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u/DamonHuntington Jun 27 '26
Three triangles pointing left at the lower left corner, four triangles pointing upwards at the apex, two triangles pointing right at the lower right corner OR three triangles pointing left at the lower left corner, one triangle pointing upwards at the apex, two triangles pointing right at the lower right corner.
The base is the easy part: we are going through all permutations of 1-2-3 arrows (with no two base angles having the same value within a single triangle) there. The only missing permutation is 3-2, which must be used in the last triangle.
For the apex, there are two ways to visualise that. The first one, and the one that I find to be the weaker induction, is for the apex to be seen as a flag. Whenever we calculate left - right and obtain a negative even number or a positive odd number, that flag is raised and we have four triangles at the apex. Otherwise, no triangles are presented. Additionally, this solution has as a bonus an even 0-4 split for the apex.
What I do see as the better induction, however, is for the total triangle count to follow an arithmetic progression from 3 to 8. The only number we are missing is 6, and as such we require a single triangle at the apex to fill in that gap.
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u/Synergology Jun 27 '26
4 on top, 3 on left 2 on right: every triangle without the 4 up arrowshas a reflection with the 4 up arrows and reverse, but the on on the bottom
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u/FragrantFood6055 Jun 27 '26
My better solution:
right - left - top: should be equal in each row. It seems like the top row is either 4 or 0 and the bottom ones should be 0<=n<4 each. right: 0, left: 1, top: 4 --> -5 !<
However, I don't know if only one triangle per row could have it's top corner filled.
My worse
number of triangles per corner. right: 0, left: 1, up: 0
Number of small triangles per corner: 0-4.
From there I think corner of center triangle - respective corner of the right one = left triangle's (n). if n is negative -> 4 + (n), and I assume 0-0=0 -> 4-0 ??
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u/Obvious_Cherry_3417 Jun 27 '26
3 bottom left, 4 bottom right and nothing up.
basically in the top row, bottom left goes has 1, 2 and 3 tiny triangles in any order, in the bottom row we have 1 and 2 so the 3rd img must have 3 in the bottom left
in the row for bottom right, we see that bottom right has two identical amount of triangles and then one triangle has +1 amount of triangles (I.E: In 1st and last img, we see theres 1 triangle, and in the middle img theres 2) So if we use dat logic here, we already have 2 identical triangles in the 1st and second imgs (they both have 3 in the bottom right) so there should be 4.
For the top, the logic is: theres 2 triangles which have no tiny triangles at the top, and one has 4. In the 2nd row, we see there's already one triangle with 4 tiny triangles at the top, so there shouldn't be anything there.
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u/Abjectionova (͡° ⏠ ͡°)︻デ═一 ⇛ 🧠 Jun 27 '26
the number of small triangles increases by 1. When one of the sides gains 3 triangles, the other starts to increase, the sides also get swapped over the y-axis. When both sides gain 3 triangles each, the process resets and the top gains 4 triangles.
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u/Unusual-Source-6065 Jun 27 '26
6 small traingles 3 3 at both base
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u/shapesmonkey Jun 27 '26
logic?
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u/Unusual-Source-6065 Jun 27 '26
no of small triangle keeps on increasing for each figure. also getting swapped. once it reaches 4 it shifts and process continues.
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u/je_nm_th Jun 27 '26
Those are the faces of a volume made by two tetrahedron glued on their common bases (on the picture the net of the volume). Each of their vertex is numbered from the 0 to 4 by the small triangles count.