Every point has a partner across the bisectors

A centre is three weights wrote every classical centre of a triangle as a weighted average of its corners, and noticed a pattern in passing: the symmedian point’s weights are $a^2 : b^2 : c^2$, the incentre’s weights squared, and the circumcentre’s and orthocentre’s weights multiply to the squares of the sides. That essay said the catalogue was a small vocabulary used many times. This one is about one word of that vocabulary — an operation on points that explains the pattern and pairs the classical centres off with each other.

Take a triangle and any point $P$ inside it. From each corner draw the line through $P$. Now reflect each of those three lines in the bisector of the angle at its own corner — the line that splits the corner’s angle into two equal halves. Three new lines, one from each corner. There is no obvious reason they should meet at a point, and yet they always do.

The figure below is the construction. The dashed blue lines run from the corners through $P$; the faint grey lines are the bisectors; the orange lines are the reflections, and they meet at a single point $P^*$, the isogonal conjugate of $P$. The figure finds $P^*$ by reflecting and intersecting, checks that the third reflected line passes through the same point, and then checks the answer against a formula that uses no reflection at all.

Why three reflections meet #

The reason is a symmetric version of the reason three lines from the corners meet in the first place, and it is cleanest in the language of angles. That two of the reflected lines meet is automatic, since any two lines that are not parallel do; the content of the theorem is entirely in the third.

A line from corner $A$ through $P$ splits the angle at $A$ into two pieces, $\alpha_1$ and $\alpha_2$. Reflecting it in the bisector swaps them: the reflected line splits the angle into $\alpha_2$ and $\alpha_1$. Do the same at $B$ and $C$. Ceva’s theorem in its trigonometric form says three lines from the corners meet at one point exactly when

$$\frac{\sin\alpha_1}{\sin\alpha_2}\cdot\frac{\sin\beta_1}{\sin\beta_2}\cdot\frac{\sin\gamma_1}{\sin\gamma_2} = 1.$$

The original three lines meet at $P$, so this product is one. Swapping every pair of angles turns the product into its reciprocal — which is also one. So the reflected lines meet too. The reflection inverts every factor, and the inverse of one is one.

The same argument, read in barycentric coordinates, gives a formula. If $P$ has barycentric weights $x : y : z$, the weights are proportional to the areas of the three small triangles $P$ cuts off, and reflecting the lines in the bisectors replaces each weight by the square of the opposite side divided by it:

$$x : y : z \;\longmapsto\; \frac{a^2}{x} : \frac{b^2}{y} : \frac{c^2}{z}.$$

The figure computes $P^*$ both ways — by reflecting and intersecting, and from this formula — and they agree to a millionth of a unit. Applying the formula twice returns $x : y : z$, so the operation is its own inverse: $P$ is the conjugate of $P^*$.

The classical centres, in pairs #

The formula makes the conjugates of the classical centres a matter of reading off weights.

The centroid has weights $1 : 1 : 1$, and its conjugate has weights $a^2 : b^2 : c^2$ — the symmedian point, the one the weights made easy to find without saying what it was. Now it has a construction: the medians, reflected in the bisectors, are the symmedians, and they meet there. Émile Lemoine studied it in 1873, and it is often called the Lemoine point; it is the same Lemoine who, fifteen years later, priced every movement of the compass.

The orthocentre and the circumcentre are partners. The altitude from $A$ makes an angle with side $AB$ equal to the angle the radius $OA$ makes with side $AC$ — both are $90°$ minus the angle at the opposite corner — so reflecting the altitudes in the bisectors gives exactly the lines from the corners to the circumcentre. Two of the four points on the Euler line are conjugates of each other.

The incentre is its own partner. Its three lines are the bisectors, and a line reflected in itself does not move. The same is true of the three excentres, whose lines are also bisectors, of the outer angles. Those four points are the only ones a conjugation fixes, and the formula agrees: $x^2 = a^2$ and its relatives have exactly the four sign choices $\pm a : \pm b : \pm c$ up to an overall sign.

The symmedian point, found by least squares #

The symmedian point has a characterisation that makes no mention of reflections, and it is the kind of coincidence that shows an operation is natural rather than contrived.

Measure the distances from a point to the three sides of the triangle, square them, and add. That sum is smallest at exactly one point inside the triangle, and the point is the symmedian point. The proof is the nearest point of a flat thing in disguise: the three distances, for points of the plane, sweep out a flat plane in three-dimensional space, since they are linear in position and tied together by one linear relation — $a\,d_a + b\,d_b + c\,d_c$ is twice the triangle’s area for every point inside. The point of that plane nearest the origin is the one whose distances are proportional to the coefficients, $d_a : d_b : d_c = a : b : c$, and distances in that ratio are exactly the symmedian point’s.

A second fact follows. The feet of the perpendiculars from the symmedian point to the three sides form a small triangle — its pedal triangle — and the symmedian point is that triangle’s centroid. So the conjugate of the centroid is the point that is the centroid of its own feet. Neither characterisation is obvious from the other, and neither is obvious from the construction by reflecting medians. The three agree because the operation that produced the point is the one that exchanges lengths with the squares of the sides, and least squares is a statement about squares.

Lines become conics #

Conjugation is not a symmetry of the plane in the ordinary sense: it does not take lines to lines.

The points of a straight line have weights satisfying a linear equation, $ux + vy + wz = 0$. Their conjugates’ weights $X, Y, Z$ satisfy $x \propto a^2/X$ and so on, and substituting turns the linear equation into $u a^2 YZ + v b^2 ZX + w c^2 XY = 0$ — a quadratic, a conic. Every term contains two of the three weights, so the conic vanishes wherever two weights are nought, which is at the three corners. Every line is carried to a conic through the corners, and every conic through the corners comes from a line.

That makes conjugation a map of a kind classical geometry did not usually consider: one that bends lines. It is a quadratic transformation, the simplest of the Cremona transformations of the plane, and it has the strange behaviour such maps have at their exceptional points. A point on side $BC$ has weight $x = 0$, so its conjugate has an infinite first weight — every point of the side, whatever its position, is carried to the single corner $A$. A whole side collapses onto the opposite corner, and the corner is blown up into the whole side. The formula is undefined at exactly the places where the construction is: at a corner, the line “through the corner and $P$” is not determined.

Applied to the Euler line, this produces a curve with a name. The Euler line holds the circumcentre, the centroid and the orthocentre, so its conjugate conic holds their conjugates — the orthocentre, the symmedian point and the circumcentre — as well as the three corners. That conic is the Jerabek hyperbola. Six notable points on one conic, and the reason for the coincidence is one line and one map.

A map that bends lines, beside one that bends circles #

There is one other familiar map of the plane that carries straight lines to curves and is its own inverse: inversion in a circle, which swaps the inside and outside of a circle and carries lines that miss the centre to circles through it. The two are close relatives, and seeing them side by side explains both.

Inversion is quadratic: in coordinates it sends $(x, y)$ to $(x, y)/(x^2 + y^2)$, and the square in the denominator is what bends lines. It is undefined at one point, the centre, which it blows up into the whole of the line at infinity, and it fixes the points of the circle. Isogonal conjugation is quadratic too, undefined at three points, the corners, each blown up into the opposite side, and it fixes four points, the incentre and excentres. Inversion turns problems about circles through a point into problems about lines, which is why it solves the problem of eight circles touching three. Conjugation turns problems about conics through three points into problems about lines, which is why the Jerabek hyperbola’s six points are one line’s worth of information.

Both belong to the family of Cremona transformations — the maps of the plane given by ratios of polynomials that have inverses of the same kind — and a theorem of Max Noether and Guido Castelnuovo says that every such map is built from the simplest quadratic one and ordinary projective maps. In that sense isogonal conjugation, read in the right coordinates, is not one example among many: it is the building block.

The circumcircle goes to infinity #

The most striking consequence concerns the triangle’s own circle.

For a point $P$ on the circumcircle, the three reflected lines are parallel. The figure checks it: the three directions agree to the last digit. Parallel lines meet only at infinity, so the conjugate of a point on the circumcircle is a point at infinity — a direction rather than a position. As $P$ moves round the circle, its conjugate’s direction turns, and every direction is reached. Conjugation carries the circumcircle to the line at infinity, the whole of it, and brings the line at infinity back to the circle.

The reason fits the pattern. The circumcircle is a conic through the three corners, so it must be the conjugate of some line, and the line turns out to be the one with equation $x + y + z = 0$ in weights — the points whose weights sum to nothing, which are the points at infinity. In the language of one circle touching four, the circumcircle and the line at infinity are the same curve seen through a quadratic lens.

The practical form of this is a statement about points near the circle. A point at ninety per cent of the radius from the centre has its conjugate some 1,800 units away, at ninety-nine per cent some 19,000, in a drawing whose triangle is a few hundred units across. The conjugate of a point just inside the circle is far outside the picture, in the direction the reflected lines are nearly parallel.

The circle two partners share #

One more property ties a point to its partner by a circle instead of by lines.

Drop perpendiculars from $P$ to the three sides; their feet are three points, and three points determine a circle — $P$’s pedal circle. Do the same from $P^*$. The six feet lie on one circle, centred at the midpoint of $P$ and $P^*$. The figure checks that the six distances from the midpoint agree to a millionth of the radius.

This is a characterisation, not just a coincidence: $P^*$ is the one other point whose three feet lie on the circle through $P$’s. And it contains a theorem met earlier. For the orthocentre and the circumcentre, the feet from $H$ are the feet of the altitudes and the feet from $O$ are the midpoints of the sides; the shared pedal circle through all six is the nine-point circle, centred halfway between $H$ and $O$. The nine-point circle is the pedal circle of an isogonal pair.

How many partners a centre has #

It is worth asking what the operation does to the catalogue as a whole. Every triangle centre has a conjugate, and the conjugate of a centre is a centre — it is defined by the triangle alone, symmetrically, as a centre must be. So the thousands of catalogued centres come in pairs, apart from the four fixed points, and a large fraction of the catalogue’s entries are conjugates of other entries.

The two Brocard points are conjugates of each other. The first is the point $\Omega$ from which the three lines to the corners make equal angles with the three sides taken in cyclic order — the angle $\Omega AB$ equals $\Omega BC$ equals $\Omega CA$ — and the second does the same in the other cyclic order. Reflecting the first point’s lines in the bisectors reverses the order, which is why the two are partners, and the common angle, the Brocard angle, satisfies $\cot\omega = \cot A + \cot B + \cot C$ and is never more than thirty degrees, reaching it only for the equilateral triangle. The Gergonne point and the Nagel point, both built from touch points of circles with the sides, have conjugates with names of their own. And the pairing respects the division a centre drawn as three weights drew between the rational family of centres, whose weights are polynomial in the squared side lengths, and the irrational family that needs the sides themselves: the formula multiplies by squares and divides, so it carries each family to itself. The symmedian point is in the rational family because the centroid is.

What the figures take on trust #

Every figure checks its claims on one triangle, or on a handful of points along one line; the algebra above proves them for every triangle and every point. The conic in the line figure is fitted through five conjugates and checked at the rest, which is evidence that the conjugates lie on a conic; the substitution into $ux + vy + wz = 0$ is the proof. The parallel lines in the circumcircle figure are checked to agree in direction at one point of the circle; the equation $x + y + z = 0$ is why they do everywhere.

The figures also stay inside the triangle, where the construction is clean. For a point outside the triangle the lines through the corners still exist and still reflect, but some of the weights are negative and the reflected lines meet on the far side of a corner; the formula handles that without comment, and the pictures do not show it. And none of them shows the collapse of a side onto a corner, because a side is where the construction breaks and a drawing of a breakdown is a drawing of nothing.

Still open: which transformations of the catalogue there are #

Isogonal conjugation is one of several involutions on triangle centres. Isotomic conjugation, $x : y : z \mapsto 1/x : 1/y : 1/z$, reflects the lines’ feet on each side about the side’s midpoint instead of reflecting the lines in the bisectors; it pairs the Gergonne point with the Nagel point and the centroid with itself. Others are built from reflections in other lines, inversions in circles attached to the triangle, and compositions of these.

What is not known is how many genuinely different operations of this kind the catalogue supports, in a sense that could be made precise, or whether a small list of them generates every incidence among catalogued centres from a few. Each operation carries lines to curves of a fixed degree and centres to centres, and each explains a family of coincidences at once — the Jerabek hyperbola’s six points are one line under one map. Whether the whole catalogue is explained that way, by a finite vocabulary of lines, conics and maps, is the structural question the weights of a centre left open, now with a candidate for part of its answer.