The non-domability conjecture for the planar rhombus ρ(1/2,15/2)\rho(1/2,\sqrt{15}/2)

Let A\mathcal A be the set of a0a\geq 0 for which the planar rhombus ρ(a,4a2)\rho(a,\sqrt{4-a^2}) can be domed, and write ρ:=ρ(1/2,15/2)\rho_\lozenge:=\rho(1/2,\sqrt{15}/2). The ρ\rho_\lozenge conjecture. One has 1/2A1/2\notin\mathcal A; equivalently, the planar unit rhombus ρ\rho_\lozenge cannot be domed. The conjecture is presented as an analogue of the (2,2,1)(2,2,1) triangle conjecture, and the paper notes that non-domability of the integral rhombus 2ρ2\rho_\lozenge would imply both claims.

Sources & referencesView supporting material

Primary source

Alexey Glazyrin and Igor Pak, “Domes over curves”, arXiv:2005.02555 (2021).

Progress summary

Never refreshed

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Solutions 0

No solutions have been posted yet.