We begin by proving the claims made by Blue in the edit to the question.
Let $I$ be the incentre of $ABC$, and let $R$ be its circumradius. Since $O$ and $H$ lie inside $ABC$, the triangle must be acute.
The incircle is divided into three arcs by its points of contact with the sides of $ABC$. At least one of these arcs, say the one nearest vertex $A$, contains neither $O$ nor $H$. Thus when the rays $AO$ and $AH$ meet the incircle at $O$ and $H$, respectively, each of these rays is intersecting the incircle for the second time. Moreover, as in any triangle, $AI$ bisects angle $OAH$. It follows that the points $O$ and $H$ are symmetric about $AI$. In particular, $AH = AO = R$.
In any triangle, $\overrightarrow{AH} = 2\overrightarrow{OA'}$, where $A'$ is the midpoint of $BC$. Hence $OA' = R/2$. It follows from this that $\angle BOC = 120^{\circ}$, hence that $\angle BAC = 60^{\circ}$.
If we introduce $O'$ as in the figure (the reflection of $O$ through $A'$), then $O$, $B$ and $C$ belong to the circle with radius $R$ centred at $O'$. Since $\overrightarrow{AH} = \overrightarrow{OO'}$, the quadrilateral $OAHO'$ is a rhombus with side $R$. Consequently, $H$ also belongs to circle $BOC$.
If $J$ is the point halfway along arc $OH$ on circle $BOC$, then $BJ$ bisects $\angle OBH$, hence $J$ lies on $BI$. Similarly, $J$ lies on $CI$. Hence $J = I$, and $I$ lies on circle $BOC$. Since $AI$ bisects $\angle BAC$, it meets the circumcircle of $ABC$ again at $O'$, which is midway between $B$ and $C$.
Conversely, we carry out a construction corresponding to the above requirements. Start with a circle centred at $O$ with radius $1$. Mark two points $B$ and $C$ on the circle so that $\angle BOC = 120^{\circ}$. Let $O'$ be the reflection of $O$ through $BC$. Then $O'$ is on the circle. Now let $I$ be any point on circle $BOC$, on the same side of $BC$ as $O$. (We will specify $I$ further below.) Let $O'I$ cut $BO'C$ again at $A$. Let $H$ be the reflection of $O$ through $O'I$. Then reversing the arguments above, we find that $H$ is the orthocentre and $I$ the incentre of triangle $ABC$, and that $IH = IO$.
The only question that remains is how to choose $I$ on circle $BOC$ so that $IO$ is equal to the inradius of $ABC$. If we let $x$ be the inradius, then $x$ is the distance from $I$ to line $BC$. We also have $IO^2 = (1/2 - x)^2 + 1 - (x+1/2)^2 = 1-2x$. The condition $OI = x$ is equivalent to $x^2 = 1 - 2x$, or $x = \sqrt{2} - 1$.
Thus the construction can be completed by letting $I$ be a point of intersection of circle $BOC$ with a circle centred at $O$ with radius $\sqrt{2}-1$.
I'm not sure how to motivate this last step geometrically.
Summary of my construction Given two points $O$ and $O'$, write $R = OO'$. Construct the circles $K$ and $K'$ of radius $R$ centred at $O$ and $O'$, respectively. Let $B$ and $C$ be the points of intersection. Let $I$ be a point of intersection of $K'$ with the circle of radius $(\sqrt{2}-1)R$ centred at $O$. Then let $A$ be the point of intersection of $O'I$ with $K$.
Alternative construction (using $IA = 2IO$, proved by dxiv below) Instead of constructing $I$, construct $A$ directly by intersecting $K$ with the circle of radius $(2\sqrt{2} - 1)R$ centred at $O'$.
Summary of dxiv's construction Construct a triangle $AIO$ with $IO= r$, $IA = 2r$, $OA = (\sqrt{2}+1)r$. Let $K$ be the circle centred at $O$ passing through $A$. Construct angles of $30^{\circ}$ on either side of $AI$. Let $B$ and $C$ be the intersections with $K$ of the outer sides of these angles.
It can't be done.
Suppose you start with the triangle $(0,0),(1,0),(0,1)$. It can be proven by induction that any point you can mark has rational coordinates (henceforth called rational points) and any line you can draw can be written in the form $ax+by=c$ with $a,b,c$ rational (call such a line rational line). Since the target incenter has irrational coordinates, it's impossible to mark.
To elucidate on the induction, suppose after $n$ steps we have only rational points and rational lines drawn. At the $n+1$-th step, we can do one of these three things:
- Mark the intersection of two rational lines. Since this amounts to solving a linear system with rational coefficients, the resulting point will have rational coordinates.
- Draw the line through two rational points. This amounts to solving the system $y_1=mx_1+c,y_2=mx_2+c$ for $m,c$ given $y_1,y_2,x_1,x_2$ rational (there's a special case when $x_1=x_2$, but that's easy to handle), but this clearly has a rational solution.
- Draw a line perpendicular to $y=Mx+c$ through a rational point $(a,b)$, where $M,c$ are rational. Now this line will of the form $y=-x/M+c_1$, where $-1/M$ is clearly rational, and $c_1$ can be determined by solving the equation $b=-a/M+c_1$, which will again be rational. Again, there's an edge case to consider when $M=0$, but that can be dealt with similarly.
Note that the key idea used in this proof is that given an uniquely solvable linear system of equations with rational coefficients, the solution will be rational. This can be easily seen, from Cramer's rule, for example.
Best Answer
It seems that although this problem is solvable (i.e. constructable), there is no simple construction.
The paper Wernickâs List: A Final Update, which is a survey whether the problems in Wernick's list are solvable or not explains:
In Example 2, this is explained for Problem 108:
So in theory, it is possible to construct Problem 82, but this would involve using a ruler and compass for arithmetic operations and square radicals, which seems out of the spirit of synthetic constructions.
They don't rule out the possibility of an elegant construction, but based on what they say I'd advise you to not spend a lot of time looking for one because even automated geometry solvers have come up empty handed so far.
There is a compendium of automatically generated constructions for Wernick's list at http://poincare.matf.bg.ac.rs/~vesnap/animations/compendiums.html. The construction for $O, T_a, I$ is notably non-existent.