Revisiting Arc Midpoints in Complex Numbers

Power Overwhelming 2018-03-12

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)

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 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 ${-yz}$ , ${-zx}$ , ${-xy}$ .

4. A version for quadrilaterals

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.