There is an engineering question that occurred to me several years ago, but I could not solve it
Assuming $H$ is the point of intersection of the heights, $I$ is the point of intersection of the bisectors, and $G$ is the point of intersection of the averages in triangle $∆ABC$, and the points $H$, $I$, and $G$ are only data, is it possible that by relying on them we can create the vertices $A$, $B$, and $C$ using the construction of the ruler and compass?
Please do not ask me to give my attempt to solve this question because it really does not show any insight or thread to reach the answer. I will be happy if someone can answer this question in the comments.
Edit: I chose these three centers because they do not lie on a single line; Because I know that there would be no hope in the general case of encrypting a triangle uniquely with three collinear centers. Yes, George Cantor's theory states that there is a correspondence between a line and a plane, but this opposition is not connected on any field However, the centers of the triangle follow the movement of the vertices in continuous paths, which prevents the possibility of a solution in cases such as taking the centers from Euler’s line.
We have G and H, the mid point N of GH is the center of the nine point circle, let's denote it's radius as $r_n$, so:
$r_n= \frac {GH}2\space\space\space\space\space\space\space\space\space(1)$
Also we have I , let's denote $IG=d$, due to Euler's theorem we have:
$d^2=R(R-2r)\space\space\space\space\space\space\space\space\space(2)$
where R and r are the radii of circumcircle and incircle respectively.
Now we use this fact that nine point circle and incircle are tangent at point M and I, N and M are colinear , hence we have:
$IN=r-r_n\space\space\space\space\space\space\space\space\space(3)$
In equations (1), (2) and (3) d and IN are known so we can find :
$r_n=\frac 14 \frac{d^2}{IN}$
In this way $R=2r_n$ and r can also be found. Now we draw three circles and use their propertis to draw the triangle. It must not be hard.
Update, how to costruct the triangle:
We use this fact that the radii of circles passing the vertexes of triangle and orthocenter have equal radii to that of circumcircle> the centers of these circles are the vertexes of triangle OPU which is equal to triangle ABC. As can be seen in figure OP is perpendicular bisector of AH. Circle pasiins vertexes A and B also passes the poin M(the tangency of nine point circle and incircle) so the center os this circle is on perpendicular bisector of HM.So the cnter of the ciecle ic the intersection of a circle center at H with radius R, which is point O. OP is perpendicular bisector of AH at T.So T is intersection of nine point circle and a circle on diameter OH. T is mid point of OP, so location of P is clear.G and E(which is the mid point of BC) are the reflects of H and T about N and BC is parallel with OP and also B and C are where this line intersect the circle centered at O and P respectively. In this way vertexes B and C can be found.