Problem:[br]Given are three parallel lines [math]a, b,[/math] and [math]c[/math]. Construct an equilateral triangle [math]ABC[/math] such that each of its vertices is on one of the lines.[br][br]Solution:[br]For any equilateral triangle a rotation with center one of the vertices and angle [math]60^{\circ} [/math] will map one of the remaining vertices onto the third vertex.[br]Let [math] B[/math] be an arbitrary point on line [math]b[/math]. [br][list][br][*]Drag the slider to the end to rotate line [math]c[/math] on angle [math]60^{\circ} [/math] around point [math]B[/math].[br][/list][br]If vertex [math]C[/math] is on line [math]c[/math], then the third vertex [math]A[/math] of the triangle should be on the image [math]c'[/math] of line [math]c[/math], but it also has to be on the third line [math]a[/math]. Therefore, vertex [math]A[/math] is the intersection point of the lines [math]c'[/math] and [math]a[/math]. This way we find two vertices and determine the side of the equilateral triangle.[br][list][br][*]Click on the Construction button to finish the construction.[br][/list][br]There will be two different solutions corresponding to the two possible directions of the rotation, at [math]60^{\circ} [/math] and [math] -60^{\circ} [/math] degrees. [br][list][br][*]Drag the gray points to change the positions of the lines. [br][*]What do you notice when line [math]b[/math] is above line [math]c[/math] or below line [math]a[/math]?[br][/list][br][br]A geometric construction using this transformation was first described by I. M. Yaglom, in [i]Geometric Transformations[/i] I, MAA, 1962, Chapter 2, Problem 18