When a Triangle is Equilateral?

The applet below is an attempt to illustrate a criterion for a triangle to be equilateral expressed with complex numbers. (We used this criterion on two occasions: in a proof of Napoleon's Theorem and in the proof of Morley's theorem by A. Connes.)

In the plane of ΔABC choose a Cartesian coordinate system, such that it becomes possible to identify points in the plane with complex numbers. (In the applet, the origin of the system is at the point O, while the axes are not shown.) In particular, we shall consider the vertices A, B, C of ΔABC as such. We also assume that ΔABC is oriented positively. Let j denote the rotation through 120° in the positive (counterclockwise) direction. This means that j3 = 1, or, since j ≠ 1,

(1)

1 + j + j2 = 0

Then ΔABC is equilateral iff

(2)

A + jB + j2C = 0

The proof is very simple. The triangle is equilateral iff each side could be obtained from another side by a rotation through 60° around their common vertex. A rotation through 120° in the opposite direction, brings one side in the direction exactly opposite to the other making their sum equal to 0. In fact, either of the assertions for just one pair of the sides is sufficient to claim regularity of the triangle. In particular, A - C is the complex number corresponding to the side AC, while B - C corresponds to the side BC. j(B - C) is the complex number that "looks" in the direction opposite to that of A - C:

A - C + j(B - C) = 0

or

(3)

A + jB - (1 + j)C = 0.

From (1), 1 + j = -j2, which reduces (3) to (2).

The applet illustrates the geometric interpretation of (2). B1 is the rotation of B through 120° in the positive direction. C1 is the rotation of C through 120° in the negative direction - exactly what j2 is. The sum of two complex numbers is determined by the parallelogram rule. According to (2), ΔABC is equilateral iff the sum of B1 and C1 is exactly -A.


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