One perimeter of 300, spent five ways

• the 3-gon obeys Bonnesen's inequality ×10
• 2 pieces against a wall do exactly as well as a 4-gon does in the open ×7
• the 3-gon encloses less than a circle of the same perimeter ×7
• the 3-gon is drawn at the same perimeter as the rest ×7
• regular 3-gon does not beat the circle ×5
• the arc in the 60° corner is 300 long ×4
• 6 reflected copies of the 60° sector make a whole disc ×3
• the 2-piece fence is 300 long ×3
• the 2-piece fence's share of the half-disc is the 4-gon's quotient ×3
• 2:1 rectangle does not beat the circle ×1
• a pentagon symmetrised has more than five corners ×1
• a quotient from 0.9 to just below 1 ×1
• a regular polygon of 3 to 12 sides ×1
• a rhombus is higher the further its angle is from a right angle ×1
• a triangle symmetrised about a side's perpendicular is a triangle ×1
• a unit cube has six units of face ×1
• a unit square cell needs two units of wall of its own ×1
• after 60 it is nearly a disc ×1
• an L does not beat the circle ×1
• and every one holds less than the half-disc ×1
• and holds twice the area ×1
• and is still short of the circle's ×1
• and it is the circle ×1
• and its area is a quarter of the perimeter, squared ×1
• and never lengthens the boundary ×1
• and shortens the boundary ×1
• and the area is strictly larger ×1
• and the best angle between the pieces is a right angle ×1
• and the circle's quotient is exactly one ×1
• and the ring's square like the inverse fourth power ×1
• and the sector holds the fence squared over twice the angle ×1
• between 10 and 80 cells ×1
• between 50 and 400 random shapes ×1
• between one and four fences of 2 to 12 pieces ×1
• each more regular sides comes closer to the disc and never reaches it ×1
• each round of symmetrisation brings the shape nearer round ×1
• each structure improves on the one before ×1
• even the best rectangle falls short of the circle ×1
• every extra side encloses strictly more ×1
• every nudged 5-gon is higher than the regular one ×1
• every random convex shape obeys Bonnesen's inequality ×1
• every shape with that quotient has the same allowance ×1
• every step keeps the area ×1
• exactly one shape in the list reaches one, and it is the circle ×1
• fences of up to between 4 and 12 pieces ×1
• five steps bring the triangle close to equilateral ×1
• four corner angles from 30 to 180 degrees ×1
• hexagons need the least wall, triangles the most ×1
• more pieces hold more ×1
• more sides always need less wall ×1
• no cell beats the regular polygon with its number of sides ×1
• no triangle on the grid is lower than the equilateral one ×1
• reflecting the dent leaves the perimeter exactly as it was ×1
• Reuleaux triangle does not beat the circle ×1
• so the fence holds less than its length squared over 2π ×1
• so the quotient rises ×1
• so their average wall is at least the hexagon's ×1
• symmetrisation keeps the area ×1
• the 12-gon's circumradius is found ×1
• the 12-gon's inradius is found ×1
• the 12-gon's ring is no wider than its deficit allows ×1
• the averaged pair is never longer ×1
• the best split is an even one ×1
• the bump is tuned to hold 0.99 of the circle's area ×1
• the bump's ring is no wider than its deficit allows ×1
• the cells average six sides ×1
• the cells cover the square exactly once ×1
• the circle does not beat the circle ×1
• the circle of that perimeter encloses more than any of the polygons drawn ×1
• the computed value for a rectangle matches π²(r + 1/r) ×1
• the deficit falls like the inverse square of the number of sides ×1
• the dent is between nothing and 160 units deep ×1
• the doubled curve is twice the fence ×1
• the doubled curve obeys the closed-curve inequality ×1
• the ellipse is tuned to hold 0.99 of the circle's area ×1
• the ellipse's circumradius is found ×1
• the ellipse's inradius is found ×1
• the ellipse's ring is no wider than its deficit allows ×1
• the equilateral triangle's exact value, 4π²/√3 ×1
• the excess falls as 4ζ(3)/n³ of the disc's value ×1
• the excess grows as the square of the nudge, the mark of a minimum rather than a corner ×1
• the hexagon's circumradius is found ×1
• the hexagon's inradius is found ×1
• the hexagon's ring is no wider than its deficit allows ×1
• the largest rectangle in the sweep is the square ×1
• the lowest grid triangle is the one nearest equilateral ×1
• the lowest mode keeps one sign throughout ×1
• the polygons drawn have between 3 and 60 sides ×1
• the regular pentagon is lowest in both measures ×1
• the reuleaux's circumradius is found ×1
• the reuleaux's inradius is found ×1
• the reuleaux's ring is no wider than its deficit allows ×1
• the shared perimeter is between 60 and 900 units ×1
• the sides of the cells add to exactly six per cell ×1
• the square's circumradius is found ×1
• the square's exact value, 2π² ×1
• the square's inradius is found ×1
• the square's ring is no wider than its deficit allows ×1
• the stadium is tuned to hold 0.99 of the circle's area ×1
• the stadium's circumradius is found ×1
• the stadium's inradius is a root of the quadratic ×1
• the stadium's inradius is found ×1
• the stadium's ring is no wider than its deficit allows ×1
• the sweep takes between 20 and 2000 samples ×1
• the triangle's circumradius is found ×1
• the triangle's inradius is found ×1
• the triangle's ring is no wider than its deficit allows ×1
• the view is one the family draws ×1
• turning directions approach a disc ×1
• two fixed directions stop short, at a shape both leave alone ×1
• two pentagons with almost the same quotient differ clearly in eigenvalue ×1
• two to four shapes ×1
• which holds an eighth of the fence squared — half of a square ×1
• whose two sides are equal ×1
• πρ² − Lρ + A is not positive at the 12-gon's two radii ×1
• πρ² − Lρ + A is not positive at the bump's two radii ×1
• πρ² − Lρ + A is not positive at the ellipse's two radii ×1
• πρ² − Lρ + A is not positive at the hexagon's two radii ×1
• πρ² − Lρ + A is not positive at the reuleaux's two radii ×1
• πρ² − Lρ + A is not positive at the square's two radii ×1
• πρ² − Lρ + A is not positive at the stadium's two radii ×1
• πρ² − Lρ + A is not positive at the triangle's two radii ×1

The whole library · What the figures prove