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 $k?$

Answer 1

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 $k = \boxed{\frac{1}{2}}.$