Let $ A,B,C$ be the vertices of wanted triangle, given its incenter $ I,$ the C-excenter $ I_c$ and the foot $ H$ of the C-altitude. Let $V$ be the foot of its C-angle bisector. Since $ (C,V,I,I_a)=-1$ $\Longrightarrow$ circle with diameter $\overline{CV}$ passing through $ H$ and the circle with diameter $ \overline{II_c}$ (which is known) are orthogonal. It remains to construct the cirle $ \mathcal{Q}$ passing through $C,H,V,$ which belows to the orthogonal pencil defined by the cirle $(K)$ with diameter $ II_c$ and whose center lies on $ II_c.$ Using that the power of $ K$ to $ \mathcal{Q}$ equals the square of the radius $ KI,$ then draw line $ KH$ and construct point $ P$ on the ray $ KH$ such that $ KP \cdot KH = KI^2$ $\Longrightarrow$ intersection of the perpendicular bisector of $\overline{PH}$ with the line $ II_c$ gives the center of $ \mathcal {Q}.$ Once the circle $ \mathcal{Q}$ is drawn, we get easily the poins $C,V$ and subsequently the line $ VH$ whose intersections with $ (K)$ are the desired vertices $ A,B.$

Let $D,I,J$ be the foot of the altitude, the incenter and excenter. Consider the line parallel to $AB$ and also tangent to the incircle and let its intersection with $CD$ be $E$. Then it is easy to see that $IE$=$ID$ and also $DJ \parallel IE$. So given, $D,I,J$, we can construct $E$. Then $C$ is simply $DE \cap IJ$. The rest of the construction follows easily.

This problem proves that you can't beat Russians when it comes to making sufficiently elegant yet decently hard problems. Here is my solution. Construction Let $I,J,H$ be the given incenter, excenter, and feet of altitude. Now, we can construct the midpoint $M$ of $IJ$. Then, we can construct a point $Y$ on the ray $MH$ such that $MH.MY=MI^2$. We also construct the circle with diameter $IJ$ and call it $\Gamma$ Now, draw tangents from $Y$ to $\Gamma$ and construct the midline of these two tangents. Let this midline intersect $IJ$ at $X$. Draw circle with centre $X$ and radius $XY$ and let it meet $IJ$ again at $A,A'$ with $A$ farther from $M$ than $A'$. Then, draw a line perpendicular to $AH$ at $H$ and let it meet $\Gamma$ at $B,C$. We claim that triangle $ABC$ is the desired triangle. Proof Notice that $XY^2=XI.XJ$ and so $(A,A';I,J)=-1$ and also that $\angle AYA'=90$ so we must have $A$ to be the requested vertex. Then, $B,C$ are also the requested ones.

Let $I,I_A,A,E$ be the incenter,excenter,vertex,and the foot of $A$-altitude and let $AI\cap BC=A'$. Let $E_1$ be the image of $E$ after inversion in $(II_A)$.We have that $AA'EE'$ is cyclic $\implies$ $\angle A'EA=\angle A'E'A=\frac{\pi}{2}$ and so let the perpendicular bisector of $EE'$ cut $II_A$ in $R$.Now $\odot(R,RE)\cap II_A=\{A',A\}$ so we have $A'$ and $E$ $EA'\cap (II_A)=\{B,C\}$ so we are done.