Question Number 13891 by Tinkutara last updated on 24/May/17
$$\mathrm{Consider}\:{n}\:\mathrm{red}\:\mathrm{and}\:{n}\:\mathrm{blue}\:\mathrm{points}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{plane},\:\mathrm{no}\:\mathrm{three}\:\mathrm{of}\:\mathrm{them}\:\mathrm{being}\:\mathrm{collinear}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{one}\:\mathrm{can}\:\mathrm{connect}\:\mathrm{each}\:\mathrm{red} \\ $$$$\mathrm{point}\:\mathrm{to}\:\mathrm{a}\:\mathrm{blue}\:\mathrm{one}\:\mathrm{with}\:\mathrm{a}\:\mathrm{segment} \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{no}\:\mathrm{two}\:\mathrm{segments}\:\mathrm{intersect}. \\ $$
Commented by mrW1 last updated on 25/May/17
$${When}\:{I}\:{image}\:{that}\:{these}\:{segments} \\ $$$${were}\:{matches}\:{of}\:{different}\:{length} \\ $$$${and}\:{the}\:{red}\:{points}\:{were}\:{the}\:{heads}\:{of} \\ $$$${the}\:{matches},\:{I}\:{know}\:{it}\:{is}\:{always} \\ $$$${possible}\:{to}\:{place}\:{the}\:{matches}\:{on}\:{a} \\ $$$${table}\:{such}\:{that}\:{they}\:{don}'{t}\:{lie}\:{over}\:{each} \\ $$$${other}. \\ $$