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Question Number 198647 by ajfour last updated on 22/Oct/23
Commented by ajfour last updated on 22/Oct/23
Q is center of smaller circle of radius a. Find maximum area of △ABC in terms of a, b.
$${Q}\:{is}\:{center}\:{of}\:{smaller}\:{circle}\:{of} \\ $$$${radius}\:{a}.\:{Find}\:{maximum}\:{area}\:{of}\: \\ $$$$\bigtriangleup{ABC}\:\:{in}\:{terms}\:{of}\:{a},\:{b}. \\ $$
Answered by mr W last updated on 22/Oct/23
Commented by mr W last updated on 23/Oct/23
let μ=(b/a) ((sin δ)/a)=((sin (α−δ))/b)=((sin α)/(PA)) ((sin α)/(tan δ))−cos α=(b/a)=μ ⇒tan δ=((sin α)/(cos α+μ)) PA=((a sin α)/(sin δ)) Δ=((b^2 sin (β+γ))/2)+((b PA [sin (π+δ−β)+sin (π−δ−γ)])/2) Δ=((b^2 sin (β+γ))/2)+((ba sin α [sin (β−δ)+sin (γ+δ)])/(2 sin δ)) Δ=((b^2 sin (β+γ))/2)+((ba sin α [sin β cos δ−cos β sin δ +sin γ cos δ+cos γ sin δ])/(2 sin δ)) Δ=((b^2 sin (β+γ))/2)+((ba sin α [(sin β+sin γ)cot δ−cos β +cos γ])/2) Δ=((b^2 sin (β+γ))/2)+((ba sin α [(sin β+sin γ)((cos α+μ)/(sin α))−cos β +cos γ])/2) Φ=((2Δ)/(ab))=μ sin (β+γ)+(sin β+sin γ)(cos α+μ)−sin α (cos β −cos γ) such that area Δ is maximum, (∂Φ/∂α)=(sin β+sin γ)(−sin α)−cos α (cos β−cos γ)=0 sin α (sin β+sin γ)+cos α (cos β−cos γ)=0 ...(iii) ⇒tan α=((cos γ−cos β)/(sin β+sin γ)) (⇒this proves that QA⊥BC.) (∂Φ/∂β)=μ cos (β+γ)+cos β (cos α+μ)+sin α sin β=0 ...(i) (∂Φ/∂γ)=μ cos (β+γ)+cos γ (cos α+μ)−sin α sin γ=0 ...(ii) (i)−(ii): (cos β−cos γ) (cos α+μ)+sin α (sin β+sin γ)=0 μ(cos β−cos γ)+cos α (cos β−cos γ)+sin α (sin β+sin γ)=0 due to (iii): μ(cos β−cos γ)=0 ⇒cos β=cos γ ⇒β=γ i.e. points B and C are symmetric about x−axis! tan α=((cos γ−cos β)/(sin β+sin γ))=0 ⇒α=0 i.e. point A lies on x−axis! insert into (i): μ cos 2β+cos β (1+μ)=0 μ(2 cos^2 β−1)+cos β (1+μ)=0 2μ cos^2 β+(1+μ) cos β−μ=0 ⇒cos β=((−(1+μ)+(√(9μ^2 +2μ+1)))/(4μ)) i.e. cos β=((−(a+b)+(√(a^2 +9b^2 +2ab)))/(4b)) ■
$${let}\:\mu=\frac{{b}}{{a}} \\ $$$$\frac{\mathrm{sin}\:\delta}{{a}}=\frac{\mathrm{sin}\:\left(\alpha−\delta\right)}{{b}}=\frac{\mathrm{sin}\:\alpha}{{PA}} \\ $$$$\frac{\mathrm{sin}\:\alpha}{\mathrm{tan}\:\delta}−\mathrm{cos}\:\alpha=\frac{{b}}{{a}}=\mu \\ $$$$\Rightarrow\mathrm{tan}\:\delta=\frac{\mathrm{sin}\:\alpha}{\mathrm{cos}\:\alpha+\mu} \\ $$$${PA}=\frac{{a}\:\mathrm{sin}\:\alpha}{\mathrm{sin}\:\delta} \\ $$$$\Delta=\frac{{b}^{\mathrm{2}} \:\mathrm{sin}\:\left(\beta+\gamma\right)}{\mathrm{2}}+\frac{{b}\:{PA}\:\left[\mathrm{sin}\:\left(\pi+\delta−\beta\right)+\mathrm{sin}\:\left(\pi−\delta−\gamma\right)\right]}{\mathrm{2}} \\ $$$$\Delta=\frac{{b}^{\mathrm{2}} \:\mathrm{sin}\:\left(\beta+\gamma\right)}{\mathrm{2}}+\frac{{ba}\:\mathrm{sin}\:\alpha\:\left[\mathrm{sin}\:\left(\beta−\delta\right)+\mathrm{sin}\:\left(\gamma+\delta\right)\right]}{\mathrm{2}\:\mathrm{sin}\:\delta} \\ $$$$\Delta=\frac{{b}^{\mathrm{2}} \:\mathrm{sin}\:\left(\beta+\gamma\right)}{\mathrm{2}}+\frac{{ba}\:\mathrm{sin}\:\alpha\:\left[\mathrm{sin}\:\beta\:\mathrm{cos}\:\delta−\mathrm{cos}\:\beta\:\mathrm{sin}\:\delta\:+\mathrm{sin}\:\gamma\:\mathrm{cos}\:\delta+\mathrm{cos}\:\gamma\:\mathrm{sin}\:\delta\right]}{\mathrm{2}\:\mathrm{sin}\:\delta} \\ $$$$\Delta=\frac{{b}^{\mathrm{2}} \:\mathrm{sin}\:\left(\beta+\gamma\right)}{\mathrm{2}}+\frac{{ba}\:\mathrm{sin}\:\alpha\:\left[\left(\mathrm{sin}\:\beta+\mathrm{sin}\:\gamma\right)\mathrm{cot}\:\delta−\mathrm{cos}\:\beta\:+\mathrm{cos}\:\gamma\right]}{\mathrm{2}} \\ $$$$\Delta=\frac{{b}^{\mathrm{2}} \:\mathrm{sin}\:\left(\beta+\gamma\right)}{\mathrm{2}}+\frac{{ba}\:\mathrm{sin}\:\alpha\:\left[\left(\mathrm{sin}\:\beta+\mathrm{sin}\:\gamma\right)\frac{\mathrm{cos}\:\alpha+\mu}{\mathrm{sin}\:\alpha}−\mathrm{cos}\:\beta\:+\mathrm{cos}\:\gamma\right]}{\mathrm{2}} \\ $$$$\Phi=\frac{\mathrm{2}\Delta}{{ab}}=\mu\:\mathrm{sin}\:\left(\beta+\gamma\right)+\left(\mathrm{sin}\:\beta+\mathrm{sin}\:\gamma\right)\left(\mathrm{cos}\:\alpha+\mu\right)−\mathrm{sin}\:\alpha\:\left(\mathrm{cos}\:\beta\:−\mathrm{cos}\:\gamma\right) \\ $$$${such}\:{that}\:{area}\:\Delta\:{is}\:{maximum}, \\ $$$$\frac{\partial\Phi}{\partial\alpha}=\left(\mathrm{sin}\:\beta+\mathrm{sin}\:\gamma\right)\left(−\mathrm{sin}\:\alpha\right)−\mathrm{cos}\:\alpha\:\left(\mathrm{cos}\:\beta−\mathrm{cos}\:\gamma\right)=\mathrm{0} \\ $$$$\mathrm{sin}\:\alpha\:\left(\mathrm{sin}\:\beta+\mathrm{sin}\:\gamma\right)+\mathrm{cos}\:\alpha\:\left(\mathrm{cos}\:\beta−\mathrm{cos}\:\gamma\right)=\mathrm{0}\:\:\:…\left({iii}\right) \\ $$$$\Rightarrow\mathrm{tan}\:\alpha=\frac{\mathrm{cos}\:\gamma−\mathrm{cos}\:\beta}{\mathrm{sin}\:\beta+\mathrm{sin}\:\gamma} \\ $$$$\left(\Rightarrow{this}\:{proves}\:{that}\:{QA}\bot{BC}.\right) \\ $$$$\frac{\partial\Phi}{\partial\beta}=\mu\:\mathrm{cos}\:\left(\beta+\gamma\right)+\mathrm{cos}\:\beta\:\left(\mathrm{cos}\:\alpha+\mu\right)+\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta=\mathrm{0}\:\:\:…\left({i}\right) \\ $$$$\frac{\partial\Phi}{\partial\gamma}=\mu\:\mathrm{cos}\:\left(\beta+\gamma\right)+\mathrm{cos}\:\gamma\:\left(\mathrm{cos}\:\alpha+\mu\right)−\mathrm{sin}\:\alpha\:\mathrm{sin}\:\gamma=\mathrm{0}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$$\left(\mathrm{cos}\:\beta−\mathrm{cos}\:\gamma\right)\:\left(\mathrm{cos}\:\alpha+\mu\right)+\mathrm{sin}\:\alpha\:\left(\mathrm{sin}\:\beta+\mathrm{sin}\:\gamma\right)=\mathrm{0} \\ $$$$\mu\left(\mathrm{cos}\:\beta−\mathrm{cos}\:\gamma\right)+\mathrm{cos}\:\alpha\:\left(\mathrm{cos}\:\beta−\mathrm{cos}\:\gamma\right)+\mathrm{sin}\:\alpha\:\left(\mathrm{sin}\:\beta+\mathrm{sin}\:\gamma\right)=\mathrm{0} \\ $$$${due}\:{to}\:\left({iii}\right): \\ $$$$\mu\left(\mathrm{cos}\:\beta−\mathrm{cos}\:\gamma\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{cos}\:\beta=\mathrm{cos}\:\gamma\: \\ $$$$\Rightarrow\beta=\gamma\: \\ $$$${i}.{e}.\:{points}\:{B}\:{and}\:{C}\:{are}\:{symmetric}\:{about}\:{x}−{axis}! \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{cos}\:\gamma−\mathrm{cos}\:\beta}{\mathrm{sin}\:\beta+\mathrm{sin}\:\gamma}=\mathrm{0} \\ $$$$\Rightarrow\alpha=\mathrm{0} \\ $$$${i}.{e}.\:{point}\:{A}\:{lies}\:{on}\:{x}−{axis}! \\ $$$${insert}\:{into}\:\left({i}\right): \\ $$$$\mu\:\mathrm{cos}\:\mathrm{2}\beta+\mathrm{cos}\:\beta\:\left(\mathrm{1}+\mu\right)=\mathrm{0} \\ $$$$\mu\left(\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \beta−\mathrm{1}\right)+\mathrm{cos}\:\beta\:\left(\mathrm{1}+\mu\right)=\mathrm{0} \\ $$$$\mathrm{2}\mu\:\mathrm{cos}^{\mathrm{2}} \beta+\left(\mathrm{1}+\mu\right)\:\mathrm{cos}\:\beta−\mu=\mathrm{0} \\ $$$$\Rightarrow\mathrm{cos}\:\beta=\frac{−\left(\mathrm{1}+\mu\right)+\sqrt{\mathrm{9}\mu^{\mathrm{2}} +\mathrm{2}\mu+\mathrm{1}}}{\mathrm{4}\mu} \\ $$$${i}.{e}.\:\mathrm{cos}\:\beta=\frac{−\left({a}+{b}\right)+\sqrt{{a}^{\mathrm{2}} +\mathrm{9}{b}^{\mathrm{2}} +\mathrm{2}{ab}}}{\mathrm{4}{b}} \\ $$$$\blacksquare \\ $$
Answered by mr W last updated on 23/Oct/23
for the triangle with maximum area the tangents at the vertexes are parallel to the coresponding opposite sides. due to symmetry, this triangle must be symmetric about the x−axis. that means following red triangle has maximum area.
$${for}\:{the}\:{triangle}\:{with}\:{maximum}\:{area} \\ $$$${the}\:{tangents}\:{at}\:{the}\:{vertexes}\:{are}\: \\ $$$${parallel}\:{to}\:{the}\:{coresponding}\:{opposite} \\ $$$${sides}. \\ $$$${due}\:{to}\:{symmetry},\:{this}\:{triangle}\:{must} \\ $$$${be}\:{symmetric}\:{about}\:{the}\:{x}−{axis}. \\ $$$${that}\:{means}\:{following}\:{red}\:{triangle} \\ $$$${has}\:{maximum}\:{area}. \\ $$
Commented by mr W last updated on 23/Oct/23
b cos θ=(b sin θ+b+a) tan θ b(1−sin^2 θ)=(b sin θ+b+a) sin θ 2b sin^2 θ+(a+b) sin θ−b=0 ⇒sin θ=((−(a+b)+(√(a^2 +9b^2 +2ab)))/(4b)) A_(max) =(b sin θ+b+a)b cos θ =(((3a+3b+(√(a^2 +9b^2 +2ab)))(√(2(a+3b)(b−a)+2(a+b)(√(a^2 +9b^2 +2ab)))))/(16))
$${b}\:\mathrm{cos}\:\theta=\left({b}\:\mathrm{sin}\:\theta+{b}+{a}\right)\:\mathrm{tan}\:\theta \\ $$$${b}\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\theta\right)=\left({b}\:\mathrm{sin}\:\theta+{b}+{a}\right)\:\mathrm{sin}\:\theta \\ $$$$\mathrm{2}{b}\:\mathrm{sin}^{\mathrm{2}} \:\theta+\left({a}+{b}\right)\:\mathrm{sin}\:\theta−{b}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{−\left({a}+{b}\right)+\sqrt{{a}^{\mathrm{2}} +\mathrm{9}{b}^{\mathrm{2}} +\mathrm{2}{ab}}}{\mathrm{4}{b}} \\ $$$${A}_{{max}} =\left({b}\:\mathrm{sin}\:\theta+{b}+{a}\right){b}\:\mathrm{cos}\:\theta \\ $$$$\:\:\:\:=\frac{\left(\mathrm{3}{a}+\mathrm{3}{b}+\sqrt{{a}^{\mathrm{2}} +\mathrm{9}{b}^{\mathrm{2}} +\mathrm{2}{ab}}\right)\sqrt{\mathrm{2}\left({a}+\mathrm{3}{b}\right)\left({b}−{a}\right)+\mathrm{2}\left({a}+{b}\right)\sqrt{{a}^{\mathrm{2}} +\mathrm{9}{b}^{\mathrm{2}} +\mathrm{2}{ab}}}}{\mathrm{16}} \\ $$
Commented by ajfour last updated on 22/Oct/23
Commented by mr W last updated on 23/Oct/23
Commented by ajfour last updated on 22/Oct/23
Sir, which (△ or △) seems to be of greater area?
$${Sir},\:{which}\:\left(\bigtriangleup\:{or}\:\bigtriangleup\right)\:{seems}\:{to}\:{be}\:{of} \\ $$$${greater}\:{area}? \\ $$
Commented by mr W last updated on 23/Oct/23
actually i just believe that the maximum triangle must also be symmetric about x−axis, but it′s not proved.
$${actually}\:{i}\:{just}\:{believe}\:{that}\:{the} \\ $$$${maximum}\:{triangle}\:{must}\:{also}\:{be} \\ $$$${symmetric}\:{about}\:{x}−{axis},\:{but}\:{it}'{s}\:{not} \\ $$$${proved}. \\ $$
Commented by mr W last updated on 22/Oct/23
Commented by ajfour last updated on 22/Oct/23
ha ha, guess m not in my senses, thanks for everything! i will follow ur sol^n till finish.
$${ha}\:{ha},\:{guess}\:{m}\:{not}\:{in}\:{my}\:{senses},\:{thanks} \\ $$$${for}\:{everything}!\:{i}\:{will}\:{follow}\:{ur}\:{sol}^{{n}} \:{till}\: \\ $$$${finish}. \\ $$
Commented by mr W last updated on 23/Oct/23
my hypothesis above is now proved, see generall method below. please help checking if i made any mistakes. thank you!
$${my}\:{hypothesis}\:{above}\:{is}\:{now}\:{proved}, \\ $$$${see}\:{generall}\:{method}\:{below}.\: \\ $$$${please}\:{help}\:{checking}\:{if}\:{i}\:{made}\:{any}\: \\ $$$${mistakes}.\:{thank}\:{you}! \\ $$
Commented by ajfour last updated on 23/Oct/23
Thanks sir. You were correct indeed, and very detailed everything you still present.
Answered by ajfour last updated on 23/Oct/23
Commented by ajfour last updated on 23/Oct/23
△=(1/2)(2bcos θ)(bsin θ+bsin φ+acos δ) (∂△/∂δ)=−bcos θsin δ=0 clearly we choose δ=0 as (a<b) (∂△/∂φ)=bcos θ(bcos φ)=0 ⇒ φ=(π/2) (∂△/∂θ)=bcos θ(bcos θ+0) −bsin θ(bsin θ+bsin φ+acos δ)=0 now since δ=0 , and φ=(π/2) we get b^2 cos^2 θ=bsin θ(bsin θ+a+b) cos^2 θ=sin θ(sin θ+1+(a/b)) ⇒ 2sin^2 θ+(1+(a/b))sin θ−1=0 sin θ is obtained from here. Now △=b^2 cos θ(sin θ+1+(a/b))
$$\bigtriangleup=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{b}\mathrm{cos}\:\theta\right)\left({b}\mathrm{sin}\:\theta+{b}\mathrm{sin}\:\phi+{a}\mathrm{cos}\:\delta\right) \\ $$$$\frac{\partial\bigtriangleup}{\partial\delta}=−{b}\mathrm{cos}\:\theta\mathrm{sin}\:\delta=\mathrm{0} \\ $$$${clearly}\:{we}\:{choose}\:\delta=\mathrm{0}\:\:\:{as}\:\:\left({a}<{b}\right) \\ $$$$\frac{\partial\bigtriangleup}{\partial\phi}={b}\mathrm{cos}\:\theta\left({b}\mathrm{cos}\:\phi\right)=\mathrm{0}\:\:\Rightarrow\:\:\phi=\frac{\pi}{\mathrm{2}} \\ $$$$\frac{\partial\bigtriangleup}{\partial\theta}={b}\mathrm{cos}\:\theta\left({b}\mathrm{cos}\:\theta+\mathrm{0}\right) \\ $$$$\:\:\:\:\:\:\:\:−{b}\mathrm{sin}\:\theta\left({b}\mathrm{sin}\:\theta+{b}\mathrm{sin}\:\phi+{a}\mathrm{cos}\:\delta\right)=\mathrm{0} \\ $$$${now}\:{since}\:\:\delta=\mathrm{0}\:\:,\:{and}\:\phi=\frac{\pi}{\mathrm{2}} \\ $$$${we}\:{get}\:\:{b}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta={b}\mathrm{sin}\:\theta\left({b}\mathrm{sin}\:\theta+{a}+{b}\right) \\ $$$$\mathrm{cos}\:^{\mathrm{2}} \theta=\mathrm{sin}\:\theta\left(\mathrm{sin}\:\theta+\mathrm{1}+\frac{{a}}{{b}}\right) \\ $$$$\Rightarrow\:\:\mathrm{2sin}\:^{\mathrm{2}} \theta+\left(\mathrm{1}+\frac{{a}}{{b}}\right)\mathrm{sin}\:\theta−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{sin}\:\theta\:\:{is}\:{obtained}\:{from}\:{here}.\:{Now} \\ $$$$\bigtriangleup={b}^{\mathrm{2}} \mathrm{cos}\:\theta\left(\mathrm{sin}\:\theta+\mathrm{1}+\frac{{a}}{{b}}\right) \\ $$
Commented by mr W last updated on 23/Oct/23
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