Question Number 14365 by RasheedSindhi last updated on 31/May/17
$$\mathrm{Related}\:\mathrm{to}\:\mathrm{Q}#\mathrm{14157} \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{ab}=\alpha^{\mathrm{2}} \\ $$$$\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} −\mathrm{bc}=\beta^{\mathrm{2}} \\ $$$$\mathrm{c}^{\mathrm{2}} +\mathrm{d}^{\mathrm{2}} −\mathrm{cd}=\gamma^{\mathrm{2}} \\ $$$$\mathrm{d}^{\mathrm{2}} +\mathrm{e}^{\mathrm{2}} −\mathrm{de}=\delta^{\mathrm{2}} \\ $$$$\mathrm{e}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} −\mathrm{ea}=\xi^{\mathrm{2}} \\ $$
Commented by mrW1 last updated on 01/Jun/17
$${The}\:{solution}\:{is}\:{a}\:{pyramid}\:{with}\:{a} \\ $$$${pentagon}\:{base}\:{whose}\:{other}\:\mathrm{5}\:{side} \\ $$$${areas}\:{are}\:{triangles}\:{with}\:{a}\:{top} \\ $$$${angle}\:{of}\:\mathrm{60}°. \\ $$