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Question Number 16803 by ajfour last updated on 26/Jun/17

Commented by ajfour last updated on 26/Jun/17

$$\mathrm{Prove}\mathrm{that}\mathrm{in}\mathrm{an}\mathrm{equilateral}$$$$\mathrm{triangle}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{distances}$$$$\mathrm{from}\mathrm{any}\mathrm{point}\mathrm{to}\mathrm{the}\mathrm{three}\mathrm{sides}$$$$\mathrm{is}\mathrm{a}\mathrm{constant}.$$

Answered by mrW1 last updated on 26/Jun/17

$$\mathrm{Area}\mathrm{of}\mathrm{\Delta }\mathrm{ABC}=\mathrm{\Delta }$$$$\mathrm{AB}=\mathrm{BC}=\mathrm{AC}=\mathrm{a}$$$$\mathrm{\Delta }=\frac{1}{2}\mathrm{AB}\times \mathrm{PN}+\frac{1}{2}\mathrm{AC}\times \mathrm{PM}+\frac{1}{2}\mathrm{BC}\times \mathrm{PL}$$$$=\frac{1}{2}\mathrm{a}(\mathrm{PL}+\mathrm{PM}+\mathrm{PN})$$$$\Rightarrow \mathrm{PL}+\mathrm{PM}+\mathrm{PN}=\frac{2\mathrm{\Delta }}{\mathrm{a}}=\mathrm{constant}$$

Commented by ajfour last updated on 26/Jun/17

$$\mathrm{Thank}\mathrm{you}\mathrm{Sir},\mathrm{didnot}\mathrm{realise}\mathrm{it}$$$$\mathrm{was}\mathrm{as}\mathrm{straight}.$$

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