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Use ImageCrop. It does exactly what you want, at least in this case... except in the case where "empty" space exists only on one side of the image, which it refuses to crop. To work around that, you c …

A solution just to show Solve can also be used directly on basis of the problem specification: With[{max = 4}, With[{coords = Table[{x[n], y[n]}, {n, 2 max - 1}]}, coords /. Solve[ (* th …

Element] reg, u, v}]; VoronoiMesh[ Join[centers[Rectangle[{-3, -3}, {5 + 2/3, 14}], -\[Pi]/24], centers[Rectangle[{5 + 1/3, -3}, {14, 14}]]], {{0, 11}, {0, 11}}] // MeshPrimitives[#, 1] & // Graphics …

It is already quite much faster than either, at least for polygons: With[{reg = Polygon@RandomReal[{0, 10}, {3, 2}]}, Graphics[{reg, Blue, Opacity[1/2], Rectangle /@ Select[Flatten[CoordinateBoundsArray …

This is a rewrite of the code in a style more natural to Mathematica: Manipulate[ Graphics[ {Opacity[.2], MapThread[Translate, {{{Yellow, Disk[{0, 0}, 10, {0, 2 Pi/3}]}, {Orange, Disk[ … 3]@RotationTransform[x][{10, 0}], {i, 0, 2}]}]}, PlotRange -> All, Axes -> True], {x, 0, 2 Pi}] This could be further simplified to a MapThread over colors and sector angles: Manipulate[ Graphics …

I want to omit parts of Graphics from scenes by defining a geometric shape where details are visible, a "keyhole" of sorts, and fill all the rest as one would do to insides of the regular graphics primitives … Polygon[pts_] :> {Line[pts]}]}], Sequence @@ opts]] showWithHole[ Graphics[{Blue, Disk[{1, 0}, 3/2]}], Directive[Opacity[1/2], White], Red, Graphics[{Disk[{0, 0}, 1], FilledCurve[{{Line@CirclePoints …

Polygon[pts_] :> {Line[pts]}, {Line[ Tuples[scaledbounds][[{1, 2, 4, 3}]]]}]]}], Sequence @@ opts]] showWithHole[ Graphics[{Red, Disk[{1, 0}, 3/2]}], Directive[Opacity[9/10], White], Black … , Graphics[{Disk[{0, 0}, 1], FilledCurve[{{Line@CirclePoints[{2, 1}, 1/2, 6]}, {Line@CirclePoints[{2, 1}, 1/4, 6]}}]}], MaxCellMeasure -> 0.001] …

Based on your image, but not your statement on lengths of lines: With[{eqn = a - a x / (1 - a)}, Show[ Quiet@Plot[Table[eqn, {a, 0, 1, 1 / 20}], {x, 0, 1}, Evaluated -> True, AspectRatio -> Aut …

I have constructed a method that produces correct results: With[{reg = Polygon@RandomReal[{0, 10}, {3, 2}]}, Graphics[{reg, Blue, Opacity[1/2], Rectangle[{x, y}] /. …

points per circle: With[{pts = RandomReal[{-10, 10}, {10000, 2}], disks = Table[Disk[RandomReal[{-5, 5}, 2], RandomReal[{1, 2}]], {3}]}, Show[ListPlot[(Select@RegionMember@#)[pts] & /@ disks], Graphics …

A method based on RegionDilation introduced in v12.3: With[{reg = Polygon@RandomReal[{0, 10}, {3, 2}]}, Graphics[{reg, Blue, Opacity[1/2], Rectangle /@ Select[Flatten[CoordinateBoundsArray@Floor@RegionBounds … This is about ten times slower for triangles than the method above, but still faster than the original: With[{reg = Polygon@RandomReal[{0, 10}, {3, 2}]}, Graphics[{reg, Blue, Opacity[1/2], Rectangle …

Consider, for instance: RegionPlot[ Element[{x, y}, Annulus[{0, 0}, {0.8, 1}]], {x, -1, 1}, {y, -1, 1}, Frame -> False, BoundaryStyle -> None, ColorFunctionScaling -> False, ColorFunction -> (Hue …

Your Line has a constant $y$ value... so I picked a bit more adventurous example: With[{line = Interpolation[ {{-Pi, Cos[-Pi]}, {Pi/4, Cos[Pi/4]}}, InterpolationOrder -> 1]}, Plot[{Cos[x …

derive the tiling from some more abstract definition or just to reproduce it, but reproducing it is not particularly hard: With[ {gridbasis = a {1, 0} + b RotationTransform[60 Degree][{1, 0}]}, Graphics … tiling. *) countregion = Rectangle[{-5/2, -5/2}, {5/2, 5/2}]; (* Visualise 12-gons (polys) which are inside in countregion. *) Cases[polys, poly_RegularPolygon /; RegionWithin[countregion, poly]] // Graphics …

.}}; (* Convert coordinate-lists to two collections of lines which can be used as primitives in both in graphics and new geometric computation. *) {lines1, lines2} = Line[Partition[#, 2]]& /@ {p1, … Solve[{x, y} \[Element] lines1 && {x, y} \[Element] lines2, {x, y}]; (* Represent all these as Graphics. *) Graphics[{Blue, lines1, Red, lines2, Black, PointSize …