 Doubled rectangle extremal-length duality

Glue two copies of $[0,w]\times [0,h]$ along their left, bottom and right sides. Their unglued top sides form the outer boundary of a disk; the two bottom corners are marked interior points. Let $\Gamma$ be the loops surrounding both points, and ${\Gamma }^{\ast }$ the boundary-to-boundary crosscuts separating them. Then $\lambda (\Gamma )\lambda ({\Gamma }^{\ast })=4$.

Each vertical seam joins a marked point to the outer boundary, so the Jordan curve theorem forces a loop surrounding both points to meet both seams. In the flat conformal metric, its total horizontal travel is therefore at least $2w$. Doubled horizontal segments attain this length. Each separating crosscut meets the bottom seam and starts and ends at height $h$, so its total vertical travel is at least $2h$; doubled vertical segments attain it. The total area is $2wh$.

For an arbitrary density, integrate the lengths of the doubled horizontal segments over their height. The Cauchy-Schwarz inequality yields $L_{\rho }(\Gamma {)}^{2}h\le 2w{A}_{\rho }$. Doubled vertical segments similarly give $L_{\rho }({\Gamma }^{\ast }{)}^{2}w\le 2h{A}_{\rho }$. These upper bounds equal the flat-metric ratios, proving both displayed formulas. Mapping the upper and lower halves of a disc to these rectangles gives the corresponding duality for a simply connected domain with two marked interior points.