Neat idea today, an elegance to the word TRIANGLES breaking perfectly into three groups of three. There's something so pretty to that. ...
read moreNeat idea today, an elegance to the word TRIANGLES breaking perfectly into three groups of three. There's something so pretty to that. I also liked Dani's natural progression, from a SCALENE (no sides of equal length) to ISOCELES (two sides of equal length) to EQUILATERAL (all three sides of equal length). I really enjoy seeing constructor's background shine through in a puzzle.
It took me a while to figure out what the circled letters were spelling, because I have a propensity to read in a clockwise fashion. I found it to be a bit odd that the TRI / ANG / LES go counterclockwise. Perhaps that's a cultural difference? In any case, I did appreciate the extra layer. Three themers + three triangles + TRI ANG LES fixed into place = a lot of depth.
Speaking of extra layers, I would have loved to see a triangle hinted at in the black squares. If you squint really hard, you can almost see the start of a triangle in the "hill" of black squares (above FLO). It would be very hard to do, especially at the tips of the triangles, but what a neat effect that could be.
Mirror symmetry can be a great tool to have in your arsenal. It's really the only way to pull off today's puzzle, because SCALENE 7 / ISOSCELES 9 / EQUILATERAL 11 can't work with regular crossword symmetry. What a fortuitous coincidence that they're all of odd length, allowing mirror symmetry to work! Another progression that would have worked: SCALENE 7 / RIGHT 5 / EQUILATERAL 11. Hard to leave out the poor right triangle in all its Pythagorean glory, but what can you do.
Quite a few constraints today, and they force a few compromises. Rich Norris at the LAT limits partials to a maximum of two, so A TAB / OH TO / OF A is an unfortunate way to open the puzzle. I understand the constraints of the EQUILATERAL triangle — you can't move around just one of the three vertices — but the SCALENE and ISOSCELES both have enough flexibility that it feels like some of the partials plus the PSAS / ARIS / OST stuff could have been cleaned up by shifting the position of circles.
All in all, an elegant idea with some compromises in execution.
ADDED NOTE: an astute reader, Jean Cranmer, was confused because the bottom triangle didn't look equilateral to her. I smugly sent her definitions of "isosceles" and "equilateral" and then... realized that the sides of the bottom triangle are 5.83 (using the Pythagorean theorem to calculate) whereas the base is 6. I imagine this is the closest Dani could come to an approximate equilateral triangle in a 15x square grid. Thanks for the catch, Jean! (And a good reminder that I'm frequently wrong.)