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 computer-aided 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 non-degenerate 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 non-degenerate 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 computer-aided 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.

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).