one is tangent to the three sides at their midpoints

the other goes through the vertices, the tangents in these vertices being parallel to the opposite sides.

First note that choosing any two points a and b on sides BC and AC,
there is a conic section tangent to BC in a, tangent to AC in b and tangent to AB in some point c.

The Brianchon theorem gives an easy construction of c :
abc is a Cevian triangle, that is Aa Bb and Cc are concurrent, considering hexagon AbCaBc.

Even if this is theoretically enough to draw the conic section (we even have 6 points), in practice we get
intersection points rejected at infinity, for in our case (a,b,c midpoints of segments) ab is parallel to AB,
otherwise we get collapsing points.

However, we easily get from Brianchon a new tangent and its contact point :

Any tangent is obtained by choosing any point M on BC,
then constructing the corresponding point N on AB :
ac and AM intersect in J, CJ intersects AB in N. (hexagon ACaMNc)

The contact point on this tangent is obtained as in the first step :
lines AM, Na and CT are concurrent. (hexagon ACaMTN)

We have then enough distinct points to draw the conic section, for instance from
points c,c,b,a,T (line cc being the tangent AB).

What is remarkable here is that when a and b are the midpoints of BC and AC,
then c is also the midpoint of AB.
A first proof directly comes from the previous construction : the cevian lines are the medians.

But we could even conclude "at once" with the help of an affine transform
of triangle ABC into an equilateral triangle, the ellipse being transformed into the incircle.

The affine transform holds the inline property, ratio of inline segments (midpoints) and tangencies.
It transforms a conic section into a conic section of same
"gender", here the incircle into an ellipse.

We deduce immediately that the center of this ellipse is the centroid (from inline points on the medians).

A dilation from center G and ratio -2 transforms this ellipse
into an ellipse tangent in A,B,C, midpoints of sides of triangle of which ABC is the midpoint triangle.
The tangents in A,B,C are then parallel to the sides of ABC.

The interesting property here is

a,b,c,a',b',c' are on a same ellipse

And even more : this ellipse is similar to the Steiner ellipses.

The dilation center being of course the centroid G.

A,B,C and m can be moved, m defines the ratio Bm/BC

The property is obvious in an equilateral triangle : the 6 points are concyclic,
on a circle centered in G.

The radius of this circle r is easily obtained for the affine transform holds the ratio of areas :

In the equilateral triangle, our triangle abc has area 1/7 of ABC.

An equilateral triangle inscribed in the incircle of ABC is the midpoints triangle, and has an area 1/4 of ABC.

The circumcircle to abc and the incircle of ABC have then radii in ratio_{ }√(1/7)/(1/4) = 2/√7

The ellipse abca'b'c' is then a dilation of the inner Steiner ellipse in that same ratio.

The outer Steiner ellipse is a dilation of the inner one in ratio 2 ( or -2), hence a dilation of ellipse abca'b'c' in ratio_{ }√7

In the equilateral triangle, Gm² = GM² + Mm² = s²/12 + (s/6)²,
that is GM = s/3.

We then get another ellipse going through the 6 points dividing the sides in 3,
similar to the Steiner ellipse in ratio _{ }2/√3

Remains to get the dilation ratio as a function of k, but I'm too lazzy...

An interesting case is whith a ratio 1.

That is when abca'b'c' are on the Steiner ellipse.

Let's transform ABC into an equilateral triangle by the previously discussed affine transform.

In this equilateral triangle Ac'GcB are on a circle for any value of k :

Angle cAB = aBC = α gives in triangle AcB the angle c = π - α - (π/3 - α) = 2π/3

Point c is then on the circle centered in G', tangent in A and B to sides AC and BC.

As we want the ellipse to be the Steiner ellipse, that is in the equilateral triangle
the circle abca'b'c' to be the incircle,
point c is the intersect point of the incircle and of this circle tangent in A and B.

Let θ = arc cG. Arc cB is then π/3 - θ, hence
angle BAc is (π/3 - θ)/2.

Angle AmB is then π - π/3 - (π/3 - θ)/2 = π/2 + θ/2
and we get k = Bm/BA = sin(π/6 - θ/2) / sin(π/2 + θ/2) =_{ }1/2 - √3/2 tan(θ/2)

Line cc' is the radical axis of these circles, hence intersects the common tangent MB in its midpoint i,
that is at 1/4 of BC.
Hence Nh = NC/4, then G'h = (7/8)G'G and θ = arccos(7/8)

We get then tan(θ/2) =_{ }1/√15 and

k = _{ }(5 - √5)/10 ≈ 0.2763932... |

Note : the area of abc is then the same as the midpoint triangle, that is 1/4 of ABC.

The ratio on the sides is then the one found in "non trivial" shares of ABC into 4 triangles of equal area.

We got there, from a direct method, BC/Bm =_{ }(5 + √5)/2,
the inverse of which is Bm/BC =_{ }2/(5 + √5) = 2(5 - √5)/(5² - 5) = _{ }(5 - √5)/10 = k

The vertices of the triangle with 1/4 ABC area are on the Steiner ellipse |