Pappus
Example Bankoff circle.
\emph{ The points of tangency of the incircle of the arbelos with the semicircles $(\beta)$ and $(\gamma)$, together with $(\alpha)$, are the points of tangency of the incircle $(\epsilon)$ of triangle $O_1O_3O_2$ with the sides of the triangle.}
View TeX code
% !TEX TS-program = lualatex
% Author Alain Matthes 2025
\directlua
View Lua code
init_elements() z.A = point(0, 0) z.B = point(10, 0) L.AB = line(z.A, z.B) z.C = L.AB : gold_ratio () L.AC = line(z.A, z.C) L.CB = line(z.C, z.B) z.O_0 = L.AB.mid z.O_1 = L.AC.mid z.O_2 = L.CB.mid C.O0B = circle(z.O_0, z.B) C.O1C = circle(z.O_1, z.C) C.O2B = circle(z.O_2, z.B) z.M_0 = C.O1C:external_similitude(C.O2B) z.M_1 = C.O0B:internal_similitude(C.O1C) z.M_2 = C.O0B:internal_similitude(C.O2B) z.Q = C.O2B.north z.P = C.O1C.north z.O = C.O0B.south z.W = C.O0B.north L.O1Q = line(z.O_1, z.Q) L.O2P = line(z.O_2, z.P) z.O_4 = intersection(L.O1Q, L.O2P) L.CO4 = line(z.C, z.O_4) z.D = intersection(L.CO4, C.O0B) L.CD = line(z.C, z.D) L.PQ = line(z.P, z.Q) L.CQ = line(z.C, z.Q) z.O_5 = intersection(L.CD, L.PQ) C.QC = circle(z.Q, z.C) C.PA = circle(z.P, z.A) z.P_0 = intersection(C.QC, C.O0B) z.P_1 = intersection(C.QC, C.O1C) _, z.P_2 = intersection(C.PA, C.O2B) T.P = triangle(z.P_0, z.P_1, z.P_2) z.O_3 = T.P.circumcenter L.AQ = line(z.A, z.Q) L.BP = line(z.B, z.P) z.T = intersection(L.BP, L.AQ) z.O_6 = L.CD.mid C.DC = circle(z.D, z.C) z.U, z.V = intersection(C.DC, C.O0B) L.UV = line(z.U, z.V) z.R, z.S = L.UV:projection(z.O_1, z.O_2)