Question Number 104101 by ajfour last updated on 19/Jul/20

Commented byajfour last updated on 19/Jul/20

If the outer triangle is equilateral,  and the two ellipses of same size  and shape, find ratio of area of  outer triangle to inner(blue)  triangle.

Commented bymr W last updated on 19/Jul/20

only if blue ellipse=yellow  ellipse=incircle of triangle  ⇒((small triangle)/(big triangle))=(1/4)

Answered by ajfour last updated on 19/Jul/20

let eq. of blue ellipse be    (x^2 /a^2 )+(((y−b)^2 )/b^2 )=1  Let right side of outer triangle has  eq:   y=x(√3)+c      and as this is  tangent to both ellipses,  first    b^2 x^2 +a^2 (x(√3)+c−b)^2 −a^2 b^2 =0  or    (3a^2 +b^2 )x^2 +2(√3)(a^2 )(c−b)x                            +a^2 (c−b)^2 −a^2 b^2 =0  has double root;  ⇒  12a^4 (c−b)^2 =4a^2 (3a^2 +b^2 )[(c−b)^2 −b^2 ]  ⇒ 3a^2 (c−b)^2 =(3a^2 +b^2 )[(c−b)^2 −b^2 ]  ⇒  (c−b)^2 =3a^2 +b^2      .....(I)  Now eq. of the yellow ellipse is        (x^2 /b^2 )+(((y−a)^2 )/a^2 )=1  y=x(√3)+c   is tangent to this ellipse  too,  hence    a^2 x^2 +b^2 (x(√3)+c−a)^2 −a^2 b^2 =0  (a^2 +3b^2 )x^2 +2(√3)(b^2 )(c−a)x                         +b^2 [(c−a)^2 −a^2 ]=0  has also a double root; ⇒  12b^4 (c−a)^2 =4b^2 (a^2 +3b^2 )[(c−a)^2 −a^2 ]  ⇒  3b^2 (c−a)^2 =(a^2 +3b^2 )[(c−a)^2 −a^2 ]  ⇒  (c−a)^2 = a^2 +3b^2      .....(II)  Now eliminating c  among (I)&(II)     b+(√(3a^2 +b^2 )) = a+(√(a^2 +3b^2 ))  let   (b/a) = μ  ⇒  μ+(√(3+μ^2 ))=1+(√(1+3μ^2 ))    ....(i)  ⇒    μ=1   ★  ⇒   c−a = c−b = 2a =2b  ⇒   a = b = (c/3)  ⇒   (△_(small) /△_(large) ) = (1/4) ■  mrW Sir  you are perfectly right!