# Zak's Triangle

Posted 5 years ago
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 At squaring.net, I learned of a triangle with sides based on powers of the real positive root of $d^6-d^2-1=0$. $$d = \sqrt{\frac{1}{3} \sqrt{\frac{27}{2}-\frac{3 \sqrt{69}}{2}}+\frac{\sqrt{\frac{1}{2} \left(9+\sqrt{69}\right)}}{3^{2/3}}} = 1.150963925257758035680601218461375863666...$$ d = Root[-1 - #1^2 + #1^6 &, 2]; Here's a picture I assembled. The numbers on the sides of the triangle are powers of $d$. The smallest triangle has sides $d^0, d^1, d^4$. The outside triangle has sides $d^{10}, d^{11}, d^{14}$. We can get point positions for the triangle vertices with uses of Solve and EuclideanDistance. zakpts = FullSimplify[{{-d^8, 0}, {0, 0}, {1, 0}, {1 + d^8, 0}, {Root[-5 + 16 #1 - 20 #1^2 + 8 #1^3 &, 1], 1/2 Root[-43 - 18 #1^2 + #1^4 + #1^6 &, 2]}, {Root[-5 + 16 #1 - 20 #1^2 + 8 #1^3 &, 1] + d^2, 1/2 Root[-43 - 18 #1^2 + #1^4 + #1^6 &, 2]}, {Root[-1 + 2 #1 - 8 #1^2 + 8 #1^3 &, 1], Root[-43 + 52 #1^2 - 416 #1^4 + 64 #1^6 &, 2]}}]; From there, it's useful to convert to barycentric coordinates, which give the coordinates of a point in relation to a triangle. ToBarycentrics[{{x1_, y1_}, {x2_, y2_}, {x3_, y3_}}, {p_, q_}] := Solve[{m*x1 + n*x2 + (1 - m - n)*x3, m*y1 + n*y2 + (1 - m - n)*y3} == {p, q}]; FromBarycentrics[{m_, n_}, {{x1_, y1_}, {x2_, y2_}, {x3_, y3_}}] := {m*x1 + n*x2 + (1 - m - n)*x3, m*y1 + n*y2 + (1 - m - n)*y3}; outsidezak = {{-d^8, 0}, {1 + d^8, 0}, {Root[-1 + 2 #1 - 8 #1^2 + 8 #1^3 &, 1], Root[-43 + 52 #1^2 - 416 #1^4 + 64 #1^6 &, 2]}}; FullSimplify[({m, n} /. ToBarycentrics[outsidezak, #])[] & /@ zakpts] Cleaning that up, we can get the barycentric version of each triangle. zakbary = With[{ r1 = Root[-1 + 2 #1 - #1^2 + #1^3 &, 1], r2 = Root[-1 + 3 #1 - 2 #1^2 + #1^3 &, 1], r3 = Root[-1 + 5 #1 + 2 #1^2 + #1^3 &, 1], r4 = Root[-5 + 14 #1 - 3 #1^2 + #1^3 &, 1]}, {{0, 0}, {0, 1}, {1, 0}, {0, r1}, {r1, r2}, {r2, r1}, {r3, r4}}]; zaktriangles = zakbary[[#]] & /@ {{7, 5, 6}, {4, 6, 7}, {1, 4, 7}, {5, 1, 7}, {6, 2, 4}, {3, 1, 5}}; And from that, we can make a triangle with the edges labeled by their powers. d = Root[-1 - #1^2 + #1^6 &, 2]; triangles = Map[FromBarycentrics[#, outsidezak] &, zactriangles, {2}]; edges = Partition[N[#], 2, 1, 1] & /@ triangles; powers = Map[Round[Log[d, EuclideanDistance[#[], #[]]]] &, edges, {2}]; edgetext = Table[{ToString[powers[[a, b]]], N[ ((powers[[a, 1]] - 2) Mean[edges[[a, b]]] + Mean[triangles[[a]]])/(powers[[a, 1]] - 1)]}, {a, 1, 6}, {b, 1, 3}]; Clear[d]; Column[{Graphics[{FaceForm[White], EdgeForm[Black], Polygon /@ triangles, Style[Map[Text[#[], #[]] &, edgetext, {2}], 24]}, ImageSize -> {800, Automatic}], Text@Style[Row[{"Values are powers of ", Style["d", Italic], ", where ", TraditionalForm[d^6 - d^2 - 1 == 0]}], 18]}, Alignment -> Center] We can also make a fractal out of it, by subdividing the largest triangles.This is where the barycentric coordinates really become useful. The previous figure could have been made from the originally calculated points. seqoftri = NestList[With[{list = #}, Sort[Flatten[{Drop[list, -1],Transpose[{First[Last[list]] - {10, 9, 8, 7, 6, 3}, Chop[Map[N[FromBarycentrics[#, Last[Last[list]]]] &, zactriangles, {2}]]}]}, 1]]] &, Transpose[{Min /@ powers, Chop[N[triangles]]}], 50]; Manipulate[Graphics[{FaceForm[White],EdgeForm[Black], {Hue[Mod[First[#] + huemod, 10]/10], Polygon[Last[#]]} & /@ seqoftri[[index]]}, ImageSize -> {800, 300}], {index, 1, 50, 1}, {huemod, 0, 10}, SaveDefinitions -> True] Do other simple polynomials lend themselves to weird similar triangle dissections? Attachments: Answer
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Posted 5 years ago - another post of yours has been selected for the Staff Picks group, congratulations !We are happy to see you at the tops of the "Featured Contributor" board. Thank you for your wonderful contributions, and please keep them coming! Answer
Posted 5 years ago
 Ed are this triangle dissections unique? Meaning do pieces assemble in a triangle in a single way? I am thinking if this is 3D printed it could make a wonderful puzzle. Nice posts, thanks for sharing and keep them coming! Answer
Posted 5 years ago
 The outside triangle has sides d10,d11,d14.Sorry - but why that? I see d(6+4) = d10, d11 and d(8+0+8) = d16 ? Answer
Posted 5 years ago
 I've done some research on these, and compiled my results in Wheels of Powered Triangles. For example: That solution can be extended to make a triangle in two ways. If anyone can piece some of these other triangles together in an interesting way, please post here. Answer