Wernick's list problem number 69: We want to find, with straightedge and compass, the vertices of triangle $\triangle ABC$ but we're only given:
- its incenter, $I$
- its circumcenter, $O$
- the midpoint of side $a$, $M_a$
What I've done:
draw half line $OM_a$, $r$
draw $s$ perpendicular to $r$ passing through $M_a$ (this line contains side $BC$)
draw $t$:a parallel to $r$ passing through $I$
$Z = t \cap s$
draw circle $c$ centered at $I$ passing thorugh $Z$ (this is the incircle of $\triangle ABC$)
$Q = c \cap t \neq Z $
reflect $Z$ at the point $M_a$ to find $T$.
draw line $QT$ ($A$ is on this line).
I don't know how to finish. I suspect we can draw the radius of the circumcircle with Steiner porisms and that would end the problem.
You can indeed find the circumcircle:
draw $IP \perp IO$ such that $P$ is in the incircle already drawn (there are two possible points for $P$ but it doesn't matter which one you pick).
let line $OP$ meet the circle centered at $P$ passing through $I$ at point $X$ ($X$ the point most distant to $O$).
The circle centered at $O$ with radius $OX$ is the circumcircle of $\triangle ABC$.
This is a result that is based on the distance $OI^2 = R^2 -2rR$.
From this, we easily get $\triangle ABC$.