Given a point P0 in the plane of the triangle A1A2A3. Define As = As-3 for all s ≥ 4. Construct a set of points P1, P2, P3, ... such that Pk+1 is the image of Pk under a rotation center Ak+1 through 120o clockwise for k = 0, 1, 2, ... . Prove that if P1986 = P0, then the triangle A1A2A3 is equilateral.
Solution
The product of three successive rotations about the three vertices of a triangle must be a translation (see below). But that means that P1986 (which is the result of 662 such operations, since 1986 = 3 x 662) can only be P0 if it is the identity, for a translation by a non-zero amount would keep moving the point further away. It is now easy to show that it can only be the identity if the triangle is equilateral. Take a circle center A1, radius A1A2 and take P on the circle so that a 120o clockwise rotation about A1 brings P to A2. Take a circle center A3, radius A3A2 and take Q on the circle so that a 120o clockwise rotation about A3 takes A2 to Q. Then successive 120o clockwise rotations about A1, A2, A3 take P to Q. So if these three are equivalent to the identity we must have P = Q. Hence ∠A1A2A3 = ∠A1A2P + ∠A3A2P = 30o + 30o = 60o. Also A2P = 2A1A2cos 30o and = 2A2A3cos 30o. Hence A1A2 = A2A3. So A1A2A2 is equilateral. Note in passing that it is not sufficient for the triangle to be equilateral. We also have to take the rotations in the right order. If we move around the vertices the opposite way, then we get a net translation.
It remains to show that the three rotations give a translation. Define rectangular coordinates (x, y) by taking A1 to be the origin and A2 to be (a, b). Let A3 be (c, d). A clockwise rotation through 120 degrees about the origin takes (x, y) to (-x/2 + y√3/2, -x√3/2 - y/2). A clockwise rotation through 120 degrees about some other point (e, f) is obtained by subtracting (e, f) to get (x - e, y - f), the coordinates relative to (e, f), then rotating, then adding (e, f) to get the coordinates relative to (0, 0). Thus after the three rotations we will end up with a linear combination of x's, y's, a's, b's, c's and d's for each coordinate. But the linear combination of x's and y's must be just x for the x-coordinate and y for the y-coordinate, since three successive 120 degree rotations about the same point is the identity. Hence we end up with simply (x + constant, y + constant), in other words, a translation.
[Of course, there is nothing to stop you actually carrying out the computation. It makes things slightly easier to take the triangle to be (0, 0), (1, 0), (a, b). The net result turns out to be (x, y) goes to (x + 3a/2 - b√3/2, y - √3 + a√3/2 + 3b/2). For this to be the identity requires a = 1/2, b = √3/2. So the third vertex must make the triangle equilateral (and it must be on the correct side of the line joining the other two). This approach avoids the need for the argument in the first paragraph above, but is rather harder work.]
Solutions are also available in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
25 Oct 1998