ – Gerry Myerson Sep 10 '11 at 5:48
  • $\begingroup$ Yes, the negation of (c) is (c'), "not all opposite sides have equal lengths". Your comment re "if opposite sides are parallel then opposite angles are equal" becomes obviously true via another comment's suggestion to "draw the diagonal" (i.e. call upon Euclid's Proposition 29, "A straight line falling on parallel straight lines makes alternate angles equal to one another..."). $\endgroup$ – James Waldby - jwpat7 Sep 10 '11 at 9:04
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    $\begingroup$ You don't have to draw any diagonals, you can just extend the four sides in question (ignoring any other sides) until you get a parallelogram. $\endgroup$ – Gerry Myerson Sep 10 '11 at 10:29

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