Let us start with metric on $\mathbb R^4=\mathbb R^2\times \mathbb R^2$. Defined by norm $\|{*}\|$ defined by $$\|(x,y)\|=\tfrac12(|x|+|y|),$$ $\|(x,y)\|=\int_0^1|t\cdot x+(1-t)\cdot y|\,dt,$$where |{ * }| denotes Euclidean norm on \mathbb R^2. This norm, is not strictly strongly convexand generically , so you have a lot of should expect many geodesics connectiong two between close points. Now, your metric is the intrisic metric induced by on the hypersurface \Sigma described by |x-y|=1. It often happens that one of the geodesic in (\mathbb R^4,\|{*}\|) lies in \Sigma. For example the geodesic between segments [(0,0),(0,1)] and [(1,0),(0,0)]. Maybe this is a closed geodesic, but at least it is a closed line with 4 geodesic segments. Maybe it already makes you happy (?). If notSo, you have a Finsler metric on \Sigma=\mathbb S^1\times \mathbb R^2. The unit ball for some coordiantes (u,v,w) of in the tangent plane can be described by is isometric the following inequality:$$\sqrt{u^2+w^2}+\sqrt{v^2+w^2}\le 2.$$This intersection of the ball in the above norm with 3-dimensional subspace in one special direction. It seems that this intersection is strongly convex. The metric is smooth (sinse there is a transitive isometric group action on \mathbb S^1\times \mathbb R^2). It reamins to write differential equasion for geodesic; this should be in any book on Finsler geometry. (It should be pain, but it might help.) 4 added 92 characters in body; added 99 characters in body Let us start with metric on \mathbb R^4=\mathbb R^2\times \mathbb R^2. Defined by norm \|{*}\| defined by$$\|(x,y)\|=\tfrac12(|x|+|y|),$$where |{ * }| denotes Euclidean norm. This norm, is not strictly convex and generically you have a lot of geodesics connectiong two points. Now, your metric is the intrisic metric induced by on the hypersurface \Sigma described by |x-y|=1. It often happens that one of the geodesic in (\mathbb R^4,\|{*}\|) lies in \Sigma. For example the geodesic between segments [(0,0),(0,1)] and [(1,0),(0,0)]. Maybe this is a closed geodesic, but at least it is a closed line with 4 geodesic segments. Maybe it already makes you happy already. (?). If not, you have a Finsler metric on \Sigma=\mathbb S^1\times \mathbb R^2. The unit ball for some coordiantes (u,v,w) of the tangent plane can be described by the following inequality:$$\sqrt{u^2+w^2}+\sqrt{v^2+w^2}\le 2.$$This ball is strongly convex. The metric is smooth (sinse there is a transitive isometric group action on \mathbb S^1\times \mathbb R^2). It reamins to write differential equasion for geodesic; this should be in any book on Finsler geometry. 3 added 254 characters in body Let us start with metric on \mathbb R^4=\mathbb R^2\times \mathbb R^2. Defined by norm \|{*}\| defined by$$\|(x,y)\|=\tfrac12(|x|+|y|),$$where |{ * }| denotes Euclidean norm. This norm, is not strictly convex and generically you have a lot of geodesics connectiong two points. Now, your metric is the intrisic metric induced by on the hypersurface \Sigma described by |x-y|=1. So It often happens that one of the geodesic in (\mathbb R^4,\|{*}\|) lies in \Sigma. For example the geodesic between segments [(0,0),(0,1)] and [(1,0),(0,0)]. Maybe it is makes you happy already. If not, you have a Finsler metric on \mathbb \Sigma=\mathbb S^1\times \mathbb R^2. The unit ball for some coordiantes (u,v,w) of the tangent plane can be described by the following inequality:$$\sqrt{u^2+w^2}+\sqrt{v^2+w^2}\le 2.$$This ball is strongly convex. The metric is smooth (sinse there is a transitive isometric group action on$\mathbb S^1\times \mathbb R^2\$). It reamins to write differential equasion for geodesic; this should be in any book on Finsler geometry.