Fillomino 6: Checkered

Consider the lower-right two. Of the four ways in which it can be placed, two force the six region to grow far too large (so that no T-intersections are created), and one more clashes with the lower-middle two and three. The fourth way forces the six and a large part of the ten.

Consider, then, the two in the middle right. There are three ways in which it can be placed, and two of them force the ten region to grow far too large. The third way forces a little more of the ten.

By this point, there are two distinct ways in which the ten region can be completed. Bifurcation rules out one of them (both branches fall in place rather neatly once you make the initial assumption), and the solve flows smoothly from this point on.

Complete solution: