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. d^{3} = -1 by Euler's formula. Let ABC_{1}, BCA_{1} and CAB_{1} be the Napoleonean triangles. We have
(2) |
A_{1} = B + d(C - B), B_{1} = C + d(A - C), C_{1} = A + d(B - A) |
Taking the arithmetic average of B, C, and A_{1} me get the centroid O_{A} of the triangle BCA_{1}. Similarly we compute the other two centroids O_{B} and O_{C} of the triangles CAB_{1} and ABC_{1}, respectively:
(3) |
O_{A} = (2B + C + d(C - B))/3, O_{B} = (2C + A + d(A - C))/3, O_{C} = (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_{A}O_{B}O_{C} is obviously at the origin. As is the centroid of ΔA_{1}B_{1}C_{1}. (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_{A}O_{B}O_{C} and prove that they are of equal length. The expressions are a little awkward:
(4) |
O_{B}O_{C} = O_{C} - O_{B} = (A + B - 2C + d(B + C - 2A))/3, O_{C}O_{A} = O_{A} - O_{C} = (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) |
O_{A}O_{B} = X_{B} + dX_{C}, O_{B}O_{C} = X_{C} + dX_{A}, O_{C}O_{A} = X_{A} + dX_{B} |
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_{A}O_{B}|, |O_{B}O_{C}|, and |O_{C}O_{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_{A}X_{B}X_{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 d^{2}.
Let's check whether d^{2}(X_{B} + dX_{C}) = X_{A} + dX_{B}. Remember two things: d^{3} = -1 and X_{A} + X_{B} + X_{C} = 0 (this is analogous to what happened before.) What we have to show simplifies first to
d^{2}X_{B} - X_{C} = X_{A} + dX_{B} |
and then to
d^{2}X_{B} + X_{A} + X_{B} = X_{A} + dX_{B} |
which is implied by the quadratic identity
d^{2} - 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_{A}X_{B}X_{C}. Please see if you can complete the proof directly for ΔO_{A}O_{B}O_{C}. The same idea will also work for ΔA_{1}B_{1}C_{1}.
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