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I feel it’s
not possible.
Define of a proof:
Observe: Because the satan is within the element I am solely 50% assured that is right. Particularly the final half remains to be quite squishy. Anyway, simply wish to put it on the market earlier than the query will get chilly.
Step 1:
Present that there have to be two reverse triangle corners. That is comparatively simple. Depend boundary squares (136). There being solely 49 tiles there have to be many who cowl 3 or 4 boundary factors and such tiles minimize off a triangle nook.
Performing some mildly annoying case bashing we will then present {that a} triangle nook should comprise both the monomino or the L-tromino, so there can solely be two of them.
Inspecting the numbers extra carefully we discover that the triangles should cowl greater than half the boundary squares and should subsequently be in reverse corners. The variety of leftover odd-sized tiles is at most 4 and they don’t seem to be brief ones.
Utilizing checkerboard parity we will additionally simply present that the smaller triangle have to be a clear stack of all 4n+1-sized or all 4n-1-sized tiles. (The opposite triangle might in precept have a couple of composite outer “onion shells” however this may be dominated out utilizing a 4-way colouring (121212 // 343434 // 121212 …) of the board.)
Step 2:
With the triangular caps established mentally lengthen the stripe sample of the caps to your entire board, i.e. cut up it into pairs of diagonals (or “ziggies” I suppose). Domino positions on the board are actually cut up into two equal dimension teams “on grid”, the domino lies fully inside one ziggy, and “off grid”, the domino straddles two ziggies.
Let’s designate one triangle the “north pole” and the opposite one the “south pole”, the min diagonal separating them could be the “equator”. Allow us to quantity northern ziggies 1,2,… ranging from the equator and southern ziggies -1,-2,… additionally ranging from the equator.
Assuming now we have managed to cowl the hexagon left between the triangle caps. Reduce the protecting tiles into dominoes. That is unambiguous for the even sized tiles, for the handful if any odd-sized tiles we minimize off a Monomino at one finish and minimize the remainder into dominoes. Observe that for any tile regardless its orientation both all its dominoes are on-grid or all off-grid.
Subsequent, take into account the dominoes and monominoes protecting a given ziggy ok. Off-grid dominoes can straddle both northwards N_k or southwards S_k. We now have N_k = S_k+1 and N_k + MN_k= S_k + MS_k +/- 1 the place the + is legitimate on the Southern hemisphere and the – within the northern hemisphere and MN_k and MS_k are the numbers of monominoes on the northern an southern components of the ziggy. By conference additionally N_-1 = S_1 on the equator and N_max = S_min = 0 on the triangle edges. Additionally, recall from step 1 that there are at most 4 monominoes whole.
These equations mandate a small however nonzero variety of off-grid dominoes. Apart from at most 4 ziggy boundaries the delta in straddlers throughout this and neighbouring boundaries is 1. The tiles that had been minimize into the off-grid dominoes contribute both a stack of 1 domino per layer crossing or all dominoes to the identical layer crossing. In a totally or predominantly grid parallel orientation there should not sufficient brief tiles accessible. Within the orthogonal orientation there should not sufficient “free” (that means non boundary touching) tiles accessible and we can’t connect the tiles to the boundary due to symmetry w.r.t. the equator this may minimize one other triangle (FIXME: Can the monominoes screw us right here?).
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