Construct a triangle given A (the angle), m_a (a-median) and
r (the in-radius). This triangle construction problem repeatedly
resurfaces in the group. Here's my attempt for the solution.
If we fix the vertex A, then locus of Ma (the foot of a-median)
is a circle A(m_a).
If we also fix the lines AB, AC, then the in-circle and in-center
"I" are also fixed, and another locus of Ma is a hyperbola whose
vertices are the intersections of AI and the in-circle, and
asymptotes parallel to AB, AC. As the axis of the hyperbola
passes through A, the construction is possible.
Using the coordination so that "I" is at the origin, and AI is
along the y-axis, the two loci are (X = A/2):
x^2 + (y + cosecX)^2 = m_a^2 [circle]
y^2 - x^2 * cotX^2 = r^2 [hyperbola]
Eliminating x, the y-coordinate of Ma may be constructed.
Draw a given angle A, its legs being rays d, d'.
Inscribe a circle with radius r in it, "I" its center,
and D the point of contact with d.
On a line AI mark point M so that AM = m_a.
Construct P1 - the orthogonal proj. of M on d,
E - the orthogonal proj. of E on AI.
On a line MP1, lay P1P2 = IE. On AP2 as diameter, draw
semi-circle $G1$. Draw a circle A(AD). It intersects $G1$
at the point P3.
Draw a circle $G2$ A(AE). Prolong DE and lay the segment
EP4 = P2P3 on it. Draw segment AP4, it will intersect $G2$
at a point P5.
Draw a circle $G3$ A(AM). Draw a line perpendicular to AI at
a distance = P4P5 from "I". It will intersect $G3$ at a sought
point Ma. The line through Ma tangent to incircle intersects
d, d' at the vertices B, C.
The problem has one solution (up to reflection in AI), if:
m_a >= r * (1 + cosec A/2).
Best regards from very hot Israel,