Steiner Inellipse Problem

We can get midpoint of triangle , write it in the form of a+ib , these three are root of cubic equation. Diff cubic we get quadratic equated with zero we get focii of ellipse (marden theorem) . We know the point of contact rest is calculation

The implicit equation of this Steiner Inellipse is

$16{ y }^{ 2 }-20xy+25{ x }^{ 2 }-56y-140x+224=0$

which can be partially checked by plugging in the coordinates of the midpoints of the triangle. Thus, the sum being $49$ , the answer is $7$ .

An affine transform

$x'={ a }_{ 00 }x+{ a }_{ 01 }y+{ a }_{ 02 }$
$y'={ a }_{ 10 }x+{ a }_{ 11 }y+{ a }_{ 12 }$

preserves ratios of lengths and areas. Hence, since the incircle of an equilateral triangle has the maximum area of all possible inscribed ellipses, we work out the coefficients

$\left( { a }_{ 00 },{ a }_{ 01 },{ a }_{ 02 },{ a }_{ 10 },{ a }_{ 11 },{ a }_{ 12 } \right) =$

$\left( -\frac { \sqrt { 3 } }{ 7 } ,\frac { \sqrt { 3 } }{ 7 } ,0,-\frac { 3 }{ 14 } ,-\frac { 3 }{ 35 } ,\frac { 7 }{ 5 } \right)$

such that it transforms the coordinates of this triangle into the equilateral triangle with vertex coordinates

$\left( 0,2 \right) ,(\sqrt { 3 } ,-1),(-\sqrt { 3 } ,-1)$

that has the incircle with the implicit equation of

${ y' }^{ 2 }+{ x' }^{ 2 }-1=0$

Once we have those coefficients, we can then apply the affine transform to the implicit equation of the incircle to come up with the implicit equation of the Steiner Inellipse of the other triangle.

The area of the Steiner Inellipse is always $\dfrac { \pi }{ 3\sqrt { 3 } }$ that of the triangle it is inscribed in.

Is using a determinant to work out the implicit equation, as suggested by Calvin Lin's post, useless for this problem, because there are only 3 known points on the Steiner Inellipse? No, it still can be used, if one recognizes that while it takes 5 points to determine a conic, it only takes 3 points if we knew the tangents of 2 of the points. So consider a point which is midpoint between a pair of vertices

$\left( p_{ 12 },{ q }_{ 12 } \right) =\left( \frac { 1 }{ 2 } \left( { x }_{ 1 }+{ x }_{ 2 } \right) ,\frac { 1 }{ 2 } \left( { y }_{ 1 }+{ y }_{ 2 } \right) \right)$

and then consider another next to it on the line through both vertices

$\left( p'_{ 12 },{ q' }_{ 12 } \right) =\left( { \left( \frac { 1 }{ 2 } +\epsilon \right) { x }_{ 1 } }+\left( \frac { 1 }{ 2 } -\epsilon \right) { x }_{ 2 },\left( \frac { 1 }{ 2 } +\epsilon \right) { y }_{ 1 }+\left( \frac { 1 }{ 2 } -\epsilon \right) { y }_{ 2 } \right)$

so that as $\epsilon \rightarrow 0$ the points merge. This defines a tangent. We do the same for another midpoint, so that we now have the 5 points we need to use a determinant. At this point, it's decidedly more convenient to use a computer, because then, after expanding and factorizing it, we can see that we can eliminate one of the factors, which is ${ \epsilon }^{ 2 }$ . Then we let $\epsilon =0$ , and after simplifying the resulting expression, we end up with the implicit equation of the Steiner Inellipse as given above.