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

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 $$L = \frac{M_B M_C}{A} = \frac{(-xz)(-xy)}{x^2} = +yz.$$ Thus $$M_A = -L = -yz$$ 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 $$\frac{z^2}{-yz} \cdot (+yt) = -zt.$$ 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.