APP009 · STEM
Archimedes Pi Explorer — Blue Team vs. Red Team Polygon Squeeze
V05 · Unit Circle (r = 1, Area = π) · Constructive Isosceles Triangle Summation
π Method:
📄Spec V05
Polygon (n):
Sides (Blue = Red):
🔵 BLUE TEAM · Inside (Inscribed)
Vertices Touch · A_Blue < π
6 / 6 Isosceles △
2.598076211
6 × 0.433012702 = 2.598076211
Blue Isosceles Triangle Basis (Legs R = r = 1)
Apex Δθ = 60.00°
Altitude h = cos(π/n)
0.866025404
Base b = 2·sin(π/n)
1.000000000
1 △ Area = ½·b·h
0.433012702
⚖️ CENTER · Average Area vs. Real π
n = 6 · Hexagon
A_Avg = (2.598076211 + 3.464101615) / 2
3.0310889132
Real π = 3.1415926536
|
|Error| = 0.110503740 (3.5174%)
Digit-by-Digit Comparison vs. Real π
1 Significant Digit Matched (“3.”)
A_Avg(n):
Real π:
Blue Lower Bound ≤ π ≤ Red Upper Bound
Spread Δ = 0.866025404
Blue: 2.598076211
π = 3.141592654
Red: 3.464101615
| Sides (n) | Blue (n·A_△) | Average | Red (n·A_△) | 1st Mismatch |
|---|
🔴 RED TEAM · Outside (Circumscribed)
Sides Tangent · A_Red > π
6 / 6 Isosceles △
3.464101615
6 × 0.577350269 = 3.464101615
Red Isosceles Triangle Basis (From Center r = 1)
Apex Δθ = 60.00°
Altitude (h = r)
1.000000000
Base b = 2·tan(π/n)
1.154700538
1 △ Area = ½·b·h
0.577350269
📐 Constructive Isosceles Triangle Formulation of π (APP009 · STEM Knowledge Graph)
Premise: A circle of radius r = 1 has exact area
A = π·r² = π ≈ 3.141592653589793.... Rather than treating polygon area as an abstract formula,
we decompose both the inscribed (Blue, Left) and circumscribed (Red, Right) regular n-sided polygons into
n congruent isosceles triangles emanating from the center (0,0),
each with central apex angle Δθ = 2π / n (half-angle α = π / n).
1. Blue Team (Left — Inside / Inscribed Equilateral Polygon — Touching Vertices):
Each vertex sits on the unit circle, so the two equal legs of each isosceles triangle have length
• Single Blue Isosceles Triangle Area:
• Sum of n Blue Triangles:
Each vertex sits on the unit circle, so the two equal legs of each isosceles triangle have length
R = r = 1,
the altitude is h_Blue = cos(π/n), and the base is b_Blue = 2·sin(π/n).• Single Blue Isosceles Triangle Area:
A_△,Blue = ½ · b_Blue · h_Blue = sin(π/n)·cos(π/n) = ½·sin(2π/n)• Sum of n Blue Triangles:
A_Blue(n) = Σ(k=1..n) A_△,Blue = (n / 2) · sin(2π/n) < π
2. Red Team (Right — Outside / Circumscribed Equilateral Polygon — Tangential Sides):
Each side is tangent to the unit circle at its midpoint, so the altitude (height) of each isosceles triangle is the circle radius
• Single Red Isosceles Triangle Area:
• Sum of n Red Triangles:
Each side is tangent to the unit circle at its midpoint, so the altitude (height) of each isosceles triangle is the circle radius
h_Red = r = 1, and the base is b_Red = 2·tan(π/n).• Single Red Isosceles Triangle Area:
A_△,Red = ½ · b_Red · h_Red = ½ · [2·tan(π/n)] · 1 = tan(π/n)• Sum of n Red Triangles:
A_Red(n) = Σ(k=1..n) A_△,Red = n · tan(π/n) > π
3. Center Average (Area Mode) & Asymptotic Convergence across {3, 4, 5, 6, 8, 10, 20, 50, 100, 200, 500, 1000, 5000}:
• Average Area:
• By Taylor series expansion,
• Stepping across
• Average Area:
A_Avg(n) = (A_Blue(n) + A_Red(n)) / 2• By Taylor series expansion,
A_Blue(n) ≈ π − 2π³/(3n²) and A_Red(n) ≈ π + π³/(3n²),
so A_Avg(n) ≈ π − π³/(6n²).• Stepping across
3 → 4 → 5 → 6 → 8 → 10 → 20 → 50 → 100 → 200 → 500 → 1000 → 5000 locks in 3. at n=3,
3.1 at n=8, 3.141 at n=100..200,
3.1415 at n=500..1000, and 3.141592 (7 significant digits, A_Avg(5000) = 3.1415924470, within 0.0000002 of π) at n=5000!
4. V04 Perimeter ÷ 2 Mode (C = 2π·r = 2π ⇒ Half-Perimeter = π) — Why Perimeter is 2× More Accurate:
• Since a unit circle (
– Blue Half-Perimeter:
– Red Half-Perimeter:
– Average Half-Perimeter:
• Key Mathematical Discovery: In Area Mode, the Blue inscribed triangle suffers a double shrinkage—both its base
• Since a unit circle (
r = 1) has circumference C = 2π·r = 2π, dividing the polygon perimeter P(n) = n · b by 2 yields the direct π estimator:– Blue Half-Perimeter:
½·P_Blue(n) = (n · b_Blue)/2 = n · sin(π/n) ≈ π − π³/(6n²) < π– Red Half-Perimeter:
½·P_Red(n) = (n · b_Red)/2 = n · tan(π/n) ≈ π + π³/(3n²) > π (identical to A_Red(n) since h_Red = r = 1)– Average Half-Perimeter:
P_Avg(n) = (½·P_Blue(n) + ½·P_Red(n)) / 2 ≈ π + π³/(12n²)• Key Mathematical Discovery: In Area Mode, the Blue inscribed triangle suffers a double shrinkage—both its base
b_Blue = 2·sin(π/n) AND its altitude h_Blue = cos(π/n) shrink inside the circle, making A_Blue underestimate π by 4× as much as ½·P_Blue! Consequently, Perimeter ÷ 2 Mode has exactly HALF the error of Area Mode at every n (+π³/(12n²) vs. −π³/(6n²)), locking in 3.14 at n=20, 3.14159 at n=500, and 3.141592 at n=1000!
5. V05 Comparator Mode (Area vs. Perimeter ÷ 2 Head-to-Head on Blue & Red Graphs):
• Left Column (Blue Team — Inscribed): Superimposes both the shaded interior triangle wedges (
• Right Column (Red Team — Circumscribed): Superimposes both the shaded interior triangle wedges (
• Center Comparator & Super-Bracket: Displays a 3-row digit-by-digit tape (
• Left Column (Blue Team — Inscribed): Superimposes both the shaded interior triangle wedges (
A_Blue(n) = (n/2)·sin(2π/n)) and the outer chord segments (½·P_Blue(n) = n·sin(π/n)), showing that A_Blue(n) = cos(π/n) · ½·P_Blue(n).• Right Column (Red Team — Circumscribed): Superimposes both the shaded interior triangle wedges (
A_Red(n)) and outer tangent segments (½·P_Red(n)), proving that A_Red(n) ≡ ½·P_Red(n) = n·tan(π/n) (exact 0.000000000 difference) because every Red triangle has altitude h_Red = r = 1!• Center Comparator & Super-Bracket: Displays a 3-row digit-by-digit tape (
A_Avg(n) vs. Real π vs. P_Avg(n)) and proves that for n ≥ 5, A_Avg(n) < π < P_Avg(n) with |Error_Area| / |Error_Perim| → 2.000×. Moreover, the weighted combination (A_Avg + 2·P_Avg)/3 cancels the O(1/n²) error completely!