= Doubled rectangle extremal-length duality
{title2=$\lambda(\Gamma)=2w/h,\quad\lambda(\Gamma^*)=2h/w$}
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 $\Gamma^*$ the boundary-to-boundary crosscuts separating them. Then $\lambda(\Gamma)\lambda(\Gamma^*)=4$.
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 $L_\rho(\Gamma)^2h\leq2wA_\rho$. Doubled vertical segments similarly give $L_\rho(\Gamma^*)^2w\leq2hA_\rho$. 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.
\Image[/doubled-rectangle-extremal-length.png]
{title=Two rectangles glued along three sides give loop length 2w, crosscut length 2h and total area 2wh}
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