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Question Number 122111 by ajfour last updated on 14/Nov/20

Commented by ajfour last updated on 14/Nov/20

Find sides of maximum area  right angled triangle inscribed  in an ellipse.

$${Find}\:{sides}\:{of}\:{maximum}\:{area} \\ $$$${right}\:{angled}\:{triangle}\:{inscribed} \\ $$$${in}\:{an}\:{ellipse}. \\ $$

Commented by MJS_new last updated on 14/Nov/20

it′s part of the problem to find the general  triangle with maximum area. I solved it,  cannot find the old question.  all triangles with centroid in the center of  the ellipse have equal area. we have to find  the rectangular triangles out of these

$$\mathrm{it}'\mathrm{s}\:\mathrm{part}\:\mathrm{of}\:\mathrm{the}\:\mathrm{problem}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{general} \\ $$$$\mathrm{triangle}\:\mathrm{with}\:\mathrm{maximum}\:\mathrm{area}.\:\mathrm{I}\:\mathrm{solved}\:\mathrm{it}, \\ $$$$\mathrm{cannot}\:\mathrm{find}\:\mathrm{the}\:\mathrm{old}\:\mathrm{question}. \\ $$$$\mathrm{all}\:\mathrm{triangles}\:\mathrm{with}\:\mathrm{centroid}\:\mathrm{in}\:\mathrm{the}\:\mathrm{center}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{ellipse}\:\mathrm{have}\:\mathrm{equal}\:\mathrm{area}.\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{rectangular}\:\mathrm{triangles}\:\mathrm{out}\:\mathrm{of}\:\mathrm{these} \\ $$

Commented by MJS_new last updated on 14/Nov/20

I only have the formulas for the axes of the  smallest ellipse around a given triangle with  sides a, b, c which is  (1/3)(√(a^2 +b^2 +c^2 ±2(√(a^2 b^2 +a^2 c^2 +b^2 c^2 −δ^2 ))))  where δ=(√((a+b+c)(a+b−c)(a+c−b)(b+c−a)))  if we let c=(√(a^2 +b^2 )) we get  ((√2)/3)(√(a^2 +b^2 ±(√(a^4 −a^2 b^2 +b^4 ))))  we have to solve this for a, b with given  axes α, β:   { ((α=((√2)/3)(√(a^2 +b^2 +(√(a^4 −a^2 b^2 +b^4 )))))),((β=((√2)/3)(√(a^2 +b^2 −(√(a^4 −a^2 b^2 +b^4 )))))) :}  sorry I have got no time right now

$$\mathrm{I}\:\mathrm{only}\:\mathrm{have}\:\mathrm{the}\:\mathrm{formulas}\:\mathrm{for}\:\mathrm{the}\:\mathrm{axes}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{smallest}\:\mathrm{ellipse}\:\mathrm{around}\:\mathrm{a}\:\mathrm{given}\:\mathrm{triangle}\:\mathrm{with} \\ $$$$\mathrm{sides}\:{a},\:{b},\:{c}\:\mathrm{which}\:\mathrm{is} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \pm\mathrm{2}\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} −\delta^{\mathrm{2}} }} \\ $$$$\mathrm{where}\:\delta=\sqrt{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}+{c}−{b}\right)\left({b}+{c}−{a}\right)} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{let}\:{c}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\mathrm{we}\:\mathrm{get} \\ $$$$\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \pm\sqrt{{a}^{\mathrm{4}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{4}} }} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{for}\:{a},\:{b}\:\mathrm{with}\:\mathrm{given} \\ $$$$\mathrm{axes}\:\alpha,\:\beta: \\ $$$$\begin{cases}{\alpha=\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\sqrt{{a}^{\mathrm{4}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{4}} }}}\\{\beta=\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\sqrt{{a}^{\mathrm{4}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{4}} }}}\end{cases} \\ $$$$\mathrm{sorry}\:\mathrm{I}\:\mathrm{have}\:\mathrm{got}\:\mathrm{no}\:\mathrm{time}\:\mathrm{right}\:\mathrm{now} \\ $$

Commented by MJS_new last updated on 14/Nov/20

found it!  with a≤b≤c the formula is  a=(√((3/8)(3(α^2 +β^2 )−(√(3(3α^4 −10α^2 β^2 +3β^4 ))))  b=(√((3/8)(3(α^2 +β^2 )+(√(3(3α^4 −10α^2 β^2 +3β^4 ))))  c=(3/2)(√(α^2 +β^2 ))

$$\mathrm{found}\:\mathrm{it}! \\ $$$$\mathrm{with}\:{a}\leqslant{b}\leqslant{c}\:\mathrm{the}\:\mathrm{formula}\:\mathrm{is} \\ $$$${a}=\sqrt{\frac{\mathrm{3}}{\mathrm{8}}\left(\mathrm{3}\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \right)−\sqrt{\mathrm{3}\left(\mathrm{3}\alpha^{\mathrm{4}} −\mathrm{10}\alpha^{\mathrm{2}} \beta^{\mathrm{2}} +\mathrm{3}\beta^{\mathrm{4}} \right.}\right.} \\ $$$${b}=\sqrt{\frac{\mathrm{3}}{\mathrm{8}}\left(\mathrm{3}\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \right)+\sqrt{\mathrm{3}\left(\mathrm{3}\alpha^{\mathrm{4}} −\mathrm{10}\alpha^{\mathrm{2}} \beta^{\mathrm{2}} +\mathrm{3}\beta^{\mathrm{4}} \right.}\right.} \\ $$$${c}=\frac{\mathrm{3}}{\mathrm{2}}\sqrt{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} } \\ $$

Commented by ajfour last updated on 14/Nov/20

Thanks a lot Sir, a glimpse of  these formulas proves  that the  proof isn′t an easy one, though..

$${Thanks}\:{a}\:{lot}\:{Sir},\:{a}\:{glimpse}\:{of} \\ $$$${these}\:{formulas}\:{proves}\:\:{that}\:{the} \\ $$$${proof}\:{isn}'{t}\:{an}\:{easy}\:{one},\:{though}.. \\ $$

Commented by MJS_new last updated on 14/Nov/20

my idea had been to start with a unit circle  and an inscribed equilateral triangle  then first rotate the triangle by ϕ and second  compress the whole thing to get the desired  ellipse. this way it′s obvious that the centroid  stays in the center and all triangles you get  have the same area and there′s no triangle  with a greater area.  I wanted to find the smallest ellipse around  a given triangle. I kept the formulas but not  the path. will send them again but not now

$$\mathrm{my}\:\mathrm{idea}\:\mathrm{had}\:\mathrm{been}\:\mathrm{to}\:\mathrm{start}\:\mathrm{with}\:\mathrm{a}\:\mathrm{unit}\:\mathrm{circle} \\ $$$$\mathrm{and}\:\mathrm{an}\:\mathrm{inscribed}\:\mathrm{equilateral}\:\mathrm{triangle} \\ $$$$\mathrm{then}\:\mathrm{first}\:\mathrm{rotate}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{by}\:\varphi\:\mathrm{and}\:\mathrm{second} \\ $$$$\mathrm{compress}\:\mathrm{the}\:\mathrm{whole}\:\mathrm{thing}\:\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{desired} \\ $$$$\mathrm{ellipse}.\:\mathrm{this}\:\mathrm{way}\:\mathrm{it}'\mathrm{s}\:\mathrm{obvious}\:\mathrm{that}\:\mathrm{the}\:\mathrm{centroid} \\ $$$$\mathrm{stays}\:\mathrm{in}\:\mathrm{the}\:\mathrm{center}\:\mathrm{and}\:\mathrm{all}\:\mathrm{triangles}\:\mathrm{you}\:\mathrm{get} \\ $$$$\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{area}\:\mathrm{and}\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{triangle} \\ $$$$\mathrm{with}\:\mathrm{a}\:\mathrm{greater}\:\mathrm{area}. \\ $$$$\mathrm{I}\:\mathrm{wanted}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{smallest}\:\mathrm{ellipse}\:\mathrm{around} \\ $$$$\mathrm{a}\:\mathrm{given}\:\mathrm{triangle}.\:\mathrm{I}\:\mathrm{kept}\:\mathrm{the}\:\mathrm{formulas}\:\mathrm{but}\:\mathrm{not} \\ $$$$\mathrm{the}\:\mathrm{path}.\:\mathrm{will}\:\mathrm{send}\:\mathrm{them}\:\mathrm{again}\:\mathrm{but}\:\mathrm{not}\:\mathrm{now} \\ $$

Commented by MJS_new last updated on 14/Nov/20

there′s no rectangular triangle for ellipses  with α≥β ∧ β>(α/( (√3))). for β=(α/( (√3))) the triangle  is isosceles  plus I corrected a typo in the formulas I gave  before. −10a^2 b^2  not −20a^2 b^2

$$\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{rectangular}\:\mathrm{triangle}\:\mathrm{for}\:\mathrm{ellipses} \\ $$$$\mathrm{with}\:\alpha\geqslant\beta\:\wedge\:\beta>\frac{\alpha}{\:\sqrt{\mathrm{3}}}.\:\mathrm{for}\:\beta=\frac{\alpha}{\:\sqrt{\mathrm{3}}}\:\mathrm{the}\:\mathrm{triangle} \\ $$$$\mathrm{is}\:\mathrm{isosceles} \\ $$$$\mathrm{plus}\:\mathrm{I}\:\mathrm{corrected}\:\mathrm{a}\:\mathrm{typo}\:\mathrm{in}\:\mathrm{the}\:\mathrm{formulas}\:\mathrm{I}\:\mathrm{gave} \\ $$$$\mathrm{before}.\:−\mathrm{10}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \:\mathrm{not}\:−\mathrm{20}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$

Answered by ajfour last updated on 14/Nov/20

let eq.of line PR  be  (x/h)+(y/k)=1  solving it with ellipse eq.  (x^2 /a^2 )+(y^2 /b^2 )=1  (x^2 /a^2 )+(k^2 /b^2 )(1−(x/h))^2 =1   ⇒   ((1/a^2 )+(k^2 /(b^2 h^2 )))x^2 −((2k^2 x)/(b^2 h))−(1−(k^2 /b^2 ))=0  (x_1 −x_2 )^2 =(x_1 +x_2 )^2 −4x_1 x_2      = (((2a^2 k^2 h)/(b^2 h^2 +a^2 k^2 )))^2 +((4a^2 h^2 )/(b^2 −k^2 ))  y_1 −y_2 =k(1−(x_1 /h))(1−(x_2 /h))              =k(1+((x_1 x_2 )/h^2 )−((x_1 +x_2 )/h))       =k(1−(a^2 /(b^2 −k^2 ))−((2a^2 k^2 )/(b^2 h^2 +a^2 k^2 )))  4△^2 ={(x_1 −x_2 )^2 +(y_1 −y_2 )^2 }p^2   .....

$${let}\:{eq}.{of}\:{line}\:{PR}\:\:{be}\:\:\frac{{x}}{{h}}+\frac{{y}}{{k}}=\mathrm{1} \\ $$$${solving}\:{it}\:{with}\:{ellipse}\:{eq}. \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{k}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\left(\mathrm{1}−\frac{{x}}{{h}}\right)^{\mathrm{2}} =\mathrm{1}\:\:\:\Rightarrow \\ $$$$\:\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{{k}^{\mathrm{2}} }{{b}^{\mathrm{2}} {h}^{\mathrm{2}} }\right){x}^{\mathrm{2}} −\frac{\mathrm{2}{k}^{\mathrm{2}} {x}}{{b}^{\mathrm{2}} {h}}−\left(\mathrm{1}−\frac{{k}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$\left({x}_{\mathrm{1}} −{x}_{\mathrm{2}} \right)^{\mathrm{2}} =\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{x}_{\mathrm{1}} {x}_{\mathrm{2}} \\ $$$$\:\:\:=\:\left(\frac{\mathrm{2}{a}^{\mathrm{2}} {k}^{\mathrm{2}} {h}}{{b}^{\mathrm{2}} {h}^{\mathrm{2}} +{a}^{\mathrm{2}} {k}^{\mathrm{2}} }\right)^{\mathrm{2}} +\frac{\mathrm{4}{a}^{\mathrm{2}} {h}^{\mathrm{2}} }{{b}^{\mathrm{2}} −{k}^{\mathrm{2}} } \\ $$$${y}_{\mathrm{1}} −{y}_{\mathrm{2}} ={k}\left(\mathrm{1}−\frac{{x}_{\mathrm{1}} }{{h}}\right)\left(\mathrm{1}−\frac{{x}_{\mathrm{2}} }{{h}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:={k}\left(\mathrm{1}+\frac{{x}_{\mathrm{1}} {x}_{\mathrm{2}} }{{h}^{\mathrm{2}} }−\frac{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} }{{h}}\right) \\ $$$$\:\:\:\:\:={k}\left(\mathrm{1}−\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} −{k}^{\mathrm{2}} }−\frac{\mathrm{2}{a}^{\mathrm{2}} {k}^{\mathrm{2}} }{{b}^{\mathrm{2}} {h}^{\mathrm{2}} +{a}^{\mathrm{2}} {k}^{\mathrm{2}} }\right) \\ $$$$\mathrm{4}\bigtriangleup^{\mathrm{2}} =\left\{\left({x}_{\mathrm{1}} −{x}_{\mathrm{2}} \right)^{\mathrm{2}} +\left({y}_{\mathrm{1}} −{y}_{\mathrm{2}} \right)^{\mathrm{2}} \right\}{p}^{\mathrm{2}} \\ $$$$..... \\ $$

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