Go to National Library of New Zealand Te Puna Mātauranga o Aotearoa
Volume 43, 1910
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– 669 –

Art. LVII.—On Centroidal Triangles.

[Read before the Philosophical Institute of Canterbury, 7th December, 1910.]

1. Let the side BC of any triangle ABC be divided internally in the point X′ and externally in the point A′ in the ratio p : q; let CA, AB be similarly divided in the points Y′, B′ and Z′, C′ respectively. The triangles X′Y′Z′, A′B′C′ are termed “centroidal” triangles, inasmuch as the centroids of these triangles are coincident with that of the triangle of reference ABC.

– 670 –

The co-ordinates of the points X′, Y′, Z′ are respectively

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(o, λ q/b, λp/c), (λ p/a, o, λq/c), (λ q/a, λp/b), o)

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where λ = 2Δ/p + q, Δ being the area of the triangle of reference.

The co-ordinates of the points A′, B′, C′ are respectively

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(o, μq/b, μp/c), (μp/a, μq/b, o), (μq/a, μp/b, o),

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where μ = 2Δ/pq.

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It follows that the co-ordinates of the centroids of the triangles X′Y′Z′, A′B′C′ are (2Δ/3a, 2Δ/3b, 2Δ/3c), which are those of the centroid of the triangle ABC. Since this result is independent of p and q we see that all triangles formed in this manner are co-centroidal with the triangle ABC.

Let now the sides Y′Z′, Z′X′, X′Y′ be divided in X0, Y0, Z0 so that Y′X0 : X0Z′ = Z′Y0 : Y0X′ = X′Z0 : Z0Y′ = p : q, then the co-ordinates of the points X0, Y0, Z0 are respectively

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(2νpq/a, νp2/b, νq2/c), (νp2/a, 2νpq/b, νp2/c), (νp2/a, νq2/b, 2νpq/c),

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Where ν = λ/p + q = 2Δ/(p + q)2.

Hence the triangle X0Y0Z0 is also co-centroidal with the triangle ABC. This result also holds for a triangle similarly formed by dividing the sides of the triangle A′B′C′, and the process may evidently be continued indefinitely.

2. The following simple relations may be easily proved :—

If Δ, Δ′, Δ, be the areas of the triangles ABC, X′Y′Z′, A′B′C′ respectively, then.

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If Δ′ = p2pq + q2/(p + q)2Δ

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Δ, = p2 + pq + q2/(p − q)2Δ.

Δ′ = nΔ, the minimum value of n is ¼, in which case p = q, and the triangle X′Y′Z′ is the medial triangle of the triangle ABC.

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The triangles AY′Z′, BZ′X′, CX′Y′ are equal in area, the common value being pq/(p + q)2Δ.

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The triangles AB′C′, BC′A′, CA′B′ are equal in area, the common value being pq/(p − q)2Δ.

The centroids of the triangles AY′Z′, AB′C′, as the ratio p : q varies, lie on straight lines parallel to BC and bisecting AG, where G is the centroid of the triangle ABC.

The middle points of the sides of all centroidal triangles lie on the sides of the medial triangle of the triangle ABC.

– 671 –

If the sides of the triangles X′Y′Z′, A′B′C′ be respectively x′, y′, z′ and a′, b′, c′

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Σ(x2) = p2pq + q2/(p + q)2 Σ (a2)

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Σ(a2) = p2 + pq + q2/(p − q)2 Σ (a2).

If p1, p2, p3 be the perpendiculars from G on the sides of a centroidal triangle, and p′, p″, p‴ the perpendiculars on those sides from A, B, C, then

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p1/p′ = p2/p″ = p3/p

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The equation of the circle circumscribing any centroidal triangle is (λ - 1)(λ3 - 1) abc (aβγ + bγα + cαβ) + λ (aα + bβ + cγ) [aα (a2λ + b2λ2 + c2) + bβ (a2 + b2λ + c2λ2) + cγ (a2λ2 + b2 + c2λ)] = 0,

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and the radical axis of this circle and of the circle ABC envelops, as λ varies, the conic (a4 - 4b2c2) a2α2 + (b4 - 4c2a2) bβ2 + (c4 - 4a2b2) c2γ2 - 2 (2a4 + b2c2) bcβγ - 2 (2b4 + c2a2) caγα - 2 (2c4 + a2b2) abαβ = 0.

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The locus of the symmedian point of the triangle AY′Z′ is the curve 2bc (cβ3 + bγ3) - c (c2 - a2) β2γ + b (a2 - b2) βγ2 = abcαβγ.

3. The equations of the lines B′C′, C′A′, A′B′ are respectively

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L′ ≡ pqaα + q2bβ + p2cγ = 0

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M′ ≡ p2aα + pq bβ + q2cγ = 0

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N′ ≡ q2aα + p2bβ + pqcγ = 0,

while those of Y′Z′, Z′X′, X′Y′ are respectively

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L1 = - pq aα + q2bβ + p2cγ = 0

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M1 = p2aα - pq bβ + q2cγ = 0

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N1 = q2aα + p2bβ - pqcγ = 0.

Hence as the ratio p : q varies, the lines L′, L1; M′, M1; N′, N1 envelop respectively the parabolas

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S′ ≡ a2α2 - 4bcβγ = 0

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S″ ≡ b2β2 - 4caγα = 0

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S‴ ≡ c2γ2 - 4abαβ = 0.

The points of contact of L′ and L1 with S′ are respectively

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(− 2pq/a, p2/b, q2/c), (2pq/a, p2/b, q2/c),

hence the sides of any centroidal triangle are divided internally and externally in the same ratio at the points in which they touch their enveloping parabolas.

– 672 –

The lines L′, L1 intersect on BC; M′, M1 intersect on CA; and N′, N1 intersect on AB. Calling these points of intersection A″, B″, C″ respectively, we have for the equations of the lines B″C″, C″A″, A″B″

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L″ ≡ p2q2aα + p4bβ + q4cγ = 0

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M″ ≡ q4aα + p2q2bβ + p4cγ = 0

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N″ ≡ p4aα + q4bβ + p2q2cγ = 0.

Comparing the equations L″, M″, N″ with those of L′, M′, N′ we see that the triangle A″B″C″ is a centroidal triangle formed by dividing the sides of the triangle ABC in the ratio q2 : p2.

The area of the triangle A″B″C″ is given by

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Δ″ = p4 + p2q2 + q4/(p2q2)2 Δ,

and therefore the areas of the triangles A″B″C″, A′B′C′, X′Y′Z′, and ABC are connected by the relation

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Δ″. Δ = Δ′. Δ,.

If BC, CA, AB be divided internally in X″, Y″, Z″ so that (A″BX″C), (B″CY″A), (C″AZ″B) form harmonic ranges, we have a fourth centroidal triangle X″Y″Z″, inscribed in the triangle ABC, the equations of whose sides, L2, M2, N2, may be formed from L″, M″, N″ by writing −q2 for q2 in the latter equations.

4. Let P′Q′R′, P″Q″R″, P1Q1R1, P2Q2R2 be respectively the poles of L′M′N′, L″M″N″, L1M1N1, L2M2N2 with regard to the triangles.

The co-ordinates of P′Q′R′ are proportional to

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(1/apq, 1/bq2, 1/cp2), (1/ap, 1/bpq, 1/cq2), (1/aq2, 1/bp2, 1/cpq).

These points are the vertices of the triangle formed by the lines AX′, BY′, CZ′; as the ratio p : q varies the loci of these points are the ellipses

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S1a2α2 - bcβγ = 0

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S2b2β2 - caγα = 0

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S2c2γ2 - abαβ = 0.

The lines AP′, BQ′, CR′ will meet S1, S2, S3 respectively in P1, Q1, R1. The position of P1 may be found by observing that (B. AP′CP1) is an harmonic pencil. Q1 and R1 may be found in a similar manner.

The lines BB′, CC′ meet in P2; CC′, AA′ in Q2; AA′, BB′ in R2. P″, Q″, R″ may be found from P2, Q2, R2 in the manner employed to determine P1Q1R1.

The four triangles P′Q′R′, P″Q″R″, P1Q1R1, and P2Q2R2 have their centroids at the point G.

5. The lines L′, L1 may be respectively written

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L′ ≡ pq (aα + bβ + cγ) - (p - q)(qbβ - pcγ) = 0

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L1 ≡ - pq (aα + bβ + cγ) + (p + q)(qbβ + pcγ) = 0.

The equations of AX′, AA′ are respectively

pbβ - qcγ = 0

pbβ + qcγ = 0.

– 673 –

Let lines drawn through A, B, C parallel respectively to L′M′N′ meet the opposite sides of ABC in D′E′F′. The equation of the line-pair AX′, AD′ will be

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(pbβ - qcγ)(qbβ - pcγ) = 0

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pq (b2β2 + c2γ2) - (p2 + q2) bcβγ = 0.

Hence the six points X′Y′Z′, D′E′F′ lie on the conic

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S0pq (a2α2 + b2β2 + c2γ2) - (p2 + q2)(bcβγ + caγα + abαβ) = 0.

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We now proceed to show that this is the Steiner ellipse of the triangle X′Y′Z′—i.e., the locus of points whose polars with respect to the triangle X′Y′Z′ pass through the point G. If 1/a, 1/b, 1/c, be substituted for α, β, γ in L1, M1, N1, these quantities have the common value p2 - pq + q2; hence the equation of the Steiner ellipse of the triangle X′Y′Z′ may be written

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1/L1 + 1/M1 + 1/N1 = 0.

This on multiplying out and dividing by the common factor p2 - pq + q2 reduces to S0.

Let lines through ABC parallel to L1M1N1 respectively meet the opposite sides of that triangle in D1E1F1. Then the six points A′B′C′, D1E1F1 lie on the conic

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S0′ ≡ pq (a2α2 + b2β2 + c2γ2) + (p2 + q2)(bcβγ + caγα + abαβ) = 0,

and this conic is at once shown to be the Steiner ellipse of the triangle A′B′C′.

The envelope of the Steiner ellipses of centroidal triangles, as the ratio p : q varies, is

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(aα + bβ + cγ)2 (√aα + √bβ + √cγ) = 0.

6. The circum-circle of the triangle AY′Z′ will for all values of the ratio p : q pass through a fixed point. The equation of the circle in question is

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p [(a2- b2) βγ + caαβ - bcγ2] - q [(c2 - a2) βγ - abγα + bcβ2] = 0.

Hence this circle passes through the intersection of two fixed circles, which may be written

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aS - bγL = 0

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aS - cβL = 0

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where S ≡ aβγ + bγα + cαβ and L = aα + bβ + cγ.

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The former of these circles touches AC at A and passes through B; the latter circle touches AB at A and passes through C. The radical axis of these circles is cβ - bγ - 0, and this line meets the circles again in the point H′ (2bc cos A/a, b, c).

Hence since Y′Z′ envelops a parabola S′ which touches AY′ and AZ′, and the circum-circle of AY′Z′ always passes through a fixed point H′, that point must be the focus of S′ = 0.

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Similarly it may be shown that the foci of the parabolas S′, S″ are at the points H″ (a2ca cos B/b, c), H‴ (a, b 2ab cos C/c) respectively.

– 674 –

It is seen by inspection that if K be the symmedian point of the triangle ABC, then H′, H″, H‴ lie on AK, BK, CK respectively.

The trilinear ratios of the middle point of the chord of the circle ABC drawn from A through K are

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2bc cos A/a, b, c).

Hence we see that the three foci H′, H″, H‴ lie on the Brocard circle or circle having as its diameter the line joining the centre of the circle ABC to the symmedian point K.

If F be the centre of the circle ABC, then H′ lies on the circle whose diameter is AF—viz., the circle

(cβ + bγ) L - 2a S = 0.

It may also be shown that the four points BCFH′ are concyclic.

If the tangent to the circle ABC at A meets BC in T, then H′ is the foot of the perpendicular from A on FT.

The equation of the Brocard circle may be written

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abc (α/a + β/b + γ/c) L - Σ(a2). S = 0,

and it is easily seen that it is satisfied by the co-ordinates of H′, H″, and H‴.

7. If δ1, δ2, δ3 be the medians drawn from A, B, C respectively to the middle points of the opposite sides of the triangle ABC, then the semilatera recta of S′, S″, S‴ are respectively

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Δ213, Δ223, Δ233,

where Δ is the area of the triangle ABC.

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Also, AH′ = bc/2δ1, BH′ = c2/2δ1, CH′ = b2/2δ1,

whence AH′2 = BH′. CH′.

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We have AK = bc/Σ(a2)· 2δ1,

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and therefore AH′. AK = b2c2/Σ(a2).

Hence if T′, T″, T‴ be the lengths of the tangents from A, B, C respectively to the Brocard circle,

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aT′ = bT″ = cT‴ = abc/√Σ(a2).

The directrices of the parabolas S′, S″, S‴ are respectively D1a cos Aα - cβ - bγ = 0

D2 ≡ - cα + b cos Bβ - aγ = 0

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D3 ≡ - bα - aβ + c cos Cγ = 0.

From the form of D1 it is seen that it passes through the point in which the tangent at A to the circle ABC meets BC.

Let the vertices of the triangle formed by D1, D2, D3 be V1, V2, V3.

The equations of the lines AV1, BV2, CV3 are respectively

β (ca + b2 cos B) - γ (ab + c2 cos C) = 0

γ (ab +c2 cos C) - α (bc + a2 cos A) = 0

α (bc + a2 cos A) - β (ca + b2 cos B) = 0.

– 675 –

Hence the triangle V1V2V3 is in perspective with the triangle ABC, the centre of perspective being the isogonal conjugate of the point

[(bc + a2 cos A), (ca + b2 cos B), (ab + c2 cos C)],

which is the centre of the Brocard circle.

If p1, p2, p3 be the lengths of the perpendiculars from V1, V2, V3, then

p1δ1 = p2δ2 = p3δ3,

the common value of these products being

⅓ [Σ(a2) + D2 (1 + cos A cos B cos C)].

If a1, b1, c1 be the lengths of the sides of the triangle V1V2V3, then

p1a1 = p2a2 = p3a3.

Hence the sides of the triangle formed by the directrices of the three parabolas S′, S″, S‴ are proportional to the medians of the triangle ABC.

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8. Writing λ for p/q we have as the equations of the sides of a centroidal triangle

λ2 cγ + λ aα + bβ = 0

λ2 aα + λ bβ + cγ = 0

λ2 bβ + λ cγ + aα = 0.

Let the condition now be determined that a triangle (A2B2C2), the parameter of whose sides is λ2, may be inscribed in a triangle (A1B1C1) whose sides have the parameter λ1.

Solving for the equation of the sides C2A2, A2B2 for the co-ordinates of the vertex A2 we have

aα : bβ : cγ = 0 : - 1 : λ2.

Hence if A2 lie on B1C1 we have the condition

λ12 λ2 = 1.

The same condition holds that B2 and C2 lie on C1A1 and A1B1 respectively.

We may now find the locus of the intersection of the corresponding sides B1C1, B2C2

λ1 aα + bβ + λ12 cγ = 0

λ2 aα + bβ + λ22 cγ = 0,

and therefore

aα : bβ : cγ = - (λ1 + λ2) : λ1λ2 : 1.

Eliminating λ1 and λ2 between the above and the equation λ12λ2 = 1, we see that the intersection of corresponding sides (BC) lies on the cubic

b3β3 + c3γ3 + abcαβγ = 0.

Similarly the intersections of corresponding sides (CA) and (AB) lie on the cubics

c3γ3 + a3α3 + abcαβγ = 0

a3α3 + b3β3 + abcαβγ = 0

respectively.

– 676 –

9. Let D′E′F′, D″E″F″ be two centroidal triangles, and let the two corresponding (BC) sides—i.e., those which touch S′ = 0—meet in the point D (α0β0γ0).

The equations of E′F′, E″F″ will be

λ′2cγ + λ′aα + bβ = 0

λ″2cγ + λ″aα + bβ = 0    (i)

when λ′, λ″ are the roots of the equation

λ2cγ0 + λaα0 + bβ0 = 0.

The sides F′D′, F″D″ will be

λ′2aα + λ′bβ + cγ = 0

λ″2aα + λ″bβ +cγ 0    (ii),

and the sides D′E′, D″E″ will be

λ′2bβ + λ′cγ + aα = 0

λ″2bβ + λ″cγ + aα = 0    (iii).

Let F′D′, F″D″ meet in E, and D′E′, D″E″ in F. The co-ordinates of E are given by

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aα : bβ : cγ = 1 : - (λ′ + λ″) : λ′λ″,

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but λ′ + λ″ = - aα0/cγ0, λ′λ″ = bβ0/cγ0,

and therefore

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α : β : γ = c/aγ0 : a/bα0 : b/cβ0.

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In the same way it may be shown that the co-ordinates of the point F are (b/aβ0, c/bγ0, b/cα0).

The equations of DE, EF, FD are respectively

aα P0 + bβ Q0 + cγ R0 = 0

aα R0 + bβ P0 + cγ Q0 = 0

aα Q0 + bβ R0 + cγ P0 = 0

where P0a2α02 - bcβ0γ0, Q0b2β02 - caγ0α0, R0c2γ02 - abα0β0.

Hence if the point (α0β0γ0) lie on any of the conics S1, S2, S3, the sides of the triangle DEF will pass through the vertices of the triangle ABC.

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We also see that if (α0β0γ0) lie on the curve f (αβγ) = 0, then E and F will lie on the curves f (cγ/a, aα/b, bβ/c) = 0 and f (bβ/a, cγ/b, aα/c) = 0.

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If the point D (α0β0γ0) lie on the Steiner ellipse 1/aα + 1/bβ + 1/cγ = 0, then the points E, F will also lie on this ellipse. In this case we have

P0 : Q0 : R0 = aα0 : bβ0 : cγ0,

and it may be easily shown that the sides of the triangle DEF touch this conic

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aα + √ bβ + √ cγ = 0.

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If D lie on the conic √ aα + √ bβ + √ cγ = 0, so also will the points E and F, and the sides of the triangle DEF will envelop the conic

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√U + √V + √W = 0,

where U ≡ - aα + bβ + cγ

V ≡ aα - bβ + cγ

W ≡ aα + bβ - cγ.

– 677 –

10. The equations of the sides of the secondary centroidal triangle PQR formed by joining the points of contact of L′, M′, N′ with S′S″S‴ respectively are

3p2q2 aα + (p4 + 2pq3) bβ + (q4 + 2p3q) cγ = 0

and two othes.

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Writing λ = p/q we have

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λ4bβ + 2λ3cγ + 3λ2aα + 2λbβ + cγ = 0.

The invariants I and J of this quartic are

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I ≡ 3a2α2/4, J ≡ ⅛ (6abcαβγ - 2b3β3 - 2c3γ3 - a3α3),

whence the envelope of the above line is

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(a3α3 + b3β3 + c3γ3 - 3abcαβγ)(b3β3 + c3γ3 - 3abcαβγ) = 0,

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whence we may infer that the envelope of the sides of the triangle PQR consists of the line at infinity, the point (1/a, 1/b, 1/c), and the system of cubics

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b3β3 + c3γ3 - 3abc αβγ = 0

c3γ3 + a3α3 - 3abc αβγ = 0

a3α3 + b3β3 - 3abc αβγ = 0.

11. The four common tangents of the conic S′ ≡ a2α2 - 4bcβγ = 0 and the circle S ≡ aβγ + bγα + cαβ = 0 form a cyclic quadrilateral.

The equation of the locus of the pole of the line ·λ2cγ + λaα + λaα + bβ = 0 with respect to the circle S is the conic

bc (cβ + bγ)2 - a2 (aγ + cα)(bβ + aα) = 0,

which may be written

a3S + bc l1l2 = 0

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where l1aα + cβ + bγ + 0

≡ L - (b - c)(β - γ) = 0

l2aα - cβ - bγ = 0

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≡ L - (b + c)(β + γ) = 0,

L being the line at infinity. The first forms of l1 and l2 show that they each pass through the point T in which the tangent to the circle ABC at A meets BC. The second form shown that l1 and l2 are parallel respectively to the internal and external bisectors of the angle A.

The tangents to S at the points in which it is met by l1 and l2 are the common tangents of S and S′.

The line l2 will always meet the circle S in real points; the ine l1 will meet it in real points if a2 > 4bc.

If two chords of a circle are at right angles, the tangents at their extremities form a cyclic quadrilateral; hence since l1 and l2 are at right angles to each other, it follows that the common tangents to S and S′ form a cyclic quadrilateral which is real if a2 > 4bc.

If a2 = 4bc, the parabola S′ and the circle S touch each other, the line l1 being the tangent at the point of contact.

The equations of the four common tangents of S and S′ are

[a (a2 - 4bc) S + bc l12] [a (a2 + 4bc) S + bcl22] = 0.