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Question-97250

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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
a/Apply the formula of the median in the triangle m_a =(1/2)(√(2(b^2 +c^2 )−a^2 ))⇔b^2 +c^2 =2m_a ^2 +(a^2 /2) Sin MI is the median of the triangle BMC we get MB^2 +MC^2 =2MI^2 +((BC^2 )/2)=2MI^2 +4(∗)(as BC=2(√2) ) we have IB=IC=(√2) .AI=(√(AB^2 −IB^2 )) =(√2) .Let K be the point symetry to J through BC then AJ=JK=(2/3)AI=((2(√2))/3),AK=((4(√2))/3) Since MJ is the median of ΔMAK, MA^2 +MK^2 =2MJ^2 +((AK^2 )/2)=2MJ^2 +((16)/9)(1) Since MI is the median of ΔMJK, MJ^2 +MK^2 =2MI^2 +((JK^2 )/2)=2MI^2 +(4/9)(2) Substract two equalities (1),(2) we get MA^2 −MJ^2 =2MJ^2 −2MI^2 +(4/3).Hence, MA^2 +2MI^2 =3MJ^2 +(4/3)(3). b/From (∗) and (3) we get MA^2 +MB^2 +MC^2 =3MJ^2 +((16)/3)(∗∗) c/Since the point J is fixed ,from (∗∗) and the hypothesis MA^2 +MB^2 +MC^2 =8 we follow that 3MJ^2 +((16)/3)=8⇔MJ=((2(√2))/3) which shows that set (T) of the points M satisfying MA^2 +MB^2 +MC^2 =8 lie on the circle has the center be the point J and the radius equal to ((2(√2))/3)
$$\mathrm{a}/\mathrm{Apply}\:\mathrm{the}\:\mathrm{formula}\:\mathrm{of}\:\mathrm{the}\:\mathrm{median}\:\mathrm{in}\:\mathrm{the}\:\mathrm{triangle} \\ $$$$\mathrm{m}_{\mathrm{a}} =\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}\left(\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \right)−\mathrm{a}^{\mathrm{2}} }\Leftrightarrow\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} =\mathrm{2m}_{\mathrm{a}} ^{\mathrm{2}} +\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{Sin}\:\mathrm{MI}\:\mathrm{is}\:\mathrm{the}\:\mathrm{median}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{BMC} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{MB}^{\mathrm{2}} +\mathrm{MC}^{\mathrm{2}} =\mathrm{2MI}^{\mathrm{2}} +\frac{\mathrm{BC}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{2MI}^{\mathrm{2}} +\mathrm{4}\left(\ast\right)\left(\mathrm{as}\:\mathrm{BC}=\mathrm{2}\sqrt{\mathrm{2}}\:\right) \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{IB}=\mathrm{IC}=\sqrt{\mathrm{2}}\:.\mathrm{AI}=\sqrt{\mathrm{AB}^{\mathrm{2}} −\mathrm{IB}^{\mathrm{2}} }\:=\sqrt{\mathrm{2}} \\ $$$$.\mathrm{Let}\:\mathrm{K}\:\mathrm{be}\:\mathrm{the}\:\mathrm{point}\:\mathrm{symetry}\:\mathrm{to}\:\mathrm{J}\:\mathrm{through}\:\mathrm{BC} \\ $$$$\mathrm{then}\:\mathrm{AJ}=\mathrm{JK}=\frac{\mathrm{2}}{\mathrm{3}}\mathrm{AI}=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}},\mathrm{AK}=\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$$$\mathrm{Since}\:\mathrm{MJ}\:\mathrm{is}\:\mathrm{the}\:\mathrm{median}\:\mathrm{of}\:\Delta\mathrm{MAK}, \\ $$$$\mathrm{MA}^{\mathrm{2}} +\mathrm{MK}^{\mathrm{2}} =\mathrm{2MJ}^{\mathrm{2}} +\frac{\mathrm{AK}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{2MJ}^{\mathrm{2}} +\frac{\mathrm{16}}{\mathrm{9}}\left(\mathrm{1}\right) \\ $$$$\mathrm{Since}\:\mathrm{MI}\:\mathrm{is}\:\mathrm{the}\:\mathrm{median}\:\mathrm{of}\:\Delta\mathrm{MJK}, \\ $$$$\mathrm{MJ}^{\mathrm{2}} +\mathrm{MK}^{\mathrm{2}} =\mathrm{2MI}^{\mathrm{2}} +\frac{\mathrm{JK}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{2MI}^{\mathrm{2}} +\frac{\mathrm{4}}{\mathrm{9}}\left(\mathrm{2}\right) \\ $$$$\mathrm{Substract}\:\mathrm{two}\:\mathrm{equalities}\:\left(\mathrm{1}\right),\left(\mathrm{2}\right)\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{MA}^{\mathrm{2}} −\mathrm{MJ}^{\mathrm{2}} =\mathrm{2MJ}^{\mathrm{2}} −\mathrm{2MI}^{\mathrm{2}} +\frac{\mathrm{4}}{\mathrm{3}}.\mathrm{Hence}, \\ $$$$\mathrm{MA}^{\mathrm{2}} +\mathrm{2MI}^{\mathrm{2}} =\mathrm{3MJ}^{\mathrm{2}} +\frac{\mathrm{4}}{\mathrm{3}}\left(\mathrm{3}\right). \\ $$$$\mathrm{b}/\mathrm{From}\:\left(\ast\right)\:\mathrm{and}\:\left(\mathrm{3}\right) \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{MA}^{\mathrm{2}} +\mathrm{MB}^{\mathrm{2}} +\mathrm{MC}^{\mathrm{2}} =\mathrm{3MJ}^{\mathrm{2}} +\frac{\mathrm{16}}{\mathrm{3}}\left(\ast\ast\right) \\ $$$$\mathrm{c}/\mathrm{Since}\:\mathrm{the}\:\mathrm{point}\:\mathrm{J}\:\mathrm{is}\:\mathrm{fixed}\:,\mathrm{from}\:\left(\ast\ast\right)\:\mathrm{and}\:\mathrm{the}\: \\ $$$$\mathrm{hypothesis}\:\:\mathrm{MA}^{\mathrm{2}} +\mathrm{MB}^{\mathrm{2}} +\mathrm{MC}^{\mathrm{2}} =\mathrm{8}\:\mathrm{we} \\ $$$$\mathrm{follow}\:\mathrm{that}\:\mathrm{3MJ}^{\mathrm{2}} +\frac{\mathrm{16}}{\mathrm{3}}=\mathrm{8}\Leftrightarrow\mathrm{MJ}=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$$$\mathrm{which}\:\mathrm{shows}\:\mathrm{that}\:\mathrm{set}\:\:\left(\mathrm{T}\right)\:\mathrm{of}\:\mathrm{the}\:\mathrm{points}\:\mathrm{M}\: \\ $$$$\mathrm{satisfying}\:\mathrm{MA}^{\mathrm{2}} +\mathrm{MB}^{\mathrm{2}} +\mathrm{MC}^{\mathrm{2}} =\mathrm{8}\:\mathrm{lie}\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{circle}\:\mathrm{has}\:\mathrm{the}\:\mathrm{center}\:\mathrm{be}\:\mathrm{the}\:\mathrm{point}\:\mathrm{J} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{equal}\:\mathrm{to}\:\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$$$ \\ $$
Commented by mathocean1 last updated on 07/Jun/20
Thank you sir....great.
$${Thank}\:{you}\:{sir}….{great}. \\ $$
Commented by 1549442205 last updated on 08/Jun/20
you are wellcome!
$$\mathrm{you}\:\mathrm{are}\:\mathrm{wellcome}! \\ $$

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