The slides a compass could have made

A marked ruler is a straightedge with two scratches on it, and its one new move is the verging: lay the edge through a given point, the pole, and slide and turn it until one scratch lies on one given line or circle and the other scratch on another. The mark that changes what is reachable used one such slide to trisect any angle, and a quintic a sliding mark reaches used more of them to reach a number of degree five. Both were arguments that the slide does something compass and straightedge cannot.

The Greek geometers did not think every slide was that powerful. Pappus of Alexandria, writing around 320, sorted problems into three kinds: plane problems, which straight lines and circles solve; solid problems, which need the conic sections; and linear problems, which need stranger curves still. He called it no small error to solve a plane problem by solid means. And he described two lost books by Apollonius of Perga, the Neuseis, given over entirely to vergings that turn out to be plane — slides that look as though they need the marked ruler and do not, because a compass and a straightedge could have found the same position.

That raises a question with a sharp answer in one simple setting. Take the two lines to be perpendicular, the pole to be the point $P = (a, b)$ measured from them, and the scratches to be $d$ apart. Which of these slides could a compass have made?

Four positions from one quartic #

Put the lines on the two axes. A placement of the ruler through $P$ with one mark at $(u, 0)$ and the other at $(0, v)$ has to satisfy two conditions: the line through the marks passes through $P$, which is $a/u + b/v = 1$, and the marks are $d$ apart, which is $u^2 + v^2 = d^2$. The first gives $v = bu/(u - a)$, and putting that into the second leaves one equation in the mark on the horizontal line:

$$F(u) = u^4 - 2a\,u^3 + (a^2 + b^2 - d^2)\,u^2 + 2a d^2\,u - a^2 d^2 = 0.$$

A slide between two lines is a quartic equation, and each of its real roots is one way to lay the ruler.

For the pole $(1, 2)$ and marks five apart the quartic is $u^4 - 2u^3 - 20u^2 + 50u - 25 = 0$, and it has four real roots. Two of them, $u = 1.745$ and $u = 4.263$, put the pole between the marks: these are the placements anyone would draw first, the ruler leaning across the corner like a beam propped against a wall. The other two, $u = -4.720$ and $u = 0.712$, put the pole outside the segment, one mark on the far side of the vertical line or below the horizontal one, and the edge carried on to reach $P$. All four are legitimate vergings, since the rule asks only that the edge pass through the pole and that the marks lie on the lines.

A quartic is not automatically beyond a compass. Its roots can always be reached by square and cube roots together, and three ways to pair four roots showed where the cube roots come from: a cubic built from the quartic’s coefficients, Ferrari’s resolvent. If that cubic is kind, the cube roots disappear and the quartic is plane. Whether the slide is plane is therefore a question about one cubic, and the rest of this essay is about which poles and which lengths make it kind.

Where four placements become two #

Before asking which slides are plane it is worth asking how many there are, because the count changes and the place where it changes is a famous curve.

Slide a segment of fixed length with one end on each axis and it sweeps out a family of lines, the motion of a beam slipping down a wall. The family folds along an envelope, and the envelope is the astroid $x^{2/3} + y^{2/3} = d^{2/3}$, a star with four cusps on the axes at distance $d$ from the corner. Inside the star, four members of the family pass through each point; outside it, two. The algebra says the same thing in its own language. The discriminant of $F$, the quantity that vanishes exactly when two roots meet, factors as

$$\Delta = -16\,a^2 b^2 d^2 \left[(a^2 + b^2 - d^2)^3 + 27 a^2 b^2 d^2\right],$$

and the bracket set to zero is the astroid’s own equation, cleared of fractional powers. The discriminant of the slide is the astroid. Where the bracket is negative, the pole is inside and $F$ has four real roots; where it is positive, two. A sweep over every whole-number pole up to 12 and every length up to 30 — 4,320 slides — found the count of real placements agreeing with the sign of the bracket every time.

The same shape of statement appeared in four perpendiculars to an ellipse, where a point inside a stretched astroid sends four normals to an ellipse and a point outside it sends two. The resemblance is an identity. The normal to the ellipse $x^2/A^2 + y^2/B^2 = 1$ at the point of angle $\varphi$ is the line $Ax/\cos\varphi - By/\sin\varphi = A^2 - B^2$. Stretch the plane, multiplying horizontal distances by $A$ and vertical ones by $B$, and that line becomes $X/\cos\varphi - Y/\sin\varphi = A^2 - B^2$, which crosses the axes at $(A^2 - B^2)\cos\varphi$ and $-(A^2 - B^2)\sin\varphi$: a segment of length exactly $A^2 - B^2$ with one end on each axis. So after the stretch, every normal of the ellipse is a placement of a ruler with marks $A^2 - B^2$ apart, and the normals through a point are the placements through the stretched point. For that essay’s ellipse, with semi-axes 2 and 1.2, the marks are 2.56 apart, and its evolute $(2x)^{2/3} + (1.2y)^{2/3} = 2.56^{2/3}$ is this astroid, stretched. Apollonius’s problem of the normals in the fifth book of the Conics is a slide between two perpendicular lines, and the boundary he found there is the edge of the vergings in his Neuseis.

The sliding segment has a second classical life. Every point of it except its ends traces an ellipse as it moves, which is the drawing instrument traditionally called the trammel of Archimedes; the astroid is the curve that instrument’s bar is always touching.

The corner a rod cannot turn #

The two placements outside the astroid never disappear, and the reason is easiest to see by trading the ruler’s position for its angle.

A line through $P$ making an angle $\theta$ with the horizontal crosses the axes, and the piece between the crossings has length $|a/\cos\theta + b/\sin\theta|$. A placement of the ruler is a crossing of that curve with the horizontal line at height $d$. For angles between $0°$ and $90°$ the line leans across the corner with $P$ between the crossings, and the length is large at both ends and has one minimum. For angles between $90°$ and $180°$ the line passes outside the corner, the two terms have opposite signs, and the length falls all the way to zero where the line runs straight through the origin — so whatever $d$ is, the right branch is crossed exactly twice. Those are the two placements that never go away.

The minimum on the left branch is the classic problem of carrying a rod flat round a corner. Two corridors of widths $a$ and $b$ meet at a right angle; the longest rod that can be turned from one into the other is the shortest segment through the inner corner $P$ that touches both outer walls, and calculus gives it as

$$d_{\min} = \left(a^{2/3} + b^{2/3}\right)^{3/2},$$

reached where $\tan\theta = \sqrt[3]{b/a}$. For corridors 1 and 2 wide it is $4.1619$, at $51.56°$. This length is exactly where the pole sits on the astroid, since the astroid for marks $d_{\min}$ apart passes through $(a, b)$; and it is not, in general, a length a compass can draw. For $(1, 2)$ it is a root of $x^6 - 15x^4 - 33x^2 - 125$, an irreducible polynomial of degree six, and six is not a power of two. It needs a cube root of 2, which the cube that will not double showed no construction reaches. Only when $b/a$ is the ratio of two cubes does the cube root become rational and the length plane — for corridors 1 and 8, it is $5\sqrt5$.

The square Apollonius did with a compass #

Now the slides themselves. Start with a case Pappus records among Apollonius’s plane vergings: a rhombus with one side extended, and a ruler laid through the opposite corner so that a given length fits in the angle outside. The two lines are a side and the extension of its neighbour, and the opposite corner lies on the diagonal, which bisects the angle between them. A square is the simplest such rhombus, and in the coordinates here it puts the pole on the bisector of the two perpendicular lines, $a = b$.

The picture shows why. Reflect any placement in the line $y = x$ and it is still a placement: the reflection swaps the two axes, fixes the pole, and keeps the distance between the marks. So the four placements come in mirror pairs, and in each pair the mark on one axis of one placement is the mark on the other axis of its twin. Write $s = u + v$ for the sum of the two marks’ distances in a placement. The condition through the pole becomes $uv = as$, and the length condition becomes $s^2 - 2uv = d^2$, which together say

$$s^2 - 2a\,s - d^2 = 0.$$

That is a quadratic for $s$, and once $s$ is known, $u$ and $v$ are the two roots of $t^2 - s\,t + a\,s = 0$, another quadratic. Two square roots in a row, and the slide is done by compass and straightedge. For $(1, 1)$ and marks 3 apart, $s = 1 \pm \sqrt{10}$, and the four placements fall out: $s = 1 + \sqrt{10}$ gives the pair whose marks sit at $1.670$ and $2.492$, $s = 1 - \sqrt{10}$ the pair at $0.744$ and $-2.906$.

This is the same calculation as the recreational puzzle of a beam leaning on a wall and just touching the far top edge of a cube-shaped box at its foot: the box’s corner is the pole, the floor and wall are the lines, and the puzzle has a quartic answer that collapses to two quadratics because the box is a cube. Make the box a cuboid and the collapse goes away — the general slide between perpendicular lines with $a \ne b$ was understood in antiquity to be solid, and Nicomedes’ conchoid, a curve built to perform slides between a line and a line, was used for the doubling of the cube. Every step is a square root is the rule that decides it: a compass reaches what a tower of quadratics reaches and nothing else, and the mirror is what builds the tower here.

Ferrari’s cubic, with one coefficient missing #

Off the bisector there is no mirror, but there might still be a tower. Ferrari’s method looks for one by pairing the four roots $r_1, r_2, r_3, r_4$ in the three possible ways and forming $r_1 r_2 + r_3 r_4$ for each; the three numbers are the roots of the resolvent cubic, and if one of them is rational the quartic splits over a quadratic field into two quadratics, so every root is two square roots away. The quartics whose roots stay paired followed that criterion through a box of millions of quartics; it is exactly the condition that the quartic’s group of symmetries has order a power of two.

For the slide’s quartic the resolvent is unusually clean. For a monic quartic $u^4 + c_3 u^3 + c_2 u^2 + c_1 u + c_0$, it is $y^3 - c_2 y^2 + (c_1 c_3 - 4c_0)\,y - (c_1^2 + c_0 c_3^2 - 4c_0 c_2)$, and here the middle coefficient is

$$c_1 c_3 - 4c_0 = (2ad^2)(-2a) + 4a^2 d^2 = 0.$$

The constant term collapses too, to $-4a^2 b^2 d^2$, so the resolvent is $y^3 - (a^2 + b^2 - d^2)\,y^2 - 4a^2 b^2 d^2$. Put $y = 2abd/w$ and multiply through, and it becomes a cubic with only three terms:

$$w^3 + (a^2 + b^2 - d^2)\,w - 2abd = 0.$$

The slide is plane exactly when this companion cubic has a rational root. For whole-number data the test is finite: a rational root of a monic cubic with whole coefficients is a whole number, and it must divide $2abd$. On the bisector the root is always there — put $a = b$ and $w = d$, and $d^3 + (2a^2 - d^2)d - 2a^2 d = 0$ — which is the mirror argument again, now in algebra. For $(1, 1)$ and marks 3 apart the cubic is $w^3 - 7w - 6 = (w + 1)(w + 2)(w - 3)$, with all three roots rational, the case where the four placements can be paired off in every one of the three ways. For the hero’s $(1, 2)$ and marks 5 apart it is $w^3 - 20w - 20$, whose only candidates are the divisors of 20, and none of them is a root. That slide is solid.

A pairing no mirror explains #

The criterion admits slides off the bisector, and they are the surprising ones, because nothing in the picture says why they should be plane.

For the pole $(1, 7)$ with marks 11 apart the companion cubic is $w^3 - 71w - 154 = (w + 7)(w^2 - 7w - 22)$, with the single rational root $w = -7$. So $y = 2abd/w = -22$, and some pairing of the four marks on the horizontal axis must have $r_1 r_2 + r_3 r_4 = -22$. It does: the placement with its mark at $-9.016$ pairs with the one at $2.945$, and the one at $7.460$ with the one at $0.611$, and $(-9.0164)(2.9453) + (7.4603)(0.6107) = -26.556 + 4.556 = -22$. The pairing is not a symmetry of the drawing. Each pair joins a placement with the pole between its marks to one with the pole outside them, two placements that look nothing alike, and the only thing they share is that their marks’ product belongs to a quadratic field. The group of the quartic here has eight elements, the symmetries of a square, which is a power of two and so within a compass’s reach.

A tower whose degrees multiply is the reason a rational root of the resolvent is enough and nothing weaker would do: the first square root splits the quartic into two quadratics over a field of degree two, the second solves each, and the degrees stay powers of two at every step. Without the rational root the splitting field has degree divisible by three, and no tower of square roots contains it. What makes the slide through $(1, 7)$ plane is arithmetic, not symmetry, and there is no way to read it off the figure.

Counting the slides a compass could make #

How common are such slides? The companion cubic makes the census exact and quick, since each slide needs only a check of the divisors of $2abd$, and of $a^2 d^2$ for a rational placement.

The diagonal of the grid is full, as the mirror requires: thirty plane lengths out of thirty for every pole on the bisector. Off it, the grid is almost empty. Counting each shape once — $a < b$, and no common factor of $a$, $b$ and $d$, so that a slide and its enlargement are not counted twice — there are 1,694 slides with $b \le 12$ and $d \le 30$. Eighteen of them are plane. Seventeen more have one rational placement and three that are not, and the remaining 1,659, which is 97.9%, reach nothing a ruler and compass could. A larger box, $b \le 20$ and $d \le 40$, gives 39 plane slides out of 6,474, and 43 with a single rational placement.

The eighteen are scattered with no obvious rule: $(1, 4)$ at lengths 3 and 7, $(1, 7)$ at 3, 9, 11 and 17, $(1, 9)$ at 7 and 13, $(4, 9)$ at 7 and 19, $(4, 11)$ at 3, 23 and 27, and a few singles such as $(5, 9)$ at 20 and $(7, 10)$ at 30. Some lengths are below the corridor minimum, where the plane slide has two real placements and the pairing joins them to two complex ones.

One placement within reach and three beyond it #

The dots in the census mark a different and stranger case: a slide with exactly one placement a compass can find.

For the pole $(3, 4)$ and marks 10 apart, the ruler can be laid from $(6, 0)$ to $(0, 8)$: the line $x/6 + y/8 = 1$ passes through $(3, 4)$, since $\tfrac36 + \tfrac48 = 1$, and $6^2 + 8^2 = 10^2$. That placement is a right triangle with whole sides, and a straightedge draws it at once. But the quartic factors as

$$u^4 - 6u^3 - 75u^2 + 600u - 900 = (u - 6)(u^3 - 75u + 150),$$

and the cubic has no rational root — its candidates are the divisors of 150, and none works — so it is irreducible, and its three roots $-9.526$, $2.129$ and $7.397$ have degree three. No compass reaches the other three placements of a slide whose first placement is a Pythagorean triple. The cubic’s discriminant is $1{,}080{,}000$, positive, so all three roots are real, and this is the case three real roots and no real radicals is about: Cardano’s formula for them passes through complex cube roots, while the marked ruler sets each of them by sliding along real lines.

The rational placements come from Pythagorean triples whose hypotenuse passes through the pole, and a tree that holds every triple lists all of those. Every one of the 17 single-rational slides in the census is one of these: its rational placement is a whole-number right triangle with its legs on the axes, from $(6, 8, 10)$ for the pole $(3, 4)$ to $(5, 12, 13)$ for $(10, 12)$, where the mark on the vertical line sits below the corner. The companion cubic does not see them, because the resolvent asks whether all four placements can be paired into quadratics, and one rational root beside an irreducible cubic is the opposite of that.

The lengths that work lie on a curve #

Fix the pole and let the length vary, and the plane slides acquire a geometry of their own. The companion cubic can be solved for $d$ instead of $w$: it is a quadratic in $d$,

$$w\,d^2 + 2ab\,d - w\,(w^2 + a^2 + b^2) = 0, \qquad d = \frac{z - ab}{w}, \quad z^2 = (w^2 + a^2)(w^2 + b^2).$$

The other root gives $d = (z + ab)/w$ at $-w$. A rational length with a plane slide is a rational point on the curve $z^2 = (w^2 + a^2)(w^2 + b^2)$, and every rational point gives two such lengths, whose product is $w^2 + a^2 + b^2$.

That curve is a quartic in $w$ with four distinct roots, $\pm ia$ and $\pm ib$, so it has genus one: it is an elliptic curve, the same kind of object that the areas a right triangle can have used to decide which whole numbers are the areas of rational right triangles. Since it has a rational point, it can be rewritten as a plane cubic, and then its rational points form a group under the rule that the three points where a line meets the cubic add to zero. The group is either finite or infinite. Barry Mazur proved in 1977 that a finite one has at most sixteen elements.

For the pole $(1, 7)$, a search of every $w = p/q$ with $p$ and $q$ up to 1,200 finds six positive values: $w = 1, 7, \tfrac{12}{5}, \tfrac{35}{12}, \tfrac{1081}{119}$ and $\tfrac{833}{1081}$. They come in pairs under the map $w \mapsto ab/w$, which is a symmetry of the curve: 1 with 7, $\tfrac{12}5$ with $\tfrac{35}{12}$, and $\tfrac{1081}{119}$ with $\tfrac{833}{1081}$. Counting both signs of $w$ and of $z$, the two points at $w = 0$ and the two at infinity, the curve has at least 28 rational points. Twenty-eight is more than sixteen, so the group is infinite, and the pole $(1, 7)$ has infinitely many rational lengths at which every placement is reachable by compass — among them $3$, $9$, $11$, $17$, $\tfrac{51}{10}$ and $\tfrac{337}{60}$.

Other poles behave quite differently. For $(1, 4)$ the same search finds only $w = 2$, fixed by the map $w \mapsto 4/w$, and the lengths 3 and 7; with the trivial points that makes eight, which is consistent with a finite group. For $(1, 2)$ — the pole on the first figure — the search finds nothing at all, and a further search with numerator and denominator up to 2,000 finds nothing either. If that curve’s group is finite, which the search suggests and does not prove, then no rational length makes the slide through $(1, 2)$ plane, and the hero’s solid verdict at length 5 is the verdict at every rational length.

Located by iteration, decided in whole numbers #

The placements are located numerically and the verdicts are exact. The positions of the ruler in every drawing are roots of a quartic found by iteration and checked against both conditions, through the pole and $d$ apart. Whether a slide is plane, solid or has one rational placement is decided in whole numbers, by testing every divisor in the finite lists the rational-root theorem allows, and it was checked against a full computation of the quartic’s group on every pole up to 8 and length up to 15.

The census is a box. Eighteen plane slides among 1,694 describes the box, not a density: the plane lengths for one pole lie on an elliptic curve and can be infinitely many, so no finite count says how often a slide is plane in general.

The verdict for $(1, 2)$ is a search, not a proof. Deciding it would mean computing the rank of an elliptic curve — by a descent, which is a calculation of its own and not one made here.

Only perpendicular lines are treated. A slide between two lines at another angle, or between a line and a circle, has a quartic of its own, and nothing above transfers without being redone; the mirror argument does transfer to a pole on the bisector of any two lines, since the reflection still swaps them. And the history is Pappus’s. The Neuseis is lost, and what it contained is known from his summary, which names the problems rather than the solutions.

Still open: which vergings between a line and a circle are plane #

Archimedes’ trisection is a slide between a line and a circle, with the pole on the circle, and it is solid for almost every angle. Pappus lists line-and-circle vergings among Apollonius’s plane problems too, so the same question can be asked there: for which poles, circles and lengths does that slide’s equation split into quadratics, and is there a companion cubic as clean as $w^3 + (a^2 + b^2 - d^2)w = 2abd$? The perpendicular case was clean because two coefficients of Ferrari’s resolvent vanished; whether anything similar happens for the circle has not been worked out here.

For the perpendicular slide itself, two questions remain open. Which poles have infinitely many plane lengths? That is the question of which of the curves $z^2 = (w^2 + a^2)(w^2 + b^2)$ have positive rank, a question about a family of elliptic curves rather than about rulers, and like most such questions it has no general answer. And how many marked-ruler moves does a solid slide save? The fewest moves to draw it counted the cheapest compass constructions by exhaustive search; run with a slide allowed as a move, the same search would price a plane slide in circles, where a solid one has no price at all.

Plane, solid, and the slide between #

The distinction Pappus drew is older than algebra, and it survives translation into algebra intact. A plane problem is one whose answer sits in a tower of square roots; a solid one needs a cube root somewhere. For the slide between two perpendicular lines the dividing line is a single cubic with three terms, $w^3 + (a^2 + b^2 - d^2)w - 2abd$, and a rational root of it is the whole difference between a compass and a marked ruler.

Most slides fail it, 97.9% of a small box. A pole on the bisector passes at every length, because the mirror pairs its placements; a few poles off it pass at isolated lengths for no visible reason, and $(1, 7)$ passes at infinitely many, because its plane lengths are the points of an elliptic curve. A verging that needs the marked ruler and a verging that only looks as if it does are told apart by arithmetic the drawing does not show, which is what Pappus meant by calling it an error to solve a plane problem by solid means: the error is invisible on the page.