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Let $ABCD$ be any quadrilateral, and let $r, s, t, u$ be its four angle bisectors. If the $r,s,t,u$ are not concurrent (as in the image below on the left), then they intersect to form the vertices of another quadrilateral $PQRS$ (as in the image below on the right). enter image description here

I don't know if there is a standard name for this construction; let's call $PQRS$ the "angle bisector quadrilateral" or abq of $ABCD$.

Here are some fairly simple properties:

  • The angle bisectors are concurrent if and only if $ABCD$ is a quadrilateral in which $AB + CD = BC + AD$.
  • As a special case, if $ABCD$ is a kite (including the special case of a rhombus) then the angle bisectors are concurrent.
  • If the angle bisectors are not concurrent, then the abq is a cyclic quadrilateral -- i.e. one in which the opposite angles are supplementary.
  • If $ABCD$ is a parallelogram (but not a rhombus) then its abq is a rectangle.
  • If $ABCD$ is a rectangle (but not a square) then its abq is a square.

In particular if we start with a parallelogram and iterate the construction we have the following sequence: $$\textrm{parallelogram} \to \textrm{rectangle} \to \textrm{square} \to \textrm{point}$$

This suggests the following questions:

  1. Is it true that for any quadrilateral, iterating the construction of abqs leads eventually to a single point after finitely many iterations?
  2. If so, is there an upper bound on the number of iterations?
  3. If no, is it possible to have a sequence of iterations that is periodic (up to scale) in the sense that the $n$th iterate is similar to the origin figure?

I'd be interested in thoughts on these questions, even if they are inconclusive.

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    $\begingroup$ By observation: take the bisectors through, say, $A$ and $B$. These meet at an angle that is the average of the interior angles at $C$ and $D$. Similar for each of the other angles of the abq. This suggests that for arbitrary initial angles, the angles of the successive abqs will approach $90^\circ$, but that the iteration will continue indefinitely. $\endgroup$ – nickgard Mar 26 '17 at 9:38
  • $\begingroup$ That's helpful, @NickG. But it's the lengths of the sides that determines whether the iteration terminates, not the angles. $\endgroup$ – mweiss Mar 27 '17 at 14:19
  • $\begingroup$ All the termination cases rely on special relationships, either between the lengths of the sides ($AB+CD=BC+AD$), or the sizes of the angles (all parallelograms lead to rectangles regardless of their side lengths - it's the opposite angles that matter). When sides and angles don't fit these special cases, I think the abqs go on forever. $\endgroup$ – nickgard Mar 27 '17 at 18:20
  • $\begingroup$ Incidentally, periodic iterations cannot happen. The only case where all angles remain the same is when they all start as right angles, and we know that leads to termination. In all other cases the angles average out and the smallest abq angle is larger than the original smallest angle. The abq angles can never get smaller again and hence can't return to the original figure. $\endgroup$ – nickgard Mar 27 '17 at 18:31

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