One circle, and a straightedge

The rung below shows that a compass alone constructs everything two instruments do — the straightedge, it turns out, buys nothing. The natural question is the other one: what does a straightedge alone construct?

Almost nothing. A straightedge draws the line through two given points and finds where two drawn lines cross, and that is all. Starting from two points it produces one line and stops; from three it produces three lines and their crossings, which are the three original points again. From four points in general position it produces six lines and three new crossings, and from those the process continues — but everything it will ever produce lies in the rational field generated by the coordinates of the starting points, so no length involving a square root is ever reached. There is no way to produce a midpoint, no way to produce a perpendicular, no way to produce a circle’s worth of anything.

That looks like the end of the subject and it is the beginning of a better theorem.

Why a straightedge fails #

The reason is a symmetry argument and it is worth having, because it explains what is missing rather than listing what cannot be done.

Every operation a straightedge performs — joining two points, crossing two lines — is preserved by any projective transformation of the plane. Those transformations move points around a great deal: they take lines to lines, and they can take a midpoint to any other point of the segment.

So if a straightedge construction produced the midpoint of $AB$ from $A$ and $B$, applying a projective transformation fixing $A$ and $B$ would take that construction to itself and the midpoint somewhere else. A quantity that a symmetry moves cannot be constructed by operations the symmetry preserves.

The argument is worth recognising as the standard one for every impossibility in this field. To show an instrument cannot reach something, find a transformation the instrument commutes with and which moves the target: the construction would have to move with it and stay put at the same time. The impossibility of trisection is the same argument with a field automorphism in place of a projective map, and the impossibility of squaring the circle is the same again with a transcendence statement doing the work.

That single argument settles the matter and identifies the missing ingredient precisely. What a straightedge lacks is not “circles” but metric information — anything that distinguishes the midpoint from any other point on the line.

What one midpoint buys #

Now the reversal. Suppose a segment $AB$ is given together with its midpoint $M$. A straightedge can then draw, through any point $P$, the line parallel to $AB$.

The reason is projective. Four points on a line have a cross ratio, and the complete quadrilateral construction produces, from three of them, the fourth in harmonic position. The harmonic conjugate of $M$ with respect to $A$ and $B$ is the point at infinity exactly when $M$ is the midpoint — so the construction hands back the direction of $AB$, which is what a parallel is.

A midpoint is a point at infinity in disguise, and the whole of the following depends on that sentence.

It is worth carrying out the cross-ratio arithmetic once, because “harmonic” is otherwise a word.

Four points $A$, $B$, $C$, $D$ on a line have cross ratio $\dfrac{(AC/CB)}{(AD/DB)}$, and they are harmonic when that ratio is $-1$. Put $A = 0$, $B = 1$ and $C = \tfrac12$: then $AC/CB = 1$, so harmonicity requires $AD/DB = -1$, which means $D$ divides externally in the ratio one to one — and there is no such finite point. The only place it can be is infinity.

So “the fourth harmonic point of a midpoint is at infinity” is a one-line computation, and the complete quadrilateral is the classical straightedge construction of the fourth harmonic. Putting the two together is the whole construction.

The reverse reading is the useful one. A straightedge can construct the fourth harmonic of any three points on a line; what it cannot do is produce a third point from two. Hand it a midpoint and it produces the direction; hand it a direction — a parallel — and it produces the midpoint. The two pieces of information are interchangeable, and a straightedge has neither.

And one parallel buys everything #

With parallels available the rest follows in a chain, each step short.

Parallels give midpoints. Given any segment and a parallel to it, the complete quadrilateral run backwards produces the segment’s midpoint. Concretely: draw the two diagonals of the trapezium formed by the segment and any two points on the parallel, join their crossing to the crossing of the two sides, and that line meets the segment at its middle.

Midpoints give division into any number of parts. Mark equal steps along an auxiliary line through one end — equal because each is constructed from the previous by a parallelogram, and a parallelogram needs only parallels — then join the last mark to the far end and draw parallels to that join through the others.

Parallels and division give multiplication and division of lengths. Put the two lengths on two rays from a common point, join the ends, draw the parallel to that join through a unit mark, and where it meets the other ray is the product or the quotient depending on which ray carried the unit. That is the field arithmetic that prices what is constructible, and it needs parallels and nothing else.

And the circle supplies square roots. This is the step that genuinely needs the circle rather than the centre. Given lengths transferred onto a line through the circle, the intersection of that line with the circle satisfies a quadratic in the transferred lengths — the power of a point — so the intersection’s position is a square root of a rational combination of what was given. That is the one operation a straightedge cannot perform, and one circle performs it as often as it is asked.

So the reachable set is the same field: rationals closed under square roots. A straightedge plus one circle with its centre reaches exactly what a straightedge and compass reach, which is Poncelet’s conjecture of 1822 and Steiner’s proof of 1833.

The direction of the theorem worth emphasising is the easy one, which is often skipped. That the weaker instruments reach no more is immediate — they are a subset of the stronger pair’s operations. The content is entirely that they reach no less, and that is what the chain above establishes, one construction at a time.

Why the centre is not optional #

The theorem needs the circle’s centre to be marked, and the requirement is not a technicality.

A circle without its centre is projectively a conic, and any conic can be taken to any other by a projective transformation. So a straightedge plus an unmarked circle is projectively equivalent to a straightedge plus any other conic, and the symmetry argument above still applies to everything that does not involve the conic’s interior structure.

There is a neat way of seeing why. A circle in the projective plane is a conic, and a conic determines its centre only once a line at infinity is chosen — the centre is the pole of that line. Projective transformations move the line at infinity around, and moving it moves the centre. So the centre is not a property of the circle; it is a property of the circle together with a choice of what counts as infinity, and that choice is precisely the metric structure a straightedge does not have.

The centre is what breaks the projective symmetry. A projective transformation preserving a circle need not preserve its centre — it takes the circle to itself and moves the centre — so a marked centre is a genuinely metric piece of data, and it is the minimum such piece.

That is the sharp form of the theorem: one metric fact, supplied once, is enough for all of Euclidean construction. Everything else is projective bookkeeping.

The independence of the free choices is worth pausing on, because it is the sign that a projective construction is well defined. A construction whose answer depended on where the apex was put would be producing a point that is not a function of the given data. Here the answer is the fourth harmonic, which is determined by the three given points and by nothing else, and every quadrilateral through them finds it.

What has been sharpened since #

The theorem has been strengthened repeatedly and the strengthenings are all about how little of the circle is needed.

An arc suffices. Steiner’s proof uses the whole circle; later work shows a single arc, however short, together with the centre, is enough — because the constructions only ever need finitely many intersections with it, and an arc supplies them after a projective adjustment. That is a considerable strengthening: a short arc carries almost no information about where the rest of the circle would be, and the centre supplies the difference.

Two circles without centres suffice, if they intersect, since their radical axis and the line of centres can be constructed and give the centres. Three circles without centres suffice even if none of them meets, by the radical centre of the three.

A circle without its centre does not suffice, which was long suspected and proved by Cauer in 1912 — the projective argument above, made precise. The proof exhibits a projective transformation taking the circle to itself and moving the centre, and observes that any construction reaching the centre would have to be moved to itself as well.

There is also a strengthening in the other direction, about how much of the plane is needed: the constructions can be confined to any region containing the circle and the given points, which matters because a construction that requires going far off the page is not a construction anybody performs. The pattern of the strengthenings is worth noticing. Each one identifies a smaller piece of metric information and shows it is still enough, and the limit is exactly the point at which the projective symmetry is no longer broken.

What it costs in practice #

Nothing here makes anything easier to draw, and it is worth saying so.

The straightedge-only constructions are long. Bisecting a segment with compass and straightedge takes three moves; doing it with a straightedge and a distant circle takes a dozen, each of which introduces a crossing that has to be accurate. The theorem is about reach and not about effort.

There is a second practical cost and it is numerical. A construction that finds a point as the crossing of two nearly parallel lines locates it badly: a small error in either line moves the crossing a long way. The straightedge-only constructions are full of such crossings, because the parallels they produce are found by intersecting lines through distant points. A construction that is exact in principle can be useless on paper, and the classical preference for the compass is partly this rather than mere habit.

Its value is entirely in what it says about the operations. Knowing that one circle is enough means the compass’s role in the whole subject is exactly “supplies square roots and one midpoint”, which is a much more precise statement than “draws circles”. Every impossibility on this ladder — the trisection, the duplication, the squaring of the circle — is then a statement about a field, with the instruments entirely accounted for.

Who asked, and why the question is a good one #

Poncelet made the conjecture in 1822, in the Traité des propriétés projectives des figures — a book written, famously, while he was a prisoner of war in Russia with no references to hand.

The setting explains the question. Poncelet’s subject is projective geometry, whose whole point is to work with the operations a straightedge performs and to discard the metric structure. Having built a subject on that discarding, the natural question is what has been lost, and the answer turned out to be: one number’s worth.

Steiner’s proof of 1833 is a chain of constructions of exactly the kind the middle sections above set out, and it is long. What makes the result memorable is not the proof but the statement, which is a precise measurement of the compass’s contribution.

The general form of the question has been asked of every instrument since. What does a marked ruler add — which is the neusis, and the answer is cube roots. What does folding paper add — the same answer, by a different route. What does a compass alone lose — nothing, which is the rung below. Each is a statement about a field, and the field is the invariant that makes the questions comparable.

What the pictures cannot show #

The construction is drawn once and the theorem is about all constructions. The figure produces one parallel from one midpoint; the theorem is that every compass-and-straightedge construction can be replayed. What makes it a theorem is the chain of the middle sections, and each link of that chain is a construction not drawn here.

The projective argument has no picture at all. “A transformation fixing two points moves the third” is a statement about a group acting on the plane, and drawing one transformation shows one instance of a claim about all of them.

The chain from parallels to arithmetic is described and not drawn. Four separate constructions are named in three sentences, and each is a figure’s worth of lines. Drawing them would take the rung’s whole length and would show four standard classical constructions, which is why the prose carries them.

And the circle is drawn near the work. In a real straightedge-only construction the circle can be anywhere — across the page, unrelated to the figure — and the constructions reach it by long chains of lines. Drawing it beside the segment makes the picture readable and understates how remote the single circle is allowed to be.

Where the ladder goes next #

The next rung weakens the compass rather than the straightedge: a compass whose opening never changes, which turns out to be no restriction either, for a reason that is arithmetic rather than projective.

Named here as a debt: the arc version of this theorem, and Cauer’s proof that an unmarked circle is not enough — both stated above and neither drawn.

Sideways, the field of constructible lengths is priced by square roots, the compass-only theorem is the rung below, and the inversion that powers that one is the map trading circles for lines.

Setting the two weakenings side by side makes the shape of the subject visible. There are two instruments; each can be removed entirely or crippled; and in every case examined the reach is unchanged. The pair of instruments is enormously redundant, and the fact that Euclid’s constructions use both freely is a matter of convenience rather than of necessity.

That redundancy is itself the finding. A set of operations whose reach is unchanged by removing most of it is a set whose reach is determined by something other than the operations — here, by the field of numbers they can express — and the whole modern treatment of the subject follows from taking that seriously.

What is worth carrying away #

When an instrument cannot do something, the useful question is which symmetry it is preserving.

A straightedge preserves projective transformations, and every quantity those move is beyond it. A midpoint is such a quantity, and that is the whole diagnosis: not that the straightedge lacks curves, but that it lacks any way of distinguishing points that a projective map can interchange.

The habit worth taking is to identify the group an operation set commutes with. Its invariants are what the operations can compute, and adding one object that breaks the symmetry — here, a marked centre — can be enough to recover everything at once.