Interactive Math & Science Lessons
Experiment 1 · Geometry · rigorous bounds
For a circle of diameter 1, the inner polygon is too short and the outer polygon is too long. Their half-perimeters therefore bracket π.
Conclusion: Because both bounds approach the same limit, the squeeze theorem identifies that limit as π.
Experiment 2 · Measurement · circumference
π is defined by C/d. A wheel with diameter 1 carries exactly one circumference of rim through one complete rotation.
Conclusion: One rotation gives distance C; with d = 1, the measured distance equals π.
Experiment 3 · Calculus · area
A thin ring at radius x has length 2πx. Cutting and stacking all such rings orders their lengths from 0 to 2πr.
Conclusion: The triangle has base 2πr and height r, so its area is ½(2πr)r = πr².
Experiment 4 · Physics · phase space
Momentum and kinetic energy constrain the two velocities to an ellipse. After rescaling, each collision becomes a reflection around a circle.
Conclusion: For M/m = 100^d, the count is floor(π·10^d), revealing successive decimal digits of π.
Experiment 5 · Probability · orientation
A needle crosses a line only when its center is close enough and its projected half-length reaches the line. Random orientation introduces an integral of sin θ.
Conclusion: When needle length equals line spacing, P(cross) = 2/π, hence π ≈ 2N/C.
Experiment 6 · Probability · area
Uniform points land in a region with probability equal to that region's share of the square's area.
Conclusion: Multiplying the observed fraction by four gives the Monte Carlo estimate π ≈ 4H/N.
Experiment 7 · Analysis · convergence
Infinite series can encode π without drawing a circle. Different identities approach the same limit at very different speeds.
Conclusion: A partial sum is an approximation, and the displayed absolute error measures how far it remains from π.
Each section isolates one idea, derives the relevant equation, and explains what the picture is actually proving.
Every Euclidean circle has the same circumference-to-diameter ratio.
C / d = π
Scaling a circle by a factor k multiplies both C and d by k, so C/d is unchanged. That invariant is π: C = πd = 2πr. The circle does not contain a hidden preferred unit; π survives every resizing.
Similarity, not a particular measurement, makes π universal.
Archimedes used one figure below the circumference and another above it.
n sin(π/n) < π < n tan(π/n)
For diameter 1, an inscribed regular n-gon gives n sin(π/n), while a circumscribed one gives n tan(π/n). The inequality n sin(π/n) < π < n tan(π/n) is guaranteed geometrically; increasing n narrows a certified interval.
A bound tells you both an approximation and its worst possible error.
The wheel's center moves uniformly, but a marked rim point speeds up and slows down.
x = r(θ − sin θ), y = r(1 − cos θ)
With rolling angle θ, the center advances rθ while the rotating radius subtracts r sin θ horizontally and r cos θ vertically. Thus x = r(θ − sin θ) and y = r(1 − cos θ). At each cusp, the marked point is instantaneously at rest on the ground.
Rolling without slipping links linear distance to angle through s = rθ.
Each thin ring contributes approximately circumference × thickness.
∫₀ʳ 2πx dx = πr²
A ring of radius x and thickness dx has area dA = 2πx dx. Adding every ring from 0 to r gives A = ∫₀ʳ 2πx dx = πr². The unrolled triangle is a geometric picture of that integral.
More, thinner rings make the staircase boundary approach a straight line.
Energy conservation turns the velocity state into motion on an ellipse.
N = floor(π√(M/m)), M/m = 100ᵈ
Using coordinates (√m·v₁, √M·v₂) changes the energy ellipse into a circle. A block collision or wall collision reflects the point across a line. The wedge angle is controlled by √(m/M), so a ratio M/m = 100^d creates roughly π·10^d reflections before escape.
The digits arise from circular geometry in velocity space, not from mysterious forces.
The crossing probability averages a trigonometric projection over all angles.
P = 2L/(πD)
For L ≤ D, a needle at angle θ crosses when its center lies within (L/2)|sin θ| of a line. Averaging over uniform center positions and θ ∈ [0, π] yields P = 2L/(πD). The simulator uses L = D, so P = 2/π.
Randomness estimates π only because the underlying probability has been derived exactly.
More points improve the estimate, but random sampling has a square-root speed limit.
π ≈ 4H/N
Each point is a Bernoulli trial with success probability p = π/4. The standard error of H/N is √(p(1−p)/N), so gaining one extra decimal digit typically needs about 100 times as many points.
A noisy estimate is expected; convergence does not mean every new batch gets closer.
Equal limits can hide dramatically different convergence rates.
Σ 1/n² = π²/6
Leibniz follows arctan(1) and needs enormous term counts for many digits. Nilakantha uses products of three consecutive integers and converges much faster. Euler's Basel identity links Σ1/n² to π²/6, exposing π inside the arithmetic of integers.
The graph compares partial sums; it does not imply that π itself changes.
The ancient construction problem was settled by the algebraic nature of π.
π ∉ Q̄
A square equal in area to a unit circle would need side √π. Straightedge-and-compass constructions produce only constructible algebraic numbers. Lindemann proved in 1882 that π is transcendental, so √π cannot be constructed by those rules.
Numerical approximations are easy; the prohibited exact construction is the impossible part.
A hexagon traps π between 3 and 3.4641. That is a spread of 0.46 — useless. Work out how many doublings it takes to make it good, and what that says about Archimedes.
The trick needs no π at all: a polygon inside the circle has a shorter perimeter, one outside has a longer one, and both perimeters are computable from trigonometry alone.
Six sides is where it starts, because a regular hexagon's side equals the radius — the inscribed perimeter is exactly 3 diameters, with no approximation anywhere. The circumscribed one is 2√3.
So π is trapped in a band 0.464 wide, which is a real proof and a bad number. Nothing about it is wrong; there is simply not enough polygon.
Doubling the sides tightens it fast. The width falls roughly fourfold each time — 0.110, 0.0270, 0.00674, 0.00168 — because the error is second order in the side length.
Four doublings reaches 96 sides, and the band is 3.141032 to 3.142715.
Answer
The tool prints 3.000000, 3.464102 and a width of 0.464102. Four doublings later a 96-gon gives π to two decimals with certainty — and that is exactly where Archimedes stopped, around 240 BC, with 3¹⁰⁄₇₁ < π < 3¹⁄₇. What makes it a landmark is not the accuracy but the bracket: he did not estimate π, he proved two numbers it must lie between, and the method improves without limit. Push the slider to its maximum of 128 sides and the bounds read 3.141277 and 3.142224 — still only two decimals certain, after more than four doublings. That is the measure of both the achievement and the method's limit: the width falls as 1/n², so every further decimal costs ten times the sides.