It is known that the tangents to a hyperbola determine on the
asymptotes segments whose product is constant, and that the point of
tangencia T is the midpoint of the segment that the asymptotes
determine in the tangent.

Therefore the segments with ends in the sides to and b, that bisect to
the triangle, are tangent in their midpoints to a hyperbola that it
has to the sides to and b like asymptotes.

Calling X, Y and Z to medium the midpoints of the medians m_a, m_b and
m_c, such hyperbola is tangent to m_a in X and m_b in and.

The problem has 3 solutions if point R is inside tricuspoid XYZ
composed by three arcs of hyperbola that they have to each pair of
sides like asymptotes, 2 solutions on the own tricuspoid, excepted
points X, and and Z, and 1 in all the rest of the plane.

That is an affine problem, the ratio of the areas is invariant under
affine transformations. Therefore, it can be solved in any triangle
with full generallity. In an equilateral triangle, by example, or in
the triangle with vertices A(0, 0), B(1, 0) and C(0, 1). In any case,
it is easy, by
integration, to calculate the area of that tricuspoid.

Its area small is compared with the one of the triangle, exactly
(3*ln(2)-2)/4=0.01986... of the one of the triangle, less of a 2%.

P.S.: This is an automatic translate from Spanish. Let be indulgent
with the grammar ...

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Saludos,