Points on a plane and triangle

Question

Given any three non-collinear points on a plane. Can there be an equilateral triangle such that the points lie on the perimeter of the equilateral triangle ?

Answer 1

Of course, just draw the line AB, draw a CX line at 60 degree from line AB, and finally close the triangle with a YZ line at 60 degree from both AB and CX.

Answer 2

Let $X$, $Y$ and $Z$ be our points.

Let $\hat{XY}=\hat{XZ}=240^{\circ}$ and let $C\in\smile{XY}$, $B\in\smile{XZ}$, such that $X\in BC$.

Now let $CY\cap BZ=\{A\}$.

Thus, since $\measuredangle XBZ=\measuredangle XCY=60^{\circ}$, we see that $\Delta ABC$ is an equilateral triangle.

Answer 3

Think in reverse.

Take an arbitrary equilateral triangle.

From an outer point A', draw lines through two vertices, that form the same angle as CAB.

Then draw a third line that forms the angles ABC and BCA with the former, and translate it to the third vertex, giving the intersections B' and C'.

Now you have a triangle which is similar to the desired triangle and circumscribed to an equilateral triangle. Apply the similarity transform that maps A'B'C' to ABC, and you have it.