## Task 14/81

In a plane, let equilateral triangles $D_{i}$ with side lengths $a_{i}=2 i-1, i=1 ; 2 ; 3 ; \ldots$ be arranged along a line $g$ such that the "right" vertex of triangle $D_{k}$ coincides with the "left" vertex of triangle $D_{k+1}$ and that the third vertices all lie in the same half-plane generated by $g$. Determine the curve on which the third vertices lie!
If one places a right-angled Cartesian coordinate system in the plane such that the abscissa axis (x-axis) coincides with the line $g$ and the ordinate axis (y-axis) passes through the left vertex of the triangle $D 1$, then the third triangle points $P_{k}$ have the coordinates

$$
P_{k}\left[x_{k} ; y_{k}\right]=P_{k}\left[k(k-1)+\frac{1}{2} ; \frac{1}{2} \sqrt{3}(2 k-1)\right]
$$

Obviously, there is a relationship between the abscissas and the ordinates of the form $y^{2}=f(x)$. A corresponding approach

$$
y^{2}=\left[\frac{1}{2} \sqrt{3}(2 k-1)\right]^{2}=3\left(k^{2}-k+\frac{1}{4}\right)=f\left(k^{2}-k+\frac{1}{4}\right)
$$

leads to the equation $y^{2}=3\left(x-\frac{1}{4}\right)$, which is the equation of a parabola with focus coordinates $(1 ; 0)$, whose axis coincides with the abscissa axis.