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Question-219318

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Question Number 219318 by ea last updated on 23/Apr/25
Answered by mr W last updated on 23/Apr/25
Commented by ea last updated on 23/Apr/25
Sir, this is perfect, but I will appreciate if you can provide detailed explanation! showing some workings would be appreciated!
Commented by mr W last updated on 23/Apr/25
((BD)/(AB))=((CE)/(AC))=r=tan θ, say ⇒((AD)/(AB))=(1/(cos θ))=((AE)/(AC)) ∠DAE=∠BAC ⇒ΔDAE∼ΔBAC say P is midpoint of DE. ((AP)/(AM))=((AD)/(AB))=(1/(cos θ)) ⇒AP=((AM)/(cos θ)) that means the locus of P is a perpendicular to AM with M being the midpoint of BC.
$$\frac{{BD}}{{AB}}=\frac{{CE}}{{AC}}={r}=\mathrm{tan}\:\theta,\:{say} \\ $$$$\Rightarrow\frac{{AD}}{{AB}}=\frac{\mathrm{1}}{\mathrm{cos}\:\theta}=\frac{{AE}}{{AC}} \\ $$$$\angle{DAE}=\angle{BAC} \\ $$$$\Rightarrow\Delta{DAE}\sim\Delta{BAC} \\ $$$${say}\:{P}\:{is}\:{midpoint}\:{of}\:{DE}. \\ $$$$\frac{{AP}}{{AM}}=\frac{{AD}}{{AB}}=\frac{\mathrm{1}}{\mathrm{cos}\:\theta} \\ $$$$\Rightarrow{AP}=\frac{{AM}}{\mathrm{cos}\:\theta} \\ $$$${that}\:{means}\:{the}\:{locus}\:{of}\:{P}\:{is}\:{a}\: \\ $$$${perpendicular}\:{to}\:{AM}\:{with}\:{M}\:{being} \\ $$$${the}\:{midpoint}\:{of}\:{BC}. \\ $$
Commented by mr W last updated on 23/Apr/25
after i have posted a diagram, i need some time to write down my solution. this can take some time! so you don′t need to give your comment or ask for more explanation as soon as you see a diagram. perhaps at this time i was just writing down my solution or the explanation you want. please have a little more petience! thanks!
$${after}\:{i}\:{have}\:{posted}\:{a}\:{diagram},\:{i}\:{need} \\ $$$${some}\:{time}\:{to}\:{write}\:{down}\:{my} \\ $$$${solution}.\:{this}\:{can}\:{take}\:{some}\:{time}! \\ $$$${so}\:{you}\:{don}'{t}\:{need}\:{to}\:{give}\:{your} \\ $$$${comment}\:{or}\:{ask}\:{for}\:{more}\: \\ $$$${explanation}\:{as}\:{soon}\:{as}\:{you}\:{see}\:{a} \\ $$$${diagram}.\:{perhaps}\:{at}\:{this}\:{time}\:{i}\: \\ $$$${was}\:{just}\:{writing}\:{down}\:{my}\:{solution}\: \\ $$$${or}\:{the}\:{explanation}\:{you}\:{want}.\: \\ $$$${please}\:{have}\:{a}\:{little}\:{more}\:{petience}! \\ $$$${thanks}! \\ $$
Commented by ea last updated on 23/Apr/25
Well noted, Prof!
Commented by ea last updated on 12/May/25
Sir, your solution is correct only if you can show that Triangle AMP and Triangle ABD are similar. Can you please show that? Thank you.
Commented by mr W last updated on 13/May/25
Commented by ea last updated on 15/May/25
Prof, the picture does not clear my doubt. So your cos theta argument is true only if AM and MC is perpendicular. The ratio between 2 sides can give you any value. It is meaningful as cos only if you can show it is a right triangle.
Commented by mr W last updated on 15/May/25
the picture above is a physical interpretation: a soild object ΔABC rotates about point A. but it seems not to be obvious to you, no problem. we can also treat this using vectors. but i think you seem to need a direct proof using basic geometry. i have provided such a proof in Q220546. i hope this can convince you.
$${the}\:{picture}\:{above}\:{is}\:{a}\:{physical} \\ $$$${interpretation}:\:{a}\:{soild}\:{object}\:\Delta{ABC} \\ $$$${rotates}\:{about}\:{point}\:{A}. \\ $$$${but}\:{it}\:{seems}\:{not}\:{to}\:{be}\:{obvious}\:{to}\:{you}, \\ $$$${no}\:{problem}.\:{we}\:{can}\:{also}\:{treat}\:{this}\: \\ $$$${using}\:{vectors}.\:{but}\:{i}\:{think}\:{you}\:{seem}\: \\ $$$${to}\:{need}\:{a}\:{direct}\:{proof}\:{using}\:{basic}\: \\ $$$${geometry}.\:{i}\:{have}\:{provided}\:{such}\:{a}\: \\ $$$${proof}\:{in}\:{Q}\mathrm{220546}.\:{i}\:{hope}\:{this}\:{can}\: \\ $$$${convince}\:{you}. \\ $$
Commented by mr W last updated on 15/May/25
vector method:
$${vector}\:{method}: \\ $$
Commented by mr W last updated on 15/May/25
Commented by mr W last updated on 16/May/25
m^→ =(1−μ)b^(→) +μc^(→) proof see below.
$$\overset{\rightarrow} {\boldsymbol{{m}}}=\left(\mathrm{1}−\mu\right)\overset{\rightarrow} {\boldsymbol{{b}}}+\mu\overset{\rightarrow} {\boldsymbol{{c}}} \\ $$$${proof}\:{see}\:{below}. \\ $$
Commented by ea last updated on 15/May/25
Perfect, Prof!
Commented by mr W last updated on 16/May/25
Commented by mr W last updated on 16/May/25
OM^(→) =r=p+μ(q−p)=(1−μ)p+μq B′C′^(→) =q+c−(p+b) OM′^(→) =p+b−μ[q+c−(p+b)] =(1−μ)(p+b)+μ(q+c) m=OM^(→) ′−OM^(→) =(1−μ)(p+b)+μ(q+c)−(1−μ)p+μq =(1−μ)b+μc
$$\overset{\rightarrow} {{OM}}=\boldsymbol{{r}}=\boldsymbol{{p}}+\mu\left(\boldsymbol{{q}}−\boldsymbol{{p}}\right)=\left(\mathrm{1}−\mu\right)\boldsymbol{{p}}+\mu\boldsymbol{{q}} \\ $$$$\overset{\rightarrow} {{B}'{C}'}=\boldsymbol{{q}}+\boldsymbol{{c}}−\left(\boldsymbol{{p}}+\boldsymbol{{b}}\right) \\ $$$$\overset{\rightarrow} {{OM}'}=\boldsymbol{{p}}+\boldsymbol{{b}}−\mu\left[\boldsymbol{{q}}+\boldsymbol{{c}}−\left(\boldsymbol{{p}}+\boldsymbol{{b}}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{1}−\mu\right)\left(\boldsymbol{{p}}+\boldsymbol{{b}}\right)+\mu\left(\boldsymbol{{q}}+\boldsymbol{{c}}\right) \\ $$$$\boldsymbol{{m}}=\overset{\rightarrow} {{OM}}'−\overset{\rightarrow} {{OM}}=\left(\mathrm{1}−\mu\right)\left(\boldsymbol{{p}}+\boldsymbol{{b}}\right)+\mu\left(\boldsymbol{{q}}+\boldsymbol{{c}}\right)−\left(\mathrm{1}−\mu\right)\boldsymbol{{p}}+\mu\boldsymbol{{q}} \\ $$$$\:\:\:\:\:=\left(\mathrm{1}−\mu\right)\boldsymbol{{b}}+\mu\boldsymbol{{c}} \\ $$

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