Oryx: B goes in the center of the formation at . W could move at , or similarly at a, b, or c -- they're all the same for our purposes here.
B now goes in any limb that W didn't, say . The best that W can do is d or e, whereupon B takes whichever of those W didn't take.
If W goes , then B connects , and whallah two eyes, and 4 safe (but kinda useless) B2Bombers.
It's interesting to note, that B used 27 stones to capture 2 - count 'em - points. 2/27 = .07 points per stone. Duh.
Alex Weldon: This looks right to me. I just wanted to say that when the author of the problem (unkx80, presumably) moves this to the solution page, he should make sure to add a link to Play on the point of symmetry.
unkx80: Thanks for the pointer. And by the way, it is definitely not a good idea to send four B2 bombers all into a collision course!