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

Commented by ajfour last updated on 14/Nov/20

$F i n d s i d e s o f m a x i m u m a r e a$ $r i g h t a n g l e d t r i a n g l e i n s c r i b e d$ $i n a n e l l i p s e .$
Commented by MJS_new last updated on 14/Nov/20

${it}^{'} s part of the problem to find the general$ $triangle with maximum area . I solved it,$ $cannot find the old question .$ $all triangles with centroid in the center of$ $the ellipse have equal area . we have to find$ $the rectangular triangles out of these$
Commented by MJS_new last updated on 14/Nov/20
$I only have the formulas for the axes of the smallest ellipse around a given triangle with sides a, b, c which is (1/3)(√(a^2 +b^2 +c^2 ±2(√(a^2 b^2 +a^2 c^2 +b^2 c^2 −δ^2 )))) where δ=(√((a+b+c)(a+b−c)(a+c−b)(b+c−a))) if we let c=(√(a^2 +b^2 )) we get ((√2)/3)(√(a^2 +b^2 ±(√(a^4 −a^2 b^2 +b^4 )))) we have to solve this for a, b with given axes α, β: { ((α=((√2)/3)(√(a^2 +b^2 +(√(a^4 −a^2 b^2 +b^4 )))))),((β=((√2)/3)(√(a^2 +b^2 −(√(a^4 −a^2 b^2 +b^4 )))))) :} sorry I have got no time right now$
$I only have the formulas for the axes of the$ $smallest ellipse around a given triangle with$ $sides a, b, c which is$ $\frac{1}{3} \sqrt{a^{2} + b^{2} + c^{2} \pm 2 \sqrt{a^{2} b^{2} + a^{2} c^{2} + b^{2} c^{2} - δ^{2}}}$ $where δ = \sqrt{(a + b + c) (a + b - c) (a + c - b) (b + c - a)}$ $if we let c = \sqrt{a^{2} + b^{2}} we get$ $\frac{\sqrt{2}}{3} \sqrt{a^{2} + b^{2} \pm \sqrt{a^{4} - a^{2} b^{2} + b^{4}}}$ $we have to solve this for a, b with given$ $axes α, β :$ ${\begin{cases} α = \frac{\sqrt{2}}{3} \sqrt{a^{2} + b^{2} + \sqrt{a^{4} - a^{2} b^{2} + b^{4}}} \\ β = \frac{\sqrt{2}}{3} \sqrt{a^{2} + b^{2} - \sqrt{a^{4} - a^{2} b^{2} + b^{4}}} \end{cases}$ $sorry I have got no time right now$
Commented by MJS_new last updated on 14/Nov/20

$found it!$ $with a ⩽ b ⩽ c the formula is$ $a = \sqrt{\frac{3}{8} (3 (α^{2} + β^{2}) - \sqrt{3 (3 α^{4} - 10 α^{2} β^{2} + 3 β^{4}}}$ $b = \sqrt{\frac{3}{8} (3 (α^{2} + β^{2}) + \sqrt{3 (3 α^{4} - 10 α^{2} β^{2} + 3 β^{4}}}$ $c = \frac{3}{2} \sqrt{α^{2} + β^{2}}$
Commented by ajfour last updated on 14/Nov/20

$T h a n k s a l o t S i r, a g l i m p s e o f$ $t h e s e f o r m u l a s p r o v e s t h a t t h e$ $p r o o f {i s n}^{'} t a n e a s y o n e, t h o u g h . .$
Commented by MJS_new last updated on 14/Nov/20

$my idea had been to start with a unit circle$ $and an inscribed equilateral triangle$ $then first rotate the triangle by φ and second$ $compress the whole thing to get the desired$ $ellipse . this way {it}^{'} s obvious that the centroid$ $stays in the center and all triangles you get$ $have the same area and {there}^{'} s no triangle$ $with a greater area .$ $I wanted to find the smallest ellipse around$ $a given triangle . I kept the formulas but not$ $the path . will send them again but not now$
Commented by MJS_new last updated on 14/Nov/20

${there}^{'} s no rectangular triangle for ellipses$ $with α ⩾ β \land β > \frac{α}{\sqrt{3}} . for β = \frac{α}{\sqrt{3}} the triangle$ $is isosceles$ $plus I corrected a typo in the formulas I gave$ $before . - 10 a^{2} b^{2} not - 20 a^{2} b^{2}$
	Answered by ajfour last updated on 14/Nov/20
	$let eq.of line PR be (x/h)+(y/k)=1 solving it with ellipse eq. (x^2 /a^2 )+(y^2 /b^2 )=1 (x^2 /a^2 )+(k^2 /b^2 )(1−(x/h))^2 =1 ⇒ ((1/a^2 )+(k^2 /(b^2 h^2 )))x^2 −((2k^2 x)/(b^2 h))−(1−(k^2 /b^2 ))=0 (x_1 −x_2 )^2 =(x_1 +x_2 )^2 −4x_1 x_2 = (((2a^2 k^2 h)/(b^2 h^2 +a^2 k^2 )))^2 +((4a^2 h^2 )/(b^2 −k^2 )) y_1 −y_2 =k(1−(x_1 /h))(1−(x_2 /h)) =k(1+((x_1 x_2 )/h^2 )−((x_1 +x_2 )/h)) =k(1−(a^2 /(b^2 −k^2 ))−((2a^2 k^2 )/(b^2 h^2 +a^2 k^2 ))) 4△^2 ={(x_1 −x_2 )^2 +(y_1 −y_2 )^2 }p^2 .....$
	$l e t e q . o f l i n e P R b e \frac{x}{h} + \frac{y}{k} = 1$ $s o l v i n g i t w i t h e l l i p s e e q .$ $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $\frac{x^{2}}{a^{2}} + \frac{k^{2}}{b^{2}} {(1 - \frac{x}{h})}^{2} = 1 \Rightarrow$ $(\frac{1}{a^{2}} + \frac{k^{2}}{b^{2} h^{2}}) x^{2} - \frac{2 k^{2} x}{b^{2} h} - (1 - \frac{k^{2}}{b^{2}}) = 0$ ${(x_{1} - x_{2})}^{2} = {(x_{1} + x_{2})}^{2} - 4 x_{1} x_{2}$ $= {(\frac{2 a^{2} k^{2} h}{b^{2} h^{2} + a^{2} k^{2}})}^{2} + \frac{4 a^{2} h^{2}}{b^{2} - k^{2}}$ $y_{1} - y_{2} = k (1 - \frac{x_{1}}{h}) (1 - \frac{x_{2}}{h})$ $= k (1 + \frac{x_{1} x_{2}}{h^{2}} - \frac{x_{1} + x_{2}}{h})$ $= k (1 - \frac{a^{2}}{b^{2} - k^{2}} - \frac{2 a^{2} k^{2}}{b^{2} h^{2} + a^{2} k^{2}})$ $4 △^{2} = {{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}} p^{2}$ $. . . . .$
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