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Edward writes,

I can't get started with this question from Step III 2006:

Show that the distinct complex numbers a, b and c represent the vertices of an equilateral triangle if and only if


I'm not normally that bad with complex numbers but I always find these geometry based questions impossible! If I start defining a, b and c in polar form I think I just complicate matters. It might have something to do with cube roots of unity only I don't see any cubes coming in anywhere. Any advice on how to proceed with this question, or any general advice about this kind of question would be gratefully received!

When I'm asked a question like this -- to show something is true -- I suppose it to be true, and then think of some interesting consequences.  In this case, an interesting consequence is that some complex offset, x, can be added to each of a, b, and c, and the result will still be true. The expression


indeed simplifies to


The "if" direction: a, b and c are vertices of an equilateral triangle if a�+b�+c�-ab-bc-ca=0

This led me to try letting x=-a, resulting in the fact that


is equivalent to


Then, I let B=b-a and C=c-a to get


Then I solved this for C,

C=(1/2 � i sqrt(3)/2)B,

or, more elegantly,

C=-ωB or C=-ω�B,

where ω=(-1/2 + i sqrt(3)/2) is that cube root of unity which you suspected was involved in this puzzle!  This shows that B and C, which represent in complex numbers the length and orientation of two consecutive sides, have the same magnitude, and their arguments are offset from one another by 60 degrees.

To show this is true for sides A and B (i.e. that sides A and B also meet at a 60 degree angle), and for sides A and C, all we need do is start over with x=-c and x=-b, respectively, or, alternatively, observe the statement is symmetrical with respect to a, b, and c, and thus true regardless of which two sides we used.

The "only if" direction: if a, b and c are vertices of an equilateral triangle then a�+b�+c�-ab-bc-ca=0

Let a and b be arbitrary.  Then c is fixed by virtue of being the third vertex of the equilateral triangle as

c = a-ω(b-a)  or  c = a-ω�(b-a)

Since ω is a cube root of unity, 1+ω+ω�=0.  It follows, then, that


Since (c-a) = -ω(b-a), we can substitute (c-a)� in place of ω�(b-a)�, and -(c-a) in place of ω(b-a), giving


which we saw earlier is equivalent to


which completes the proof when c = a-ω(b-a).  To prove it when c = a-ω�(b-a), simply use ω� everywhere in place of ω, and the result follows.

More about this puzzle

It was pointed out by Geoff a.k.a. unkseven that a2+b2+c2-ab-bc-ca can be factorized as


This struck me as surprising, so I checked the result by multiplying everything out, treating it as a polynomial in ω, and then simplifying the result using the identities ωnn-3 and ω+ω�=-1.

bcω4 + (b�+c�)ω³ + (ab+ac+bc)ω� + (ab+ac)ω + a�
(ab+ac+bc)ω� + (ab+ac+bc)ω + a�+b�+c�


Even more amazing, I think, is that a, b, and c form an equilateral triangle iff a+bω�+cω=0 or a+bω+cω�=0.  To understand this, I visualized a,b,c as the original equilateral triangle, and then aω,bω,cω as an image of the original triangle rotated 120 degrees about the origin, and then aω�,bω�,cω� as yet another image.  Since the a,b,c triangle is moved a third of the way around the origin and also rotated 120 degrees, the image looks as if the original triangle were translated, and its vertices were rotated.  In other words, in the first image, bω (or cω) bears the same relationship to the origin as a.  In the second image, cω� (or bω�) bears the same relationship to the origin as a.  In other words, due to the apparent translation of triangle a,b,c and rotation of its vertices, a,bω,cω� (or a,bω�,cω) are equidistant from the origin with arguments exactly 120 degrees apart.

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