What two points build in two rounds

Two points and two operations are enough to describe everything a compass and straightedge can do: a line through two known points, and a circle centred at one known point through another, with every crossing of two such curves added to what is known. The essays that followed answered the question which numbers this reaches — every step is a square root, so the reachable numbers are the ones built from the rationals by square roots, and nothing else — and that answer settled the doubling of the cube, the trisection of the angle and the squaring of the circle.

None of them asked how fast. The answer to “which” is an infinite set described by a rule; the operations themselves reach it in stages, and the stages can be counted. Do everything possible at once — draw every line and every circle the known points allow, and add every crossing — and call that a round. From two points, the first round gives 6 points, the second gives 203, and the third gives 1,723,816,861. The figure below is the second round, every point computed exactly and coloured by the square roots its coordinates need.

One round, then an explosion #

The first round is small enough to do by hand. Two points give one line, through both, and two circles, each centred at one point and passing through the other. The line meets each circle in two points, one of which is a starting point; the circles meet each other above and below the line. Six points in all: 0, 1, −1 and 2 on the line, and $(\tfrac12, \pm\tfrac{\sqrt3}2)$ off it — the two apexes of the equilateral triangles on the unit segment, the first construction in Euclid’s Elements.

The second round starts from those six points. Six points give fifteen pairs, but four of the points lie on one line, so the fifteen pairs give only ten different lines; they give thirty ordered pairs for circles, but several share a centre and a radius, leaving sixteen different circles. Ten lines and sixteen circles cross in many places, many crossings coincide, and the distinct points number 203.

The third round draws every line through two of 203 points and every circle about one through another — about twenty thousand lines and forty thousand circles — and their crossings number 1,723,816,861. That count was published in 2020 after a computation with exact arithmetic, and it is cited here rather than repeated, because checking coincidences among nearly two billion points needs care that floating point cannot give. Each round’s curves number about the square of its points, and its crossings about the square of its curves, so the points grow roughly like a fourth power: $203^4$ is about 1.7 billion, startlingly close to the actual count, while $6^4 = 1{,}296$ overshoots 203 by far, because among so few curves so many crossings coincide.

The coincidences are where the geometry shows. Of the fifteen pairs of first-round points, six lie on the starting line and give one line between them; the other nine pairs give nine different lines, ten in all. The circles coincide more: the circle about the origin through 1 also passes through −1 and through both apexes, so four ordered pairs give one circle, and the thirty ordered pairs give sixteen circles. Then the crossings coincide. The ten lines and sixteen circles cross one another 497 times, but those crossings land on only 203 distinct points, because many points lie on three, four or more of the curves at once. A point where three curves meet is counted three times among the crossings; the 203 is what is left after Euclid’s geometry has been allowed to coincide.

The hexagon is finished, the square is not #

Some familiar figures are complete after two rounds. The six points at distance one from the origin in the directions of a regular hexagon — $(\pm1, 0)$ and $(\pm\tfrac12, \pm\tfrac{\sqrt3}2)$ — are all among the 203: two were starting points or first-round points, and the other four are crossings of the unit circle about the origin with the unit circles about $\pm1$ and the apexes. A regular hexagon inscribed in a circle is the construction every compass makes almost by accident, stepping its own radius round the circle, and the rounds make it in two stages.

The square on the unit segment is not there. Its other two corners, $(0, 1)$ and $(1, 1)$, need a perpendicular at an endpoint and then a quarter-circle, and the perpendicular through the origin exists only once the second round has produced points above and below it; its crossing with the unit circle is a third-round point. Nor is the centre of the first equilateral triangle, at height $\sqrt3/6$, though its three vertices are known after one round: the centre is the crossing of two medians, and the medians need the midpoints of the sides, which also wait for the third round. The point $(0, \sqrt3)$, by contrast, is already there — the apex of a larger triangle.

Euclid’s own order of propositions follows the same logic. The first proposition of the Elements constructs the equilateral triangle, the first round exactly; the construction of a square comes forty-five propositions later, after perpendiculars and parallels have been built from the triangle. The rounds measure the same dependency mechanically: what the triangle gives at once, what needs one more stage, and what needs two.

Exact coordinates, and the tower made visible #

Floating point finds the 203 points and draws them. It cannot say which numbers they are. For that the second round is done again with exact arithmetic, and the description is short enough to carry out completely.

Every first-round point has coordinates of the form $a + b\sqrt3$ with $a$ and $b$ rational, so every line and circle of the second round has its equation in the field $\mathbb{Q}(\sqrt3)$. Two lines cross where a pair of linear equations is solved, which stays inside $\mathbb{Q}(\sqrt3)$. A line and a circle, or two circles, cross where a quadratic equation is solved, and its solutions are $u \pm v\sqrt{D}$ for some $u$, $v$ and $D$ in $\mathbb{Q}(\sqrt3)$. So every second-round coordinate has the form

$$a + b\sqrt3 + (c + d\sqrt3)\sqrt{D},$$

and the computation records $a$, $b$, $c$, $d$ and $D$ as exact fractions. That is the theorem that every step is a square root, carried out on every crossing at once.

The point needs no new square root exactly when $D$ is already a square in $\mathbb{Q}(\sqrt3)$, and that is a finite check: $D = p + q\sqrt3$ is the square of $x + y\sqrt3$ when $x^2 + 3y^2 = p$ and $2xy = q$, which leads to a quadratic for $x^2$ whose solutions must be squares of fractions. Running the check on every crossing sorts the 203 points into three kinds. Eleven are rational — the integers from −4 to 5 on the line, and one half. Seventy-six more need $\sqrt3$ and nothing else. And 116 need a square root that is not in $\mathbb{Q}(\sqrt3)$ at all.

Those 116 do not each need a different one. Two values of $D$ give the same new field when their ratio is a square in $\mathbb{Q}(\sqrt3)$, and grouping them that way leaves six new square roots: $\sqrt2$, $\sqrt5$, $\sqrt7$, $\sqrt{11}$, $\sqrt{13}$ and $\sqrt{35}$, where each is named by a square-free integer and 3 can be dropped from any of them because it is already a square. Each gives a field of degree four over the rationals, $\mathbb{Q}(\sqrt3, \sqrt D)$, whose degree is the product of the two steps. The first round climbed one step of the tower, to degree two; the second round climbs to degree four, in six different directions at once. No point needs a cube root, or a square root of a square root, because a single round solves only one quadratic beyond what the previous round knew.

Fifteen points on the line #

The starting line is where the numbers live in their most familiar form, and after two rounds it holds fifteen points.

They are the integers from −4 to 5, one half, and the four numbers $\pm1 \pm \sqrt3$ shifted by one: $-1 - \sqrt3$, $2 - \sqrt3$, $\sqrt3 - 1$ and $2 + \sqrt3$. Every one of them is a crossing of the starting line with a second-round line or circle. One half arrives as the foot of the perpendicular through the two apexes; $2 \pm \sqrt3$ arrive where a circle of radius $\sqrt3$ about the point 2 crosses the line.

The fifteen are symmetric about one half — −4 pairs with 5, $-1 - \sqrt3$ with $2 + \sqrt3$ — because reflecting the plane in the perpendicular bisector of the starting segment swaps the two starting points and so carries every round onto itself. The symmetry is a free check on the computation: a point found on one side without its partner on the other would mean an error.

Missing are numbers as plain as one third, one quarter and minus one half, and $\sqrt2$. All of them are constructible: a third, for instance, follows from the intercept theorem in a handful of steps. They are simply not reached by two rounds of doing everything at once, because the steps that build them require curves that only exist after the second round’s points are known. The difference between a number being constructible and being reached quickly is the whole subject of this essay, and the line shows it in the simplest place.

Reached, not yet, never #

The line is one line. Distances between points are a fuller test of what two rounds have reached: the 203 points have a little over three thousand distinct distances between pairs of them, and familiar lengths can be looked for among those.

The square roots of 2, 3, 5 and 7 are all distances after two rounds, and so is one third — two points a third apart exist even though the line does not hold the number itself. The golden ratio, which needs $\sqrt5$ combined with further arithmetic, is not yet a distance, and neither is the side of the regular pentagon, $2\sin 36°$, though both are constructible and the pentagon is among the polygons that can be drawn. Neither is $\sqrt{2 + \sqrt2}$, a nested square root, which needs two levels of tower above the rationals in the same direction.

Two lengths will never appear, in any round. The cube root of 2 has degree three, and three does not divide any power of two, so it lies in no field the tower reaches; π is transcendental, so it lies in no finite tower at all. Those were the classical impossibilities, and the rounds put them in context: a list of what has been reached after each round grows explosively, and these numbers are not on any of the lists.

Where each new root appears #

The six new square roots are not scattered at random over the second round’s points.

Each set is symmetric about the starting line and about the perpendicular bisector of the starting segment. Both reflections fix the pair of starting points, so they carry the whole construction onto itself, round by round, and a reflection of that kind changes no coordinate’s field. The $\sqrt{11}$ points are the most numerous, thirty-six of them, and the $\sqrt{35}$ points the rarest, four. Every one of the 116 comes from a crossing that involves at least one circle, since two lines cross without any square root; which circles produce which root can be read off the exact computation, crossing by crossing.

How the count could be checked further #

The third round is where exact arithmetic becomes a serious undertaking, and the reason is coincidence. Two different pairs of curves can cross at the same point, and in floating point two crossings that differ by $10^{-12}$ may be one point or two. At 203 points rounding to eight decimal places separates every distinct crossing, as the exact computation confirms here. At 1.7 billion points the gaps between genuinely different points shrink so far that rounding can merge distinct points or split equal ones, and every coincidence must be decided exactly, by comparing numbers that each lie in a field of degree up to eight.

At 203 points the danger is remote, and it can be measured. The closest two of the 203 points are about 0.024 apart — one at $(\tfrac54, \tfrac{\sqrt{39}}4)$, where $\sqrt{39} = \sqrt3\sqrt{13}$, and one a crossing that needs $\sqrt{11}$ — which is more than a million times the rounding used to merge crossings, so no two distinct points could have been merged and no point split. That makes the third-round count a different kind of fact from the second-round one. The 203 is computed here twice, and the two computations agree point for point. The 1,723,816,861 rests on a single published computation of a much larger size. Nothing suggests it is wrong, and it is the kind of number that would be worth recomputing independently; in the meantime it is reported here as cited, with that status, rather than as checked.

What the figures cannot show #

The figures show two rounds, and two rounds are a very small part of the constructible plane. Every constructible point is reached in some finite number of rounds, so the union of all rounds is the whole set of constructible points, dense in the plane; any picture of finitely many rounds shows a scattering that will eventually fill everything, and the eye cannot tell from the 203 points which regions fill first.

The exact classification also stops at the field each point needs, not at its full description. A point needing $\sqrt2$ might have coordinates like $1 + \sqrt6/2$ or $\sqrt2 - \sqrt3$, and the figures do not distinguish them. They record the step of the tower, which is what matters for the impossibility theorems, and leave the arithmetic within each field to the exact computation behind them.

And rounds are one way to measure speed; there are others. Counting the number of individual lines and circles drawn — the measure used by construction puzzles — gives a different and much finer notion, in which a third can be made in a few strokes while the rounds need three full stages. A single marked ruler can reach further than both, and a straightedge with one circle drawn once reaches everything the pair does, at a very different cost per point. Each measure gives a different answer to “how fast”, and only the rounds give a single number per stage.

Still open: the fourth round #

How many points does the fourth round reach? Nobody knows. The third round’s count needed exact arithmetic in fields of degree up to eight; the fourth round would draw every line and circle among 1.7 billion points — on the order of $10^{18}$ curves — and decide every coincidence among their crossings exactly. Even its order of magnitude has not been computed.

The growth has a cleaner form of question that is also open: is there a formula, or even an asymptotic estimate, for the number of points after $n$ rounds? The rounds are defined by a simple rule, and their counts are 2, 6, 203 and 1,723,816,861; no rule is known that predicts the next term, and the four known terms are too few to guess one.

The closure in stages #

From two points, one round of every line and circle gives 6 points and the next gives 203; the third gives 1,723,816,861. Done exactly, the second round shows the tower of square roots as it grows: eleven rational points, seventy-six needing $\sqrt3$, and 116 needing one of six new roots — $\sqrt2$, $\sqrt5$, $\sqrt7$, $\sqrt{11}$, $\sqrt{13}$ and $\sqrt{35}$ — each a field of degree four and nothing beyond. Familiar numbers arrive at different speeds: $\sqrt5$ and one third are distances after two rounds, the golden ratio is not, and $\sqrt[3]2$ and π never will be.