Emperor of the French
Karen Van Hoey
Napoleon, a mathematician ?
Napoleon is best known as a military genius
and Emperor of France but he was also an outstanding mathematics student.
He was born on the island of Corsica and died in exile on the island of Saint-Hélène after being defeated in Waterloo. He attended school at Brienne in France where he was the top maths student. He took algebra, trigonometry and conics but his favorite was geometry. After graduation from Brienne, he was interviewed by Pierre Simon Laplace(1749-1827) for a position in the Paris Military School and was admitted by virtue of his mathematics ability. He completed the curriculum, which took others two or three years, in a single year and subsequently he was appointed to the maths section of the French National Institute.
During the Egyptian military campaign of 1798-1799, Napoleon was accompanied by a group of educators, civil engineers, chemists, mineralogists and mathematicians, including Gaspard Monge (1746-1818) and Joseph Fourier(1768-1830). On his return from Egypt, Napoleon led a successful coup d'état and became head of France. As emperor, he instituted a number of juridical, economical and educational reforms and placed men such as Laplace, Monge and Fourier in government positions with the commission to establish new educational institutions, recruit teachers and revamp the curriculum to emphasize maths. As his career began to rise, he continued his study of maths and he assembled a group of mathematicians, including Lorenzo Mascheroni (1750-1800), Pierre Simon Laplace and Joseph Louis Lagrange (1736-1813), to discuss mathematics.
Napoleon was most proud of having solved
one of Mascheroni's problems, the construction of a circle into fourths. As he began to
explain his solution to the assembled mathematicians, Laplace(
denominated as his chief military engineer), commented :
"We expect all things from you, General, except a lesson in geometry".
A theorem is named after him as well, although he was not the first to discover it but apparently found it independently (some dispute this claim) and so it bears his name.The famous Napoleon Theorem is stated by Coxeter and Greitzer as follows:
If equilateral triangles are erected externally on the sides of any triangle, their centers form an equilateral triangle. (the following portrait is made by Anne-Louis Girodet-Trioson )
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 form an equilateral triangle.
Proof #1 ("Hammer and Tongs" trigonometry)
In the following A denotes both the vertex A and the corresponding angleA, a is both BC and its length. In addition, let G denote the centroid of the equilateral triangle on side AB, and I denote the centroid of the equilateral triangle on side AC, etc. Let s denote the length of segment GI, t the length of segment AG, and u the length of segment AI.
Since IÂC = GÂB = 30°, we may apply the Law of Cosines to compute
s² = u² + t² - 2ut.cos(A + 60°) (1)
Since the centroid of an equilateral triangle lies along each median, 2/3 of the distance from the vertex to the midpoint of the opposite side, we have
t = (2/3). sqrt(3)/2 . c = c/sqrt(3)
u = (2/3). sqrt(3)/2 . b = b/sqrt(3)
and (1) becomes
3. s2 = b2 + c 2 - 2bc. cos(A + 60°) (2)
Expanding the cosine of the sum, and recalling that cos(60°) = 1/2 , sin(60°) = sqrt(3)/2, we have
cos(A + 60°) = cos(A)/2 - sin(A). sqrt(3)/2 (3)
Substituting (3) into (2) yields
3. s² = b² + c² - bc. cos(A) + sqrt(3).bc. sin(A) (4)
Now apply the Law of Cosines to triangle ABC :
a² = b² + c² - 2bc. cos(A) (5)
and recall, as in the derivation of the Law of Sines :
2. (Area of triangle ABC) = bc. sin(A) (6)
Substituting (5) and (6) into (4) gives
3.s² = (1/2)(a² + b² + c²) + 2. sqrt(3). (Area of triangle ABC) (7)
Since (7) is symmetrical in a, b, and c, it follows that the triangle GIH connecting the three centroids is equilateral.
Proof #2 (an argument by symmetrization)
|Notations are the same as before: let triangle ABC be the original triangle. Choose D, E, and F exterior to triangles ABC so that the triangles ADB, BEC, and AFC are equilateral triangles, with centroids G, H, and I respectively. Rotation(center I and 120° to the right) around I results in BB, DD, EE, GG, HH being the images of points B, D, E, G, and H, respectively. Connect D to EE and G to HH. By the rigidity of the rotation, triangle GHI = triangle GG.HH.I. In particular, GH = GG.HH.Now consider the six triangles that converge on point A. Three of them (ABD, ACF, and A.EE.BB) are equilateral. Recollect that the angles of a triangle sum to 180°, while the angles around a point sum to 360°. Since triangle BB.A.F is a copy of triangle BCA, it follows that D.Â.EE = A^BC. Now consider the six triangles that converge on point A. Three of them (ABD, ACF, and A.EE.BB) are equilateral. Finally, triangle D.A.EE = triangle ABC, and the pentagon A.BB.EE.D.B is congruent to the pentagon BECAD. It follows that G.HH = GH. And thus G.HH = GH = GG.HH.|
|Repeating the rotation by 120° once more, and connecting the tips of the equilateral triangles as above, we obtain the figure on the right. Arguing as above, it is clear that the central hexagon is equilateral, and that the six triangles which meet at the center of rotation are congruent. Therefore, 6 . HÎG = 360°, and HÎG = 60°. Since (among the points G,H,I) the choice of the centroid I was arbitrary, we have shown that triangle GHI is equiangular, hence equilateral.|
A Generalization of Napoleon's Theorem
It was already thought that the theorem allows several generalizations. In particular, equilateral triangles can be replaced with similar triangles of arbitrary shape. This sounds even more surprising than the Napoleon's theorem itself.
|Start with two similar triangles (black). On each of the lines connecting their corresponding vertices (white), construct triangles (red) similar to each other and similarly oriented. Then three free vertices of these triangles form a fourth triangle similar to the original two.|
|(1) In a special case where two vertices of the given similar
triangles coincide, only one (white) line is needed to connect vertices of the two
triangles. The other two pairs are connected by sides of the triangles.
Three similar isosceles triangles are constructed on the vertex connecting lines.
|Changing the viewpoint, note that on the sides of D ABC we only discern two similar triangles (thick black, on the lower two sides of D ABC.) To remove this lopsidedness we construct an additional (and spurious) triangle (thin black) similar to the other two (thick black.)|
The latter illustrates the most general reformulation of the Napoleon's Theorem. The three similar triangles may be of various shapes and, in addition, one is permitted to connect any three corresponding points (and not just centroids) in order to obtain a fourth similar triangle. (On the diagram, we took the apices of similar isocseles triangles as the three corresponding points.)
(1) has a more immediate appeal than such a generalization. After all, all we did was constructing nothing but similar triangles. Having constructed a sequence of similar triangles, it seems quite "natural" to expect, as a result, another similar triangle. What else might it be? Fortunately for (1) there exists a proof, based on the theory of spiral similarities.
Geschiedenis van de wiskunde, D.J.Struik, Uitgeverij Het Spectrum (-Aula Paperback;178)
Classic Math : History Topics for the Classroom, Art Johnson, Dale Seymour Publications
(Palo Alto, CA 94303)
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