Menu Close

Question-39404

Tinku Tara 0 Comments
Pin




Question Number 39404 by ajfour last updated on 06/Jul/18
Commented by ajfour last updated on 06/Jul/18
Q.39365 solution..
$${Q}.\mathrm{39365}\:\:\:{solution}.. \\ $$
Commented by MrW3 last updated on 06/Jul/18
let′s say the angles of isosceles triangles are θ, in our case θ=(π/6). Are there any other values for θ such that ΔGJN is equilateral?
$${let}'{s}\:{say}\:{the}\:{angles}\:{of}\:{isosceles}\:{triangles} \\ $$$${are}\:\theta,\:{in}\:{our}\:{case}\:\theta=\frac{\pi}{\mathrm{6}}. \\ $$$${Are}\:{there}\:{any}\:{other}\:{values}\:{for}\:\theta\:{such} \\ $$$${that}\:\Delta{GJN}\:{is}\:{equilateral}? \\ $$
Commented by behi83417@gmail.com last updated on 06/Jul/18
AG=(b/2)secθ,AN=(c/2)secθ NG^2 =(b^2 /4)sec^2 θ+(c^2 /4)sec^2 θ−((bc)/2)sec^2 θ.cos(60+2θ)= =(1/4)sec^2 θ[b^2 +c^2 −2bc.cos(60+2θ)] NJ^2 =(1/4)sec^2 θ[a^2 +c^2 −2ac.cos(60+2θ)] ⇒a^2 +c^2 −2ac.cos(60+2θ)=b^2 +c^2 −2bc.cos(60+2θ) ⇒a^2 −b^2 =2c(a−b)cos(60+2θ) ⇒^(a≠b) cos(60+2θ)=((a+b)/(2c)) ⇒θ=(1/2)cos^(−1) [((a+b)/(2c))]−30 .
$${AG}=\frac{{b}}{\mathrm{2}}{sec}\theta,{AN}=\frac{{c}}{\mathrm{2}}{sec}\theta \\ $$$${NG}^{\mathrm{2}} =\frac{{b}^{\mathrm{2}} }{\mathrm{4}}{sec}^{\mathrm{2}} \theta+\frac{{c}^{\mathrm{2}} }{\mathrm{4}}{sec}^{\mathrm{2}} \theta−\frac{{bc}}{\mathrm{2}}{sec}^{\mathrm{2}} \theta.{cos}\left(\mathrm{60}+\mathrm{2}\theta\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{sec}^{\mathrm{2}} \theta\left[{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}.{cos}\left(\mathrm{60}+\mathrm{2}\theta\right)\right] \\ $$$${NJ}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}{sec}^{\mathrm{2}} \theta\left[{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{ac}.{cos}\left(\mathrm{60}+\mathrm{2}\theta\right)\right] \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{ac}.{cos}\left(\mathrm{60}+\mathrm{2}\theta\right)={b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}.{cos}\left(\mathrm{60}+\mathrm{2}\theta\right) \\ $$$$\Rightarrow{a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{2}{c}\left({a}−{b}\right){cos}\left(\mathrm{60}+\mathrm{2}\theta\right) \\ $$$$\overset{{a}\neq{b}} {\Rightarrow}{cos}\left(\mathrm{60}+\mathrm{2}\theta\right)=\frac{{a}+{b}}{\mathrm{2}{c}} \\ $$$$\Rightarrow\theta=\frac{\mathrm{1}}{\mathrm{2}}{cos}^{−\mathrm{1}} \left[\frac{{a}+{b}}{\mathrm{2}{c}}\right]−\mathrm{30}\:. \\ $$
Commented by MrW3 last updated on 06/Jul/18
AG=(b/(2 cos θ)) AN=(c/(2 cos θ)) NG^2 =((b/(2 cos θ)))^2 +((c/(2 cos θ)))^2 −2((b/(2 cos θ)))((c/(2 cos θ))) cos (A+2θ) =((b^2 +c^2 −2bc (cos A cos 2θ−sin A sin 2θ))/(4 cos^2 θ)) =((b^2 +c^2 −2bc cos A cos 2θ+2bc sin A sin 2θ)/(4 cos^2 θ)) =((b^2 +c^2 −(b^2 +c^2 −a^2 )cos 2θ+4Δ sin 2θ)/(4 cos^2 θ)) =((a^2 cos 2θ+(b^2 +c^2 )(1−cos 2θ)+4Δ cos 2θ)/(4 cos^2 θ)) cos 2θ=1−cos 2θ ⇒cos 2θ=(1/2) ⇒θ=(π/6) ⇒there is only one value for θ.
$${AG}=\frac{{b}}{\mathrm{2}\:\mathrm{cos}\:\theta} \\ $$$${AN}=\frac{{c}}{\mathrm{2}\:\mathrm{cos}\:\theta} \\ $$$${NG}^{\mathrm{2}} =\left(\frac{{b}}{\mathrm{2}\:\mathrm{cos}\:\theta}\right)^{\mathrm{2}} +\left(\frac{{c}}{\mathrm{2}\:\mathrm{cos}\:\theta}\right)^{\mathrm{2}} −\mathrm{2}\left(\frac{{b}}{\mathrm{2}\:\mathrm{cos}\:\theta}\right)\left(\frac{{c}}{\mathrm{2}\:\mathrm{cos}\:\theta}\right)\:\mathrm{cos}\:\left({A}+\mathrm{2}\theta\right) \\ $$$$=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}\:\left(\mathrm{cos}\:{A}\:\mathrm{cos}\:\mathrm{2}\theta−\mathrm{sin}\:{A}\:\mathrm{sin}\:\mathrm{2}\theta\right)}{\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$$=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}\:\mathrm{cos}\:{A}\:\mathrm{cos}\:\mathrm{2}\theta+\mathrm{2}{bc}\:\mathrm{sin}\:{A}\:\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$$=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)\mathrm{cos}\:\mathrm{2}\theta+\mathrm{4}\Delta\:\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$$=\frac{{a}^{\mathrm{2}} \:\mathrm{cos}\:\mathrm{2}\theta+\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta\right)+\mathrm{4}\Delta\:\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$$\mathrm{cos}\:\mathrm{2}\theta=\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta \\ $$$$\Rightarrow\mathrm{cos}\:\mathrm{2}\theta=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\theta=\frac{\pi}{\mathrm{6}} \\ $$$$\Rightarrow{there}\:{is}\:{only}\:{one}\:{value}\:{for}\:\theta. \\ $$
Commented by ajfour last updated on 06/Jul/18
marvelous Sir!
$${marvelous}\:{Sir}! \\ $$
Commented by behi83417@gmail.com last updated on 06/Jul/18
dear master!for what? cos2θ=1−cos2θ
$${dear}\:{master}!{for}\:{what}? \\ $$$${cos}\mathrm{2}\theta=\mathrm{1}−{cos}\mathrm{2}\theta \\ $$
Commented by MrW3 last updated on 06/Jul/18
the term a^2 cos 2θ+(b^2 +c^2 )(1−cos 2θ) must be the same for all three sides including b^2 cos 2θ+(c^2 +a^2 )(1−cos 2θ) c^2 cos 2θ+(a^2 +b^2 )(1−cos 2θ) but this is true only when cos 2θ=1−cos 2θ then a^2 cos 2θ+(b^2 +c^2 )(1−cos 2θ) =b^2 cos 2θ+(c^2 +a^2 )(1−cos 2θ) =c^2 cos 2θ+(a^2 +b^2 )(1−cos 2θ) =(a^2 +b^2 +c^2 )cos 2θ and it′s the same for all three sides.
$${the}\:{term}\:{a}^{\mathrm{2}} \:\mathrm{cos}\:\mathrm{2}\theta+\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta\right) \\ $$$${must}\:{be}\:{the}\:{same}\:{for}\:{all}\:{three}\:{sides} \\ $$$${including} \\ $$$${b}^{\mathrm{2}} \:\mathrm{cos}\:\mathrm{2}\theta+\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta\right) \\ $$$${c}^{\mathrm{2}} \:\mathrm{cos}\:\mathrm{2}\theta+\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta\right) \\ $$$${but}\:{this}\:{is}\:{true}\:{only}\:{when} \\ $$$$\mathrm{cos}\:\mathrm{2}\theta=\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta \\ $$$${then}\: \\ $$$${a}^{\mathrm{2}} \:\mathrm{cos}\:\mathrm{2}\theta+\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta\right) \\ $$$$={b}^{\mathrm{2}} \:\mathrm{cos}\:\mathrm{2}\theta+\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta\right) \\ $$$$={c}^{\mathrm{2}} \:\mathrm{cos}\:\mathrm{2}\theta+\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta\right) \\ $$$$=\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\mathrm{cos}\:\mathrm{2}\theta \\ $$$${and}\:{it}'{s}\:{the}\:{same}\:{for}\:{all}\:{three}\:{sides}. \\ $$
Answered by ajfour last updated on 06/Jul/18
AG=(b/2)sec (π/6) = (b/( (√3))) AN=(c/2)sec (π/6)= (c/( (√3))) NG^2 =AG^2 +AN^2 −2(AG)(AN)cos (A+(π/3)) = ((b^2 +c^2 )/3)−((2bc)/3)((1/2)cos A−((√3)/2)sin A) =((b^2 +c^2 −bccos A)/3) + ((bcsin A)/( (√3))) Now since sin A = (a/(2R)) (R being circumradius of △ABC) and a^2 = b^2 +c^2 −2bccos A NG^2 =((2b^2 +2c^2 −2bccos A)/6)+((abc)/(2(√3)R)) = ((a^2 +b^2 +c^2 )/6)+((abc)/(2(√3)R)) similarly NJ^2 and GJ^( 2) result in the same expression. Hence △NJG is equilateral.
$${AG}=\frac{{b}}{\mathrm{2}}\mathrm{sec}\:\frac{\pi}{\mathrm{6}}\:=\:\frac{{b}}{\:\sqrt{\mathrm{3}}} \\ $$$${AN}=\frac{{c}}{\mathrm{2}}\mathrm{sec}\:\frac{\pi}{\mathrm{6}}=\:\frac{{c}}{\:\sqrt{\mathrm{3}}} \\ $$$${NG}^{\mathrm{2}} ={AG}^{\mathrm{2}} +{AN}^{\mathrm{2}} −\mathrm{2}\left({AG}\right)\left({AN}\right)\mathrm{cos}\:\left({A}+\frac{\pi}{\mathrm{3}}\right) \\ $$$$\:\:\:\:=\:\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{3}}−\frac{\mathrm{2}{bc}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:{A}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:{A}\right) \\ $$$$\:\:\:=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{bc}\mathrm{cos}\:{A}}{\mathrm{3}}\:+\:\frac{{bc}\mathrm{sin}\:{A}}{\:\sqrt{\mathrm{3}}} \\ $$$$\:{Now}\:{since}\:\:\mathrm{sin}\:{A}\:=\:\frac{{a}}{\mathrm{2}{R}} \\ $$$$\left({R}\:{being}\:{circumradius}\:{of}\:\bigtriangleup{ABC}\right) \\ $$$${and}\:\:{a}^{\mathrm{2}} =\:{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}\mathrm{cos}\:{A} \\ $$$$\:\:{NG}^{\mathrm{2}} =\frac{\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{c}^{\mathrm{2}} −\mathrm{2}{bc}\mathrm{cos}\:{A}}{\mathrm{6}}+\frac{{abc}}{\mathrm{2}\sqrt{\mathrm{3}}{R}} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{6}}+\frac{{abc}}{\mathrm{2}\sqrt{\mathrm{3}}{R}} \\ $$$${similarly}\:\:{NJ}\:^{\mathrm{2}} \:{and}\:{GJ}^{\:\mathrm{2}} \:{result} \\ $$$${in}\:{the}\:{same}\:{expression}. \\ $$$${Hence}\:\:\bigtriangleup{NJG}\:{is}\:{equilateral}. \\ $$
Commented by behi83417@gmail.com last updated on 06/Jul/18
nice proof sir Ajfour.thanks for drawing the perfect figure.
$${nice}\:{proof}\:{sir}\:{Ajfour}.{thanks}\:{for}\: \\ $$$${drawing}\:{the}\:{perfect}\:{figure}. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *