Explain circle formulas by polygon bounds

Try

A circle is not a polygon, so adding a few chord lengths does not give its exact circumference. Yet polygons can trap its measures. Draw a square inside a unit circle and another square tangent outside it. The inner square has diagonal 2, area 2 and perimeter 4√2. The outer square has side 2, area 4 and perimeter 8. These are starting bounds, not the values we hope to prove.

Predict

Predict what happens when every central angle is bisected, producing 8, then 16, then 32 sides. Inner chord paths get longer while remaining inside; outer tangent polygons lose their corners and get shorter. Why should the gap approach zero? “Many sides look round” is not a mathematical reason. We will use a right triangle and a bounded perimeter to control that gap.

Experiment

In the equal-scale unit-circle diagram, the inscribed and circumscribed squares have parallel corresponding sides. Label r=1, inner apothem a=1/√2 and half-side s/2=1/√2 in the right triangle. An apothem is the perpendicular distance from the center to a polygon side.

Compute each polygon area by joining its center to its vertices. There are n triangles with total base length equal to the polygon perimeter. Their common perpendicular heights are a for the inner polygon and r for the tangent polygon. Thus the areas are pa/2 and qr/2 respectively.

Understand

For the nested regular polygons with n=4,8,16,…, let p be the inner perimeter, q the corresponding outer perimeter, s=p/n the inner side and a its apothem. Similar central triangles give q=pr/a. Pythagoras gives a²=r²−(s/2)².

Here is the control on the limit. Each inner perimeter is at most 8r: an edge length is at most its absolute horizontal change plus its absolute vertical change. Along a convex closed polygon inside the bounding square, total horizontal variation is at most 4r and total vertical variation at most 4r. Adding these bounds controls the whole perimeter. Consequently s=p/n≤8r/n, which tends to zero. The right-triangle identity then forces a to approach r, without a trigonometric or calculus assumption. Since a≥r/√2 from the initial square onward, division by a stays bounded. Hence q−p=p(r/a−1) tends to zero.

Triangle inequality makes the refined inner perimeter increase. Removing an outer corner and replacing its two pieces by the new tangent segment makes the outer perimeter decrease. Both remain bounded and the gap tends to zero, so they approach a common number C. We use this standard polygon-limit definition of circle circumference; it is not a claim that every finite polygon has perimeter C.

Similarity multiplies every polygon length by a positive factor k, so it multiplies C by k. Therefore C/(2r) is independent of radius. Call this dimensionless constant π: C=2πr. This defines the constant through circumference; it does not calculate its decimal expansion.

The circle lies between the two polygons. Their areas pa/2 and qr/2 both approach Cr/2 because a approaches r and p,q approach C. Monotonicity of area then traps the circle area at Cr/2=πr². The vanishing difference, rather than a rearrangement of finitely many curved wedges into a perfect rectangle, supplies the justification.

Apply

Use the practice to derive bounds at a new radius and explain the scale argument in your own words. If a numerical perimeter is needed, report an inner/outer interval or a justified rounding tolerance. A midpoint of a valid bracket has error at most half the bracket width.

The argument relies on ordinary Euclidean similarity, Pythagoras, polygon area and the completeness principle that nested bounds with vanishing gap determine a number. It needs no derivatives or integrals, but it does need that limiting reasoning; merely substituting π≈3.14 does not explain the formula.

Continue

Disk area now has a justified expression. Extruding such a disk through a perpendicular height produces a cylinder, while comparing entire families of section areas leads to Cavalieri’s principle and later to a sphere-volume argument.