Six tangents and the point nobody drew

The ellipse the drawing office draws established that the four-centre construction is not an ellipse and measured how far out it is: the drawn curve reaches 2√2/3 of the true semi-major axis, so a hole drawn this way is 5.72 per cent short across the direction the drawing is measured in and 3.53 per cent wide across the other.

Every number in that sentence needed the truth. The projected circle had to be computed, the drawn curve sampled, and the two compared — which is a measurement a reader with a drawing in front of them cannot make, because they do not have the circle.

This essay is the test they can make.

Brianchon’s theorem #

Six tangent lines of a conic, taken as a hexagon. Consecutive tangents meet in six vertices; the three lines joining opposite vertices are the hexagon’s main diagonals; they pass through one point.

It is the dual of Pascal’s theorem, which says that six points of a conic, taken as a hexagon, have their three pairs of opposite sides meeting in three collinear points. Exchange point and line throughout and one becomes the other, which is the mechanism a point and a line are one object is about and the same exchange that turns Desargues into its own converse.

Both are joins and meets and nothing else, so both survive a projection and both run on a photograph with the camera unknown.

Why the tangent version is the one a reader can run #

A point on a drawn curve is a guess. A tangent to a drawn curve is a straightedge laid against it until it touches, which is a thing a hand does well and an eye checks easily.

More to the point here, the curve under test is defined by tangency. The four-centre construction’s four arcs are each fixed by the requirement that they touch two sides of the projected square at their midpoints; nothing in the recipe ever names a point of the curve except those four points of contact. Testing such a curve with its own tangents is asking it a question in the language it was built in.

The reading #

Six tangents taken across three of the four arcs, three diagonals built, and the miss measured: 1.693 per cent of the figure’s own width.

The instrument’s own floor is what makes that a rejection rather than a number. Run on the true projected circle, sampled at the same density and tested with the same six parameters, the diagonals meet to 3.1 × 10⁻⁵ of the width — which is the finite differencing that produces the tangents and nothing else. The rejection stands 546 times clear of it.

Nothing in that comparison is available to a reader with only the drawing. The control is here because this is a site that measures its own instruments; the test needs only the drawing, and on the drawing it answers.

The part that is not a defect #

Six tangents taken from one arc give a residual of 4.9 × 10⁻⁶ of the width — at the floor, indistinguishable from the true conic.

That is the instrument telling the truth. A circular arc is part of a conic, so six of its tangents are six tangents of a circle and Brianchon holds exactly. The four-centre curve is not one conic; it is four conics glued at four points, and a test that looks at only one of them finds a conic.

So the reading is a function of how far the six tangents are spread, and reporting one number would have been reporting the sampling:

• six from one arc: 0.000 per cent of the width;
• across two arcs: 0.119 per cent;
• across three: 1.693 per cent;
• across all four: 0.889 per cent.

The fall at the last row is not noise. Six tangents spread symmetrically over all four arcs sample the curve’s two-fold symmetry, and a symmetric arrangement is closer to a conic’s than an asymmetric one — which is exactly the trap the cube that is a box fell into with a symmetric taught construction reporting a side ratio of exactly one, and the trap the third point put where it looks right meets again in a different costume.

Why a symmetric hexagon reads zero on anything #

The all-four-arcs row is explained above as symmetry buying back half the reading, and the statement is milder than the truth. A symmetric set of six tangents makes the test vacuous, exactly, on any centrally symmetric curve whatever — and the argument is three lines with no arithmetic in it.

Let $\rho$ be the half-turn about the curve’s own centre, and suppose the six tangents are chosen so that $t_{4} = \rho(t_{1})$, $t_{5} = \rho(t_{2})$, $t_{6} = \rho(t_{3})$ — which is what “spread them symmetrically” produces on a curve with two-fold symmetry. Then the vertex $V_{1} = t_{1} \cap t_{2}$ has $V_{4} = t_{4} \cap t_{5} = \rho(V_{1})$, so the main diagonal $V_{1}V_{4}$ joins a point to its own half-turn image and therefore passes through the centre. The same holds for $V_{2}V_{5}$ and $V_{3}V_{6}$.

All three diagonals pass through the centre, and they do so whatever the curve is. A hexagon of tangents to a rounded rectangle, a superellipse, a stadium shape or a four-arc approximation returns a perfect Brianchon concurrence if the six are laid out symmetrically. The theorem has not failed; the test has been handed a configuration in which concurrence is forced by the symmetry rather than by the conic.

That reframes the advice from a refinement into a precondition. The instruction is not spread the six widely and preferably unevenly; it is break the curve’s symmetry, and on a four-arc ellipse — which has two mirror axes and a centre — that means the six tangents must not be arranged in opposite pairs about the centre or in mirror pairs about either axis.

It also explains the shape of the four readings without appealing to noise. Six from one arc read zero because an arc is a conic. Six over two arcs read 0.119 per cent because the set is small and crosses one junction. Six over three read 1.693 per cent, which is the honest maximum: three arcs cannot be arranged symmetrically about a two-fold centre, so nothing is cancelling. And six over all four fall back to 0.889 per cent because a set spread evenly over four arcs is close to the symmetric arrangement the paragraph above describes, and the closer it gets the more of the residual is cancelled by construction.

So the widest spread is not the best test, and the natural instinct — sample the whole curve evenly — is the worst thing to do. That is an unusually clean instance of a pattern this collection keeps meeting: a test evaluated at a symmetric configuration is a test evaluated where it cannot fail, which is the same fault as the conformality measurement taken along one tangent basis and the cross-ratio checked on four equally spaced marks. The cure is the same each time, and it is not more care. It is a deliberately awkward sample.

What a reader does with it #

Lay a straightedge against the drawn oval at six places, spread as widely as the curve allows and deliberately not symmetrically. Mark the six vertices where consecutive tangents cross. Draw the three lines joining opposite vertices. If they meet, the curve is a conic; if they miss by more than the drawing’s own line width, it is not.

That is a complete procedure with no arithmetic in it, and it is the thing this collection means by an instrument a reader owns. It also answers a question about their own drawing that no amount of comparing against a computed ellipse can, because they do not have the computed ellipse — which is the same reason what a panel says about its maker reads a finished panel rather than the recipe behind it.

The one discipline it requires is the spread. Six tangents clustered on one side of a four-arc curve will report a conic, correctly and uselessly.

Why the construction is refused off the isometric plane #

Worth pairing with the test, because the two failures are different and a reader will meet both.

The four-centre recipe is only a recipe on a 60°/120° rhombus. An arc is tangent to a side at its midpoint exactly when the line from its centre to that midpoint is perpendicular to the side, and in the isometric case that line runs to the opposite obtuse vertex — which is what makes the two large arcs centred on the vertices themselves. On a dimetric plane it does not, and the construction is refused rather than quietly producing a slightly worse curve.

So there are two things a drawn oval can be doing wrong: it can be a four-arc approximation on a plane where the recipe exists, which Brianchon catches; or it can be a four-arc approximation on a plane where the recipe does not exist, which is not a curve at all. The test catches the first. The second is caught by the construction refusing to run.

How large is the error, in the units of a drawing #

The construction’s departure from the true ellipse is 5.72 per cent of the semi-major axis at its worst, and the Brianchon reading is 1.69 per cent of the figure’s width. Those are different quantities and the second is the one a reader measures.

On a hole drawn 40 millimetres across, the diagonals miss by about two thirds of a millimetre — several line widths, visible without a magnifier, and comfortably outside anything a hand would produce on a genuine conic. On a hole drawn 6 millimetres across it is a tenth of a millimetre, which is a line width, and the test stops being usable. That is the honest limit and it is a limit of the drawing, not of the theorem: the residual scales with the figure and the pen does not.

Why five tangents will not do #

A natural first attempt is to fit a conic to five of the curve’s tangents and see how far the sixth is from touching it. That works, and it is arithmetic: it needs the adjugate, the nullspace and a normalisation, all of which five tangents name the same conic sets out.

Brianchon needs none of them, and the reason is worth stating because it is the reason duality is useful rather than decorative. A fit produces an object — a curve — and then compares evidence to it. A theorem produces a relation among the evidence and asks whether it holds. The second needs no object, no coordinates and no arithmetic, which is what lets it be executed with a straightedge on a drawing that has no coordinate system at all.

That is the same difference as between recovering a camera and checking a cross-ratio: one is a fit and one is an invariant, and on a picture whose provenance is unknown the invariant is the stronger instrument.

What the test does not say #

Three limits, and each one is a place a reader could over-read the result.

It does not say how wrong the curve is. A residual of 1.7 per cent of the width is a rejection and not an error estimate; the curve’s actual departure from the conic it is standing in for is 5.72 per cent of a semi-axis, and neither number is computable from the other without knowing the conic.

It does not say which conic the curve is not. Brianchon rejects; it does not fit. A reader who wants the true ellipse still has to construct it, which on an isometric plane is the projection of a circle and needs the projector rather than a compass.

And it does not distinguish a bad conic from a good non-conic. A conic drawn badly by hand and a four-arc curve drawn perfectly can produce the same reading. What separates them is the pattern of readings across different spreads: a hand’s error is spread evenly and reads roughly the same wherever the six are taken, while a four-arc curve reads zero within an arc and rises with the number of junctions crossed. That pattern is the four-arc signature, and it is what the figure’s slider walks through.

Where else the test applies #

Anywhere a curve is claimed to be a conic and was not drawn as one.

A french curve’s arc, a template’s edge, a road alignment set out as a series of circular arcs, the outline of a turned part on a shadowgraph, a fitted ellipse in a photograph that was fitted badly. The last of those matters here: the conic a circle becomes is where this collection establishes that a photographed circle is a genuine conic, so a fitted outline that fails Brianchon has been fitted wrong rather than photographed wrong. In each case the question is the same and the evidence is the same: is this one conic, or several stuck together?

And in each case the alternative — fitting a conic to points and measuring the residual — needs arithmetic, needs a decision about which points, and gives a number whose units depend on the fit. Brianchon gives a distance on the page, which is the unit the question was asked in.

What the wrong field’s rule says about this one #

The rule from this field’s first essay is never mock a method, measure it, and the four-centre construction comes out of the measurement rather well.

It is exact at four points, it is tangent to the projected square where the true ellipse is, it is drawn with a compass in four settings, and its worst departure is under six per cent — for a construction that predates any means of computing the alternative. What it is not is a conic, and it does not claim to be; what it is used as is a conic, and drawings measured off it inherit the error.

The contribution here is that a reader can now tell which they have without owning the truth. That is the same move three procedures, one panel makes for a perspective pavement and the rule that draws another room makes for a taught depth rule: the question stops being is this the right method and becomes what does this finished drawing say about itself.

What the six parameters are #

The test needs six places on the curve and nothing else, and how they are chosen is the only judgement in it.

Here they are parameters along the drawn curve’s own arc length, taken from the four sets the figure’s slider walks through: six clustered inside one arc, six spanning two, six spanning three, six spread over all four. The residual is then read against the figure’s own width, so it is a fraction rather than a length and transfers between drawings of any size.

The one thing a reader must not do is pick the six symmetrically. The four-centre curve has two-fold symmetry, and a symmetric set of six tangents inherits it — which pulls the three diagonals back toward concurrence and halves the reading, as the all-four row shows. Asymmetry is not a refinement of the method; it is the method working.

The short version #

Brianchon’s theorem is Pascal’s with points and lines exchanged, and the exchange turns an arithmetic test into a straightedge one. Six tangents, six vertices, three diagonals: they meet if and only if the curve is a conic, and the miss is a distance on the page.

Pointed at the four-centre ellipse it reads 1.693 per cent of the figure’s width where the true projected circle reads 3.1 × 10⁻⁵ — a rejection standing 546 times clear of the instrument’s floor, made with no knowledge of the circle the drawing is standing in for. Taken from a single arc it reads zero, correctly, because an arc is part of a conic; so the one thing the reader must do is spread the six.