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**On πr²/4r² = π/4 < 1 as a Constant Subunit Sum**

The areal ratio of any circle to its containing square is πr²/4r² wherein r² cancels out such that π/4 < 1 singular unit.

This means any/all circle/square comparisons sharing a common r invariably reduce into the domain of the unit square.

This is confirmed by the same relation π/4 describing the ratio of π circ. to the perimeter of its containing unit square.

Because πr²/4r² = π/4 < 1 for any/all r, it is a

*fallacy*to assume (and/or make use of)

*several units*to squarely contain

what is otherwise already contained by only a single squared unit. The smallest possible containing square

(of the diameter of the circle whose circumference is π) is 1² = 1 > π/4 with π/4 = πr²/4r² for any/all common r.

Accordingly: Above is shown the surface area of the unit square as a sum of hyperbolic space k & π/4. Importantly,

the circle simultaneously contains & reflects k in/as a hyperbolic region of its own circular squared area: Accordingly, πr²/4r² = π/4 < 1 is a

**constant subunit sum:**composed of two discretely weighted contributions (π-2)/2 < 1 & k < 1 with(π-2)/2 + k = π/4

**(π-2)/2 + 2k = 1**.

When mathematicians are surrounding a circle with a non-circular plane figure, they are crudely estimating this sum in/as a single aggregate of area.

They are assuming the area of the aggregate must be numerically greater than the weight of the circle against the square of its own diameter.

Further, they are assuming the perimeter of a surrounding regular polygon must always be greater than the circumference of the circle.

Both of these assumptions are false.

**Perimeter**

If/as a regular polygon "rolls" along a flat plane alongside a rolling circle, it is always losing translational motion to the y-axis.

If/as the number of sides is sufficiently high (n > 58), the polygon can no longer keep up with the null loss of the circle.

This means a circle can roll further than the regular polygon containing it provided it has enough sides.

The shortest distance between two points is a straight line & polygons approach this better than curves.

**Area**

As above, the unit square is composed of a subunit sum (π-2)/2 + 2k = 1 with π/4 = (π-2)/2 + k. The polygon approximates the latter.

If/as one removes the (π-2)/2 region from the polygon & doubles the remainder, it is always relatively more than (π-2)/2 + 2k = 1.

**In other words: mathematicians have been assuming a**As is self-evident, such figures incessantly overestimate the aggregate as a whole while/as assuming it must be numerically greater than π/4 per unit of 1.

*bigger unit*than*actual*.In reality, the surrounding polygon is with respect to a containing square whose surface area is greater than the surface area containing the circle.

Because the containing square of the circle is smaller, the circle weighs more against it than the polygon against its own larger containing square.