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Question Number 97250 by mathocean1 last updated on 07/Jun/20

Answered by 1549442205 last updated on 07/Jun/20

Commented by 1549442205 last updated on 07/Jun/20

$$\mathrm{a}/\mathrm{Apply}\mathrm{the}\mathrm{formula}\mathrm{of}\mathrm{the}\mathrm{median}\mathrm{in}\mathrm{the}\mathrm{triangle}$$$${\mathrm{m}}_{\mathrm{a}}=\frac{1}{2}\sqrt{2({\mathrm{b}}^2+{\mathrm{c}}^2)-{\mathrm{a}}^2}\iff {\mathrm{b}}^2+{\mathrm{c}}^2={2\mathrm{m}}_{\mathrm{a}}^2+\frac{{\mathrm{a}}^2}{2}$$$$\mathrm{Sin}\mathrm{MI}\mathrm{is}\mathrm{the}\mathrm{median}\mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{BMC}$$$$\mathrm{we}\mathrm{get}{\mathrm{MB}}^2+{\mathrm{MC}}^2={2\mathrm{MI}}^2+\frac{{\mathrm{BC}}^2}{2}={2\mathrm{MI}}^2+4(\ast )(\mathrm{as}\mathrm{BC}=2\sqrt{2})$$$$\mathrm{we}\mathrm{have}\mathrm{IB}=\mathrm{IC}=\sqrt{2}.\mathrm{AI}=\sqrt{{\mathrm{AB}}^2-{\mathrm{IB}}^2}=\sqrt{2}$$$$.\mathrm{Let}\mathrm{K}\mathrm{be}\mathrm{the}\mathrm{point}\mathrm{symetry}\mathrm{to}\mathrm{J}\mathrm{through}\mathrm{BC}$$$$\mathrm{then}\mathrm{AJ}=\mathrm{JK}=\frac{2}{3}\mathrm{AI}=\frac{2\sqrt{2}}{3},\mathrm{AK}=\frac{4\sqrt{2}}{3}$$$$\mathrm{Since}\mathrm{MJ}\mathrm{is}\mathrm{the}\mathrm{median}\mathrm{of}\mathrm{\Delta }\mathrm{MAK},$$$${\mathrm{MA}}^2+{\mathrm{MK}}^2={2\mathrm{MJ}}^2+\frac{{\mathrm{AK}}^2}{2}={2\mathrm{MJ}}^2+\frac{16}{9}\left(1\right)$$$$\mathrm{Since}\mathrm{MI}\mathrm{is}\mathrm{the}\mathrm{median}\mathrm{of}\mathrm{\Delta }\mathrm{MJK},$$$${\mathrm{MJ}}^2+{\mathrm{MK}}^2={2\mathrm{MI}}^2+\frac{{\mathrm{JK}}^2}{2}={2\mathrm{MI}}^2+\frac{4}{9}\left(2\right)$$$$\mathrm{Substract}\mathrm{two}\mathrm{equalities}\left(1\right),\left(2\right)\mathrm{we}\mathrm{get}$$$${\mathrm{MA}}^2-{\mathrm{MJ}}^2={2\mathrm{MJ}}^2-{2\mathrm{MI}}^2+\frac{4}{3}.\mathrm{Hence},$$$${\mathrm{MA}}^2+{2\mathrm{MI}}^2={3\mathrm{MJ}}^2+\frac{4}{3}\left(3\right).$$$$\mathrm{b}/\mathrm{From}(\ast )\mathrm{and}\left(3\right)$$$$\mathrm{we}\mathrm{get}{\mathrm{MA}}^2+{\mathrm{MB}}^2+{\mathrm{MC}}^2={3\mathrm{MJ}}^2+\frac{16}{3}(\ast \ast )$$$$\mathrm{c}/\mathrm{Since}\mathrm{the}\mathrm{point}\mathrm{J}\mathrm{is}\mathrm{fixed},\mathrm{from}(\ast \ast )\mathrm{and}\mathrm{the}$$$$\mathrm{hypothesis}{\mathrm{MA}}^2+{\mathrm{MB}}^2+{\mathrm{MC}}^2=8\mathrm{we}$$$$\mathrm{follow}\mathrm{that}{3\mathrm{MJ}}^2+\frac{16}{3}=8\iff \mathrm{MJ}=\frac{2\sqrt{2}}{3}$$$$\mathrm{which}\mathrm{shows}\mathrm{that}\mathrm{set}\left(\mathrm{T}\right)\mathrm{of}\mathrm{the}\mathrm{points}\mathrm{M}$$$$\mathrm{satisfying}{\mathrm{MA}}^2+{\mathrm{MB}}^2+{\mathrm{MC}}^2=8\mathrm{lie}\mathrm{on}$$$$\mathrm{the}\mathrm{circle}\mathrm{has}\mathrm{the}\mathrm{center}\mathrm{be}\mathrm{the}\mathrm{point}\mathrm{J}$$$$\mathrm{and}\mathrm{the}\mathrm{radius}\mathrm{equal}\mathrm{to}\frac{2\sqrt{2}}{3}$$$$$$

Commented by mathocean1 last updated on 07/Jun/20

$$Thankyousir....great.$$

Commented by 1549442205 last updated on 08/Jun/20

$$\mathrm{you}\mathrm{are}\mathrm{wellcome}!$$

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