Areas of Napoleon triangles
NB: I came across a survey paper (Zhao, 2011) on some plane geometry problems that a group of friends and I explored around 2000. It’s a nice trip down the memory lane. I played with computeraided proofs using Maxima recently and thought it might be fun to write down the results.
Consider triangle \(ABC\). The centers of equilateral triangles constructed outward on the sides form the outer Napoleon triangle, \(DEF\). Similarly, the centers of equilateral triangles constructed inward on the sides form the inner Napoleon triangle, \(D'E'F'\). Napoleon’s theorem says that both \(DEF\) and \(D'E'F'\) are equilateral, one counterclockwise and the other clockwise.
A particularly interesting property is that the sum of the (signed) areas of \(DEF\) and \(D'E'F'\) equals the area of \(ABC\) (Coxeter and Greitzer, 1967, §3.3). A signed area is positive if the vertices are ordered counterclockwise, and negative if clockwise.
Note that in Napoleon triangle, \(A', B', C'\), the reflections of \(A, B, C\) across \(EF, FD, DE\), respectively, coincide such that \(\mathrm{area}\ A'B'C' = 0\). A general and symmetric form of the areal property is the following:
\[\mathrm{area}\ ABC + \mathrm{area}\ A'B'C' = \mathrm{area}\ DEF + \mathrm{area}\ D'E'F'.\]This post focuses on generalizations of the areal property.
Antisimilitude and involution
First, we would like to generalize how \(D, E, F\) are chosen. What’s a necessary and sufficient condition for their reflections \(D', E', F'\) across \(BC, CA, AB\), respectively, to form a triangle inversely similar to \(DEF\) (i.e., with the same shape and opposite rotations)?
Zhonghao Ye and Gang Cao (1997) answered this question with the following theorem.
Theorem 1: Consider nondegenerate triangles \(ABC\) and \(DEF\). Let \(A', B', C', D', E', F'\) be the reflections of \(A, B, C, D, E, F\) across \(EF, FD, DE, BC, CA, AB\), respectively.
Triangles \(DEF\) and \(D'E'F'\) ‘are inversely similar if and only if the Möbius transformation that maps \(A, B, C\) to \(D, E, F\) is an involution (or simply, \(A, B, C, D, E, F\) are involutoric).
Let’s start with some definitions. A Möbius transformation is a linear fractional transformation in the following form:
\[\varphi(z) = \frac{\alpha z + \beta}{\gamma z + \delta}.\]Here \(\alpha, \beta, \gamma, \delta\) are complex numbers such that \(\alpha\delta  \beta\gamma \neq 0\). We use a lowercase letter \(z\) to represent the complex number of the corresponding point \(Z\).
A transformation \(\varphi\) is an involution if \(\varphi(\varphi(z)) = z\). For example, if an involutoric transformation maps \(A\) to \(D\), then it also maps \(D\) to \(A\).
A Möbius transformation is uniquely determined by three pairs of points (Deaux 1956, article 74). When the transformation determined by \((A, D), (B, E), (C, F)\) is an involution, these points satisfy the following (Deaux 1956, article 102):
\[\begin{vmatrix} ad & a + d & 1 \\ be & b + e & 1 \\ cf & c + f & 1 \end{vmatrix} = 0.\]We use the above definition in the proofs.
A more intuitive way is to rewrite the determinant as:
\[\frac{(a  f)(b  d)(c  e)}{(f  b)(d  c)(e  a)} = 1.\]Therefore, one may view an involution as a hexagon that satisfy the following properties:

\(\dfrac{AF}{FB} \cdot \dfrac{BD}{DC} \cdot \dfrac{CE}{EA} = 1\);

\(\angle AFB + \angle BDC + \angle CEA = 0\).
We use \(\angle XYZ\) to denote the directed angle from line \(XY\) to line \(YZ\).
With this view, it’s easy to see that below are a few examples of involution:
 A triangle \(ABC\) and its outer (or inner) Napoleon triangle \(DEF\);
 A triangle \(ABC\) and points \(D, E, F\) on \(BC, AC, AB\), respectively, such that \(AD, BE, CF\) are concurrent (Ceva’s theorem);
 A triangle \(ABC\) and points \(D, E, F\) on \(BC, AC, AB\), respectively, such that \(D, E, F\) are collinear (Menelaus’s theorem);
 The three pairs of opposite vertices of a complete quadrilateral (Deaux 1956, article 103).
Proof. \(DEF\) inversely similar to \(D'E'F'\) translates to:
\[\begin{vmatrix} d & \overline{d'} & 1 \\ e & \overline{e'} & 1 \\ f & \overline{f'} & 1 \end{vmatrix} = 0.\]Since \(D'\) is the reflection of \(D\) across \(BC\), we have \(d' = \dfrac{(b  c)\overline{d} + \overline{b}c  b\overline{c}}{\overline{b}  \overline{c}}\). We obtain \(e'\) and \(f'\) similarly. Plug \(d', e', f'\) into the above equation, which simplifies to:
\[\begin{vmatrix} ad & a + d & 1 \\ be & b + e & 1 \\ cf & c + f & 1 \end{vmatrix} \begin{vmatrix} a & \overline{a} & 1 \\ b & \overline{b} & 1 \\ c & \overline{c} & 1 \end{vmatrix} = 0.\]Since \(ABC\) is a nondegenerate triangle, the above reduces to:
\[\begin{vmatrix} ad & a + d & 1 \\ be & b + e & 1 \\ cf & c + f & 1 \end{vmatrix} = 0.\]Therefore, a necessary and sufficient condition for \(DEF\) and \(D'E'F'\) to be inversely similar is that \(A, B, C, D, E, F\) are involutoric. Q.E.D.
Note that by symmetry, this is also the necessary and sufficient condition for triangles \(ABC\) and \(A'B'C'\) to be inversely similar.
How are antisimilitude and involution related to the areal property? Below is my answer to this question in a letter to Zhonghao Ye (personal communication, 1999; Zhao, 2011).
Theorem 2: Consider points \(A, B, C, D, E, F\) in the plane. Let \(A', B', C', D', E', F'\) be the reflections of \(A, B, C, D, E, F\) across \(EF, FD, DE, BC, CA, AB\), respectively.
If \(A, B, C, D, E, F\) are involutoric, the sum of the areas of \(ABC\) and \(A'B'C'\) equals the sum of the areas of \(DEF\) and \(D'E'F'\):
\[\mathrm{area}\ ABC + \mathrm{area}\ A'B'C' = \mathrm{area}\ DEF + \mathrm{area}\ D'E'F'.\]Proof. Suppose the involution determined by \(A, B, C, D, E, F\) is:
\[\varphi(z) = \frac{\alpha z + \beta}{\gamma z + \delta}.\]This boils down to three cases.
If \(\gamma = 0\), given \(\varphi(\varphi(z)) = z\), this reduces to one of the following:

\(\alpha = \delta, \beta = 0\). In this case, \(\varphi\) reduces to the identity transformation \(\varphi(z) = z\), where \(DEF\) coincides with \(ABC\). We have \(d = a, e = b, f = c\).

\(\alpha = \delta\). In this case, \(\varphi\) reduces to \(\varphi(z) = z + {\beta}/{\delta}\), where \(DEF\) is the symmetry of \(ABC\) with respect to point \(\dfrac{\beta/\delta}{2}\). Choose that point as the origin. We have \(d = a, e = b, f = c\).
If \(\gamma \neq 0\), \(\varphi\) has two fixed points \(p_i\) such that \(\varphi(p_i) = p_i (i = 1, 2)\):
 WLOG, choose the midpoint of \(p_1, p_2\) as the origin and the unit such that \(p_1 = 1, p_2 = 1\). In this case, \(\varphi\) reduces to \(\varphi(z) = 1/z\). We have \(d = 1/a, e = 1/b, f = 1/c\).
The rest is calculation for each of the three cases. See area.mac in Maxima. Q.E.D.
Concyclicity
Involution is necessary and sufficient for antisimilitude (e.g., \(DEF\) inversely similar to \(D'E'F'\)). It is also sufficient for the areal property, but not necessary, as evidenced by the following theorem.
Theorem 3: Consider points \(A, B, C, D, E, F\) in the plane. Let \(A', B', C', D', E', F'\) be the reflections of \(A, B, C, D, E, F\) across \(EF, FD, DE, BC, CA, AB\), respectively.
If \(A, B, C, D, E, F\) are concyclic, the sum of the areas of \(ABC\) and \(A'B'C'\) equals the sum of the areas of \(DEF\) and \(D'E'F'\):
\[\mathrm{area}\ ABC + \mathrm{area}\ A'B'C' = \mathrm{area}\ DEF + \mathrm{area}\ D'E'F'.\]
Proof. Choose the circle on which \(A, B, C, D, E, F\) all lie as the unit circle.
We have \(\overline{a} = 1/a, \overline{b} = 1/b, \overline{c} = 1/c, \overline{d} = 1/d, \overline{e} = 1/e, \overline{f} = 1/f\).
The rest of the proof is similar to that of Theorem 2. See area.mac. Q.E.D.
Theorems 2 and 3 indicate that to unify the areal property for both involutoric and cyclic cases, a property “weaker” than antisimilitude is needed, as detailed next.
Hexagons with opposite sides parallel
Yong Zhao (2011) described several new results regarding the areal property.
First, Yong Zhao considered the hexagon formed by \(A_1, B_1, C_1, D_1, E_1, F_1\), the feet of the altitudes from \(A, B, C, D, E, F\) to \(EF, FD, DE, BC, CA, AB\), respectively: if \(A, B, C, D, E, F\) are involutoric, the opposite sides of hexagon \(A_1F_1B_1D_1C_1E_1\) are parallel (not necessarily equal in length): \(A_1F_1 \parallel D_1C_1, F_1B_1 \parallel C_1E_1, B_1D_1 \parallel E_1A_1\).
Second, it’s straightforward to show that if \(A, B, C, D, E, F\) are concyclic, the opposite sides of hexagon \(A_1F_1B_1D_1C_1E_1\) are also parallel.
Third, the hexagon with opposite sides parallel has the following property:
\[\mathrm{area}\ A_1B_1C_1 = \mathrm{area}\ D_1E_1F_1.\]Zhouxing Mao further derived an alternative proof of Theorem 2 using the above property, demonstrating the connection between the areal property and the hexagon with opposite sides parallel.
Therefore, the key is to find a necessary and sufficient condition for the opposite sides of hexagon \(A_1F_1B_1D_1C_1E_1\) to be parallel.
Libing Huang conjectured the following:
Theorem 4. Consider distinct points \(A, B, C, D, E, F\) in the plane. Let \(A_1, B_1, C_1, D_1, E_1, F_1\) be the feet of the altitudes from \(A, B, C, D, E, F\) to \(EF, FD, DE, BC, CA, AB\), respectively.
The opposite sides of hexagon \(A_1F_1B_1D_1C_1E_1\) are parallel if and only if \(A, B, C, D, E, F\) are either involutoric or concyclic.
Libing Huang mentioned that he had a computeraided proof but didn’t provide details. Below I’ll describe a proof in Maxima.
Proof. The proof of the sufficient condition is similar to those of theorems 2 and 3. Let’s focus on the necessary condition.
Since \(A_1\) is the midpoint of \(A\) and \(A'\) (the reflection of \(A\) across \(EF\)), we have:
\[a_1 = \frac{a + a'}{2} = \frac{a(\overline{e}  \overline{f}) + \overline{a}(e  f) + \overline{e}f  e\overline{f}} {2(\overline{e}  \overline{f})}.\]We obtain \(b_1, c_1, d_1, e_1, f_1\) similarly.
\(A_1F_1 \parallel D_1C_1\) translates to:
\[(a_1  f_1)(\overline{d_1}  \overline{c_1})  (\overline{a_1}  \overline{f_1})(d_1  c_1) = 0.\]Plug \(a_1, c_1, d_1, f_1\) into the above equation, we have:
\[\left[(a  b)(\overline{e}  \overline{f}) + (\overline{a}  \overline{b})(e  f)\right] \left[(b  c)(\overline{d}  \overline{e}) + (\overline{b}  \overline{c})(d  e)\right] p_1 = 0.\]The detail of \(p_1\) is omitted due to the complexity (120 subexpressions!). The other two expressions correspond to the degenerate cases, \(AB \bot EF\) (\(A_1\) coincides with \(F_1\)) and \(BC \bot DE\) (\(D_1\) coincides with \(C_1\)), which we don’t consider. Therefore, the above equation simplifies to \(p_1 = 0\).
Similarly, from \(F_1B_1 \parallel C_1E_1\) we obtain \(p_2 = 0\) (details omitted).
Eliminating \(\overline{d}\) from \(p_1 = 0, p_2 = 0\) gives the following two cases:
\[\begin{vmatrix} ad & a + d & 1 \\ be & b + e & 1 \\ cf & c + f & 1 \end{vmatrix} = 0\] \[\left(c  a\right)\left(f  b\right)\left(\overline{c}  \overline{b}\right)\left(\overline{f}  \overline{a}\right)  \left(\overline{c}  \overline{a}\right)\left(\overline{f}  \overline{b}\right)\left(c  b\right)\left(f  a\right) = 0.\]The first corresponds to the case where \(A, B, C, D, E, F\) are involutoric. The second corresponds to the case where \(A, B, C, F\) are concyclic; similarly, we obtain that \(A, B, C, D\) and \(A, B, C, E\) are concyclic, together implying that \(A, B, C, D, E, F\) are concyclic. See parallel.mac in Maxima for the complete proof. Q.E.D.
Summary
Below is a list of the Maxima files used in this post.
 Common library: geometry.mac (see Automated geometry bashing)
 Proof of Theorems 2 and 3: area.mac
 Proof of Theorem 4: parallel.mac
Acknowledgments: Zhonghao Ye provided a copy of my original proof of Theorems 2 and 3. Zhilei Xu provided feedback on a draft of this post.
References

H. S. M. Coxeter and S. L. Greitzer. (1967). Geometry Revisited. The Mathematical Association of America.

Roland Deaux. 1956. Introduction to the Geometry of Complex Numbers. Translated by Howard Eves.

Zhonghao Ye and Gang Cao. (1997, September). Solutions to Problem 165. Bulletin of Mathematics, no. 9. 问题征解栏: 165题的解答. 数学通讯.

Yong Zhao. (2011). A Survey of Involutoric Hexagons. 完美六边形研究综述 (draft).