# Some Properties of Napoleon's Configuration

*Napoleon's theorem* claims that the centers $A', B', C',$ of the equilateral triangles $A''BC,$ $AB''C,$ $ABC'',$ erected on the sides of a given triangle $ABC$ form an equilateral triangle. This is true if the triangles $A''BC,$ $AB''C,$ $ABC'',$ are constructed "outwardly":

and also when they are constructed "inwardly":

The corresponding triangles are known as the *outer* and, respectively, *inner*, *Napoleon triangles*. The properties of one parallel the properties of the other, and there are also properties they have as a pair.

### First Property

The centroid of Napoleon's triangle $A'B'C'$ coincides with that of the reference triangle ABC.

The simplest proof I am aware of makes use of complex numbers, see a separate discussion. The proof holds both for the outer and the inner Napoleon's triangles. It follows that the centroids of the two Napoleon triangles - the outer and the inner - coincide. This fact was mentioned in the classical [*Geometry Revisited*, p. 65], [*Complex Numbers and Geometry*, p. 113] or [*MathematicalGems*, p. 40]. The two books, however, failed to mention that the same point also serves as the centroid of the base triangle $ABC.$

### Second Property

Lines $AA'',$ $BB'',$ $CC''$ meet in a point - one (either *inner* or *outer*) of Fermat's points of $\Delta ABC.$

The three lines form equal angles of $60^{\circ}$ or $120^{\circ},$ depending on how one looks at the configuration. This is equivalent to the claim that the circumcircles of triangles $ABC'',$ $AB''C,$ and $A''BC$ are concurrent.

This implies the next property.

### Third Property

The sides of the Napoleon triangle serve as perpendicular bisectors of the lines joining Fermat's point to the vertices of $\Delta ABC.$

This follows from the fact that, by the construction, the vertices of Napoleon's triangle are the centers of the circumcircles of triangles $ABC'',$ $AB''C,$ and $A''BC$ while the segments $AF,$ $BF,$ $CF$ ($F$ being the Fermat point) are the common chords of the circles taken two at a time.

### Fourth Property

The circumcricles of Napoleon's triangles are concurrent; in an acute triangle they meet at Fermat's point. As such, they admit a triangle with one vertex on each of the circles whose sides pass through the vertices of the base triangle.

This is actually a *porism*: the starting point for the construction of that triangle can be chosen arbitrarily. The triangle is always equilateral.

### Fifth Property

Here we deal with both inner and outer Napoleon's triangles. This necessitates notations that differentiate between the two.

Let $ABC_{o},$ $BCA_{o},$ $CAB_{o}$ be the outer, $ABC_{i},$ $BCA_{i},$ $CAB_{i}$ the inner Napoleon's triangles.

Then the midpoints of $AA_{o}$ and $B_{i}C_{i}$ coincide as are the midpoints of $BB_{o},$ $C_{i}A_{i}$ and $CC_{o},$ $A_{i}B_{i}.$ A similar assertion holds with the exchange of the indices $_{o}$ and $_{i}$ throughout.

For proof see a separate page.

### Sixth Property

Here we return to the previous notations for Napoleon's triangles $A''BC,$ $AB''C,$ $ABC''$ and their centroids $A',$ $B',$ and $C',$ respectively.

Let $K_A,K_B,K_C$ be the centroids of triangles $AB''C'',A''BC'',A''B''C,$ respectively.

Then the hexagon $A'K_CB'K_AC'K_B$ is regular.

### Seventh Property

Form equilateral triangles $ABC'',\;$ $AB''C,\;$ $A''BC\;$ on the sides of $\Delta ABC\;$ and define $D = \frac{1}{2}(A''+B'')\;$ and $E=\frac{1}{2}(A''+C'').$

Prove that $\Delta ADE\;$ is equilateral.

For a proof see a separate page.

### References

- H. S. M. Coxeter, S. L. Greitzer,
*Geometry Revisited*, MAA, 1967 - Liang-shin Hahn,
*Complex Numbers & Geometry*, MAA, 1994 - R. Honsberger,
*Mathematical Gems*, MAA, 1973

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