Given the centroid and orthocenter of an isosceles triangle. What is the geometric locus of the endpoints of the base if the third vertex moves along the axis of symmetry?
I. Solution. We assume that the centroid $S$ and the orthocenter $M$ are different, because otherwise any line through $S$ could be the axis of symmetry, and any equilateral triangle $ABC$ with $S$ as its center would satisfy the problem's requirements, so the geometric locus would be the entire plane, except for $S$.

Since $S$ and $M$ lie on the axis of symmetry, $SM$ itself is the axis of symmetry, and the vertex $C$ moves along this line. The base is perpendicular to this and is half the distance from $S$ as $C$ is. We choose the origin of the Cartesian coordinate system to be $S$, the $Y$-axis to be the line $SM$, the ordinate of $M$ to be 1, and the coordinates of $A$ and $B$ to be $(x, y)$ and $(-x, y)$, respectively. Thus, the coordinates of $C$ are $(0, -2y)$.

The side $CA$ and the altitude $BM$ are perpendicular, which is true if and only if the product of their slopes is -1, so if $x \neq 0$, and

$$
\frac{y-(-2y)}{x} \cdot \frac{1-y}{x} = \frac{3y(1-y)}{x^2} = -1
$$

Since $x$ is not 0, we can multiply by its square and rearrange the equation. The geometric locus thus consists of points whose coordinates satisfy

$$
3y^2 - 3y - x^2 = 0, \quad x \neq 0
$$

This equation can also be written as:

$$
3\left(y - \frac{1}{2}\right)^2 - x^2 = \frac{3}{4}, \quad -\frac{x^2}{\frac{3}{4}} + \frac{\left(y - \frac{1}{2}\right)^2}{\frac{1}{4}} = 1
$$

Thus, the geometric locus of $A$ and $B$ is a hyperbola, excluding the points on the $Y$-axis, whose center is the point $(0, 1/2)$, which is the midpoint of the segment $SM$, the major axis is the $Y$-axis, i.e., the line $SM$, the semi-major and semi-minor axes are $(a =) 1/2$ and $(b =) \sqrt{3}/2$, respectively, so its vertices are $S$ and $M$ (which do not belong to the geometric locus), the distance of the foci from the center is 1 (i.e., the square root of $a^2 + b^2$), and the asymptotes form a $60^\circ$ angle with the major axis.

However, not every position of the point $C$ moving along the line $SM$ corresponds to a triangle. From (1), $x^2 = 3y(y-1)$ is negative if $0 < y < 1$, positive if $y > 1$, or $y < 0$. Therefore, the geometric locus is the part of the hyperbola where $y > 1$ or $y < 0$.

We can use the construction of two pairs of points $A, B$ in one stroke of the compass for the positions of $C$ on the line $KM$. (If $KS < KC < KM$, we get one pair of $A, B$ for one position of $C$, if $KC < KS$, we get none. If $KC = KS$, a degenerate triangle is formed, and if $KC = KM$, one position of $C$ gives a degenerate triangle.) Again, it turns out that the points $C$ on the segment $SE$ obtained by extending $SK$ four times do not form a triangle. By multiplying the pairs of points, a hyperbola appears to form, whose vertices are $S$ and $M$. If our conjecture is correct, the line connecting the midpoint $G$ of the segment $SM$ to a distant point $B^*$ on the diagram should be close to one of the asymptotes, and the intersection point $H$ of this line with the perpendicular to $SM$ at $S$ (the tangent at the vertex) should give a point close to one of the foci when rotated around $G$ by $2^2$. The rotation gives a point close to $K$, so we try to prove that the geometric locus is the hyperbola whose vertices are $M$ and $S$, and one of the foci is $K$ (thus the other focus is the reflection of $K$ over $G$, $F_2$).

Let $C$ be on the extension of $SM$ beyond $M$ (so $D$ is on the extension of $SK$ beyond $K$), and let $KC = KA = r$, and $SM = 2a$, so $F_2M = SK = a$.

$$
F_2A^2 = F_2D^2 + DA^2 = F_2D^2 + KA^2 - KD^2 = KA^2 + (F_2D + KD)F_2K
$$

Here, $F_2D = F_2S + SD$, and $KD = SD - SK$, so

$$
F_2D + KD = 2a + 2SD = 2a + CS = 2a + (r - a) = r + a \\
F_2A^2 = r^2 + (r + a)4a = (r + 2a)^2 \\
F_2A = r + 2a, \quad \text{or} \quad F_2A - KA = 2a = SM
$$

which is what we wanted to prove.

After minor intermediate differences, we arrive at the same result if $C$ is on the segment $SM$, or if $C$ is on the extension of $SE$ beyond $E$, then $KA - F_2A = 2a$.

Similarly, it can be shown that if $F_2K = 4a$, and $F_2A - KA = 2a$, or $KA - F_2A = 2a$, but $A$ is not on $F_2K$, and the point $C$ is marked on the line $KF_2$ such that it is three times the distance from the projection of $A$ on $D$ as $S$ (the first quarter point of the segment $KF_2$), and on the same side, and $B$ is the reflection of $A$ over $D$, then $KC = KA$, so $K$ is the circumcenter of the isosceles triangle $ABC$, $S$ is its centroid, and thus $M$ is its orthocenter; so $A$ belongs to the geometric locus.

László Márki (Budapest, Fazekas M. Gyak. G. IV. o. t.)[^1]

[^0]:    ${ }^{1}$ It is true for every triangle that $M, S$, and $K$ lie on the same line, the so-called Euler line of the triangle, and $SK = SM / 2$. See, for example, Gallai T.-Hódi E.-Szabó P.-Tolnai J.: Mathematics for the general secondary school, 12th edition, Tankönyvkiadó, Budapest, 1962, p. 163.

[^1]:    ${ }^{2}$ Indeed, the center of the hyperbola is the circle passing through the foci and the intersection points of the tangents at the vertices and the asymptotes.