A centre is three weights

Nine points on one circle proves its coincidence by a scaling and then names the other route: “Every claim on this page can be turned into polynomial identities in the six coordinates of the triangle, and then verified by algebra rather than by a sweep — which is what a proof by coordinates amounts to, and which is how these results are checked when a sweep is not enough.”

The coordinates that make it work are not the six. They are three weights per point, and a coincidence becomes a determinant.

Three weights, and why they are the right coordinates #

A point in the plane of a triangle can be written as $\alpha A + \beta B + \gamma C$ with $\alpha + \beta + \gamma = 1$, and the three numbers are determined by the point. They are barycentric coordinates, and the name is literal: the point is where the triangle balances with those weights at its corners.

Two things make them better than ordinary coordinates for this subject.

They are intrinsic. Moving, rotating or scaling the triangle leaves a centre’s weights unchanged, because the weights refer to the corners rather than to any axes. So a centre is a rule assigning weights to a triangle’s shape, and two triangles of the same shape have centres with identical weights — which is what being a centre means and is awkward to say in the six coordinates the nine-point circle’s proof works in.

And a collinearity is a determinant. Three points with weight triples $u$, $v$, $w$, each summing to one, are collinear exactly when

$$\det \begin{pmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{pmatrix} = 0.$$

That is the same statement as three points in the plane being collinear when a certain determinant vanishes, transported into weights, and it is the whole reason the reformulation pays: an incidence claim becomes an algebraic identity with no geometry left in it.

The classical four, written out #

The weights are computed from the side lengths, and it is worth having them because the pattern in them is the point.

The centroid is $1 : 1 : 1$. It is the average of the corners, which is why the balance-point argument works for it and for nothing else on the list.

The circumcentre is $a^2(b^2 + c^2 - a^2) : b^2(c^2 + a^2 - b^2) : c^2(a^2 + b^2 - c^2)$, which looks unpleasant and is built entirely from squared side lengths.

The orthocentre is $(a^2 + b^2 - c^2)(c^2 + a^2 - b^2)$ and its two cyclic relatives — again squares only.

The nine-point centre is the midpoint of the first two of those, so its weights are their normalised weights added, and it inherits their character.

And the incentre is $a : b : c$ — the side lengths themselves, not squared.

That last line is the whole of the essay’s second half. $a^2$ is a polynomial in the six coordinates: a difference of coordinates, squared and added. $a$ is the square root of one.

Where the weights come from #

Each of those triples is derived rather than looked up, and two derivations are short enough to have here because they show what the pattern is made of.

The centroid. It is the average of the corners, so the weights are equal. Nothing to derive.

The circumcentre. It is equidistant from the three corners. Writing a point as $\alpha A + \beta B + \gamma C$ and computing its squared distance to each corner gives three expressions; setting them equal gives two linear equations in the weights, whose solution is the triple above. The squared side lengths appear because a squared distance between two points is a sum of squared coordinate differences, and nothing else enters.

The incentre. It is equidistant from the three sides. The distance from a point to a line is a linear expression divided by the line’s length, and the line’s length is a side length — so the equations carry $a$, $b$ and $c$ in the denominators, and clearing them gives weights proportional to the side lengths themselves.

That single difference is the essay’s whole content. A distance to a point is a square root of a polynomial and squaring it removes the root; a distance to a line is a polynomial divided by a square root, and the root survives into the answer. So equidistant from points gives rational weights and equidistant from lines does not, and the two families an earlier essay names are the two kinds of equidistance.

The figures check each triple by computing the point it produces and comparing it against the point computed geometrically, so the weights are verified rather than quoted — which matters for the circumcentre and orthocentre, whose triples nobody would recognise as correct on sight.

The Euler line as one number #

Put the circumcentre, the centroid and the orthocentre into the determinant. Every entry is a rational function of the squared side lengths, so the determinant is one too, and computing it gives identically nought — not nought for this triangle, but nought as a rational function.

That is the Euler line, proved. The figures compute the determinant on the drawn triangle and across a sweep of a couple of hundred, and it never exceeds a part in a hundred million; the identity itself is an algebraic calculation that a computer does in a moment and that nobody needs to read.

The nine-point centre goes in the same way and the determinant vanishes again, which is the statement the nine-point circle’s essay proves by a halving towards the orthocentre. Two proofs, and they are not the same proof: the scaling explains why, and the determinant establishes that, for every triangle at once and with no case distinction.

The one-to-two ratio comes out as well. The determinant says the three are collinear and the weights say where: normalising and subtracting gives the centroid at exactly a third of the way from the circumcentre to the orthocentre, which is the ratio the same nine-point argument measures to a billionth.

Why the incentre cannot be in it #

Put the circumcentre, the centroid and the incentre into the determinant. It does not vanish, and the figures report the value — which is the whole content of the statement that the incentre is off the Euler line.

The reason it cannot vanish identically is the square roots. A determinant whose rows are $a^2$-polynomials and $a$-polynomials is not a rational function of the coordinates, so it cannot be identically nought unless it vanishes for a reason visible in the square roots — and it does not.

What it can do is vanish for particular triangles, and the figures check which. Across the sweep the determinant falls to nought on some triangles, and every one of those is found to be isosceles. That is the theorem an earlier essay states without proof — the incentre is on the Euler line exactly when two sides are equal — and it arrives here as a fact about when a specific expression vanishes.

The check is also the essay’s own guard against a wrong claim. The first version required the determinant to stay away from nought across the whole sweep, and a nearly-isosceles triangle in it made the figure refuse to draw. The refusal was correct and the claim was wrong, which is the kind of error a sweep with a strong claim in it catches and a sweep with a weak one does not.

Concurrency, which is the same determinant #

The determinant has a second reading and it is worth having, because it doubles what the calculation settles.

Three points are collinear when the determinant of their weight triples vanishes. Three lines are concurrent when the determinant of their coefficient triples vanishes — and a line in barycentric coordinates is an equation $l_1\alpha + l_2\beta + l_3\gamma = 0$, so it has a triple of its own.

So the same computation decides both kinds of coincidence, and the two are exchanged by swapping points for lines. That the medians meet, that the altitudes meet, that the perpendicular bisectors meet: each is one determinant in the lines’ coefficients, and each is a polynomial identity.

An earlier essay treats each concurrency as its own small miracle — “three perpendicular bisectors meet because a point equidistant from A and B and a point equidistant from B and C is equidistant from A and C” — and that argument is better than the determinant, because it explains. What the determinant adds is that the three concurrences are three instances of one calculation, and that a fourth would be found the same way.

The exchange of points for lines is the plane’s own duality, and it is the reason barycentric coordinates are the natural ones here rather than merely convenient: the coordinates make points and lines the same kind of object, so an incidence between them is symmetric and the determinant does not care which is which.

What the mechanical version costs #

It tells nobody why. That earlier essay puts the trade plainly: a proof by coordinates “settles a whole family at once and tells nobody why any of it is true, which is the standing trade between a proof that convinces and a proof that explains.” The determinant for the Euler line is a polynomial identity with no geometric content, and reading it teaches nothing about triangles.

The expressions grow. The circumcentre’s weights are degree four in the side lengths, the orthocentre’s degree four, and a determinant of three such rows is degree twelve before simplification. For the classical centres that is manageable; for the further entries in the catalogue it is not something anybody writes down, and the calculations are done by machine.

And it does not reach the irrational family. Every identity among the rational centres is a polynomial identity and is settled by a computation. A statement joining them to the incentre — Feuerbach’s tangency, or the incentre’s distance from the Euler line — needs the square roots cleared, which squares the expressions and loses sign information, and that is exactly why the tangency theorem is hard when the collinearities are easy.

What a centre has to be #

The weights make it possible to say precisely what a triangle centre is, which an earlier essay defines informally as “a point defined from the triangle by a rule that does not depend on how the triangle is labelled or where it sits”.

In weights that becomes: a centre function is a function $f(a, b, c)$ of the side lengths, with the centre’s weights $f(a,b,c) : f(b,c,a) : f(c,a,b)$, subject to two conditions. It must be symmetric in its last two arguments — $f(a, b, c) = f(a, c, b)$ — so that relabelling the two corners other than the first does not move the point. And it must be homogeneous, so that scaling the triangle scales nothing.

Both conditions are visible in the five triples. The centroid’s $f = 1$ is symmetric and homogeneous of degree nought. The incentre’s $f = a$ is symmetric in $b$ and $c$ trivially and homogeneous of degree one. The circumcentre’s $f = a^2(b^2 + c^2 - a^2)$ is symmetric in $b$ and $c$ and homogeneous of degree six.

So the catalogue is a catalogue of functions, and its thousands of entries are thousands of choices of $f$. That is a deflating description and it is the accurate one: a great many entries are $f$ written in a slightly different form, which is why the question of how much of the catalogue is independent has no clean answer.

What the pictures cannot show #

The table is the figure, and a table of weights is a poor picture of a triangle. What the weights mean — that the point is where the triangle balances with those loads at its corners — is a physical image the figure does not draw, and it is the only intuition the coordinates come with — the same balancing a shear proof uses for areas. Nor do the figures draw a circle condition, which in weights is quadratic rather than linear and so has no determinant of its own.

The determinants are reported as numbers and the claim is that one of them is identically nought. A number of order $10^{-17}$ is consistent with an identity and does not establish one; the identity is an algebraic fact and the figure’s contribution is the sweep, which shows the number staying at that size across two hundred triangles rather than drifting.

And the isosceles exception is located rather than drawn. The figure reports how many triangles in its sweep put the incentre on the line and requires that each has two equal sides; what the configuration looks like as a triangle approaches isosceles, with the Euler line rotating into the axis of symmetry, is a motion no static picture carries.

Two centres the weights make easy to find #

The value of a mechanical apparatus is the things it produces without insight, and two of them are worth showing because they are not in the classical four.

The symmedian point is $a^2 : b^2 : c^2$ — the incentre’s weights squared. Geometrically it is where the reflections of the medians in the angle bisectors meet, which is a construction nobody would guess from the weights and which the weights settle immediately: the triple is rational in the squared side lengths, so the point is in the rational family and every collinearity involving it is a polynomial identity.

The Gergonne point is where the three lines from each corner to the incircle’s touch point on the opposite side meet. Its weights involve $1/(s-a)$ and its relatives, which after clearing denominators are polynomials in the side lengths — unsquared. So it is in the irrational family, and a statement relating it to the Euler line is a hard one.

Both classifications were made by looking at the weights and not at the geometry. That is the apparatus working as intended: a question about which theorems will be easy is answered by inspecting an expression, before any theorem is attempted.

The pattern in the squares is worth noting on its own. Squaring the incentre’s weights gives the symmedian point; the circumcentre and orthocentre have weights that are products of the quantities $b^2+c^2-a^2$; and a great many catalogued centres are built from those two ingredients. The catalogue is not a flat list but a small vocabulary used many times, which is the structural statement the next section says nobody can make precise.

Still open: what the catalogue’s structure is #

Thousands of triangle centres are catalogued and the weights of each are known. Writing them all as weight triples turns every claimed incidence into a determinant, and machine verification of the catalogue is routine. What is not available is an account of its structure.

The natural question is which identities generate the rest, and it has no accepted formulation. There is no agreed notion of a basis for the incidences, no measure of how many independent coincidences a triangle has, and no theorem saying that a given identity is not a consequence of others. So the catalogue is verified and not understood, which is an unusual state for a subject with this much computation in it.

One direction that does have structure is the distinction this essay turns on. The centres whose weights are rational in the squared side lengths form a family closed under the operations that produce new centres from old, and the ones needing a square root form another; the boundary between them is where the hard theorems are, and classifying which statements cross it is a question with a definite shape and no general answer.

What the change of coordinates bought #

Nothing about the triangle changed. The Euler line was collinear before, the incentre was off it, and Feuerbach’s tangency held. What changed is that each of those is now a statement about an expression, and an expression can be evaluated.

The gain is that a family is settled at once. A scaling argument proves one coincidence and has to be found again for the next; a determinant identity proves the same coincidence for every triangle and the calculation is reusable without thought. That is the standard reason a subject moves to coordinates, and the standard cost is that the proofs stop being readable.

The distinction the weights make visible is the part worth keeping. A centre defined by equal distances to points has rational weights and a centre defined by equal distances to lines does not, because a distance to a line brings in a square root. Which family a centre belongs to predicts which theorems about it are easy — the tangency being the standing example of a hard one, and that prediction is available from the weights before any theorem is attempted.