Volume 45, 1912
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### Art. XLV.—On certain Tripolar Relations: Part I.

[Read before the Philosophical Institute of Canterbury, 4th December, 1912.]

 § 1. [The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.] If any point P be taken in the plane of the triangle ABC, its tripolar co-ordinates are AP2, BP2, CP2. Writing X, Y, Z for these quantities we have X sin2A = β2 + γ2 + 2βγ cos A Y sin2B = γ2 + α2 + 2γα cos B Z sin2C = α2 + β2 + 2αβ cos C, where (α, β, γ) are the trilinear co-ordinates of the point P. The fundamental relation—due to Cayley–connecting the mutual distances of the points A, B, C, P is a2X2 + b2Y2 + c2Z2–2bc cos A YZ–2ca cos B ZX–2ab cos C XY–2abc (a cos A X + b cos B Y + c cos C Z) + a2b2c2 = O (1). [The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.] If X: Y: Z = λ: μ: v, then there are two points P, Q which satisfy this relation—viz., the common points of the coaxial system of circles X/λ = Y/μ = Z/v This system of circles is out orthogonally by the circle ABC, and P, Q are inverse points with respect to the circle ABC. § 2. [The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.] The equation lX + mY +nZ == K represents in general a system of concentric circles as K varies. Transforming to trilinear co-ordinates the tripolar equation lK + mY + nZ = o, we have (α sin A + β sin B + γ sin C) {α/sin A (m/sin2B + n/sin2C) + β/sin B (n/sin2C + l/sin2A) + γ/sin C (l/sin2A + m/sin2B)} = l + m + n/sin A sin B sin C (βγ sin A + γα sin B + αβ sin C). [The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.] Hence if l + m + n = o, the equation lX + mY + nZ = o reduces to the product of the line at infinity, and the line α/a (m/b2 + n/c2) + β/b (n/c2 + l/a2) + γ/c (l/a2 + m/b2) = o, which, being satisfied by (cos A, cos B, cos C), is a diameter of the circle ABC. The tripolar equation of the diameter through the point (α1β1γ1) is [(b2 - c2) aα1 + a2 (bβ1 - cγ1)] X + [(c2 - a2) bβ1 + b2 (cγ1 - aα1)] Y + [(a2 - b2) cγ1 + c2 (aα1 - bβ1)] Z = o.
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 The following particular cases may be noted:– Euler's line ∑[(b2 - c2) X] = o. Brocard diameter ∑ [a2 (b2 - c2) X] = o. Diameter through in-centre ∑ [a (b - c) X] = o. Diameter through symmedian point of medial triangle of triangle ABC ∑ [cos A cos (B - C) X] = o. Diameter parallel to BC aX - b cos C Y - c cos B Z = o. Diameter through vertex A (b2 - c2) X - b2Y + c2Z = o. Diameter parallel to tangent to circle ABC at vertex A a cos (B - C) X - b cos B Y - c cos C Z = o. § 3. If l + m + n = o, the equation lX + mY + nZ = K represents a straight line parallel to lX + mY + nZ = o. We may now find the equation in tripolar co-ordinates of the line whose trilinear equation is pα + qβ + rγ = o. Let the required equation be written L' ≡ lX + mY + nZ + 4 Δ2K = o, where l + m + n = o and Δ is the area of the triangle of reference. [The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.] Transforming L' to trilinear co-ordinates and dividing out by aα + bβ + cγ, we have α/a (m/sin2B + n/sin2C) + β/b (n/sin2C + l/sin2A) + γ/c (l/sin2A + m/sin2B) + K (aα + bβ + cγ) = o. [The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.] Hence, comparing coefficients, we have m/sin2B + n/sin2C + Ka2 = γpa n/sin2C + l/sin2A + Kb2 = γqb l/sin2A + m/sin2B + Kc2 = γrc, which give 2l + 4KΔ sin A cos A = γ sin2A (- pa + qb + rc) 2m + 4KΔ sin B cos B = γ sin2B (pa - qb + rc) 2n + 4KΔ sin C cos C = γ sin2C (pa + qb - rc), whence, by addition, we obtain [The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.] K = γR/2Δ (- p cos A + q cos B + r cos C) 2l = γa (- p + q cos C + r cos B) 2m = γb (p cos C - q + r cos A) 2n = γc (p cos B + q cos A - r). Hence the required line is L' ≡ (- p + q cos C + r cos B) aX + (p cos C - q + r cos A) bY + (p cos B + q cos A - r) cZ = o. The tripolar equations of the sides of the triangle of reference are aX - b cos C Y - c cos B Z = abc cos A - a cos C X + bY - c cos A Z = abc cos B - a cos B X - b cos A Y + cZ = abc cos C.
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• The equation of the line through (α1β1γ1) parallel to BC is - (bβ1 + cγ1) α + α1 (bβ + cγ) = o, whence, substituting - (bβ1 + cγ1), bα1, cα1, for p, q, r respectively in L' we have -aX + b cos C Y + c cos B Z + abc cos A = 4 Δα1 a cos C X - bY + c cos A Z + abc cos B = 4 Δ β1 a cos B X + b cos A Y - cZ + abc cos C = 4 Δ γ1, relations by means of which we can transform from trilinear to tripolar co-ordinates.

These three relations constitute Lucas's Theorem. I am unaware of the method by which Lucas obtained them. Casey has proved them by an application of Stewart's Theorem.

By squaring these relations and making use of the fundamental relation (i) we obtain 4a2α12 = 4YZ - (Y + Z - a2)2 4b2β12 = 4ZX - (Z + X - b2)2 4c2γ12 = 4XY - (X + Y - c2)2

Hence, since 2 Δ ∑ [(Y - Z)2 cos A] = ∑ (a2X2) - 2 ∑ (bc cos A YZ) = 8RΔ [∑ (a cos A X) - 2RΔ], we easily deduce that 4 (a2 cot A α12 + b2 cot B β12 + c2 cot C γ12) = [2 ∑ {a2 cot A (Y + Z) - ∑ (a4 cot A)}] - 4R [∑ (a cos A X) - 2RΔ], whence 8R2 (sin 2 Aα12 + sin 2 Bβ12 + sin 2 Cγ12) = [4R ∑ {a cos (B - C) X} - 2R ∑ (a3 cos A)] - 4R [∑ (a cos A X) - 2RΔ] = 8R [∑ (a cos B cos C X) - 4RΔ cos A cos B cos C] = 16R2 cos A cos B cos C [∑ (tan A X) - 2Δ].

Hence the tripolar equation of the polar circle of the triangle ABC is tan A X + tan B Y + tan C Z = 2Δ.

It will be noticed that if in the relation 4a2α12 = 4 YZ - (Y + Z - a2)2 X be substituted for α, the result is the tripolar equation of the parabola which has its focus at A and has BC for its directrix.

If the two lines lX + mY + nZ = k and l'X + m'Y + n'Z = k1 be parallel, then ∑ (l) = o and l/l' = m/m' = n/n'

The conditions that the two lines should be at right angles to each other are ∑ (l) = o, ∑ (l') = o, and a2 (mn' + m'n) + b2 (nl' + n'l) + c2 (lm' + l'm) = o.

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 § 4. If d1, d2, d3 be the lengths of the sides of the pedal triangle of the point P (α1β1γ1), then d12 = X sin2A, d22 = Y sin2B, d32 = Z sin2C. Hence if the sum of the squares of the sides of the pedal triangle of P be given, the locus of P is the circle a2X + b2Y + c2Z = κ, a circle whose centre is at the symmedian point of the triangle ABC. The minimum value of d12 + d22 + d32 is 3a2b2c2/∑ (a2) The locus of points whose pedal triangles are right-angled is a2X = b2Y + c2Z and two similar circles. The above circle passes through B and C and has its centre at the point in which the tangents to the circle ABC at B and C intersect. If the pedal triangle of P be equilateral, then a2X = b2Y = c2Z = κ. Substituting in relation (i) we obtain the quadratic equation in κ κ2 [∑ (a4) - ∑ (b2c2)] - κa2b2c2 ∑ (a2) + a4b4c4 = o. The roots of this equation are real, showing that every triangle has two points whose pedal triangles are equilateral. [The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.] It is easily deduced that the areas of the two triangles are √3 Δ2 [∑ (a2) + √3 Δ] / 2 [∑ (a4) - ∑ (b2c2)] and √3 Δ2 [∑ (a2) - √3 Δ] / 2 [∑ (a4) - ∑ (b2c2)] The coaxial circles a2X = b2Y = c2Z are the circles of Apollonius, and their common points P, P' lie on the Brocard diameter. [The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.] The distance between P and P' is √3 abc Δ / ∑ (a4) - ∑ (b2c2) § 5. [The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.] If P (α1β1γ1) and Q (κ2 / α1 κ2 / β1 κ2 / γ1) be isogonal conjugates, and if AP = P1, BP = P2, CP = P3, AQ = Q1, BQ = Q2, CQ = Q3, then P12 sin2A = β12 + γ12 + 2β1γ1 cos A Q12 sin2 A = κ4 (1 / β12 + 1 / γ12 + 2 cos A / β1γ1) κ4 P12 sin2 A / β12γ12 [The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.] hence Q1β1γ1 = ± κ2P1, Q2γ1α1 = ± κ2P2, Q3α1β1 = ± κ2P3. Multiplying these in turn by aα1, bβ1, cγ1 and adding α1β1γ1 (aQ1 + b2 + c3 = κ2 (aP1α1 ± bP2β1 ± cP3γ1) Also κ2 (a / α1 + b / β1 + c / γ1) = 2 Δ, hence (aβ1γ1 + bγ1α1 + cα1β1) (aQ1 + bQ2 +cQ3 = 2 Δ (aP1α1 ± bP2β1 ± cP3γ1). [The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.] If d be the distance between (α1β1γ1) and the circum-centre, then aβ1γ1 + bγ1α1 + cα1β1 = Δ / R (R2 - d2) Hence aQ1 + bQ2 +cQ3 = 2R / R2 - a2 (aP1α1 ± bP2β1 ± cP3γ1).
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• From the above relations we easily obtain Q1S1 = ± 2Δ P1α1 Q2S1 = ± 2Δ P2β1 Q3S1 = ± 2Δ P3γ1, where S1 = aβ1γ1 + bγ1α1 + cα1β1, therefore α1: β1: γ1 = Q1/P1: Q2/P2: Q3/P3, (ii)

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

The signs of the ratios following those of the co-ordinates Also aQ1S1/P1 = ± 2Δ aα1 bQ2S1/P2 = ± 2Δ bβ1 cQ3S1/P3 = ± 2Δ cγ1 hence S1(aQ1/P1 + bQ2/P2 + cQ3/P3 = 2 Δ (aα1 ± bβ1 ± cγ1)

If (α1β1γ1) move on a circle of radius d concentric with the circumcircle, S1 = Δ/R (R2d2), and therefore aQ1/P1 + bQ2/P2 + cQ3/P3 = 4 Δ R/R2d2

Hence the theorem “If P be any point within the triangle ABC at a distance d from the circum-centre, and if Q be its isogonal conjugate, then BC. AQ/AP + CA. BQ/BP + AB. CQ/CP = 4Δ R/R2d2 (iii)

If P lie outside the triangle ABC, the relation (iii) must be modified in accordance with relation (ii), and d2–R2 substituted for R2d2.