• If a cell beside a number has a triangle, the triangle must border the cell. See counterexample, where a white right-angle cannot be created otherwise (left). By the same corollary, cells beside the board edge containing a triangle must have the triangle border the edge. This leads to the cell below being blank:

• Because the black cell is also a border, by extension, the left cell is blank. This creates a white rectangle that demands a black triangle (left). White sections of the triangle must be followed up by two triangles either beside or diagonal from them, to either create a right-angle, or form a straight line. In this case, the diagonal is filled, so the triangle must appear on the side. We extend the same logic for the cell below:

• Extending the white section creates a white area that cannot form a square, so the left side cannot be true. Since we cannot use the diagonal, the cell beside must be filled (right):

• White areas cannot have widths equal to half the diagonal, see counterexample below. This also implies that cells with triangles must have symmetry with the neighboring cells bordering the white areas.

• The white cell to the right of the numbered cell below cannot be blank: this implies the cell sandwiched by two borders must also be blank, and this creates a non-rectangular white area, see counterexample (left). This means the right cell must be filled with a black triangle, and the bottom square is empty, including the new sandwiched area (right):

• The left side must therefore be filled with a black triangle to separate the two white rectangular regions (left). Extending the idea of symmetry leads to the bottom left solved (right):

• The blank above the zero cell (left) creates a sandwich, and this forces a rectangular region to appear (right):

The remaining solution is trivial.