Looking at
https://www.cut-the-knot.org/triangle/index.shtmlI was asking myself how to find the vertices of ABC, with
straightedge and compass, knowing the positions of O,H,I.
(Incidentally, I underline that (A,a,R) does not fix
a triangle, since the sine law holds.)
There are some remarkable facts.
By calling N the midpoint of OH we have that
1) IN = R/2 - r due to Feuerbach's theorem
2) OI^2 = 2R(R/2-r) due to Euler's theorem
so we know (R,r) and can easily draw the circumcircle, the incircle,
the 9-points circle, the polar circle, the ortic axis and so on.
Moreover, due to Poncelet's lemma, if we take a point U on
the circumcircle, draw the tangents to the incircle and intersect
them with the circumcircle, we determine a chord VW that is
also tangent to the incircle.
(Incidentally, I've noted that one of the proof of the Butterfly
theorem on this site proves Poncelet's lemma too.)
For all this UVW triangles, I,O,R and r are fixed, so, for example,
when P travels on the circumcircle, the nine-point-center N(UVW)
travels on a circumference with center I and radius (R/2-r).
However, the map that brings U into N(UVW) is not a "simple"
map between circumferences: it is 3-to-1, and its inversion
is a cubic problem, generally not solvable with straightedge
and compass.
Otherwise, we can build G (centroid), Na (Nagel point)
and F (tangency point of the incircle and the 9-point-circle)
and try to intersect the Feuerbach hyperbola (we know
its center F and three points on it: I,H,Na) with the circumcircle,
but this is a fourth (or third?!) degree problem.
This considerations raise the question of the possibility,
or, more probably, the impossibility of the construction.
Using trilinears, one may say that it is hard to determine
unsymmetric things like the coordinates of the vertices A<1,0,0>
B<0,1,0> C<0,0,1> dealing with proper (symmetric) centers of the
triangle (there are also a good number of reformulations in terms
of trigonometric functions of the angles in A,B,C, or symmetric
functions of the lengths of the sides).
So things become even more hard, and I ask you:
1) Does the (I,O,H) problem admit a simpler reverse construction
(lower degree) than mine?
2) WHEN, given three proper centers of ABC, is it possible
to find (A,B,C) with straightedge and ruler?