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 = e^iπ/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 O_A of the triangle BCA₁. Similarly we compute the other two centroids O_B and O_C 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 ΔO_AO_BO_C 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 ΔO_AO_BO_C 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 X_A = (B + C - 2A)/3, X_B = (C + A - 2B)/3, and X_C = (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 |O_AO_B|, |O_BO_C|, and |O_CO_A|, 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 ΔX_AX_BX_C 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(X_B + dX_C) = X_A + dX_B. Remember two things: d³ = -1 and X_A + X_B + X_C = 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 ΔX_AX_BX_C. Please see if you can complete the proof directly for ΔO_AO_BO_C. The same idea will also work for ΔA₁B₁C₁.

Napoleon's Theorem

• Napoleon's Theorem
• A proof with complex numbers
• A second proof with complex numbers
• A third proof with complex numbers
• Napoleon's Theorem, Two Simple Proofs
• Napoleon's Theorem via Inscribed Angles
• A Generalization
• Douglas' Generalization
• Napoleon's Propeller
• Napoleon's Theorem by Plane Tessellation
• Fermat's point
• Kiepert's theorem
• Lean Napoleon's Triangles
• Napoleon's Theorem by Transformation
• Napoleon's Theorem via Two Rotations
• Napoleon on Hinges
• Napoleon on Hinges in GeoGebra
• Napoleon's Relatives
• Napoleon-Barlotti Theorem
• Some Properties of Napoleon's Configuration
• Fermat Points and Concurrent Euler Lines I
• Fermat Points and Concurrent Euler Lines II
• Escher's Theorem
• Circle Chains on Napoleon Triangles
• Napoleon's Theorem by Vectors and Trigonometry
• An Extra Triple of Equilateral Triangles for Napoleon
• Joined Common Chords of Napoleon's Circumcircles
• Napoleon's Hexagon
• Fermat's Hexagon
• Lighthouse at Fermat Points
• Midpoint Reciprocity in Napoleon's Configuration
• Another Equilateral Triangle in Napoleon's Configuration
• Yet Another Analytic Proof of Napoleon's Theorem
• Leo Giugiuc's Proof of Napoleon's Theorem
• Gregoie Nicollier's Proof of Napoleon's Theorem
• Fermat Point Several Times Over

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