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OpenGeometry·1911

Square Peg Problem

Every simple closed curve in the plane is conjectured to contain four vertices of a square.

Formula

Conjecture (Toeplitz 1911)
γ:S1R2 continuous, injective    x1,x2,x3,x4γ forming a square\gamma:\,S^1\hookrightarrow\mathbb{R}^2\text{ continuous, injective}\;\Longrightarrow\;\exists\,x_1,x_2,x_3,x_4\in\gamma\text{ forming a square}

Every Jordan curve in the plane inscribes a square — open in full generality, proved for smooth and piecewise-linear curves.

Smooth case (proved)
γC1    γ inscribes a square\gamma\in C^1\;\Longrightarrow\;\gamma\text{ inscribes a square}

For smooth curves the result follows from topological arguments; the difficulty is purely continuous curves with bad local behavior.

Inscribed rectangle theorem
γ Jordan curve,   inscribed rectangle of every aspect ratio r(0,1]\forall\,\gamma\text{ Jordan curve},\;\exists\text{ inscribed rectangle of every aspect ratio }r\in(0,1]

Greene–Lobb (2021) proved every smooth Jordan curve inscribes rectangles of every aspect ratio, using symplectic geometry.

Summary

Draw any closed loop on paper — no matter how wild. Can you always find four points on it forming a perfect square? Toeplitz conjectured yes in 1911, and partial results cover convex, smooth, and piecewise-linear curves. The most dramatic recent progress came from Greene and Lobb (2021), who used symplectic geometry — Möbius bands in R⁴ forming a Klein bottle — to prove every smooth curve inscribes rectangles of every aspect ratio. The general case for non-smooth Jordan curves remains open.

Sources

Videos