As far as I am concerned I consider J89 being a good rule which is perfect in practice in almost 100% of the games but that does not prevent me from liking discussion on beast positions. BTW seeing the difficulty to find a wording corresponding to my own interpretation I consider, as Jann said, that J89 is much better than usually claimed.
$$
$$ -----------------------
$$ | O . O . X . . O O . O
$$ | O O O X O X O O O O O
$$ | X X X O . O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .
- Click Here To Show Diagram Code
[go]$$
$$ -----------------------
$$ | O . O . X . . O O . O
$$ | O O O X O X O O O O O
$$ | X X X O . O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .[/go]
Coming back to this interesting position it appears to me as a perfect (?) beast (thank you lightvector!) because this position adresses a number of difficulties with the wording of the rule I will try to summarize:
Let's take the main variation in order to try and find the status of the 5 white stones:
$$B
$$ -----------------------
$$ | O . O . X . . O O . O
$$ | O O O X O X O O O O O
$$ | X X X O 1 O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .
- Click Here To Show Diagram Code
[go]$$B
$$ -----------------------
$$ | O . O . X . . O O . O
$$ | O O O X O X O O O O O
$$ | X X X O 1 O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .[/go]
$$B
$$ -----------------------
$$ | O . O . X . . O O . O
$$ | O O O X a X O O O O O
$$ | X X X O X O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .
- Click Here To Show Diagram Code
[go]$$B
$$ -----------------------
$$ | O . O . X . . O O . O
$$ | O O O X a X O O O O O
$$ | X X X O X O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .[/go]
Seeing this position it is clear that black has taken a ko and white need a pass-for-ko before being allowed to retake the ko at "a". Now seeing some previous comment in this thread, we can also observe that, in order to try and capture the 5 white stone, black begins by taking another white stone for wich the status is unkown though black is not looking at the status of this white stone.
$$B
$$ -----------------------
$$ | O . O . X 2 . O O . O
$$ | O O O X . X O O O O O
$$ | X X X O X O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .
- Click Here To Show Diagram Code
[go]$$B
$$ -----------------------
$$ | O . O . X 2 . O O . O
$$ | O O O X . X O O O O O
$$ | X X X O X O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .[/go]
$$B
$$ -----------------------
$$ | O . O . X O b O O . O
$$ | O O O X a X O O O O O
$$ | X X X O X O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .
- Click Here To Show Diagram Code
[go]$$B
$$ -----------------------
$$ | O . O . X O b O O . O
$$ | O O O X a X O O O O O
$$ | X X X O X O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .[/go]
After the move

the position has completly changed:
everybody can see that a ko has been created in "b" but we can also see that "a" is no more a ko because a white move here will take two black stones. Now a question arises : have now white to use a pass-for-ko if this ko reappears in the future?
$$B
$$ -----------------------
$$ | O . O . X O 3 O O . O
$$ | O O O X a X O O O O O
$$ | X X X O X O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .
- Click Here To Show Diagram Code
[go]$$B
$$ -----------------------
$$ | O . O . X O 3 O O . O
$$ | O O O X a X O O O O O
$$ | X X X O X O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .[/go]
$$B
$$ -----------------------
$$ | O . O . X . X O O . O
$$ | O O O X a X O O O O O
$$ | X X X O X O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .
- Click Here To Show Diagram Code
[go]$$B
$$ -----------------------
$$ | O . O . X . X O O . O
$$ | O O O X a X O O O O O
$$ | X X X O X O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .[/go]
After

"a" becomes again a ko. Has this ko to be considered the same ko or is it a new ko?
If it is a new ko then white can immediatly take at "a" because no pass-for-ko is needed for a new ko.
If it is the same ko the situation is still not obvious : we are in a double ko situation and we know that, strictly speaking, the rule does not handle correctly a double ko situation. Is it really the intention of the rule to prevent white to play one of the two ko in such situation?
$$B :w4: pass for ko at "a"
$$ -----------------------
$$ | O 7 O 5 X 8 X O O . O
$$ | O O O X 6 X O O O O O
$$ | X X X O X O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .
- Click Here To Show Diagram Code
[go]$$B :w4: pass for ko at "a"
$$ -----------------------
$$ | O 7 O 5 X 8 X O O . O
$$ | O O O X 6 X O O O O O
$$ | X X X O X O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .[/go]
$$B :w4: pass for ko at "a"
$$ -----------------------
$$ | . X . X X 8 . O O . O
$$ | . . . X 6 . O O O O O
$$ | X X X O . O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .
- Click Here To Show Diagram Code
[go]$$B :w4: pass for ko at "a"
$$ -----------------------
$$ | . X . X X 8 . O O . O
$$ | . . . X 6 . O O O O O
$$ | X X X O . O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .[/go]
In this final position can we consider the capture of the 5 white stones has enabled to play

and

? In any case, I mean even if black is not trying to kill the 5 white stones, white is able to play the uncapturable stones at

and
As you can see this alone position adresses number of issues and we can be sure that a common interpretation is quite impossible to reach.
Anyway any comment are welcome.