Revisiting arc midpoints in complex numbers

1. Synopsis

One of the major headaches of using complex numbers in olympiad geometry problems is dealing with square roots. In particular, it is nontrivial to express the incenter of a triangle inscribed in the unit circle in terms of its vertices.

The following lemma is the standard way to set up the arc midpoints of a triangle. It appears for example as part (a) of Lemma 6.23.

Theorem 1 (Arc midpoint setup for a triangle)

Let ${ABC}$ be a triangle with circumcircle ${\Gamma}$ and let ${M_A}$, ${M_B}$, ${M_C}$ denote the arc midpoints of ${\widehat{BC}}$ opposite ${A}$, ${\widehat{CA}}$ opposite ${B}$, ${\widehat{AB}}$ opposite ${C}$.

Suppose we view ${\Gamma}$ as the unit circle in the complex plane. Then there exist complex numbers ${x}$, ${y}$, ${z}$ such that ${A = x^2}$, ${B = y^2}$, ${C = z^2}$, and

$\displaystyle M_A = -yz, \quad M_B = -zx, \quad M_C = -xy.$

Theorem 1 is often used in combination with the following lemma, which lets one assign the incenter the coordinates ${-(xy+yz+zx)}$ in the above notation.

Lemma 2 (The incenter is the orthocenter of opposite arc midpoints)

Let ${ABC}$ be a triangle with circumcircle ${\Gamma}$ and let ${M_A}$, ${M_B}$, ${M_C}$ denote the arc midpoints of ${\widehat{BC}}$ opposite ${A}$, ${\widehat{CA}}$ opposite ${B}$, ${\widehat{AB}}$ opposite ${C}$. Then the incenter of ${\triangle ABC}$ coincides with the orthocenter of ${\triangle M_A M_B M_C}$.

Unfortunately, the proof of Theorem 1 in my textbook is wrong, and I cannot find a proof online (though I hear that Lemmas in Olympiad Geometry has a proof). So in this post I will give a correct proof of Theorem 1, which will hopefully also explain the mysterious introduction of the minus signs in the theorem statement. In addition I will give a version of the theorem valid for quadrilaterals.

2. A Word of Warning

I should at once warn the reader that Theorem 1 is an existence result, and thus must be applied carefully.

To see why this matters, consider the following problem, which appeared as problem 1 of the 2016 JMO.

Example 3 (JMO 2016, by Zuming feng)

The isosceles triangle ${\triangle ABC}$, with ${AB=AC}$, is inscribed in the circle ${\omega}$. Let ${P}$ be a variable point on the arc ${BC}$ that does not contain ${A}$, and let ${I_B}$ and ${I_C}$ denote the incenters of triangles ${\triangle ABP}$ and ${\triangle ACP}$, respectively. Prove that as ${P}$ varies, the circumcircle of triangle ${\triangle PI_{B}I_{C}}$ passes through a fixed point.

By experimenting with the diagram, it is not hard to guess that the correct fixed point is the midpoint of arc ${\widehat{BC}}$, as seen in the figure below. One might be tempted to write ${A = x^2}$, ${B = y^2}$, ${C = z^2}$, ${P = t^2}$ and assert the two incenters are ${-(xy+yt+xt)}$ and ${-(xz+zt+xt)}$, and that the fixed point is ${-yz}$.

This is a mistake! If one applies Theorem 1 twice, then the choices of “square roots” of the common vertices ${A}$ and ${P}$ may not be compatible. In fact, they cannot be compatible, because the arc midpoint of ${\widehat{AP}}$ opposite ${B}$ is different from the arc midpoint of ${\widehat{AP}}$ opposite ${C}$.

In fact, I claim this is not a minor issue that one can work around. This is because the claim that the circumcircle of ${\triangle P I_B I_C}$ passes through the midpoint of arc ${\widehat{BC}}$ is false if ${P}$ lies on the arc on the same side as ${A}$! In that case it actually passes through ${A}$ instead. Thus the truth of the problem really depends on the fact that the quadrilateral ${ABPC}$ is convex, and any attempt with complex numbers must take this into account to have a chance of working.

3. Proof of the theorem for triangles

Fix ${ABC}$ now, so we require ${A = x^2}$, ${B = y^2}$, ${C = z^2}$. There are ${2^3 = 8}$ choices of square roots ${x}$, ${y}$, ${z}$ we can take (differing by a sign); we wish to show one of them works.

We pick an arbitrary choice for ${x}$ first. Then, of the two choices of ${y}$, we pick the one such that ${-xy = M_C}$. Similarly, for the two choices of ${z}$, we pick the one such that ${-xz = M_B}$. Our goal is to show that under these conditions, we have ${M_A = -yz}$ again.

The main trick is to now consider the arc midpoint ${\widehat{BAC}}$, which we denote by ${L}$. It is easy to see that:

Lemma 4 (The isosceles trapezoid trick)

We have ${\overline{AL} \parallel \overline{M_B M_C}}$ (both are perpendicular to the ${\angle A}$ bisector). Thus ${A L M_B M_C}$ is an isosceles trapezoid, and so ${ A \cdot L = M_B \cdot M_C }$.

Thus, we have

$\displaystyle L = \frac{M_B M_C}{A} = \frac{(-xz)(-xy)}{x^2} = +yz.$

Thus

$\displaystyle M_A = -L = -yz$

as desired.

From this we can see why the minus signs are necessary.

Exercise 5

Show that Theorem 1 becomes false if we try to use ${+yz}$, ${+zx}$, ${+xy}$ instead of ${-yz}$, ${-zx}$, ${-xy}$.

We now return to the setting of a convex quadrilateral ${ABPC}$ that we encountered in Example 3. Suppose we preserve the variables ${x}$, ${y}$, ${z}$ that we were given from Theorem 1, but now add a fourth complex number ${t}$ with ${P = t^2}$. How are the new arc midpoints determined? The following theorem answers this question.

Theorem 6 (${xytz}$ setup)

Let ${ABPC}$ be a convex quadrilateral inscribed in the unit circle of the complex plane. Then we can choose complex numbers ${x}$, ${y}$, ${z}$, ${t}$ such that ${A = x^2}$, ${B = y^2}$, ${C = z^2}$, ${P = t^2}$ and:

• The opposite arc midpoints ${M_A}$, ${M_B}$, ${M_C}$ of triangle ${ABC}$ are given by ${-yz}$, ${-zx}$, ${-xy}$, as before.
• The midpoint of arc ${\widehat{BP}}$ not including ${A}$ or ${C}$ is given by ${+yt}$.
• The midpoint of arc ${\widehat{CP}}$ not including ${A}$ or ${B}$ is given by ${-zt}$.
• The midpoint of arc ${\widehat{ABP}}$ is ${-xt}$ and the midpoint of arc ${\widehat{ACP}}$ is ${+xt}$.

This setup is summarized in the following figure.

Note that unlike Theorem 1, the four arcs cut out by the sides of ${ABCP}$ do not all have the same sign (I chose ${\widehat{BP}}$ to have coordinates ${+yt}$). This asymmetry is inevitable (see if you can understand why from the proof below).

Proof: We select ${x}$, ${y}$, ${z}$ with Theorem 1. Now, pick a choice of ${t}$ such that ${+yt}$ is the arc midpoint of ${\widehat{BP}}$ not containing ${A}$ and ${C}$. Then the arc midpoint of ${\widehat{CP}}$ not containing ${A}$ or ${B}$ is given by

$\displaystyle \frac{z^2}{-yz} \cdot (+yt) = -zt.$

On the other hand, the calculation of ${-xt}$ for the midpoint of ${\widehat{ABP}}$ follows by applying Lemma 4 again. (applied to triangle ${ABP}$). The midpoint of ${\widehat{ACP}}$ is computed similarly. $\Box$

In other problems, the four vertices of the quadrilateral may play more symmetric roles and in that case it may be desirable to pick a setup in which the four vertices are labeled ${ABCD}$ in order. By relabeling the letters in Theorem 6 one can prove the following alternate formulation.

Corollary 7

Let ${ABCD}$ be a convex quadrilateral inscribed in the unit circle of the complex plane. Then we can choose complex numbers ${a}$, ${b}$, ${c}$, ${d}$ such that ${A = a^2}$, ${B = b^2}$, ${C = c^2}$, ${D = d^2}$ and:

• The midpoints of ${\widehat{AB}}$, ${\widehat{BC}}$, ${\widehat{CD}}$, ${\widehat{DA}}$ cut out by the sides of ${ABCD}$ are ${-ab}$, ${-bc}$, ${-cd}$, ${+da}$.
• The midpoints of ${\widehat{ABC}}$ and ${\widehat{BCD}}$ are ${+ac}$ and ${+bd}$.
• The midpoints of ${\widehat{CDA}}$ and ${\widehat{DAB}}$ are ${-ac}$ and ${-bd}$.

To test the newfound theorem, here is a cute easy application.

Example 8 (Japanese theorem for cyclic quadrilaterals)

In a cyclic quadrilateral ${ABCD}$, the incenters of ${\triangle ABC}$, ${\triangle BCD}$, ${\triangle CDA}$, ${\triangle DAB}$ are the vertices of a rectangle.

4 thoughts on “Revisiting arc midpoints in complex numbers”

1. Here’s my preferred way to think about this: let $A_1A_2\dots A_n$ be a cyclic $n$-gon on the unit circle with $n\geq3$, and let $B_{1,2},B_{2,3},\dots,B_{n,1}$ be the midpoints of the $n$ arcs partitioning the circle.

Key Observation: if $A_1,A_2,\dots,A_n$ go counterclockwise around the circle, then the product $\prod_i B_{i,i+1} A_i^{-1}$ is $\exp(i\pi) = -1$, because the sum of the corresponding arcs is precisely half the circumference.

So, if $a_1$ is an arbitrary square root of $A_1$, then there is a **unique** square root $a_2$ of $A_2$ such that $B_{1,2} = -a_1a_2$: namely, $a_2 = -B_{1,2} a_1^{-1}$ works! Continuing in this manner uniquely specifies all $a_i$. The catch is, the choice is consistent if and only if $n$ is odd: we require $(-1)^n$ to be consistent with $-1$ from the Key Observation.

The case $n$ is even (e.g. quadrilateral as you discuss) is similar, except we can no longer take all the signs in the $B_{i,i+1} = \pm a_ia_{i+1}$ to be $-1$. An odd number of the signs must be $+1$ instead.

Liked by 3 people

• In the explanation for the cyclic inductive construction, I should have elaborated that the terms $B_{i,i+1} A_i^{-1} = \pm a_{i+1} a_i^{-1}$ form a cyclic telescoping product (I only mentioned the sign issues). In any case, an example where I needed to think through this carefully was http://artofproblemsolving.com/community/c6h404946p2271784

(Sorry I forgot about WordPress math in the previous comment, and I can’t seem to edit.)

Liked by 1 person

2. No comments on the post, but a question: what do you use for writing WordPress math? I am struggling with open-source solutions. Thank you in advance!

Liked by 1 person