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Question Number 222718 by Mingma last updated on 05/Jul/25
Commented by gabthemathguy25 last updated on 06/Jul/25
what?
$${what}? \\ $$
Commented by AntonCWX8 last updated on 06/Jul/25
Did you just so happen to have a death wish, MathematicalUser?
$${Did}\:{you}\:{just}\:{so}\:{happen}\:{to}\:{have}\:{a}\:{death}\:{wish},\:{MathematicalUser}? \\ $$
Commented by MathematicalUser2357 last updated on 07/Jul/25
NO, There was CH and I thought that this question was for check_CH. So I called check_CH for help. Additionally, I thought that the topic of this question was about check_CH. This is my last word message. I will not send anything without mathematics.
$$\boldsymbol{\mathrm{NO}},\:\mathrm{There}\:\mathrm{was}\:\mathrm{CH}\:\mathrm{and}\:\mathrm{I}\:\mathrm{thought}\:\mathrm{that}\:\mathrm{this}\:\mathrm{question}\:\mathrm{was}\:\mathrm{for}\:\mathrm{check\_CH}. \\ $$$$\mathrm{So}\:\mathrm{I}\:\mathrm{called}\:\mathrm{check\_CH}\:\mathrm{for}\:\mathrm{help}. \\ $$$$\mathrm{Additionally},\:\mathrm{I}\:\mathrm{thought}\:\mathrm{that}\:\mathrm{the}\:\mathrm{topic}\:\mathrm{of}\:\mathrm{this}\:\mathrm{question}\:\mathrm{was}\:\mathrm{about}\:\mathrm{check\_CH}. \\ $$$$\mathrm{This}\:\mathrm{is}\:\mathrm{my}\:\mathrm{last}\:\mathrm{word}\:\mathrm{message}. \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{not}\:\mathrm{send}\:\mathrm{anything}\:\mathrm{without}\:\mathrm{mathematics}. \\ $$
Answered by A5T last updated on 06/Jul/25
Commented by A5T last updated on 06/Jul/25
WLOG, let D(0,0) be the origin and affiixes of A,B,C be A(0,a) B(b,0), C(c,0) resp. Then affixes of E,F are E((b/2),0), F((c/2),0) resp. Let G(g_1 ,g_2 ) be affix of G. EG⊥AB⇒((0−a)/(b−0))=((g_1 −(b/2))/(−(g_2 −0)))⇒(a/b)=((2g_1 −b)/(2g_2 ))...(i) FG⊥AC⇒((0−a)/(c−0))=((g_1 −(c/2))/(−(g_2 −0)))⇒(a/c)=((2g_1 −c)/(2g_2 ))...(ii) (i)−(ii)⇒(a/b)−(a/c)=((−b+c)/(2g_2 ))⇒g_2 =((bc)/(2a)) ⇒c+b=2g_1 ⇒g_1 =((b+c)/2) ⇒G(((b+c)/2),((bc)/(2a))) is the affix of G. ⇒BG=(√((((b+c)/2)−b)^2 +(((bc)/(2a))−0)^2 ))=(√((((c−b)^2 )/4)+((b^2 c^2 )/(4a^2 )))) Similarly, GC=(√((((b+c)/2)−c)^2 +(((bc)/(2a))−0)^2 )) =(√((((b−c)^2 )/4)+((b^2 c^2 )/(4a^2 )))) ⇒BG=GC since (b−c)^2 =(c−b)^2 ■
$$\mathrm{WLOG},\:\mathrm{let}\:\mathrm{D}\left(\mathrm{0},\mathrm{0}\right)\:\mathrm{be}\:\mathrm{the}\:\mathrm{origin}\:\mathrm{and}\:\mathrm{affiixes}\:\mathrm{of} \\ $$$$\mathrm{A},\mathrm{B},\mathrm{C}\:\mathrm{be}\:\mathrm{A}\left(\mathrm{0},\mathrm{a}\right)\:\mathrm{B}\left(\mathrm{b},\mathrm{0}\right),\:\mathrm{C}\left(\mathrm{c},\mathrm{0}\right)\:\mathrm{resp}.\:\mathrm{Then}\:\mathrm{affixes}\:\mathrm{of}\: \\ $$$$\mathrm{E},\mathrm{F}\:\mathrm{are}\:\mathrm{E}\left(\frac{\mathrm{b}}{\mathrm{2}},\mathrm{0}\right),\:\mathrm{F}\left(\frac{\mathrm{c}}{\mathrm{2}},\mathrm{0}\right)\:\mathrm{resp}.\: \\ $$$$\mathrm{Let}\:\mathrm{G}\left(\mathrm{g}_{\mathrm{1}} ,\mathrm{g}_{\mathrm{2}} \right)\:\mathrm{be}\:\mathrm{affix}\:\mathrm{of}\:\mathrm{G}. \\ $$$$\mathrm{EG}\bot\mathrm{AB}\Rightarrow\frac{\mathrm{0}−\mathrm{a}}{\mathrm{b}−\mathrm{0}}=\frac{\mathrm{g}_{\mathrm{1}} −\frac{\mathrm{b}}{\mathrm{2}}}{−\left(\mathrm{g}_{\mathrm{2}} −\mathrm{0}\right)}\Rightarrow\frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{2g}_{\mathrm{1}} −\mathrm{b}}{\mathrm{2g}_{\mathrm{2}} }…\left(\mathrm{i}\right) \\ $$$$\mathrm{FG}\bot\mathrm{AC}\Rightarrow\frac{\mathrm{0}−\mathrm{a}}{\mathrm{c}−\mathrm{0}}=\frac{\mathrm{g}_{\mathrm{1}} −\frac{\mathrm{c}}{\mathrm{2}}}{−\left(\mathrm{g}_{\mathrm{2}} −\mathrm{0}\right)}\Rightarrow\frac{\mathrm{a}}{\mathrm{c}}=\frac{\mathrm{2g}_{\mathrm{1}} −\mathrm{c}}{\mathrm{2g}_{\mathrm{2}} }…\left(\mathrm{ii}\right) \\ $$$$\left(\mathrm{i}\right)−\left(\mathrm{ii}\right)\Rightarrow\frac{\mathrm{a}}{\mathrm{b}}−\frac{\mathrm{a}}{\mathrm{c}}=\frac{−\mathrm{b}+\mathrm{c}}{\mathrm{2g}_{\mathrm{2}} }\Rightarrow\mathrm{g}_{\mathrm{2}} =\frac{\mathrm{bc}}{\mathrm{2a}} \\ $$$$\Rightarrow\mathrm{c}+\mathrm{b}=\mathrm{2g}_{\mathrm{1}} \Rightarrow\mathrm{g}_{\mathrm{1}} =\frac{\mathrm{b}+\mathrm{c}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{G}\left(\frac{\mathrm{b}+\mathrm{c}}{\mathrm{2}},\frac{\mathrm{bc}}{\mathrm{2a}}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{affix}\:\mathrm{of}\:\mathrm{G}. \\ $$$$\Rightarrow\mathrm{BG}=\sqrt{\left(\frac{\mathrm{b}+\mathrm{c}}{\mathrm{2}}−\mathrm{b}\right)^{\mathrm{2}} +\left(\frac{\mathrm{bc}}{\mathrm{2a}}−\mathrm{0}\right)^{\mathrm{2}} }=\sqrt{\frac{\left(\mathrm{c}−\mathrm{b}\right)^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{b}^{\mathrm{2}} \mathrm{c}^{\mathrm{2}} }{\mathrm{4a}^{\mathrm{2}} }} \\ $$$$\mathrm{Similarly},\:\mathrm{GC}=\sqrt{\left(\frac{\mathrm{b}+\mathrm{c}}{\mathrm{2}}−\mathrm{c}\right)^{\mathrm{2}} +\left(\frac{\mathrm{bc}}{\mathrm{2a}}−\mathrm{0}\right)^{\mathrm{2}} } \\ $$$$=\sqrt{\frac{\left(\mathrm{b}−\mathrm{c}\right)^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{b}^{\mathrm{2}} \mathrm{c}^{\mathrm{2}} }{\mathrm{4a}^{\mathrm{2}} }} \\ $$$$\Rightarrow\mathrm{BG}=\mathrm{GC}\:\mathrm{since}\:\left(\mathrm{b}−\mathrm{c}\right)^{\mathrm{2}} =\left(\mathrm{c}−\mathrm{b}\right)^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare \\ $$
Commented by Mingma last updated on 10/Jul/25
Perfect, sir!
Answered by parthasc last updated on 06/Jul/25

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