Inscribe a square on a shape’s outline
Place four points on the shape’s outline so they form a square. As you move your cursor, the nearest outline point follows it — tap to drop a vertex. After four drops the quadrilateral is scored: closer to a perfect square (equal sides, 90° angles) means a higher score. You can drag any placed point afterwards to refine.
The Inscribed Square Problem (Toeplitz, 1911)
Otto Toeplitz asked a deceptively simple question in 1911: does every simple closed curve in the plane contain four points that form a square? More than a century later, the answer is probably yes — but it’s still one of the oldest open problems in geometry.
The problem is settled for nice classes of curves: polygons, convex curves, smooth curves, piecewise-smooth curves, and curves with bounded curvature all provably have an inscribed square. Recent work by Greene & Lobb (2020) settled the problem for smooth Jordan curves by connecting it to symplectic geometry. But for arbitrary Jordan curves — including pathological fractal ones — existence is still unproved.
On this page every generated shape is piecewise-smooth — a mix of straight segments, circular arcs, and Bezier curves. That’s well within the classes where the problem is proven, so a perfect inscribed square always exists on these shapes. Your job is to find one by eye.
Tips for spotting an inscribed square
- Start with a diameter: pick two points on roughly opposite sides of the shape — they’re candidates for the square’s diagonal.
- The two other vertices must sit on a line perpendicular to that diagonal, with the same length.
- Convex regions usually have several inscribed squares; concave shapes sometimes have just one obvious one.
- If three vertices already look good but the fourth is hard to place, adjust vertex two — the constraint often means the whole square has to rotate slightly.
Related
- Inscribed Equilateral Triangle — the solved sibling of the Toeplitz problem.
- Cut — slice shapes by area instead of placing vertices.
- Balance — find where the shape’s centroid sits.