Triangle covering problems and the Viterbo inequality in the plane

Abstract

We review a certain problem on covering triangles in the plane. Equivalently, it can be viewed as a family of 'isobilliard' inequalities in convex shapes, and as a special case of Viterbo's conjecture in symplectic geometry. We give an elementary overview of these topics and, using the optics of the covering problem, we establish several new special cases of Viterbo's conjecture, provide a simple explanation of the counterexample of Haim-Kislev and Ostrover, and state a few open questions. The main novel result is a proof of Viterbo's conjecture for lagrangian products $K \times Q$, where $Q \subset \mathbb{R}^2$ is any quadrilateral and $K \subset \mathbb{R}^2$ is any convex shape.

Figures (13)

• Figure 1: Mallée's parabolic triangle (right) is composed of six parabolic arcs. One of these arcs is shown on the left, and the full shape is obtained by reflecting it under the $D_3$ dihedral group around the central regular triangle.
• Figure 2: A periodic (generalized) $Q$-Minkowski billiard orbit in $K$ corresponds to a pair of closed polygonal curves in $K$ and $Q$ satisfying symmetric conditions: $q_{i+1} - q_i \in N_Q(p_i)$ and $p_{i-1} - p_{i} \in N_Q(q_i)$ for all $i$.
• Figure 3: $\triangle$-normed triangles define the curve $\Gamma$ (middle), and in the most symmetric placement their convex hull is the regular hexagon $K_0$ (right).
• Figure 4: Half of the symmetrized curve $\Gamma$
• Figure 5: The two-dimensional family of area-minimizing $\triangle$-covers topologically is a disk. Each point represents a relative placement of the two $\triangle$-normal triangles, whose convex hull is a tight hexagon.

Theorems & Definitions (26)

• Conjecture 1.1
• Theorem 1.2
• Theorem 2.1
• Conjecture 2.2
• Theorem 2.3
• Definition 2.4
• Theorem 2.5
• Example 3.1
• Definition 3.2
• Theorem 3.3