Napoleon's Theorem
A proof with complex numbers

  On each side of a triangle, erect an equilateral triangle, lying exterior to the original triangle. Then the segments connecting the centroids of the three equilateral triangles themselves form an equilateral triangle.

That morning I went to a garage for a routine oil change that I was promised would take about 1/2 an hour. In reality it took 1/2 a day (they also changed two hoses and a thermostat) during which more then once I mentally patted myself on the back for taking along the latest (November, 1997) issue of the American Mathematical Monthly.

So I was sitting in the garage reading some and skimming some articles in the Monthly. The journal starts with a very interesting article by G.W.Cobb and D.S.Moore on teaching Statistics and its relation to Mathematics. The authors very convincingly make a point that "In data analysis, context provides meaning." This is juxtaposed with "In mathematics, context obscures structure" which left me with ambivalent feeling and thoughts. I registered my doubts and went on reading.

In the Problem section there was a solution to the following problem

  Let A be a triangle whose centroid is at the origin. Choose a real k, k>1, and dilate one of the Napoleon triangles of A by a factor of -k and the other by a factor of k/(1-k). Prove that A is (simultaneously) perspective with both dilated triangles.

The solution used some complex numbers arithmetic. So I decided to find a similar solution to Napoleon's Theorem in terms of complex numbers.

Let A,B,C be three complex numbers that correspond to vertices of a given triangle in the counterclockwise direction. Placing the centroid of the triangle at the origin, we may assume that

(1) A + B + C = 0

Let d = eiπ/3 = (1 + 3i)/2. Multiplication by d rotates a number 60° in the counterclockwise direction. d3 = -1 by Euler's formula. Let ABC1, BCA1 and CAB1 be the Napoleonean triangles. We have

(2) A1 = B + d(C - B),
B1 = C + d(A - C),
C1 = A + d(B - A)

Taking the arithmetic average of B, C, and A1 me get the centroid OA of the triangle BCA1. Similarly we compute the other two centroids OB and OC of the triangles CAB1 and ABC1, respectively:

(3) OA = (2B + C + d(C - B))/3,
OB = (2C + A + d(A - C))/3,
OC = (2A + B + d(B - A))/3

I was staring at these three numbers for a while thinking of what to do next. For one, the centroid of ΔOAOBOC is obviously at the origin. As is the centroid of ΔA1B1C1. (You have to take the average of three numbers again. That the results are zero in both cases follows from (1) and the symmetry with respect to A,B, and C of the expressions involved.)

It's now natural to find the sides of ΔOAOBOC and prove that they are of equal length. The expressions are a little awkward:

(4) OAOB = OB - OA = (C + A - 2B + d(A + B - 2C))/3,
OBOC = OC - OB = (A + B - 2C + d(B + C - 2A))/3,
OCOA = OA - OC = (B + C - 2A + d(C + A - 2B))/3

So far it's all structural. Formally, we can do even better. Let's introduce XA = (B + C - 2A)/3, XB = (C + A - 2B)/3, and XC = (A + B - 2C)/3. Then (4) simplifies to

(5) OAOB = XB + dXC,
OBOC = XC + dXA,
OCOA = XA + dXB

Actually, in the garage, I did this just in order to do something because nothing else has crossed my mind. A natural step would be to evaluate the lengths |OAOB|, |OBOC|, and |OCOA|, and verify that they coincide. The computations ought to be tedious and uninspiring. Perhaps somebody at this point may see how to complete the solution immediately. I needed to invoke the context. It appears that ΔXAXBXC has been rotated 60° counterclockwise and the vertices of the two triangles have been added pairwise. These sums (as complex numbers of course) stand as vertices of a triangle that must be equilateral. Its centroid is at the origin, as before. Therefore, it must be possible to get from one vertex to the next by a rotation through 120°, i.e. by multiplying by d2.

Let's check whether d2(XB + dXC) = XA + dXB. Remember two things: d3 = -1 and XA + XB + XC = 0 (this is analogous to what happened before.) What we have to show simplifies first to

  d2XB - XC = XA + dXB

and then to
  d2XB + XA + XB = XA + dXB

which is implied by the quadratic identity

  d2 - d + 1 = 0

obviously satisfied by d.

I think the proof is nice. As a byproduct, it shows that several involved triangles all have the same centroid. Something that does not come for free from other proofs. It's also an example where keeping context in mind proved instrumental at least to me.

In hindsight, there was no compelling need to introduce ΔXAXBXC. Please see if you can complete the proof directly for ΔOAOBOC. The same idea will also work for ΔA1B1C1.

Napoleon's Theorem

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