The (Un)equal Tangents Problem

Abstract

The problemGiven a point A outside of a closed strictly convex plane curve γ, there are two tangent segments from A to γ, the left and the right ones, looking from point A.Problem: Does there exist a curve γ such that one can walk around it so that, at all moments, the right tangent segment is smaller than the left one?In other words, does there exist a pair of simple closed curves, γ and Γ, the former strictly convex, the latter containing the former in its interior, such that for every point A of Γ the right tangent segment to γ is smaller than the left one?Over the years, I have polled numerous colleagues, mostly as a dinner table topic. Most of them thought that the answer was negative, and quite a few tried to provide a proof, but each attempt had a flaw. I invite the reader to think about this question too before reading any further.Up until recently, I have believed that for any oval 1 γ and every closed curve Γ going around γ, there existed a point A ∈ Γ from which the tangent segments to γ were equal. In fact, I conjectured in [7] that there existed at least four such points. Figure 1 illustrates the situation for an ellipse. The extensions of the axes partition the plane into four quadrants marked + and − according to the sign of the difference between the left and the right tangent segments. The axes themselves are the locus of points from which the tangent segments * Department of Mathematics, Pennsylvania State University, University Park, PA 16802, USA; e-mail: tabachni@math.psu.edu 1 a smooth strictly convex closed curve.

The equal tangents property

“…In conclusion, we remark that Lemma 1 holds in all three classical geometries: elliptic, Euclidean, and hyperbolic; we give a proof that works in all three cases, cf. [5,7].…”

“…Next we construct pairs of curves in the plane, Γ and γ, such that Γ encloses γ and, for every point x ∈ Γ, the two tangent segments from x to γ have equal lengths. Compare with [7] where a pair of curves Γ and γ is constructed such that, for every point x ∈ Γ, the tangent segments from x to Γ have unequal lengths.…”

Fun Problems in Geometry and Beyond

“…Here is the whole process: See [30] for more details and comments. We mention a relevant result from [24], motivated by the flotation theory: given an oval γ and a fixed angle α, the locus of points from which the oval is seen under angle α contains at least four points from which the tangent segments to γ have equal lengths.…”

Configuration Spaces of Plane Polygons and a sub-Riemannian Approach to the Equitangent Problem

“…Here we construct a nested pair of curves γ ⊂ Γ such that Γ is contained in the equitangent locus of γ. Our approach is similar to that of [23].…”

“…Another, somewhat surprising, observation is that there exist nested curves γ ⊂ γ 1 such that γ 1 is disjoint from the equitangent locus of γ, see [23]. In other words, a point can make a full circuit around γ in such a way that, at all times, the two tangent segments to γ have unequal lengths.…”