4.6 TCGL Polygon Is Reconstructible from g$_{r}$ and g$_{s}$ without tail Theorem 4.6.1. For a tangent-cone graph-like polygon Ω , knowing g ( s, r ) , g$_{r}$ ( s, r ) and g$_{s}$ ( s, r ) for all s ∈ [0 , L ) and a particular r for which ∂ Ω is tangent-cone graph-like is sufficient to completely determine Ω up to translation and rotation; that is, we can recover the side lengths and angles of Ω . Proof. For a given s and r where g$_{r}$ and g$_{s}$ exist, we can use them to obtain r ( θ$_{2}$ − θ$_{1}$ ) as the length of the circular arc between the entry and exit points by Lemma 4.4.1 and r (sin θ$_{2}$ − sin θ$_{1}$ ) as the difference in heights of the entry and exit points by Lemma 4.4.2. We wish to recover θ$_{1}$ and θ$_{2}$ from these quantities. Note that if ( θ$_{1}$, θ$_{2}$ ) = ( φ$_{1}$, φ$_{2}$ ) is one possible solution, then so is ( θ$_{1}$, θ$_{2}$ ) = (2 π − φ$_{2}$, 2 π − φ$_{1}$ ) so solutions always come in pairs. We can imagine placing a circular arc with angle g$_{r}$ r on our circle and sliding it around until the endpoints have the appropriate height difference, yielding our θ$_{1}$ and θ$_{2}$ . Note that since Ω is tangent-cone graph-like, one endpoint must be on the left side of the circle and the other must be on the right and we cannot slide either endpoint to or beyond the vertical line through the center of the circle. Therefore, as we slide the right endpoint down, the left endpoint slides up so that the height difference as a function of the slide is strictly monotonic. Therefore, the slide that gives us θ$_{1}$ and θ$_{2}$ is unique for a given starting arc placement. However, there are two starting arc placements: the first calls the angle for the right endpoint θ$_{1}$ and the left endpoint θ$_{2}$ (so the interior of Ω is “up” in the circle) and the second swaps these (so the interior of Ω is “down”). Since we have adopted the convention that ∂ Ω is traversed in a counterclockwise direction (so the interior of Ω is up in the circles) we therefore pick 104