Three points lie on the graph of the parabola $y = x^2$. The three points form an equilateral triangle. One of the vertices is at the vertex of the parabola, and the other two points lie on the line $y = k$. What is the value of $k?$
Plotting the points, we are trying to find $k$ such that the three black line segments have the same length.
[asy]
size(8cm);
for (int i=0; i<=6; ++i) {
draw((-6,i)--(6,i), lightgray, EndArrow(5));
draw((i,-6)--(i,36), lightgray, EndArrow(5));
}
draw((-6,0)--(6,0),linewidth(1.25),EndArrow(5));
draw((0,-6)--(0,36),linewidth(1.25),EndArrow(5));
real a(int x) { return x^2; }
draw(graph(a,-3,3),linewidth(1.25));
xaxis("$x$"); yaxis("$y$");
label("$y = x^2$", (6,6^2), E);
dot("$(0,0)$", (0,0), NW);
dot("$(-1,1)$", (-1,1), NW);
dot("$(1,1)$", (1,1), NE);
draw((-3,1)--(3,1)^^(0,0)--(2,4)^^(0,0)--(-2,4));
label("$k$", (6,1), E);
dot((6,1));
draw((6,0)--(2,4), red);
[/asy]
Using the distance formula, each of the black line segments has length $\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} = \sqrt{(2)^2 + (k-0)^2} = \sqrt{4+k^2}.$ Also, the distance is equal to $\sqrt{(1-(-1))^2 + (k-1)^2} = \sqrt{(2)^2 + (k-1)^2} = \sqrt{4+k^2-2k+1} = \sqrt{k^2 - 2k + 5}.$ Solving $\sqrt{4+k^2} = \sqrt{k^2 - 2k + 5},$ we find $k^2 - 2k + 5 = 4 + k^2,$ so $2k = 1,$ and $k = \boxed{\frac{1}{2}}.$
Let's start by finding the coordinates of the vertex of the parabola. The vertex of the parabola $y = x^2$ has the form $(h, k)$, where $h$ and $k$ are the $x$ and $y$ coordinates of the vertex, respectively.
Since the equation is in the form $y = x^2$, we can see that $h = 0$ since the squared term is just $x$. Substituting $h = 0$ into the equation, we find that $k = 0$ as well. Therefore, the vertex of the parabola is $(0, 0)$.
One of the vertices of the equilateral triangle is given as the vertex of the parabola, which is $(0, 0)$. The other two vertices of the triangle lie on the line $y = k$.
Let's consider one of the other vertices of the triangle. The $y$-coordinate of this point is $k$, so let's call it $(a, k)$, where $a$ is the $x$-coordinate of this point. Since the point lies on the graph of the parabola, we have $k = a^2$.
Now we need to find the $a$-value that ensures the two vertices are the same distance from the vertex of the parabola.
The distance between $(0, 0)$ and $(a, k)$ is
\[\sqrt{(a - 0)^2 + (k - 0)^2} = \sqrt{a^2 + k^2}.\]The distance between $(0, 0)$ and one of the other vertices of the equilateral triangle is
\[\sqrt{(0 - a)^2 + (0 - k)^2} = \sqrt{a^2 + k^2}.\]Since the two distances are equal, we must have that $a = a^2$.
Since $a = a^2$, we can subtract $a$ from both sides to get $a^2 - a = 0$. Factoring the left side gives $a(a - 1) = 0$, so $a = 0$ or $a = 1$. If $a = 1$, then $k = a^2 = 1^2 = 1$.
Therefore, the value of $k$ is $\boxed{1}$.