Napoleon points

First Napoleon point

Let △ABC be any given plane triangle. On the sides BC, CA, AB of the triangle, construct outwardly drawn equilateral triangles △DBC, △ECA, △FAB respectively. Let the centroids of these triangles be X, Y, Z respectively. Then the lines AX, BY, CZ are concurrent. The point of concurrence N₁ is the first Napoleon point, or the outer Napoleon point, of the triangle △ABC.

The triangle △XYZ is called the outer Napoleon triangle of △ABC. Napoleon's theorem asserts that this triangle is an equilateral triangle.

In Clark Kimberling's Encyclopedia of Triangle Centers, the first Napoleon point is denoted by X(17).[3]

• The trilinear coordinates of N₁:
  $${\displaystyle {\begin{aligned}&\csc(A+{\tfrac {\pi }{6}}):\csc(B+{\tfrac {\pi }{6}}):\csc(C+{\tfrac {\pi }{6}})\\[2pt]=\ &\sec(A-{\tfrac {\pi }{3}}):\sec(B-{\tfrac {\pi }{3}}):\sec(C-{\tfrac {\pi }{3}})\end{aligned}}}$$

  
• The barycentric coordinates of N₁:
  $${\displaystyle a\csc(A+{\tfrac {\pi }{6}}):b\csc(B+{\tfrac {\pi }{6}}):c\csc(C+{\tfrac {\pi }{6}})}$$

  

Second Napoleon point

Let △ABC be any given plane triangle. On the sides BC, CA, AB of the triangle, construct inwardly drawn equilateral triangles △DBC, △ECA, △FAB respectively. Let the centroids of these triangles be X, Y, Z respectively. Then the lines AX, BY, CZ are concurrent. The point of concurrence N₂ is the second Napoleon point, or the inner Napoleon point, of △ABC.

The triangle △XYZ is called the inner Napoleon triangle of △ABC. Napoleon's theorem asserts that this triangle is an equilateral triangle.

In Clark Kimberling's Encyclopedia of Triangle Centers, the second Napoleon point is denoted by X(18).[3]

• The trilinear coordinates of N₂:
  $${\displaystyle {\begin{aligned}&\csc(A-{\tfrac {\pi }{6}}):\csc(B-{\tfrac {\pi }{6}}):\csc(C-{\tfrac {\pi }{6}})\\[2pt]=\ &\sec(A+{\tfrac {\pi }{3}}):\sec(B+{\tfrac {\pi }{3}}):\sec(C+{\tfrac {\pi }{3}})\end{aligned}}}$$

  
• The barycentric coordinates of N₂:
  $${\displaystyle a\csc(A-{\tfrac {\pi }{6}}):b\csc(B-{\tfrac {\pi }{6}}):c\csc(C-{\tfrac {\pi }{6}})}$$

  

Two points closely related to the Napoleon points are the Fermat-Torricelli points (ETC's X(13) and X(14)). If instead of constructing lines joining the equilateral triangles' centroids to the respective vertices one now constructs lines joining the equilateral triangles' apices to the respective vertices of the triangle, the three lines so constructed are again concurrent. The points of concurrence are called the Fermat-Torricelli points, sometimes denoted F₁ and F₂. The intersection of the Fermat line (i.e., that line joining the two Fermat-Torricelli points) and the Napoleon line (i.e., that line joining the two Napoleon points) is the triangle's symmedian point (ETC's X(6)).

A point on the Kiepert hyperbola.

Kiepert hyperbola of △ABC. The hyperbola passes through the vertices A, B, C, the orthocenter O and the centroid G of the triangle.

This generalization asserts the following:[4]

If the three triangles △XBC, △YCA, △ZAB, constructed on the sides of the given triangle △ABC as bases, are similar, isosceles and similarly situated, then the lines AX, BY, CZ concur at a point N.

If the common base angle is θ, then the vertices of the three triangles have the following trilinear coordinates.

$${\displaystyle {\begin{array}{rccccc}X=&-\sin \theta &:&\sin(C+\theta )&:&\sin(B+\theta )\\Y=&\sin(C+\theta )&:&-\sin \theta &:&\sin(A+\theta )\\Z=&\sin(B+\theta )&:&\sin(A+\theta )&:&-\sin \theta \end{array}}}$$

The trilinear coordinates of N are

$${\displaystyle \csc(A+\theta ):\csc(B+\theta ):\csc(C+\theta )}$$

A few special cases are interesting.

Value of θ	The point N
${\displaystyle 0}$	G, the centroid of △ABC
${\displaystyle {\frac {\pi }{2}},-{\frac {\pi }{2}}}$	O, the orthocenter of △ABC
${\displaystyle {\frac {\pi }{4}},-{\frac {\pi }{4}}}$	The Vecten points
${\displaystyle {\frac {\pi }{6}}}$	N1, the first Napoleon point X(17)
${\displaystyle -{\frac {\pi }{6}}}$	N2, the second Napoleon point X(18)
${\displaystyle {\frac {\pi }{3}}}$	F1, the first Fermat–Torricelli point X(13)
${\displaystyle -{\frac {\pi }{3}}}$	F2, the second Fermat–Torricelli point X(14)
${\displaystyle {\begin{cases}-A&{\text{if }}A<\pi /2\\\pi -A&{\text{if }}A>\pi /2\end{cases}}}$	The vertex A
${\displaystyle {\begin{cases}-B&{\text{if }}B<\pi /2\\\pi -B&{\text{if }}B>\pi /2\end{cases}}}$	The vertex B
${\displaystyle {\begin{cases}-C&{\text{if }}C<\pi /2\\\pi -C&{\text{if }}C>\pi /2\end{cases}}}$	The vertex C

Moreover, the locus of N as the base angle θ varies between −π/2 and π/2 is the conic

$${\displaystyle {\frac {\sin(B-C)}{x}}+{\frac {\sin(C-A)}{y}}+{\frac {\sin(A-B)}{z}}=0.}$$

This conic is a rectangular hyperbola and it is called the Kiepert hyperbola in honor of Ludwig Kiepert (1846–1934), the mathematician who discovered this result.[4] This hyperbola is the unique conic which passes through the five points A, B, C, G, O.

Generalization of Napoleon point: A special case

The three triangles △XBC, △YCA, △ZAB erected over the sides of the triangle △ABC need not be isosceles for the three lines AX, BY, CZ to be concurrent.[5]

If similar triangles △XBC, △AYC, △ABZ are constructed outwardly on the sides of any triangle △ABC then the lines AX, BY, CZ are concurrent.

The concurrence of the lines AX, BY, CZ holds even in much relaxed conditions. The following result states one of the most general conditions for the lines AX, BY, CZ to be concurrent.[5]

If triangles △XBC, △YCA, △ZAB are constructed outwardly on the sides of any triangle △ABC such that

$${\displaystyle \angle CBX=\angle ABZ,\quad \angle ACY=\angle BCX,\quad \angle BAZ=\angle CAY;}$$

then the lines AX, BY, CZ are concurrent.

The point of concurrency is known as the Jacobi point.

A generalization of Napoleon point