Olympiad geometry without coordinates: where to start

Most students who solve olympiad geometry with coordinates are not choosing coordinates. They are falling back on them, because they looked at the figure, did not know what the first synthetic move was, and coordinates at least give you something to write. Then the circle equation arrives, the algebra triples, and forty minutes later the page is full and the proof is not.

The synthetic route is not a matter of seeing a clever line. It has a first move and a second move like any method, and once you know them you will reach for coordinates much less often, and on purpose when you do.

Move 1: translate the goal before you touch the figure

Every olympiad geometry statement asks for one of a handful of things, and each has a short list of standard ways to prove it. Write the translation down before you draw anything else.

• Four points are concyclic. Two equal angles standing on a common segment, or opposite angles summing to $180^\circ$, or $PA \cdot PB = PC \cdot PD$ for two lines through a point $P$.
• Three points are collinear. The angles at the middle point add to $180^\circ$, or all three have equal power with respect to two circles, so they sit on the radical axis.
• Three lines are concurrent. Characterise each line by a property, show the intersection of two of them has the third property, or find three circles whose radical axes they are.
• Two lengths are equal. An isosceles triangle from two equal angles, two tangents from one point, or a pair of congruent triangles.
• Two lines are perpendicular. An angle in a semicircle, or the radical axis of two circles against the line through their centres.
• A line is tangent to a circle. The tangent–chord angle equals an inscribed angle, or $PT^2 = PA \cdot PB$.

Notice how often a circle appears in that list even when the problem never mentions one. That is the whole secret of synthetic geometry: the statement is about angles or lengths, and the proof is about a circle somebody forgot to draw.

Move 2: find every cyclic quadrilateral in the figure

Once you know what you want, go looking for circles, because circles are what convert angles into lengths and back. The angle chasing page puts it in one line: equal angles on a common segment also prove four points concyclic. Its partner is the fact behind opposite angles of a cyclic quadrilateral summing to $180^\circ$, which works in both directions.

The cheapest source of circles is right angles. Two right angles standing on the same segment put both vertices on the circle with that segment as diameter. A triangle with its three altitudes drawn therefore hands you six cyclic quadrilaterals for free, before you have had a single idea. Mark them all. A student who marks them first and chases second finishes problems that a student who chases first never does.

Watching it work

Take an acute triangle $ABC$ with altitudes $AD$, $BE$ and $CF$ meeting at the orthocentre $H$. Reflect $H$ in the side $BC$ to a point $H'$. Show that $H'$ lies on the circumcircle of $ABC$.

Move 1: "lies on the circumcircle" means $A$, $B$, $H'$, $C$ are concyclic, and the standard proof is opposite angles summing to $180^\circ$. So the target is

$$\angle BH'C = 180^\circ - \angle A.$$

Move 2: the right angles at $E$ and $F$ both stand on $AH$, so $A$, $F$, $H$, $E$ are concyclic. Opposite angles: $\angle FHE = 180^\circ - \angle A$. The angle $\angle BHC$ is vertically opposite $\angle FHE$, so it is also $180^\circ - \angle A$. A reflection keeps angles, so $\angle BH'C = \angle BHC = 180^\circ - \angle A$. Done, in four lines, with no lengths and no algebra.

Now let power of a point collect the lengths. $D$ is the midpoint of $HH'$, and $D$ lies on the two chords $BC$ and $AH'$ of the circumcircle, so

$$BD \cdot DC = AD \cdot DH' = AD \cdot DH.$$

The same move on the small circles gives $HA \cdot HD = HB \cdot HE = HC \cdot HF$: the right angles at $D$ and $E$ put $A$, $B$, $D$, $E$ on a circle with diameter $AB$, and the right angles at $E$ and $F$ put $B$, $C$, $E$, $F$ on a circle with diameter $BC$. Those are length identities nobody gave you a single length for. In coordinates each one is a page; here each one is a line, because the circle did the work.

One more falls out if you keep going. Reflecting $H$ in the midpoint of $BC$ instead lands on the point of the circumcircle diametrically opposite $A$. That fact sits underneath half the configurations involving the orthocentre, and it is another one you can reach without writing an equation.

Concurrency is a property, not an intersection

The move that coordinates make hardest to see is the concurrency argument. To show three lines meet, you do not solve for the intersection and substitute. You describe each line by what its points have in common.

Take the three angle bisectors of a triangle. The bisector of $\angle A$ is exactly the set of points inside the angle that are equally far from the lines $AB$ and $AC$. Where the bisectors from $A$ and $B$ cross, the point is equally far from $AB$ and $AC$, and from $AB$ and $BC$, so it is equally far from $AC$ and $BC$ too, and the third bisector has to pass through it. That point is the incentre, and there is no equation of a line anywhere in the argument. The perpendicular bisectors are concurrent has the same shape with a different property, and it is worth proving yourself before you read the solution.

When to give in and compute

Synthetic is not always faster, and pretending otherwise wastes contest time. Reach for coordinates, vectors or complex numbers on purpose when:

• the conclusion is a specific ratio or length with no circle anywhere near it, so there is nothing for power of a point to grab;
• the figure is all midpoints and ratios along lines, where vectors make every step linear. The Euler line is a good example: it has a synthetic proof, but $H = A + B + C - 2O$ in vectors makes it a two-line calculation;
• you have done both moves honestly and found no cyclic quadrilateral at all after ten minutes.

That last test is the one to keep. Coordinates as a decision are a tool. Coordinates as an escape from not knowing where to start are the habit this page is trying to replace.

How to train it

Take a page of olympiad geometry problems and, for each one, do only the two moves: write the translated goal in one line, and list every cyclic quadrilateral in the figure. Do not solve anything. You will get through a page in the time one solution takes, and the two moves are exactly what is missing when the first instinct is to set up axes.

The angle chasing and power of a point pages have both tools in their shortest form and where they sit in the ladder, the problem archive has the proofs above in full, and the daily problem gives you a fresh figure to practise on each day, no account needed.