Polyiamonds are a diamond-inspired construction based on polyominoes made from component equilateral triangles instead of squares. This puzzle dealt with hexiamonds, those shapes constructed from exactly six adjacent triangles. There are twelve different hexiamond shapes under rotation and reflection, which lumped together would comprise of 72 total smaller triangles. Fittingly, the grids presented in this puzzle are also composed of 72 small triangles. Considering all different types of hexiamonds are featured in the puzzle as different outlines on top of these grids, it seems likely we want to use the 12 unique hexiamonds in some way. One intuitive use would be to tile these grids with the 12 different hexiamonds, a feat which can certainly be achieved in many different ways. It seems reasonable then that the inclusion of shape outlines on each grid might restrict the number of such tilings, and in fact in every grid, looking only at red or only at blue outlines in turn, the resultant board tiling by the 12 unique hexiamonds turns out to be unique. For example, tiling the first board with the 12 unique hexiamonds such that the two blue outlines represent two pieces' fixed positions results in the following unique tiling (left): Next to that tiling is the unique red tiling also indicated by the red outlines given on the first board. Similar tilings can be found for the remaining boards - you might like to try achieving these tilings on a rainy day if you haven't already. You can find a copy of the board and pieces here. All we have achieved thus far though is a complete tiling of each board, twice. It seems reasonable to want to use the two tilings per board together in some way, and one of the most obvious ways to take an "intersection" of such tilings that doesn't trivially encompass the whole board might be to focus on individual pieces and observe their overlap between tilings. More explicitly, if a blue hexiamond and a red hexiamond of the same shape (the bar shape, say) overlap at some point, those overlapped triangles can be considered part of the total intersection we'd like to take. This is hopefully more clearly emphasised by the purple shadings in the below image, whose separate sections from top to bottom represent the overlap between the two bars, crowns, hooks, and crooks:
Now we can finally use the letter assignment provided in the first board of
each page. It's fair to assume the letters of the alphabet continue to be
assigned to triangles in a left-right, top-bottom order, and that we start
again from A after reaching Z - as has been reflected in the image above.
The intersection's letters give us the string
Promisingly, often the spare letters seem to be proximal in the alphabet. In fact upon closer inspection, in each group of six boards, their spare letters come from adjacent triangles. The spare letters seem to be representing triangles that should be cut out of each board just as they have been cut from each letter string, and combining them all together while considering the blue- or red-coloured board borders would give us two new blue and two new red hexiamonds and their positions on another unseen board. Specifically, these triangles come together to indicate the new board:
Completing the task one more time of tiling the board with unique hexiamonds
and taking the tilings' special intersection, the final string of letters
communicated give a full anagram this time of the word
Unfortunately despite many alternative approaches, I eventually realised the puzzle would never eventuate unless I forced letters into it somehow. The fairly boring alphabet assignment was designed, and even then very few words/anagrams arose after checking all intersections. FELDSPAR stood out as one of the only longer words that somewhat tied into the theme, and was picked as the answer - but now the tiling and intersection steps were only being hinted at by a single board, and neither step had a particularly convincing mid-step signal for the solver to know they were going in the right direction. Thus the recursive step was added, allowing for many more examples to be played with and hopefully provide more hints towards the intended solution mechanic. | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

The answer is: feldspar |
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