I don't think that's the issue.
For perfect right angles, the total perimeter will still be 4 no matter the length of the individual segments.
The problem is, one thinks of the length converging to the perimeter of the circle but it doesn't. The area of the inscribed polygon does converge to the area of the circle and I think that's the correct way to look at this.
Area of unit square = 1. Obviously, by inspection, the area is being reduced as more and more cuts are being made at the perimeter. One can easily count up the area removed, so the area of the circle < area of square. The area will converge to pi pi/4 as the number of cuts increases (from the definition of pi).
[edit: ack - messed up earlier]
Area of unit circle = pi * r * r = pi * 0.5 * 0.5, = pi * 0.25. By inspection area of square > area of circle, so 1 > 0.25 * pi, so pi < 4. QED.
In other words, one has to carefully think about limits and not just assume that because something is being cut into smaller and smaller segments that it converges to something else. The perimeter doesn't converge in this case. If one instead circumscribes regular polygons with increasing numbers of sides, then that perimeter will converge (the angles are changing rather than all being 90 degrees).
Cheers,
Scott.
(Who should know better than to try to do geometry without a pencil and paper.)
I don't think that's the issue.
For perfect right angles, the total perimeter will still be 4 no matter the length of the individual segments.
The problem is, one thinks of the length converging to the perimeter of the circle but it doesn't. The area of the inscribed polygon does converge to the area of the circle and I think that's the correct way to look at this.
Area of unit square = 1. Obviously, by inspection, the area is being reduced as more and more cuts are being made at the perimeter. One can easily count up the area removed, so the area of the circle < area of square. The area will converge to pi as the number of cuts increases (from the definition of pi).
Area of unit circle = pi, so pi < 4. QED.
In other words, one has to carefully think about limits and not just assume that because something is being cut into smaller and smaller segments that it converges to something else. The perimeter doesn't converge in this case. If one instead circumscribes regular polygons with increasing numbers of sides, then that perimeter will converge (the angles are changing rather than all being 90 degrees).
Cheers,
Scott.
I don't think that's the issue.
For perfect right angles, the total perimeter will still be 4 no matter the length of the individual segments.
The problem is, one thinks of the length converging to the perimeter of the circle but it doesn't. The area of the inscribed polygon does converge to the area of the circle and I think that's the correct way to look at this.
Area of unit square = 1. Obviously, by inspection, the area is being reduced as more and more cuts are being made at the perimeter. One can easily count up the area removed, so the area of the circle < area of square. The area will converge to pi as the number of cuts increases (from the definition of pi).
[edit: ack - messed up earlier]
Area of unit circle = pi * r * r = pi * 0.5 * 0.5, = pi * 0.25. By inspection area of square > area of circle, so 1 > 0.25 * pi, so pi < 4. QED.
In other words, one has to carefully think about limits and not just assume that because something is being cut into smaller and smaller segments that it converges to something else. The perimeter doesn't converge in this case. If one instead circumscribes regular polygons with increasing numbers of sides, then that perimeter will converge (the angles are changing rather than all being 90 degrees).
Cheers,
Scott.
