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 and FAB respectively. Let the centroids of these triangles be X, Y and Z respectively. Then the lines AX, BY and 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 the triangle 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:

${\displaystyle {\begin{aligned}&\left(\csc \left(A+{\frac {\pi }{6}}\right),\csc \left(B+{\frac {\pi }{6}}\right),\csc \left(C+{\frac {\pi }{6}}\right)\right)\\&=\left(\sec \left(A-{\frac {\pi }{3}}\right),\sec \left(B-{\frac {\pi }{3}}\right),\sec \left(C-{\frac {\pi }{3}}\right)\right)\end{aligned}}}$

• The barycentric coordinates of N1:

${\displaystyle \left(a\csc \left(A+{\frac {\pi }{6}}\right),b\csc \left(B+{\frac {\pi }{6}}\right),c\csc \left(C+{\frac {\pi }{6}}\right)\right)}$

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 and FAB respectively. Let the centroids of these triangles be X, Y and Z respectively. Then the lines AX, BY and CZ are concurrent. The point of concurrence N2 is the second Napoleon point, or the inner Napoleon point, of the triangle ABC.

The triangle XYZ is called the inner Napoleon triangle of the triangle 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:

${\displaystyle {\begin{aligned}&\left(\csc \left(A-{\frac {\pi }{6}}\right),\csc \left(B-{\frac {\pi }{6}}\right),\csc \left(C-{\frac {\pi }{6}}\right)\right)\\&=\left(\sec \left(A+{\frac {\pi }{3}}\right),\sec \left(B+{\frac {\pi }{3}}\right),\sec \left(C+{\frac {\pi }{3}}\right)\right)\end{aligned}}}$

• The barycentric coordinates of N2:

${\displaystyle \left(a\csc \left(A-{\frac {\pi }{6}}\right),b\csc \left(B-{\frac {\pi }{6}}\right),c\csc \left(C-{\frac {\pi }{6}}\right)\right)}$

Two points closely related to the Napoleon points are the Fermat-Torricelli points (ETC's X13 and X14). 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 X6).

A point on the Kiepert hyperbola.

Kiepert hyperbola of triangle 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 and 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 ${\displaystyle \theta }$, then the vertices of the three triangles have the following trilinear coordinates.

• ${\displaystyle X(-\sin \theta ,\sin(C+\theta ),\sin(B+\theta ))}$
• ${\displaystyle Y(\sin(C+\theta ),-\sin \theta ,\sin(A+\theta ))}$
• ${\displaystyle Z(\sin(B+\theta ),\sin(A+\theta ),-\sin \theta )}$

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
0	G, the centroid of triangle ABC
π/2 (or –π/2)	O, the orthocenter of triangle ABC
π/4 (or –π/4)	The Vecten points
π/6	N1, the first Napoleon point (X17)
– π/6	N2, the second Napoleon point (X18)
π/3	F1, the first Fermat–Torricelli point (X13)
– π/3	F2, the second Fermat–Torricelli point (X14)
–A (if A < π/2) π – A (if A > π/2)	The vertex A
–B (if B < π/2) π – B (if B > π/2)	The vertex B
–C (if C < π/2) π – C (if C > π/2)	The vertex C

Moreover, the locus of N as the base angle ${\displaystyle \theta }$ 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 and 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 and CZ are concurrent.

The concurrence of the lines AX, BY, and 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

∠CBX = ∠ABZ, ∠ACY = ∠BCX, ∠BAZ = ∠CAY,

then the lines AX, BY and CZ are concurrent.

The point of concurrency is known as the Jacobi point.

A generalization of Napoleon point