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Find-the-maximum-area-of-a-triangle-inscribed-in-an-ellipse-with-parameters-a-and-b-

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Question Number 50674 by mr W last updated on 18/Dec/18
Find the maximum area of a triangle inscribed in an ellipse with parameters a and b.
$${Find}\:{the}\:{maximum}\:{area}\:{of}\:{a}\:{triangle} \\ $$$${inscribed}\:{in}\:{an}\:{ellipse}\:{with}\:{parameters} \\ $$$${a}\:{and}\:{b}. \\ $$
Commented by ajfour last updated on 18/Dec/18
Commented by MJS last updated on 18/Dec/18
all triangles created as follows have the same (maximum) area 1. draw a circle with radius a 2. insert an equilateral triangle with 0≤α<((2π)/3) A= (((acos α)),((asin α)) ) B= (((acos (α+((2π)/3)))),((asin (α+((2π)/3)))) ) C= (((acos (α−((2π)/3)))),((asin (α−((2π)/3)))) ) 3. compress (((x′)),((y′)) ) = ((x),(((b/a)y)) )
$$\mathrm{all}\:\mathrm{triangles}\:\mathrm{created}\:\mathrm{as}\:\mathrm{follows}\:\mathrm{have}\:\mathrm{the} \\ $$$$\mathrm{same}\:\left(\mathrm{maximum}\right)\:\mathrm{area} \\ $$$$\mathrm{1}.\:\mathrm{draw}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{radius}\:{a} \\ $$$$\mathrm{2}.\:\mathrm{insert}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{0}\leqslant\alpha<\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$${A}=\begin{pmatrix}{{a}\mathrm{cos}\:\alpha}\\{{a}\mathrm{sin}\:\alpha}\end{pmatrix}\:{B}=\begin{pmatrix}{{a}\mathrm{cos}\:\left(\alpha+\frac{\mathrm{2}\pi}{\mathrm{3}}\right)}\\{{a}\mathrm{sin}\:\left(\alpha+\frac{\mathrm{2}\pi}{\mathrm{3}}\right)}\end{pmatrix}\:\:{C}=\begin{pmatrix}{{a}\mathrm{cos}\:\left(\alpha−\frac{\mathrm{2}\pi}{\mathrm{3}}\right)}\\{{a}\mathrm{sin}\:\left(\alpha−\frac{\mathrm{2}\pi}{\mathrm{3}}\right)}\end{pmatrix} \\ $$$$\mathrm{3}.\:\mathrm{compress}\:\begin{pmatrix}{{x}'}\\{{y}'}\end{pmatrix}\:=\begin{pmatrix}{{x}}\\{\frac{{b}}{{a}}{y}}\end{pmatrix} \\ $$
Commented by mr W last updated on 19/Dec/18
thank you sir!
$${thank}\:{you}\:{sir}! \\ $$
Answered by ajfour last updated on 18/Dec/18
let a > b , B(acos θ, bsin θ) A = abcos θ(1−sin θ) (dA/dθ) = 0 ⇒ cos^2 θ = sin^2 θ−sin θ ⇒ 2sin^2 θ−sin θ−1= 0 (sin θ−(1/4))^2 = (9/(16)) ⇒ sin θ = −(1/2) ⇒ θ = −(π/6) A_(max) = ((3(√3) ab)/4) .
$${let}\:\:{a}\:>\:{b}\:,\:{B}\left({a}\mathrm{cos}\:\theta,\:{b}\mathrm{sin}\:\theta\right) \\ $$$${A}\:=\:{ab}\mathrm{cos}\:\theta\left(\mathrm{1}−\mathrm{sin}\:\theta\right) \\ $$$$\frac{{dA}}{{d}\theta}\:=\:\mathrm{0}\:\:\Rightarrow\:\:\mathrm{cos}\:^{\mathrm{2}} \theta\:=\:\mathrm{sin}\:^{\mathrm{2}} \theta−\mathrm{sin}\:\theta \\ $$$$\Rightarrow\:\:\mathrm{2sin}\:^{\mathrm{2}} \theta−\mathrm{sin}\:\theta−\mathrm{1}=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\left(\mathrm{sin}\:\theta−\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} =\:\frac{\mathrm{9}}{\mathrm{16}} \\ $$$$\Rightarrow\:\:\mathrm{sin}\:\theta\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\Rightarrow\:\theta\:=\:−\frac{\pi}{\mathrm{6}} \\ $$$$\:\:\:{A}_{{max}} \:=\:\frac{\mathrm{3}\sqrt{\mathrm{3}}\:{ab}}{\mathrm{4}}\:. \\ $$
Commented by mr W last updated on 18/Dec/18
thanks! but why must A be on the axis?
$${thanks}!\:{but}\:{why}\:{must}\:{A}\:{be}\:{on}\:{the}\:{axis}? \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 19/Dec/18
△ABC A(acosα,bcosα) B(acosβ,bsinβ) C(acosγ,bsinγ) (1/2)[x_1 (y_2 −y_3 )+x_2 (y_3 −y_1 )+x_3 (y_1 −y_2 )] =(1/2)ab[cosα(sinβ−sinγ)+cosβ(sinγ−sinα)+cosγ(sinα−sinβ)] =((ab)/2)[sin(β−α)+sin(γ−β)+sin(α−γ)] =((ab)/2)[−2sin(((α−β)/2))cos(((α−β)/2))+2sin(((α−β)/2))cos(((2γ−α−β)/2))] =((ab)/2)×2sin(((α−β)/2))[2sin(((γ−β)/2))sin(((γ−α)/2))] =−2absin(((α−β)/2))sin(((β−γ)/2))sin(((γ−α)/2)) so area of △ABC is 2absin(((α−β)/2))sin(((β−γ)/2))sin(((γ−α)/2)) wait pls maximising the area... now let P corespond to A in auxilary circle Q→B and R→C area of △PQR=2a^2 sin(((α−β)/2))sin(((β−γ)/2))sin(((γ−α)/2)) ((area△ABC)/(area △PQR))=((2ab)/(2a^2 ))=(b/a) max area △ABC=(b/a)×max area△ PQR △PQR achive max area when it is eauilateral triangle.hence△ABC is also equilateral triangle... area of equilateral tria ngle inscribed in circle of rsdius a is =(1/2)(acos30^o +acos30^o )×(a+asin30^o ) =(a^2 /2)(2×((√3)/2))(1+sin30^o )=((3(√3))/4)a^2 so required area of trizngle△ABC=(b/a)×((3(√3))/4)a^2 =((3(√3))/4)ab
$$\bigtriangleup{ABC}\:\:{A}\left({acos}\alpha,{bcos}\alpha\right)\:{B}\left({acos}\beta,{bsin}\beta\right) \\ $$$${C}\left({acos}\gamma,{bsin}\gamma\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[{x}_{\mathrm{1}} \left({y}_{\mathrm{2}} −{y}_{\mathrm{3}} \right)+{x}_{\mathrm{2}} \left({y}_{\mathrm{3}} −{y}_{\mathrm{1}} \right)+{x}_{\mathrm{3}} \left({y}_{\mathrm{1}} −{y}_{\mathrm{2}} \right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ab}\left[{cos}\alpha\left({sin}\beta−{sin}\gamma\right)+{cos}\beta\left({sin}\gamma−{sin}\alpha\right)+{cos}\gamma\left({sin}\alpha−{sin}\beta\right)\right] \\ $$$$=\frac{{ab}}{\mathrm{2}}\left[{sin}\left(\beta−\alpha\right)+{sin}\left(\gamma−\beta\right)+{sin}\left(\alpha−\gamma\right)\right] \\ $$$$=\frac{{ab}}{\mathrm{2}}\left[−\mathrm{2}{sin}\left(\frac{\alpha−\beta}{\mathrm{2}}\right){cos}\left(\frac{\alpha−\beta}{\mathrm{2}}\right)+\mathrm{2}{sin}\left(\frac{\alpha−\beta}{\mathrm{2}}\right){cos}\left(\frac{\mathrm{2}\gamma−\alpha−\beta}{\mathrm{2}}\right)\right] \\ $$$$=\frac{{ab}}{\mathrm{2}}×\mathrm{2}{sin}\left(\frac{\alpha−\beta}{\mathrm{2}}\right)\left[\mathrm{2}{sin}\left(\frac{\gamma−\beta}{\mathrm{2}}\right){sin}\left(\frac{\gamma−\alpha}{\mathrm{2}}\right)\right] \\ $$$$=−\mathrm{2}{absin}\left(\frac{\alpha−\beta}{\mathrm{2}}\right){sin}\left(\frac{\beta−\gamma}{\mathrm{2}}\right){sin}\left(\frac{\gamma−\alpha}{\mathrm{2}}\right) \\ $$$${so}\:{area}\:{of}\:\bigtriangleup{ABC}\:{is} \\ $$$$\mathrm{2}{absin}\left(\frac{\alpha−\beta}{\mathrm{2}}\right){sin}\left(\frac{\beta−\gamma}{\mathrm{2}}\right){sin}\left(\frac{\gamma−\alpha}{\mathrm{2}}\right) \\ $$$${wait}\:{pls}\:{maximising}\:{the}\:{area}… \\ $$$${now}\:{let}\:{P}\:{corespond}\:{to}\:{A}\:{in}\:{auxilary}\:{circle} \\ $$$${Q}\rightarrow{B}\:\:\:{and}\:{R}\rightarrow{C} \\ $$$${area}\:{of}\:\bigtriangleup{PQR}=\mathrm{2}{a}^{\mathrm{2}} {sin}\left(\frac{\alpha−\beta}{\mathrm{2}}\right){sin}\left(\frac{\beta−\gamma}{\mathrm{2}}\right){sin}\left(\frac{\gamma−\alpha}{\mathrm{2}}\right) \\ $$$$\frac{{area}\bigtriangleup{ABC}}{{area}\:\bigtriangleup{PQR}}=\frac{\mathrm{2}{ab}}{\mathrm{2}{a}^{\mathrm{2}} }=\frac{{b}}{{a}} \\ $$$${max}\:{area}\:\bigtriangleup{ABC}=\frac{{b}}{{a}}×{max}\:{area}\bigtriangleup\:{PQR} \\ $$$$\bigtriangleup{PQR}\:{achive}\:{max}\:{area}\:{when}\:{it}\:{is}\:{eauilateral} \\ $$$${triangle}.{hence}\bigtriangleup{ABC}\:{is}\:{also}\:{equilateral}\: \\ $$$${triangle}… \\ $$$${area}\:{of}\:{equilateral}\:{tria}\:{ngle}\:{inscribed}\:{in}\:{circle} \\ $$$${of}\:{rsdius}\:{a}\:{is} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({acos}\mathrm{30}^{{o}} +{acos}\mathrm{30}^{{o}} \right)×\left({a}+{asin}\mathrm{30}^{{o}} \:\right) \\ $$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{2}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\left(\mathrm{1}+{sin}\mathrm{30}^{{o}} \right)=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}}{a}^{\mathrm{2}} \\ $$$${so}\:{required}\:{area}\:{of}\:{trizngle}\bigtriangleup{ABC}=\frac{{b}}{{a}}×\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}}{a}^{\mathrm{2}} \\ $$$$=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}}{ab} \\ $$
Commented by mr W last updated on 20/Dec/18
thanks sir!
$${thanks}\:{sir}! \\ $$

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