Napoleon's Theorem

Napoleon's theorem states that if we construct equilateral triangles on the sides of any triangle (all outward or all inward), the centers of those equilateral triangles themselves form an equilateral triangle, as illustrated below.

This is said to be one of the most-often rediscovered results in mathematics. The earliest definite appearance of this theorem is an 1825 article by Dr. W. Rutherford in "The Ladies Diary". Although Rutherford was probably not the first discoverer, there seems to be no direct evidence supporting any connection with Napoleon Bonaparte, although we know that he did well in mathematics as a school boy. According to Markham's biography,

Even after becoming First Consul he was proud of his membership in the Institute de France (the leading scientific society of France), and was close friends with several mathematicians and scientists, including Fourier, Monge, Laplace, Chaptal and Berthollet. (Oddly enough, Markham refers to Fourier as Francois, apparently confusing him with the social reformer and utopianist. The mathematician and friend of Napoleon was actually Jean Baptiste Joseph Fourier, no relation to Francois as far as I know, although they were almost exact contemporaries.) Indeed, in his grand expedition to Egypt in 1798 Napoleon brought along (in addition to 35,000 troops) over 150 experts in various fields, among them Monge, Fourier, and Berthollet, not to mention a complete encyclopedie vivante with libraries and instruments. Napoleon held long discussions with his movable Institute - much to the consternation of his generals, who had trouble staying awake during these sessions.

One result of the expedition was that Fourier served for a time as the governor of lower Egypt. Likewise Laplace (who interviewed the young Napoleon for admission to the artillery) received titles and high office as a result of his friendship with Bonaparte. However, Laplace was relieved of his duties as the Minister of the Interior after only six weeks, and Napoleon later commented that Laplace had "sought subtleties everywhere, had only doubtful ideas, and carried the spirit of the infinitely small into administration". The most famous exchange between these two men occurred after Laplace had given Napoleon a copy of his great work, the Mecanique Celeste. Napoleon looked it over, and remarked that in this massive volume about the universe there was not a single mention of God, its creator. Laplace replied "Sire, I had no need of that hypothesis".

Regarding the idea that Napoleon might actually have discovered what we now call Napoleon's Theorem, Coxeter and Greitzer have said that

(I wonder how many times that palindrome has been independently re-discovered... but I digress.) It's actually not very difficult to prove this theorem, nor to discover it, so I don't think it would have been impossible for Napoleon to have found it. Consider the drawing below, where the original triangle is shaded in gray, and we have added a 4th equilateral triangle to the left hand vertex.

Notice that if we rotate the figure counter-clockwise through an angle of p/3 about the point "c", the triangle originally centered at "b" moves to the position originally occupied by the triangle centered at "d". This proves the line segments cb and cd are of equal length and make an angle of p/3. Likewise if we rotate the figure clockwise through an angle p/3 about the point "a", the triangle centered at "b" again moves to the position of the triangle at "d", so the line segments ab and ad are of equal length and make an angle of p/3. Consequently the line ac bisects the angles at "a" and "c", so the triangle abc has an angle of p/6 at each vertex, so it is equilateral (as is acd). This is a simple enough observation that it could easily be made by a bright school boy. On the other hand, since Rutherford's article first appeared in 1825, just four years after Bonaparte died on St Helena, it's also conceivable that Rutherford just decided to name his theorem after the famous fallen Emperor.

In any case, the theorem really just expresses the nice tiling pattern, as shown below:

This makes the theorem visually obvious, and is also suggestive of various generalizations, but it doesn't provide much quantitative information. Given the edge lengths A,B,C of the original triangle, what is the edge length of "Napoleon's equilateral triangle"? It's not too difficult to give a proof of Napoleon's theorem using coordinate geometry and a little algebra that also yields the answer to this question. It turns out to be closely related to Heron's formula for the area of a triangle in terms of the edge lengths.

If we construct equilateral triangles on any two of the sides of the given triangle, the distance from their centers is the same, regardless of which two sides we choose. Hence, if we express this distance in terms of the edge lengths, it must be a symmetrical function of those lengths, i.e., any permutation of the edges should leave the result unchanged. The same is true for the area of a triangle, so we shouldn't be surprised if there turns out to be a relation between the sides of Napoleon's equilateral triangle and the area of the original triangle. On the other hand, these two quantities can't be identical, nor even proportional, because we can have a degenerate triangle with zero area but still a non-zero edge length for Napoleon's equilateral triangle.

To give another proof Napoleon's theorem, and in the process discover how to express the sides of the Napoleon's triangle in terms of the sides of the given triangle, consider the general case of a triangle with edge lengths a,b,c placed on a Cartesian coordinate system with the side c on the X axis as shown below.

The center of the equilateral triangle constructed on the side a is to be found by moving from the midpoint of that side perpendicularly outward a distance of a/(12)^1/2. The components m,n of this segment satisfy the requirements

where x,y are the coordinates of the vertex opposite the side c. Thus, noting that x² + y² = a², we find

Likewise we can find the components for the segment leading from the midpoint of the side b to the center of the equilateral triangle constructed on b. Making use of x = (a² - b² + c²)/(2c) we have

In terms of these lengths the distance s between the centers of the equilateral triangles constructed on the edges a and b can be expressed as

Expanding the squares gives

Noting that the last term is x²/3, we can combine this with y²/3 to give a²/3. Also, we can combine c²/4 and c²/12 to give c²/3. Furthermore, recalling that 2cx = a² - b² + c², we can make this substitution and collect terms to give

Now, since y² = a² - x², the quantity cy can be written as

This is just twice the area of the triangle with sides a,b,c, as discussed in the note Heron's Formula For Triangle Area, and it has the symmetrical factorization

Hence the distance between the centers of the equilateral triangles constructed on the sides a and b is

Since this is perfectly symmetrical in the three sides, it's clear that the distances between the centers of the equilateral triangles constructed on any two sides of the triangle are the same, and so the triangle formed by connecting those centers is equilateral, which proves Napoleon's theorem. The theorem can be generalized to say that the centers of regular n-gons constructed on the sides of a regular n-gon form a regular n-gon, and naturally it also applies to n-gons subjected to conformal and affine transformations.

Another proof of Napoleon's Theorem, based on a more explicit trigonometric approach, can be developed from the figure below.

Recall that the edge lengths of a triangle are proportional to the sines of the opposite angles, which implies that the lengths of the segments from the vertex to the centroid of the equilateral triangles centered on points a,b,c are proportional to sin(a), sin(b), sin(g) respectively. Therefore, noting that the angle between the segments Pb and Pc is a + p/3, the normalized squared length of segment can be expressed using the law of cosines as

Adding and subtracting sin(a)² and expanding the cosine of a + p/3 in the above expression gives

Napoleon's Theorem is true if and only if the corresponding expressions for the squares of the other two normalized lengths s_ab and s_ca equal this same value. These expressions are given by simply permuting the angles a, b, and g. Since the overall expression is obviously symmetrical under transpositions of b and g, and the first four terms are symmetrical in all three of the angles, it only remains to consider the effect of transposing a and b on the quantity in the square brackets. Hence Napoleon's Theorem is equivalent to the equality

Solving for sin(g) gives

Since a, b, and g are the three interior angles of a triangle, we have g = p - (a + b), and therefore sin(g) = sin(a + b). Making this substitution, the above relation is equivalent to the trigonometric identity

We can write this in a more familiar form by making the substitutions sin(x)² = (1-cos(2x))/2 on the right hand side, and defining u = a - b and v = a + b. This gives

This identity is easily verified by writing the sines and cosines in exponential form, and carrying out the multiplications.

Yet another approach to Napoleon's Theorem is by representing points in the plane as complex numbers. Both the premise and the conclusion of the theorem involve equilateral triangles, so it will be useful to know the necessary and sufficient on three complex numbers z₁, z₂, z₃ to be the vertices of an equilateral triangle. The centroid of these points is simply z₀ = (z₁ + z₂ + z₃)/3, and the rays from the centroid to the vertices are (z₁ - z₀), (z₂ - z₀), and (z₃ - z₀). The vertices form an equilateral triangle if and only if these rays are of equal length L and separated by the angle 2p/3. Hence the necessary and sufficient condition is

for some arbitrary angle q. It follows that

Substituting for z₀ and expanding the products, we arrive at the condition

This condition on the vertices of an equilateral triangle can be expressed in several alternate (but algebraically equivalent) forms, the most fundamental of which is

Another equivalent form is

Also, since the quantity in the left parentheses is 3z₀, this can be written as

which represents an alternate expression for the centroid of an equilateral triangle.

Given any two points, these expressions enable us to compute a third points that forms an equilateral triangle with the first two. Of course, the relation is quadratic in each of the three points, so there are two solutions, corresponding to the two possible directions that the third point can lie in with respect to the other two. Given z₁ and z₂, an equilateral triangle is formed by setting z₃ to either of the values

Therefore the centroid of this triangle is at

The root given by the + sign places the centroid such that the loop 1,2,0 proceeds in the counter-clockwise direction. Now, if we are given three arbitrary points p₁, p₂, p₃ in the complex plane, Napoleon's Theorem asserts that the centroids of the outer equilateral triangles on the bases (p₁,p₂), (p₁,p₃), and (p₂,p₃) form an equilateral triangle. Letting c₁₂, c₁₃, and c₂₃ denote these centroids, we can use the preceding formula to express these in terms of the points p₁, p₂, p₃. For example, we have

It is then straightforward to verify that

so the centroids do indeed satisfy the condition for being the vertices of an equilateral triangle. Furthermore, letting p₀ denote the centroid of the original triangle (not necessarily equilateral), the right-hand quantity is 3p₀², but we know that the middle quantity equals 3c₀² where c₀ is the centroid of the equilateral triangle formed by the three original centroids. Consequently, the centroid of Napoleon's triangle coincides with the centroid of the original three points.

We can also verify the expression for the length of the edge of Napoleon's triangle given above. The length is simply |c₁₂ - c₂₃|, i.e., the magnitude (norm) of the difference between two vertices. We have

Letting p₁ = x₁ + i y₁, etc., this becomes

The squared length of the edge is then the product of this quantity and its complex conjugate, so we have

The quantity in the first square bracket is just twice the area of the original triangle (as discussed in Net Area and Green's Theorem), and the quantity in the second square bracket is the sum of the squares of the edge lengths of the original triangle. Hence if we let A denote the area and a,b,c denote the edge lengths of the original triangle, the edge length of Napoleon's triangle is

in agreement with the previous demonstration.

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