Let's construct a triangle, given one of its vertices, the orthocenter, and the centroid.
Solution. Let the given vertex of the triangle to be constructed be $A$, the other two vertices $B$ and $C$, the orthocenter, centroid, and circumcenter of the triangle be $M, S$, and $K$, respectively, and the midpoint of side $BC$ be $F$. In the following, we assume that $A \neq S$; otherwise, the problem is clearly unsolvable.

![](https://cdn.mathpix.com/cropped/2024_05_02_d547a6d3c85650361dfeg-1.jpg?height=477&width=572&top_left_y=268&top_left_x=771)

We know that the centroid divides the medians in a $2:1$ ratio, so the point $S$ is the point on segment $AF$ that is closer to $F$ and divides it in a $2:1$ ratio. Furthermore, $S$ also divides the segment $MK$ in a $2:1$ ratio according to the properties of the Euler line. Based on this, the points $F$ and $K$ can be easily constructed. Knowing these, the vertices $B$ and $C$ can also be determined because they lie on the circle $k$ with center $K$ and radius $KA$, and also on the line $e$ perpendicular to $AM$ and passing through $F$. These are given by the two intersection points of $k$ and $e$.

The triangle $ABC$ thus constructed clearly has $k$ as its circumcircle. From the construction, it follows that the triangles $SFK$ and $SAM$ are similar by central similarity, with the center of similarity at point $S$ and the ratio being $-2$. Therefore, $FK$ is parallel to $AM$, so the line $e$ is also perpendicular to $FK$, which implies that $KF$ is the perpendicular bisector of the chord $BC$ of circle $k$, and thus $F$ is the midpoint of side $BC$. Since $S$ divides $AF$ in a $2:1$ ratio, it coincides with the centroid of triangle $ABC$, and since it also divides $KM$ in a $2:1$ ratio, it follows from the properties of the Euler line that $M$ is indeed the orthocenter of the constructed triangle $ABC$.

The discussion of the problem is more complex than the construction. If the circle $k$ and the line $e$ can be constructed uniquely in the steps described above, then there is a solution if and only if the point $F$ lies inside the circle $k$, which is true if $KA > KF = \frac{AM}{2}$. The length of segment $KA$ can be expressed in terms of the lengths of segments $AM$, $MS$, and $AS$, because $AS$ and $KA$ are medians of triangles $AMN$ and $ASN$, respectively, where $N$ is the reflection of point $M$ over $S$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_d547a6d3c85650361dfeg-1.jpg?height=325&width=409&top_left_y=1480&top_left_x=847)

Figure 2

The formula for the length of a median in a triangle, valid even in degenerate cases (with the usual notation $4s_c^2 = 2(a^2 + b^2) - c^2$),

$$
4AS^2 = 2AM^2 + 2AN^2 - MN^2 \quad \text{and} \quad 4KA^2 = 2AS^2 + 2AN^2 - SN^2
$$

Since $SN = SM$ and $MN = 2SN$, a short calculation yields

$$
4KA^2 = 6AS^2 + 3MS^2 - 2AM^2
$$

Thus, the condition $KA > \frac{AM}{2}$ is equivalent to the inequality $2AS^2 + MS^2 > AM^2$ for the given points $A$, $M$, and $S$.

If points $A$ and $M$ coincide, then every step of the construction can be carried out, but the line $e$ is not uniquely determined. In this case, $K$ and $F$ also coincide, and any line passing through $K$ and not containing $A$ can be chosen as $e$. Thus, we obtain infinitely many different solutions (each of the constructed triangles has a right angle at $A$).

Finally, if any step of the construction cannot be carried out, then there is no solution. This occurs in two main cases.

1. The circle $k$ cannot be constructed. This happens if $K$ coincides with $A$, or if the point $S$ lies on segment $MA$ and divides it in a $2:1$ ratio.
2. The circle $k$ can be constructed, and $k$ and $e$ intersect, but one of the intersection points of $k$ and $e$ is the point $A$. This happens precisely when the angle $KFS$ is a right angle, which is true if and only if the angle $MAS$ is a right angle.