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Volume 39, 1906
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– 305 –

Art. XXIX.—On Isogonal Transformations: Part I.

[Read before the Philosophical Institute of Canterbury, 5th December, 1906.]

1. “Two points P, P′, which are such that lines drawn from them to the summits of the triangle of reference are equally inclined to the bisectors of its angles are called isogonal conjugates with respect to the triangle.”—Casey.

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If the trilinear co-ordinates of P' be (α β γ), those of P′ will be (κ2/α κ2/β κ2/γ); but as in what follows trilinear ratios will be for the most part used, the co-ordinates of P′ will be (1/α 1/β 1/γ). If the co-ordinates of P′ be written (α′ β′ γ′) we have αα′ = ββ′ = γγ′ = a constant: hence an isogonal transformation is a species of inversion, and in the following paper isogonal transformations will be described in the language of inversion.

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The incentre and three excentres (1 ± 1 ± 1) of the triangle of reference ABC are the only points which invert into themselves. The four points (β ± β ± γ) forming the vertices of a harmonic quadrangle invert into four points (1/α ± 1/β ± 1/γ) forming the summits of another harmonic quadrangle.

– 306 –

It may also be noticed that according as P is within or without the triangle ABC so is its inverse point P′ within or without that triangle.

2. The line whose equation is

lα + mβ + nγ = ο

will invert into the conic having for equation

lβγ + mγα + nαβ = ο

Also, any conic circumscribed to the triangle ABC will invert, into a line: in particular the circumcircle of the triangle ABC will invert into the line at infinity.

If a point P (α1β1γ1) be determined by the intersection of the circle ABC with the conic lβγ + mγα + nαβ = ο, it may be at once shown that the lines

ββ1 − γγ1 = ο, γγ1 − αα1 = ο, αα1 − ββ1 = ο

which determine the position of the inverse of P, are all parallel to the line lα + mβ + nγ = ο.

A line passing through a vertex of the triangle ABC inverts into a line passing through the same vertex.

3. The conic lβγ + mγα + nαβ = ο will be a hyperbola, parabola, or ellipse according as

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la + √mb + √nc > = or < ο

but this is the condition that the line lα + mβ + nγ = ο shall intersect, touch, or not intersect the circle ABC: hence the theorem that a line inverts into a hyperbola, parabola, or ellipse according as it cuts, touches, or does not cut the circumcircle of the triangle of reference.

4. The asymptotes of the conic lβγ + mγα + nαβ = ο are given by

lmn (aα + bβ + cγ)2 + Δ (lβγ + mγα + nαβ) = ο

where

Δ = a2l2 + b2m2 + c2n2 − 2bcmn − 2canl − 2ablm

It is easily shown that the angle (φ) between the asymptotes is given by

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tan φ = √Δ/2R (l cos A + m cos B + n cos C)

R being the radius of the circle ABC. Hence

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cos φ = l cos A + m cos B + n cos C/Ω

where Ω2 = l2 + m2 + n2 − 2mn cos A − 2nl cos B − 2lm cos C.

– 307 –

If p be the length of the perpendicular from the centre of the circle ABC on lα + mβ + nγ = ο, then

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p = R (l cos A + m cos B + n cos C)/Ω

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therefore cos φ = p/R: but p/R is the cosine of the angle between the chord lα + mβ + nγ = ο of the circle, ABC and the tangent to the circle at the extremity of the chord, hence the angle between the asymptotes of the conic lβγ + mγα + nαβ = ο is equal to the angle at which the line lα + mβ + nγ = ο cuts the circle ABC.

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Moreover, since the excentricity (∊) of the conic is connected with φ by the relation ∊ = sec φ/2 and p = R cos φ, it follows at once that tangents to circles concentric with the circle ABC invert into similar conics.

5. Suppose that a curve S inverts into a curve S′: then to any two points P and Q on S will be two corresponding inverse points P′ and Q′ on S′. If now the point Q move up to P and become infinitely close to it, the point Q′ will become infinitely close to P′. Hence if the tangent to S′ at the point P be inverted, it will become a circumconic touching S′ at the point P′.

If the line lα + mβ + nγ = ο be inverted, then any tangent to the conic lβγ + mγα + nαβ = ο will invert into a conic touching lα + mβ + nγ = ο, and a pair of tangents to the conic lβγ + mγα + nαβ = ο will invert into a pair of circumconics intersecting in the point which is the inverse of that from which the tangents were drawn and having the line lα + mβ + nγ = ο as a common tangent.

6. Let two lines L1 = l1α + m1β + n1γ = ο and L2 = l2α + m2β + n2γ = ο be taken: these will invert into the conics

S1 = l1βγ + m1γα + nαβ = ο

S2 = l2βγ + m2γα + n2αβ = ο

Let L = λα + μβ + νγ = ο be a common tangent of S1 and S2: then L will invert into the conic

S = λγγ + μγα + ναβ = ο

which will have double contact with the line pair L1 L2: its equation therefore will be of the form

L1L2 − (pα + qβ + r)2 = ο

– 308 –

Comparing this with the form of S given above we have

p2 = l1l2, q2 = m1m2, r2 = n1n2

hence the equations of the four chords of contact with L1 and L2 of the conics which are the inverses of the common tangents of S1 and S2 are

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l1l2 α ± √m1m2 β + √n1n2 γ = ο

The inverses of the points in which these four lines meet L1 and L2 are the points of contact of the common tangents of S1 and S2.

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Let c1 = √l1l2 α + √m1m2 β + √n1n2 = ο

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c2 = √l1l2 α − √m1m2 β − √n1n2 = ο

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c3 = − √l1l2 α + √m1m2 β − √n1n2 = ο

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c4 = − √l1l2 α − √m1m2 β + √n1n2 = ο

and form the conic

T1 = L1L2c12 = ο

which is the inverse of a common tangent t1.

Now write

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P1 = √m1n2 − √m2n1  P2 = √m1n2 + √m2n1

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Q1 = √n1l2 − √n2l1  Q2 = √n1l2 + √n2l1

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R1 = √l1m2 − √l2m1  R2 = √l1m2 + √l2m1

Then the conic T1 reduces to

P12βγ + Q12γα + R12αβ = ο

On inversion we obtain the four common tangents of S1 and S2

t1 = P12α + Q12β + R12γ = ο

t2 = P12α + Q22 + R22γ = ο

t3 = P22α + Q12β + R22γ = ο

t4 = P22α + Q22β + R12γ = ο

To find the co-ordinates of the points of contact of t1 with S1 and S2, solve for α β γ between c1 and L1 and c1 and L2 and invert.

We thus find that t1 will touch S1 and S2 respectively in the points

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(√l1/P1m1/Q1n1/R1)    (√l2/P1m2/Q1n2/R1)

with similar expressions for the points of contact of t2, t3, and t4 with these conics.

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7. Any triangle circumscribing the conic

S = λβγ + μγα + ναβ = ο

will invert into three circumconics having the line

L = λα + μβ + νγ = ο

as a common tangent.

A family of n parabolas circumscribing the triangle of reference will invert into an n-sided polygon in which the circle ABC is inscribed.

The pencil of lines represented by the equation

l1α + m1β + n1γ + κ (l2α + m2β + n2γ) = ο

where κ varies, will invert into a family of conics passing through the four points of intersection of the conics

l1βγ + m1γα + n1αβ = ο

l2βγ + m2γα + n2αβ = ο

In particular a system of parallel lines will invert into a family of conics passing through four concyclic points.

Hence, as there will always be two lines, whether of the pencil or of the parallel system, which are equidistant from the centre of the circle ABC (excluding those lines of either system which are diameters of this circle), we see that all conics passing through four given points may be arranged in pairs of similar conics.

8. Two tangents drawn from a point P to the circle ABC will invert into two parabolas passing through ABC and P′—the inverse of P with respect to the triangle ABC.

Hence if four points, ABCD, be given, and if A′ B′ C′ D′ be respectively the inverses of those points with respect to the triangle formed by joining the remaining three points, we see that the two parabolas which may be drawn through four given points can be regarded as originating by inversion of the pair of tangents from the four points A′ B′ C′ D′ to the circles BCD, CDA, DAB, ABC respectively.

Now, if one of the points, say D′, fall within the circle ABC, the tangents from it to that circle are imaginary; and consequently the two parabolas through ABCD are imaginary: therefore the remaining points A′ B′ C′ must lie within the respective circles BCD, CDA, DAB.

We may state this result as follows: If any four points be taken on a parabola, the inverse of any one of the points with respect to the triangle formed by joining the remaining three points lies without the circumcircle of that triangle.

– 310 –

9. We may determine the equation of the two parabolas which can be drawn through ABC and P(α1 β1 γ1) as follows:—

The curve whose equation is

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√a/α + √b/β + √c/γ = ο

is the locus of points whose axes of homology touch the circle ABC, while the conic

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1/α1α + 1/β1β + 1/γ1γ = ο

is the locus of points whose axes of homology pass through

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(1/α1 1/β1 1/γ1)

Let these two curves cut in the point α′β′γ′: then

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α/α′ + β/β′ + γ/γ′ = ο

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will be a tangent to the circle through 1/α1 1/β1 1/γ1

We have also

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1/αα′ + 1/ββ′ + 1/γγ′ = ο

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√a/α′ + √b/β′ + √c/γ′ = ο

whence, eliminating α′β′γ′, we have the equation of the two tangents in the form

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aα11β − γ1γ) + √bβ11γ − α1α) + √cγ11α − β1β) = ο

and the equation of the pair of parabolas is

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aα(β/β1 − γ/γ1) + √bβ(γ/γ1 − α/α1) + √cγ(α/α1 − β/β1) = ο

10. Let there be four concyclic points A, B, C, D, and let the position of the point D be determined by the intersection of the circle ABC and the conic

lβγ + mγα + nαβ = ο

Then the two parabolas through the four points will be the inverses of the two tangents to the circle ABC which are parallel to the line lα + mβ + nγ = ο.

Consider the conic whose equation

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mcnb/α + nalc/β + lbma/γ =

– 311 –

It is the locus of points whose axes of homology are parallel to lα + mβ + nγ = ο.

Let this conic cut in the point (α′β′γ′) the curve

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a/α + √b/β + √c/γ = ο

Then the axis of homology of (α′β′γ′) will be a tangent to the circle ABC and parallel to the line. lα + mβ + nγ = ο. Eliminating (α′β′γ′) between the equations

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α/α′ + β/β′ + γ/γ′ = ο

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mcnb/α′ + nalc/β′ + lbma/γ′ = ο

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a/α′ + √b/β′ + √c/γ′ = ο

we have for the equation of the pair of tangents

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al/a(bβ + cγ) − (mβ + nγ) + bb/b(cγ + aα) − (nγ + lα)

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+ cn/c(aα + bβ) − (lα + mβ) = ο

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The equation of the two parabolas may be written down from the above by substituting in it 1/α 1/β 1/γ for α β γ respectively.

11. Any line parallel to α = ο will invert into a conic of the form

κβγ + aβγ + bγα + cαβ = ο

All conics of this family touch each other and the circle ABC at the vertex A of the triangle of reference.

The two tangents to the circle ABC parallel to α = ο invert into the pair of parabolas

(b ± c)2βγ + aα(bγ + cβ) = ο

The two tangents to the same circle drawn parallel to the diameter of the circle through A invert into the pair of parabolas

aβγ + bγα + cαβ ± 4 R sin B sin C α (β cos β − γ cos C) = ο

where R is the radius of the circle ABC.

– 312 –

12. Any diameter of the circle ABC will invert into a rectangular hyperbola.

Let the diameter be taken which is perpendicular to the line α = ο; the equation of its inverse is

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sin (B − C)/α + sin B/β − sin C/γ = ο

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This conic cuts the circle ABC at the extremity H of the diameter which passes through the vertex A of the triangle of reference; it also passes through the points (− 1/a 1/b 1/c) and the orthocentre of the triangle ABC: its centre is at the middle point of the line BC: the tangent to the conic at A passes through the symmedian point (abc) of the Δ ABC, while the tangent at H passes through the point (− abc).