I have the following problem :
Show that the symmetrical (ie reflection) of a triangle's vertexes by the opposite side are aligned iff the distance between the orthocenter and the circumcenter is twice the circumradius.
I made a few pictures with GeoGebra to try and find a way to solve that, but it didn't really help.
When two of them ($A'$ and $C'$) are in the same place :
When they are distinct and aligned : 
First draw the lines through $A', B'$ and $C'$ parallel to $BC, CA$ and $AB$. These make a triangle $A''B''C''$ say. See the diagram below.
1) We prove that $A''B''C''$ is an enlargement of $ABC$ scale factor $4$ from the centroid $G$.
It is easy to see that $BYZ$ is an enlargement of $BCA$ scale factor $2$ from $B$. Hence $YZ = 2CA$ and similarly for the other two sides. Also $BC=CY$ etc.
The quadrilateral $YC''XC$ is a parallelogram, so $YC''=XC$. Applying this to every side gives that $A''C''= 4AC$ etc.
Since the diagonals of a parallelogram bisect eachother, $C''C$ intersects $YX$ at its midpoint. And since $C''YX \cong CAB$ by SAS, it follows that $C''C, A''A, B''B$ intersect at the centroid, giving us our original statement that we wanted to prove above.
2) We know that $GH = 4GT$ for any triangle and that $T$ lies on circle $ABC$ (since $OH$ is twice the circumradius and because $OT = TH$). It follows from this and the enlargement that $H$ lies on circle $A''B''C''$.
But now notice that $A'B'C'$ is simply the Simson line of circle $A''B''C''$ from $H$! Hence $A', B'$ and $C'$ are collinear, as required. The converse is simply this last step applied backwards, as the converse of the Simson Theorem is also true.
Hope this helps!