I wrote some code in 1999 that defines a PTiling function that does something close to what you want. I ve included that code at the bottom of this response. Once you execute that to define PTiling, you should be able to generate a picture like so: Show[PTiling[5, 2, 3], ImageSize -> 600, PlotRange -> 1.1] Obviously, we have just a black and white picture illustrating the boundaries of the pentagons that you want but, hopefully, this is a good start. I think that one could use portions of disks, rather than circles, to get closer to what you want. A few of comments on the code are in order. The ideas are all described in Saul Stahl s excellent book, The Poincare Half-Plane - specifically, the chapter on the Poincare disk. I wrote the code to illustrate some ideas for a geometry class that I was teaching in 1999 so it must have been for version 3 or 4. I have done nothing to try to optimize the code for any subsequent version. Also, it s not the case that all combinations of n and k work so well. I think they actually need to be chosen somewhat carefully; I honestly don t recall exactly how they should be chosen. (* Generates a groovy image of the Poincare disk *) (* Not all combinations of n and k work great *) PTiling[n_Integer, k_Integer, depth_ : 2] := Module[ {aH, a, q, r, idealPoints, z1Ideal, z2Ideal, init, initCircs}, aH = ArcCosh[(Cos[Pi/n] Cos[Pi/2] + Cos[(n - 2) Pi/(2 k)])/(Sin[Pi/n] Sin[Pi/2])]; a = (Exp[aH] - 1)/(Exp[aH] + 1); q = (a + 1/a)/2; r = q - a; (* The Action *) idealPoints = {x, y} /. NSolve[{x^2 + y^2 == 1, (x - q)^2 + y^2 == r^2}, {x, y}]; {z1Ideal, z2Ideal} = toC /@ idealPoints; init = N@IdealPLine[{z1Ideal, z2Ideal}]; initCircs = NestList[RotateCircle[#, 2 Pi/n] &, init, n - 1]; Show[Graphics[{Nest[Iter, initCircs, depth], PGamma}], AspectRatio -> Automatic] ]; (* Ancillary code *) (* One step in the iteration *) Iter[PLineList_List] := Union[PLineList, Select[Flatten[ Outer[Reflect, PLineList, PLineList]], (# =!= Null &)], SameTest -> sameCircleQ ]; (* Generate the ideal Poincare line through the points z1 and z2 *) (* Should be an arc, if z1 and z2 are not on the same diameter *) IdealPLine[{z1_, z2_}] := Module[ {center}, center = Exp[I (Arg[z2] + Arg[z1])/2] / Cos[(Arg[z2] - Arg[z1])/2]; arc[{z1, z2}, center] ]; arc[{z1_, z2_}, z0_] := Module[{theta1, theta2}, theta1 = Arg[z1 - z0]; theta2 = Arg[z2 - z0]; If[Abs[N[theta1 - theta2]] > Pi, If[N[theta1] < N[theta2], theta1 = theta1 + 2 Pi, theta2 = theta2 + 2 Pi] ]; Circle[toR2[z0], Abs[z1 - z0], Sort[{theta1, theta2}, N[#1] < N[#2] &] ] ]; (* Circle operations *) Invert[Circle[c_, r1_, ___], z_] := r1^2/Conjugate[z - toC[c]] + toC[c]; Reflect[circ1_Circle, Circle[c2_, r2_, thetaList_]] := IdealPLine[ Invert[circ1, toC[c2 + r2 *{Cos[#], Sin[#]}]] & /@ thetaList ]; RotateCircle[Circle[c_, r_, thetaList_], theta_] := Circle[RotationMatrix[theta] . c, r, theta + thetaList]; cSameCircleQ = Compile[ {{c1, _Real, 1}, {r1, _Real}, {th1, _Real, 1}, {c2, _Real, 1}, {r2, _Real}, {th2, _Real, 1}}, (c1[[1]] - c2[[1]])^2 + (c1[[2]] - c2[[2]])^2 + (r1 - r2)^2 + (th1[[1]] - th2[[1]])^2 + (th1[[2]] - th2[[2]])^2 < 0.00001]; sameCircleQ[Circle[c1_, r1_, th1_], Circle[c2_, r2_, th2_]] := cSameCircleQ[c1, r1, th1, c2, r2, th2]; (* Basics *) toR2 = {Re[#], Im[#]} &; toC = #[[1]] + #[[2]] I &; PGamma = {Thickness[.008], GrayLevel[.3],Circle[{0, 0}, 1]};