Volume 40, 1907 This text is also available in PDF
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– 333 – ### Art. XXXI.—On Isogonal Transformations: Part II.

[Read before the Philosophical Institute of Canterbury, 4th December, 1907.]

1. If from a point P perpendiculars PD, PE, PF be drawn to the sides BC, CA, AB respectively of the triangle of reference ABC, it is easily shown that the perpendiculars from A, B, and C on EF, FD, and DE respectively are concurrent in a point P', and that the points P and P' are isogonal conjugates.

If now the point P be supposed to move on to the circle ABC, the point P' will move to infinity, and the pedal triangle DEF will become the Simson line of the point P. Hence we derive the important theorem—“The isogonal conjugate of a point on the circumcircle of the triangle of reference lies at infinity in a direction 0perpendicular to the Simson line of the given point.”

2. In this paper use will also be made of the following therem: “The Simson lines of the extremities of a chord of a circle intersect at an angle equal to that at which the chord cuts the circle.” This may be easily proved from the consideration that if the perpendicular drawn from any point P on the circle ABC to BC meets that circle again in the point A', then AA' isparallel to the Simson line of P.

3. It has been shown in section 4 of Part I of this paper that the asymptotic angle of the circumconic which is the isogonal transformation of a chord of the circumcircle of the triangle of reference is equal to the angle at which that chord cuts the cirlce. Combining this with sections 1 and 2 of this paper, we see that the asymptotoes of the conic which is the isogonal transformation of a chord PQ of the circle ABC are perpendicular to the Simson lines of the points P and Q.

In general, if S' be the isogonal transformation of a curve S, and if S cut the circle ABC in the points P, Q, R……, then the directions of the asymptotes of S' are perpendicular to the Simson lines of the points P, Q, R……

4. If the position of a point P be determined by the intersection of the circle ABC and the conic whose equation is

lβγ + mγα + nαβ = 0,

– 334 – then the Simson line of the point P is perpendicular to the line lα + mβ + nγ = o, and its equation may be written

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α/l(b/m-c/n)/dω/dl/α + b/m(c/n-α/l)/dω/dm/β + c/n(α/l-b/m)/dω/dn/γ = 0

where

ω ≡ l2+ m2+ n2-2mn cos A-nl cos B-2lm cos C.

5. The isogonal transformation of a tangent to the circle ABC is a parabola circumscribed to that triangle. If the tangent touch at the Point P, then thje axis of the parabola is perpendicular to the Simson line of the point P. The isogonal transformations of two tangents TP, TQ are two parabolas having their axes inclined at an angle equal to that at which PQ cuts the circle ABC. Parallel tangents transform isogonally into two parabolas passing through four concyclic points and having their axes mutually perpendicular.

6. Let four points A, B, C, D (no three of which are collinear) be taken, and let the triangles formed by omitting in turn each of the points be called Δ1, Δ2, Δ3, Δ4: let also the isogonal conjugates of A, B, C, D with regard to the triangles Δ1, Δ2, Δ3, Δ4 be called respectively A′, B′, C′, D′. If the tangents from A′, B′, C′, D′ tough the circumcircles of Δ1, Δ2, Δ3, Δ4 in P1Q1 : P2Q2 : P3Q3 : P4Q4, then the two parabolas which can be drawn through the four given points may be regarded as the isogonal transformation of any pair of tangents to the corresponding circumcircle Hence we see that the eight points of contact of the tangents may be arranged in two groups of four points such that the Simson lines of the points of each group are parallel to one another. This result may also be expressed by saying that each of the chords of contact PQ cuts its associated circle at the same angle—viz., the angle at which the axes of the parabolas are inclined to each other.

7. If the direction of the axis of a parabola circumscribing the triangle ABC be given, the line of which the parabola is the isogonal transformation may be constructed in the following manner: Draw through A a chord AA′ perpendicular to the given direction; let the perpendicular from A′ on BC meet the circle ABC in the point P, then the tangent at P to the circle ABC will isogonally transform into a parabola whose axis is perpendicular to the Simson line of P, and therefore parallel to the given direction.

8. Let a straight line L ≡ lα + mβ + nγ = o be taken, and let p be its distance from the centre of the circle ABC and φ

– 335 – [The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

the angle at which it cuts that circle. Then if P < R, the radius of the circle ABC, L will isogonally transform into the hyperbola S ≡ lβγ + mγα + nαβ = o, whose eccentricity (ε) is given by the relation ε = sec φ/2. From this we may deduce the following expression for the eccentricity:-

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ε2 = 2R/p + R

We now proceed to the case where p > R. Suppose the line L' ≡ l'α + m′β + n′γ = 0 be drawn parallel to L and passing through the pole of that line with respect to the circle ABC. Let p′ be the distance of L′ from the centre of the circle ABC, and let L′ cut that circle at the angle φ′.

The line L transforms into an ellipse, and the angle (ψ) between its equi-conjugate diameters, expressed in terms of the invariants θ, θ′, is given by

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cos2 ψ = Θ′2 - 4 Θ/Θ′2

Taking S to be 2lβγ + 2mγα + 2nαβ = 0, we have

Θ′ = - 2(l cos A + m cos B + n cos C)

Θ = - l2 sin2 A - m2 sin2 B - n2 sin2 C + 2 mn sin B sin C + 2 nl sin C sin A + 2 lm sin A sin B,

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whence cos2 ψ = ω/(l cos A + m cos B + n cos C)2

where

ω = l2 + m2 + n2 - 2mn cos A - 2nl cos B - 2lm cos C.

We also have

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p2 = R2 (l cos A + m cos B + n cos C)2

and pp′ = R2,

therefore

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cos ψ = R/p = p′/R = cos ψ′

Hence we derive the result that the angle between the equiconjugate diameters of S is equal to the angle at which L′ cuts the circle ABC.

Moreover, since L and L′ are parallel, their isogonal transformations, S and S′, will intersect in four concyclic points: the chords of the circle joining these four points will be equally inclined to the axes of each of the conics: in other words, the axes of the two conics will be parallel, and therefore the Simson lines of the two points in which L′ cuts the circle ABC

– 336 – will be perpendicular to the equi-conjugate diameters of the ellipse S. The directions of the axes of that conic may therefore be determined.

If ε and ε′ are the eccentricities of S and S′ respectively,

we have

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ε2 = 2R/p + R, ε′2 = 2R/p′+ R

Eliminating p and p′ by means of the relation pp′ = R2, we have

ε2 + ε′2 = 2.

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9. The foci of any conic inscribed in the triangle of reference are isogonal conjugates. If the trilinear co-ordinates of one focus be (α0β0γ0), then the co-ordinates of the other focus will be (k20, k20, k20) where k is the semi-minor axis of the conic.

The conic may be regarded as the envelope of a variable line lα + nβ + nγ = 0, which moves so that the product of the perpendiculars on it from the foci is equal to k2. The relation between l, m, n is easily found to be

mnα0Θ1 + nlβ0Θ2 + lmγ0Θ3 = 0,

where Θ1 ≡ β02 + γ02 + 2β0γ0 cos A

Θ2 ≡ γ2 + α02 + 2γ0α0 cos B

Θ3 ≡ α02 + β02 + 2α0β0 cos C,

and the equation of the inscribed conic is

√αα0Θ1 + √ββ0Θ2 + √γγ0Θ3 = 0

If D be the focus (α0β0γ0), then Θ1, Θ2, Θ3 are respectively (DA sin A)2, (DB sin B)2, (DC sin C)2.

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10 If we take D to be the incentre of the triangle ABC, then Θ1 = 4r2 cos2A/2, Θ2 = 4r2 cos2B/2, Θ3 = 4r2 C/2, and we obtain the equation of the incircle, viz.,

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cosA/2 √α + cosB/2 √β + cosC/2 √γ √γ = 0

In a similar manner the equations of the ex-circles may be at once determined.

Let D be the centre of the circle ABC, then

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Θ1/sin2A = Θ2/sin2 = Θ3/sin2= R2,

and we find the equation of the inscribed conic having its foc

– 337 – at the circum- and ortho-centres of the triangle of reference to be

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sin A √α cos A + sin B √β cos B + sin C √γ cos C = 0

This conic has the nine-point circle of the triangle as its auxiliary circle, and its eccentricity is

√1–8 cos A cos B cos C.

If three conics be inscribed in the triangle of reference (supposed acute), the middle points of the perpendiculars from the vertices on the opposite sides being each a focus of one conic, then the major axes of the conics all pass through the centroid of the triangle.

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11. The polar of any point with respect to a rectangular hyperbola self-conjugate with respect to the triangle of reference passes through its isogonal conjugate. Taking the equation of the hyperbola to be lα2 + mβ2 + nγ2 = 0, where l + m + n = 0, the polar of any point P (α′β′γ′) is lα′α + mβ′β + nγ′γ = 0, which passes through P' (1/α′,1/β′,1/γ′).

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Let the polars of P and P' intersect in P″, then the co-ordinates of P″ are (U′/l, V′/m, W′/n), where U = α(β2 - γ2),

V - β(γ2 - α2), W = γ(α2 - β2). Hence the P″ lies on U′/α + V′/β + W′/γ = 0,

a conic which passes through P and P′. Since its equation is independent of l, m, n, we derive the following theorem: Given a fixed tirangle and a fixed point, the locus of the intersection of the polars of the given point and its isogonal conjugate with regard to rectangular hyperbolas having a given self-conjugate triangle is a conic passing through thevertices of that triangle, the given point, and its isogonal conjugate.

The tangent at any pointof the rectangular hyperbola lα2+mβ2+nγ2 = 0 passes through its isogonal conjugate. If O be the centre of the hyperbola, and if its asymptotes meet the circle ABC again in the point X, Y, then these points are the isogonal conjugates of the points in which the hyperbola is touched by its asymptotes: hence the diameter XY of the circle ABC will isogonally transform into a rectangular hyperbola whose asymptotes are parallel to those lα2 + mβ2 + nγ2 = 0.

The equation of XY is easily found to be

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l/α(cβ + bγ) + m/b(αγ + cα) + n/c(bα+αβ) = 0

– 338 – The tangents to the above hyperbola at the in-and excentres form the standard quadrilatral

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lα ± mβ ± nγ = 0

The equation of the line joining the middle points of the diagonals of this quadrilateral is

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l2α/a + m2β/b + n2γ/c = 0

Hence, since l + m + n + = 0, the envelope of this line is the circle ABC, and the line touches its envelope at the centre of the corresponding hyperbola.

The tangents to this hyperbola, at its intersections with the conic λβγ + μγα + ναβ = 0 meet the quartic lβ2+ mγ2α2 + nα2β2 = 0 in four points lying on the strainght line λα + μβ + νγ = 0

12. Let five points A, B, C, D, E, no three of which are collinear, be taken. If any three—say, A, B, C—be taken as the vertices of the triangle of reference, and the isogonal conjugates D′ E′ of the two remaining points be consturcted, then the conic through the five given points may be regarded as the isogonal transformation of D′ E′ with respect to the triangle ABC. If, therefore, D′ E′ touch the circle ABC, then will each of the lines A′ B′, A′ C′……formed in a corresponding manner touch the circles CDE, BDE……respectively. If the line D′E′ cut the circle ABC at an angle φ, then the lines A′ B′, A′ C′……will cut the corresponding circle CDE, BDE……at the same angle φ, and the Simson lines of the points of intersection of each line with its associated circle will form two sets of parallel lines.

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13. In connection with the theory of isogonal transformation are certain curves which remain unaltered when the coordinates of any point (α,β,γ) are changed into (1/α, 1/β, 1/γ) Among such curves we have the conics of the forms

α2 ± βγ = 0

α2 ± βγ κα(β + γ) = 0

Other curves are homogeneous functions of U, V, W, such as

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U/α0 + V/β0 + W/γ0 = 0 (i)

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U/l(b/m - c/n) + V/m(c/n-a/l) + W/n(a/l-b/m) (ii)

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√α0U + √β0V + √γ0W = 0 (iii)

– 339 – If P be any point (αβγ), and P its isogonal conjugate, then

(i) expresses that PP′pases through the fixed point (α0β0γ0),

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(ii) expresses that PP′ is parallel to the line lα + mβ + nγ = 0, and (iii) expresses that PP′ touches the conic α0/α + β0/β + γ0/γ = 0.

Such curves as the above posses the property that the tangent at any point transforms isogonally into a conic touching the curve at the isogonal conjugate of the point of contact of the tangent.