Question-222062

Question Number 222062 by ajfour last updated on 16/Jun/25

Answered by fantastic last updated on 16/Jun/25

Commented by fantastic last updated on 16/Jun/25

RQ=(√(a^2 +a^2 −2.a.acos 120^0 ))=(√(2a^2 −2a^2 .−(1/2)))=a(√3) ∴AQ=a(√3) Similarly CS=b(√3) PV[V=⊥OQ I forgot to mention]=(√((a+b)^2 −(a−b)^2 ))=(√(4ab))=2(√(ab)) ∴SQ=2(√(ab)) So side length=a(√3)+b(√3)+2(√(ab)) =(√3)(a+b)+2(√(ab)) u

$${RQ}=\sqrt{{a}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}.{a}.{a}\mathrm{cos}\:\mathrm{120}^{\mathrm{0}} }=\sqrt{\mathrm{2}{a}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} .−\frac{\mathrm{1}}{\mathrm{2}}}={a}\sqrt{\mathrm{3}} \\ $$$$\therefore{AQ}={a}\sqrt{\mathrm{3}} \\ $$$${Similarly}\:{CS}={b}\sqrt{\mathrm{3}} \\ $$$${PV}\left[{V}=\bot{OQ}\:{I}\:{forgot}\:{to}\:{mention}\right]=\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −\left({a}−{b}\right)^{\mathrm{2}} }=\sqrt{\mathrm{4}{ab}}=\mathrm{2}\sqrt{{ab}} \\ $$$$\therefore{SQ}=\mathrm{2}\sqrt{{ab}} \\ $$$${So}\:{side}\:{length}={a}\sqrt{\mathrm{3}}+{b}\sqrt{\mathrm{3}}+\mathrm{2}\sqrt{{ab}} \\ $$$$=\sqrt{\mathrm{3}}\left({a}+{b}\right)+\mathrm{2}\sqrt{{ab}}\:{u} \\ $$

Answered by mr W last updated on 16/Jun/25

Commented by mr W last updated on 16/Jun/25

s=(√3)a+(√3)b+(√((a+b)^2 −(a−b)^2 )) =(√3)(a+b)+2(√(ab))≥2(√3)a

$${s}=\sqrt{\mathrm{3}}{a}+\sqrt{\mathrm{3}}{b}+\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −\left({a}−{b}\right)^{\mathrm{2}} } \\ $$$$\:\:=\sqrt{\mathrm{3}}\left({a}+{b}\right)+\mathrm{2}\sqrt{{ab}}\geqslant\mathrm{2}\sqrt{\mathrm{3}}{a} \\ $$

Commented by ajfour last updated on 16/Jun/25

Thanks sir. Now if it′s an isosceles triangle what must the semi vertical angle be such that triangle area is a maximum for given a and b?

$${Thanks}\:{sir}.\:{Now}\:{if}\:{it}'{s}\:{an}\:{isosceles}\:{triangle} \\ $$$${what}\:{must}\:{the}\:{semi}\:{vertical}\:{angle}\:{be}\:{such} \\ $$$${that}\:{triangle}\:{area}\:{is}\:{a}\:{maximum}\:{for} \\ $$$${given}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{b}}? \\ $$

Commented by mr W last updated on 16/Jun/25

Commented by mr W last updated on 16/Jun/25

maximum area doesn′t exist, since θ→0 ⇒A→∞ θ→(π/2) ⇒A→∞ we should find the minimum area.

$${maximum}\:{area}\:{doesn}'{t}\:{exist},\:{since} \\ $$$$\theta\rightarrow\mathrm{0}\:\Rightarrow{A}\rightarrow\infty \\ $$$$\theta\rightarrow\frac{\pi}{\mathrm{2}}\:\Rightarrow{A}\rightarrow\infty \\ $$$${we}\:{should}\:{find}\:{the}\:{minimum}\:{area}. \\ $$

Commented by mr W last updated on 17/Jun/25

2α+θ=(π/2) ⇒α=(π/4)−(θ/2) s=(a/(tan θ))+(b/(tan α))+(√((a+b)^2 −(a−b)^2 )) =(a/(tan θ))+((b(1+tan (θ/2)))/(1−tan (θ/2)))+2(√(ab)) A=((s^2 sin 2θ)/2)=((sin 2θ)/2)[(a/(tan θ))+((b(1+tan (θ/2)))/(1−tan (θ/2)))+2(√(ab))]^2 =((tan (θ/2)(1−tan^2 (θ/2)))/((1+tan^2 (θ/2))^2 ))[((a(1−tan^2 (θ/2)))/(2 tan (θ/2)))+((b(1+tan (θ/2)))/(1−tan (θ/2)))+2(√(ab))]^2 =((t(1−t^2 ))/((1+t^2 )^2 ))[((a(1−t^2 ))/(2t))+((b(1+t))/(1−t))+2(√(ab))]^2 example: a=3, b=c A_(min) ≈38.7933 at tan (θ/2)≈0.2812, i.e. θ≈31.412°

$$\mathrm{2}\alpha+\theta=\frac{\pi}{\mathrm{2}}\:\Rightarrow\alpha=\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}} \\ $$$${s}=\frac{{a}}{\mathrm{tan}\:\theta}+\frac{{b}}{\mathrm{tan}\:\alpha}+\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −\left({a}−{b}\right)^{\mathrm{2}} } \\ $$$$\:\:=\frac{{a}}{\mathrm{tan}\:\theta}+\frac{{b}\left(\mathrm{1}+\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\right)}{\mathrm{1}−\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}+\mathrm{2}\sqrt{{ab}} \\ $$$${A}=\frac{{s}^{\mathrm{2}} \mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2}}=\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2}}\left[\frac{{a}}{\mathrm{tan}\:\theta}+\frac{{b}\left(\mathrm{1}+\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\right)}{\mathrm{1}−\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}+\mathrm{2}\sqrt{{ab}}\right]^{\mathrm{2}} \\ $$$$\:\:=\frac{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\left(\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}\right)}{\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}\right)^{\mathrm{2}} }\left[\frac{{a}\left(\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}\right)}{\mathrm{2}\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}+\frac{{b}\left(\mathrm{1}+\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\right)}{\mathrm{1}−\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}+\mathrm{2}\sqrt{{ab}}\right]^{\mathrm{2}} \\ $$$$\:\:=\frac{{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\left[\frac{{a}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\mathrm{2}{t}}+\frac{{b}\left(\mathrm{1}+{t}\right)}{\mathrm{1}−{t}}+\mathrm{2}\sqrt{{ab}}\right]^{\mathrm{2}} \\ $$$$ \\ $$$${example}:\:{a}=\mathrm{3},\:{b}={c} \\ $$$${A}_{{min}} \approx\mathrm{38}.\mathrm{7933}\: \\ $$$${at}\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\approx\mathrm{0}.\mathrm{2812},\:{i}.{e}.\:\theta\approx\mathrm{31}.\mathrm{412}° \\ $$

Commented by mr W last updated on 17/Jun/25