Complex numbers for olympiad geometry: when to reach for them

Every geometry problem you meet has two possible attacks. One is to see the configuration: the cyclic quadrilateral hiding in it, the spiral similarity, the point that is secretly a midpoint. The other is to give up on seeing anything and compute.

The second attack has a bad reputation, and the reputation is half deserved. A student who reaches for coordinates on every problem never learns to see, and gets crushed on the problems where the computation does not close. But a student who refuses to compute has no move at all when the configuration will not reveal itself, and on a timed paper that is a zero.

So the useful question is not whether complex numbers are elegant. It is: given this problem, in front of me, right now, will the computation close? That is a recognition skill, and it is learnable, because the configurations where it closes have a shape.

The setup is the entire decision

Put the circumcircle of the figure at the origin with radius $1$. Every point of that circle is now a complex number $a$ with $|a| = 1$, and one fact does all the work:

$$\bar a = \frac{1}{a} \quad \text{whenever} \quad |a| = 1.$$

That is why the unit circle and not some other circle. Conjugation is the operation that appears in every geometric condition worth writing down, and on the unit circle it stops being a second, independent variable and becomes a reciprocal of the first one. A condition that would need $a$ and $\bar a$ as separate unknowns collapses to a rational function of $a$ alone, and rational functions of one variable are things you can actually finish.

Everything downstream follows from that single substitution. If the figure has no circle to put at the origin, you have lost most of the advantage before you start.

The dictionary

Six lines, and you want them cold, because looking them up mid-problem costs more time than the method saves. Throughout, lower-case letters are the complex numbers of the upper-case points, and $|a| = |b| = |c| = 1$.

• Perpendicular and parallel. $PQ \perp RS$ exactly when $\frac{p-q}{r-s}$ is purely imaginary; they are parallel when it is real.
• Collinear. $P$, $Q$, $R$ are collinear exactly when $\frac{p-q}{p-r}$ is real. This is the parallel test with a shared point, and it is the single most used line in the dictionary.
• Rotation. Rotating $P$ about $Q$ through $\theta$ sends $p \mapsto q + (p-q)e^{i\theta}$. Multiplication is rotation, which is why spiral similarity, the configuration that defeats most synthetic attempts, is a one-line computation here.
• Orthocentre. $h = a + b + c$. No fractions, no case analysis.
• Foot of an altitude. The foot from $C$ to chord $AB$ is $\tfrac12\left(a + b + c - \frac{ab}{c}\right)$.
• Intersection of two chords. Chords $AB$ and $CD$ meet at $\dfrac{ab(c+d) - cd(a+b)}{ab - cd}$.

The four signals that say compute

• A circle is already in the statement. A circumcircle, an incircle you can invert away, a fixed circle points move on. The setup is free, so the method starts ahead.
• The conclusion is collinearity, concurrency or a perpendicularity. These are exactly the conditions the dictionary turns into single equations. A conclusion about an inequality or an area comparison is not on that list.
• The configuration is defined by intersections rather than by angles. If the points are built by "let $X$ be where these two lines meet", you have closed forms for all of them and the problem is bookkeeping. If they are built by "let $X$ be the point with $\angle XAB = \angle XCB$", you are about to introduce an unknown that does not simplify.
• You have spent ten minutes and seen nothing. This one is not a property of the problem, it is a property of your paper. A guaranteed grind that closes in twenty minutes beats an elegant argument you do not have.

Watching it work

Take the Euler line: the orthocentre, centroid and circumcentre of a triangle are collinear, with $HG = 2\,GO$.

Put the circumcircle at the origin, so $O = 0$. The centroid is the average, $g = \tfrac{a+b+c}{3}$, which needs no setup at all. And the orthocentre is $h = a+b+c$, which you can verify in one line: $h - a = b + c$, and $\frac{b+c}{b-c}$ is purely imaginary because on the unit circle its conjugate is $\frac{1/b + 1/c}{1/b - 1/c} = \frac{c+b}{c-b}$, the negative of itself. So $AH \perp BC$, and the same computation runs at the other two vertices.

Now the theorem has already happened. $h = 3g$, so $H$, $G$ and $O = 0$ are three points on one line through the origin, and $|h - g| = 2|g|$. There is nothing left to prove. The result that takes a page of synthetic work is a consequence of the coordinates being chosen well.

The nine-point circle falls out of the same two facts. Its centre is $\tfrac{a+b+c}{2}$, the midpoint of $O$ and $H$, and every one of the nine points is at distance $\tfrac12$ from it. Each side midpoint is $\tfrac{a+b}{2}$; each vertex-to-orthocentre midpoint is $\tfrac{a+b+c}{2} + \tfrac{c}{2}$ in disguise; each altitude foot is the dictionary entry above. Three formulas, nine verifications, no diagram.

Ptolemy is an identity, not a theorem

The clearest illustration of what the algebra buys you is Ptolemy's inequality: for any four points, $AC \cdot BD \le AB \cdot CD + BC \cdot AD$.

For any four complex numbers whatsoever,

$$(a-c)(b-d) = (a-b)(c-d) + (b-c)(a-d).$$

Expand both sides and every term cancels. Take absolute values and apply the triangle inequality, and Ptolemy's inequality is proved for arbitrary points in the plane. Equality holds when the two terms on the right have the same argument, which is precisely the condition that the four points lie on a circle in that order.

Notice what happened. The geometric content became an algebraic identity you can verify by expanding brackets, and the geometric hypothesis (concyclic) became a condition on arguments. That trade is the whole method in miniature, and it is why the cyclic quadrilateral results feel less mysterious once you have coordinates: opposite angles summing to $180^\circ$ is a statement about arguments of ratios, and arguments of ratios are what complex numbers are for.

Where it stalls

The method has a hard boundary, and knowing it is worth as much as knowing the dictionary.

• The incentre needs a square root. The orthocentre is a polynomial in $a$, $b$, $c$. The incentre is not: it is built from the arc midpoints, and the midpoint of arc $BC$ is a point $m$ with $m^2 = bc$. Choosing which square root is the geometric content, and the standard fix is to start from $a = p^2$, $b = q^2$, $c = r^2$ instead of from $a, b, c$. If a problem is about the incircle, decide before you start whether you are willing to pay that cost.
• Lengths are worse than positions. $|a - b|$ carries a square root that does not cancel. Configurations stated in ratios of segments are fine; configurations stated in sums of lengths usually are not.
• Inequalities do not translate. The dictionary turns equalities into equations. It has no entry that turns a $\le$ into anything tractable.
• Sometimes vectors are enough, and cheaper. For the diagonals of a parallelogram, or anything about midpoints and centroids, plain vectors do the same job with less machinery. Reach for complex numbers when you need rotation or the unit-circle conjugate; if you need neither, you are paying for tools you will not use.

What to do with this

Do not practise this by solving. Take fifteen geometry problems you have already seen solved, and for each one write three lines only: what goes at the origin, what the unknown points are in closed form, and what single equation the conclusion becomes. Then stop.

You will be wrong on some of them, and being wrong at that stage costs you two minutes instead of forty. That is the entire skill: not the algebra, which is mechanical, but the judgement about whether the algebra will close, made early enough to change your mind.

Lemma teaches this as complex numbers and coordinates in geometry, a named technique with a place in the sequence rather than a trick you meet twice and forget. The problem archive has the statements above in full, with difficulty and expected time on each, and the daily problem posts a new one every day in three tiers, no account needed.