The point that the article is addressing (but you have to ignore the image and study the equations to see this!) is that this sort of shifting can't equalize ev

The point that the article is addressing (but you have to ignore the image and study the equations to see this!) is that this sort of shifting can't equalize everything. In the span of 3 white keys C to E at the front, you have 2 black keys at the back, so if you take r to be the ratio of back-width to white key front-width then you have 3 = 5r. But in the 4 keys F to B, you've got 3 black keys so 4 = 7r. No single ratio works! So the article investigates various compromises. The B/12 solution is what seems to me the most straightforward, divide white keys in each of the sections C to E and F to B equally at the back, and don't expect anyone to notice the difference.