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 coordinates are AP^{2}, BP^{2}, CP^{2}. Writing X, Y, Z for these quantities we have X sin^{2}A = β^{2} + γ^{2} + 2βγ cos A Y sin^{2}B = γ^{2} + α^{2} + 2γα cos B Z sin^{2}C = α^{2} + β^{2} + 2αβ cos C, where (α, β, γ) are the trilinear coordinates of the point P. The fundamental relation—due to Cayley–connecting the mutual distances of the points A, B, C, P is a^{2}X^{2} + b^{2}Y^{2} + c^{2}Z^{2}–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) + a^{2}b^{2}c^{2} = 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 coordinates the tripolar equation lK + mY + nZ = o, we have (α sin A + β sin B + γ sin C) {α/sin A (m/sin^{2}B + n/sin^{2}C) + β/sin B (n/sin^{2}C + l/sin^{2}A) + γ/sin C (l/sin^{2}A + m/sin^{2}B)} = 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/b^{2} + n/c^{2}) + β/b (n/c^{2} + l/a^{2}) + γ/c (l/a^{2} + m/b^{2}) = 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 (α_{1}β_{1}γ_{1}) is [(b^{2}  c^{2}) aα_{1} + a^{2} (bβ_{1}  cγ_{1})] X + [(c^{2}  a^{2}) bβ_{1} + b^{2} (cγ_{1}  aα_{1})] Y + [(a^{2}  b^{2}) cγ_{1} + c^{2} (aα_{1}  bβ_{1})] Z = o. 
The following particular cases may be noted:– Euler's line ∑[(b^{2}  c^{2}) X] = o. Brocard diameter ∑ [a^{2} (b^{2}  c^{2}) X] = o. Diameter through incentre ∑ [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 (b^{2}  c^{2}) X  b^{2}Y + c^{2}Z = 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 coordinates of the line whose trilinear equation is pα + qβ + rγ = o. Let the required equation be written L' ≡ lX + mY + nZ + 4 Δ^{2}K = 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 coordinates and dividing out by aα + bβ + cγ, we have α/a (m/sin^{2}B + n/sin^{2}C) + β/b (n/sin^{2}C + l/sin^{2}A) + γ/c (l/sin^{2}A + m/sin^{2}B) + 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/sin^{2}B + n/sin^{2}C + Ka^{2} = γpa n/sin^{2}C + l/sin^{2}A + Kb^{2} = γqb l/sin^{2}A + m/sin^{2}B + Kc^{2} = γrc, which give 2l + 4KΔ sin A cos A = γ sin^{2}A ( pa + qb + rc) 2m + 4KΔ sin B cos B = γ sin^{2}B (pa  qb + rc) 2n + 4KΔ sin C cos C = γ sin^{2}C (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. 

The equation of the line through (α_{1}β_{1}γ_{1}) parallel to BC is  (bβ_{1} + cγ_{1}) α + α_{1} (bβ + cγ) = o, whence, substituting  (bβ_{1} + cγ_{1}), bα_{1}, cα_{1}, for p, q, r respectively in L' we have aX + b cos C Y + c cos B Z + abc cos A = 4 Δα_{1} a cos C X  bY + c cos A Z + abc cos B = 4 Δ β_{1} a cos B X + b cos A Y  cZ + abc cos C = 4 Δ γ_{1}, relations by means of which we can transform from trilinear to tripolar coordinates.
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 4a^{2}α_{1}^{2} = 4YZ  (Y + Z  a^{2})^{2} 4b^{2}β_{1}^{2} = 4ZX  (Z + X  b^{2})^{2} 4c^{2}γ_{1}^{2} = 4XY  (X + Y  c^{2})^{2}
Hence, since 2 Δ ∑ [(Y  Z)^{2} cos A] = ∑ (a^{2}X^{2})  2 ∑ (bc cos A YZ) = 8RΔ [∑ (a cos A X)  2RΔ], we easily deduce that 4 (a^{2} cot A α_{1}^{2} + b^{2} cot B β_{1}^{2} + c^{2} cot C γ_{1}^{2}) = [2 ∑ {a^{2} cot A (Y + Z)  ∑ (a^{4} cot A)}]  4R [∑ (a cos A X)  2RΔ], whence 8R^{2} (sin 2 Aα_{1}^{2} + sin 2 Bβ_{1}^{2} + sin 2 Cγ_{1}^{2}) = [4R ∑ {a cos (B  C) X}  2R ∑ (a^{3} cos A)]  4R [∑ (a cos A X)  2RΔ] = 8R [∑ (a cos B cos C X)  4RΔ cos A cos B cos C] = 16R^{2} 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 4a^{2}α_{1}^{2} = 4 YZ  (Y + Z  a^{2})^{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 = k^{1} 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 a^{2} (mn' + m'n) + b^{2} (nl' + n'l) + c^{2} (lm' + l'm) = o.
§ 4. 
If d_{1}, d_{2}, d_{3} be the lengths of the sides of the pedal triangle of the point P (α_{1}β_{1}γ_{1}), then d_{1}^{2} = X sin^{2}A, d_{2}^{2} = Y sin^{2}B, d_{3}^{2} = Z sin^{2}C. 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 a^{2}X + b^{2}Y + c^{2}Z = κ, a circle whose centre is at the symmedian point of the triangle ABC. The minimum value of d_{1}^{2} + d_{2}^{2} + d_{3}^{2} is 3a^{2}b^{2}c^{2}/∑ (a^{2}) The locus of points whose pedal triangles are rightangled is a^{2}X = b^{2}Y + c^{2}Z 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 a^{2}X = b^{2}Y = c^{2}Z = κ. Substituting in relation (i) we obtain the quadratic equation in κ κ^{2} [∑ (a^{4})  ∑ (b^{2}c^{2})]  κa^{2}b^{2}c^{2} ∑ (a^{2}) + a^{4}b^{4}c^{4} = 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 Δ^{2} [∑ (a^{2}) + √3 Δ] / 2 [∑ (a^{4})  ∑ (b^{2}c^{2})] and √3 Δ^{2} [∑ (a^{2})  √3 Δ] / 2 [∑ (a^{4})  ∑ (b^{2}c^{2})] The coaxial circles a^{2}X = b^{2}Y = c^{2}Z 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 / ∑ (a^{4})  ∑ (b^{2}c^{2}) 
§ 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 (α_{1}β_{1}γ_{1}) and Q (κ^{2} / α_{1} κ^{2} / β_{1} κ^{2} / γ_{1}) be isogonal conjugates, and if AP = P_{1}, BP = P_{2}, CP = P_{3}, AQ = Q_{1}, BQ = Q_{2}, CQ = Q_{3}, then P_{1}^{2} sin^{2}A = β_{1}^{2} + γ_{1}^{2} + 2β_{1}γ_{1} cos A Q_{1}^{2} sin^{2} A = κ^{4} (1 / β_{1}^{2} + 1 / γ_{1}^{2} + 2 cos A / β_{1}γ_{1}) κ^{4} P_{1}^{2} sin^{2} A / β_{1}^{2}γ_{1}^{2} [The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.] hence Q_{1}β_{1}γ_{1} = ± κ^{2}P_{1}, Q_{2γ1α1} = ± κ^{2}P_{2}, Q_{3α1}β_{1} = ± κ^{2}P_{3}. Multiplying these in turn by aα_{1}, bβ_{1}, cγ_{1} and adding α_{1}β_{1}γ_{1} (aQ_{1} + b_{2} + c_{3} = κ^{2} (aP_{1}α_{1} ± bP_{2}β_{1} ± cP_{3}γ_{1}) Also κ^{2} (a / α_{1} + b / β_{1} + c / γ_{1}) = 2 Δ, hence (aβ_{1}γ_{1} + bγ_{1}α_{1} + cα_{1}β_{1}) (aQ_{1} + bQ_{2} +cQ_{3} = 2 Δ (aP_{1}α_{1} ± bP_{2}β_{1} ± cP_{3}γ_{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 d be the distance between (α_{1}β_{1}γ_{1}) and the circumcentre, then
aβ_{1}γ_{1} + bγ_{1}α_{1} + cα_{1}β_{1} = Δ / R (R^{2}  d^{2}) 

From the above relations we easily obtain Q_{1}S_{1} = ± 2Δ P_{1}α_{1} Q_{2}S_{1} = ± 2Δ P_{2}β_{1} Q_{3}S_{1} = ± 2Δ P_{3}γ_{1}, where S_{1} = aβ_{1}γ_{1} + bγ_{1}α_{1} + cα_{1}β_{1}, therefore α_{1}: β_{1}: γ_{1} = Q_{1}/P_{1}: Q_{2}/P_{2}: Q_{3}/P_{3}, (ii)
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The signs of the ratios following those of the coordinates Also aQ_{1}S_{1}/P_{1} = ± 2Δ aα_{1} bQ_{2}S_{1}/P_{2} = ± 2Δ bβ_{1} cQ_{3}S_{1}/P_{3} = ± 2Δ cγ_{1} hence S_{1}(aQ_{1}/P_{1} + bQ_{2}/P_{2} + cQ_{3}/P_{3} = 2 Δ (aα_{1} ± bβ_{1} ± cγ_{1})
If (α_{1}β_{1}γ_{1}) move on a circle of radius d concentric with the circumcircle, S_{1} = Δ/R (R^{2}–d^{2}), and therefore aQ_{1}/P_{1} + bQ_{2}/P_{2} + cQ_{3}/P_{3} = 4 Δ R/R^{2}–d_{2}
Hence the theorem “If P be any point within the triangle ABC at a distance d from the circumcentre, and if Q be its isogonal conjugate, then BC. AQ/AP + CA. BQ/BP + AB. CQ/CP = 4Δ R/R^{2}–d^{2} (iii)
If P lie outside the triangle ABC, the relation (iii) must be modified in accordance with relation (ii), and d^{2}–R^{2} substituted for R^{2}–d^{2}.