Art. XXXII.—On certain Tripolar Relations: Part III.
[Read before the Philosophical Institute of Canterbury, 1st November, 1916; received by Editors, 22nd December, 1917; issued separately, 24th June, 1918.]
The equation of the circle of radius _{p} having its centre at the point O, whose trilinear co-ordinates are (α_{o}, β_{o}, γ_{o}), is
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U ≡ a;α_{o}X + bβ_{o}Y + cγ_{o}Z − 2RS_{o} − 2δ ρ^{2} = o (1)
Let d be the distance of O from H, the centre of the circle ABC. If U pass through H, then
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R^{2} (aα_{o} + bβ_{o} + cγ_{o}) − 2RS_{o} − 2 δ d^{2} = o;
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i.e., 2RS_{o} = 2 δ (R^{2}–d^{2}): hence U may be written
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aα_{o}X + bβ_{o}Y + cγ_{o}Z = 2 δ (R^{2} + ρ^{2} − d^{2}) (ii)
If the circle U cut the circle ABC at the angle θ, then R^{2} + p^{2} — d^{2} = 2R_{p} cos θ whence
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U ≡ aα_{o}X + bβ_{o}Y + cγ_{o}Z − abc ρ cos θ = o (iii)
The equations of the circles of radius ρ and centre (α_{o}β_{o} γ_{o}) touching the circle ABC internally and externally are respectively
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aα_{o}X + bβ_{o}Y + cγ_{o}Z − abc ρ = o
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aα_{o}X + bβ_{o}Y + cγ_{o}Z + abc ρ = o
the trilinear co-ordinates of the point of contact being
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R/R ± ρ [a_{o} ± ρ cos A, β_{o} ± ρ cos B, γ_{o} ± ρ cos C]
the negative sign being taken for internal contact.
If U reduce to a point-circle, (ii) then takes the form
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aα_{o}X + bβ_{o}Y + cγ_{o}Z. = (aα_{o} + bβ_{o} + cγ_{o}) (R^{2} −d^{2});
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i.e., aα_{o} (X + d^{2} − R^{2}) + bβ_{o} (Y + d^{2} − R^{2}) + cγ_{o} (Z + d^{2} − R^{2}) = o.
Let × = ρ_{1}^{2}, Y = ρ_{2}^{2}, Z = ρ _{8}^{2}, and let the radii HA, HB, HC subtend at O the angles λ, μ, ν, respectively; we then have
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aα_{o}ρ_{1} cos λ + bβ_{o}ρ_{2} cos μ + cγ_{o}ρ_{3} cos ν = o (iv)
The following particular cases of (iv) are of interest. For the ortho-centre, in-centre, centroid, and symmedian point respectively
Sin A cos λ + sin B cos μ + sin C cos ν = o
cos A/2 cos λ + cos B/2 cos μ + cos C/2 cos ν = o
m_{1} cos λ + m_{2} cos μ + m_{3} cos ν = o
am_{1} cos λ + bm_{2} cos μ + cm_{3} cos ν = o,
where m_{1}, m_{2}, and m_{3} are the medians of the triangle ABC.
If O be either of the two points whose pedal triangles, are equiangular, then, since for these points a_{ρ1} = b_{ρ2} = c_{ρ3},
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a √X_{o}α_{o}α + b √Y_{o}β_{o}β + c √Z_{o} γ_{o}γ = o.
If O be the focus of a conic inscribed in the triangle ABC, then the equation of the conic is
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a √X_{o}α_{o}α + b √Y_{o}β_{o}β + c √Z_{o} γ_{o}γ = o.
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Comparing this with the equation of the maximum inscribed ellipse, √ aα + √bβ √γ = o, we have aX_{o}α_{o} = bY_{o}β_{o} = cZ_{o}γ_{o}, whence
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cosλ/AO + cosμ/BO + cosν/CO = o.
For the Brocard ellipse this gives
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cosλ/BC. AO + cosμ/CA. BO + cosν/AB. CO = o.
Let U^{1} be the result of substituting in U the co-ordinates (X^{1}, Y^{1}, Z^{1}) for (X, Y, Z). Suppose a circle of radius p^{1} concentric with U to pass through the point (X^{1}Y^{1}Z^{1}), then
U^{1} = aα_{o}X^{1} + bγ_{o}Y^{1} − 2RS_{o} − 2δ ρ^{2}
O = aα_{o}X^{1} + bβ_{o}Y^{1} − 2RS_{o} − 2δ ρ^{1 2}
∴ U^{1} = 2δ (ρ^{1 2} − ρ^{2}) = 2δ t^{1 2},
where t^{1} is the length of the tangent from (X^{1}Y^{1}Z^{1}) to the circle U = O.
Let t_{1}, t_{2}, and t_{3} be the lengths of the tangents to the circle U from A, B, C respectively: then
2 δ t^{2}_{1} = bβ_{o}c^{2} + cγ_{o}b^{2} − 2 δ (R^{2} − d^{2} + ρ^{2});
i.e., 2 δ (t_{1}^{2} + R^{2} − d_{2} + ρ^{2}) = bβ_{o}c^{2} + cγ_{o}b^{2}.
We have X_{o} = t_{1}^{2} + ρ^{2}, whence, if OH subtend at A, B, and C the angles θ, φ ψ respectively,
abc ρ^{1} cos θ = bβ_{o}c^{2} + cγ_{o}b^{2}
abc ρ^{2} cos φ = aα_{o}c^{2} + cγ_{o}a^{2}
abc ρ^{1} cos ψ = α_{o}b^{2} + bβ_{o}a^{2}
2 δ = aα_{o} + bβ_{o} + cγ_{o}.
Eliminating aα_{o} bβ_{o} cγ_{o} we have
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2Rρ_{1} cos θ | o | c^{2} | b^{2} | |
2Rρ_{2} cos φ | c2 | o | a^{2} | |
2Rρ_{3} cos ψ | b^{2} | a^{2} | = o, | o |
1 | 1 | 1 | 1 |
which reduces to the relation
aρ_{o} cos A θ + bρ_{2} cos B cos φ + cρ_{3} cos C cosψ = 2 δ.
From the equations
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αρ_{1} cos θ/bc = bβ_{o}/b + γ_{o}/c, bρ_{2} cosφca = γ_{o}/o + α_{o}/a, cρ_{3} cos ψ/ab = α_{o}/a + β_{o}/b,
we see that if O lie on the trilinear polar of the symmedian point BC^{2} AO cosθ + CA^{2}BO cosφ + AB^{2}CO cosψ = o.
Solving for α_{o}, β_{o}, γ_{o} from the above equations, and substituting in the equation of the circle ABC, we obtain the relation
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BC √OA cosθ + CA √OBcosφ + AB√OC cosψ = o
for any point O on the circle.
If t_{o} be the length of the tangent to any circle from the middle point of BC, then
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2 δ t^{2}_{2} = aα_{o}c^{2} + cγ_{o}a^{2}−2 (RS_{o} + δρ^{2})
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2 δ t^{2}_{3} = aα_{o}b^{2} + bβ_{o}a^{2}−2 (RS_{o} + δρ^{2})
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2 δ t^{2}_{o} = aα_{o}m^{2}_{1} + a^{2}/4(bβ_{o} + cγ_{o})−2(RS_{o} + δρ^{2})
where m_{1} is the median drawn through A.
Hence 2δ(t^{2}_{2} + t^{2}_{3}−2t^{2}_{o}
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= aα_{o}(b^{2} + c^{2}−2m^{2}_{1}) + a^{2}/2(bβ_{o} + cγ_{o})
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= a^{2}/2 δ;
i.e. t^{2}_{2} + t^{2}_{3} = 2t^{2}_{o} + a^{2}/2,
an extension of the Theorem of Apollonius.
If U be the polar circle of the triangle ABC, then
U ≡ tan AX + tan BY + tan CZ − 2 δ = o
∴ δ t^{2}_{2} = κ (c^{2} tan A + a^{2} tan C − 2δ),
where κ = 4R^{2}cos A cos B cos C.
which reduces to t_{2}^{2} = ca cos B. Also t_{3}^{2} = ab cos C, hence t_{3}_{2} + t_{3}^{2} = a^{2}.
If in the triangle ABC the angle A be obtuse, then the sum of the squares of the tangents to the polar circle from B and C is BC^{2}.
If U = o pass through the point (X_{1}Y_{1}Z_{1}), then the locus of O (α_{o}β_{o}γ_{o}) is the circle
V ≡ aαX_{1} + bβY_{1} + cγZ_{1} − 2RS −2δρ^{2} = o.
If t_{1}, t_{2}, t_{3} be the lengths of the tangents to the circle V from A, B, C respectively, then X_{1} = t_{1}^{2} + p^{2}, Y_{1} = t_{2}^{2} + p^{2}, Z = t_{3}^{2} + p^{2}, and the equation of the circle takes the form
V ≡ (t^{2}_{1}aα + t^{2}_{2}bβ + cγ) (aα + bβ + cγ) − abc S = o.
If V touch the circle ABC, then expressing that the radical axis of the two circles is a tangent to S we have
at_{1} ± bt_{2}± ct_{3} = o.
from which we obtain Ptolemy's theorem if we suppose V to reduce to a point-circle
This extension of Ptolemy's theorem may be proved geometrically as follows:—
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Suppose the circle V to touch the circle ABC at the point O on the arc BC, and let AO, BO, CO meet V in the points D, E, F respectively; then t_{2}^{2} = BO. BE and t_{3}^{2} = CO. CF. Also OE/OF = OB/OC, and therefore BE/CF = OB/OC, hence t_{2}^{2}: t_{3}^{2} = OB^{2}: OC^{2},
i.e., t_{1}: t_{2}: t_{3} = OA: OB: OC.
By Ptolemy's theorem BC . OA = AC . OB + AB . OC;
i.e., at_{1} = bt_{2} + ct_{3}.
If t be the length of the tangent to the circle V = o from any point P (α_{o}β_{o}γ_{o}), then
2 δ t^{2} = aα_{o}t_{1}^{2} + bβ_{o}t_{2}^{2} + cγ_{o}t_{8}^{2} − 2RS_{o}.
For the circle BPC, t = t_{2} = t_{3} = o.
Consider now the circles BPC, CPA, APB. We have
aα_{o}t^{2}_{1} = bβ_{o}t^{2}_{2} = cγ_{o}t^{2}_{3} = 2RS_{o}
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Hence aα_{o} + bβ_{o} − cγ_{o}/2RS_{o} = 1/t^{2}_{1} + 1/t^{2}_{2} + 1/t_{3}^{2}
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i.e. 1/t^{2}_{1} + 1/t^{2}_{2} + 1/t^{2}_{3} = 1/R^{2}−d^{2}
where d is the distance of P from the circum-centre of the triangle ABC.
If P be the symmedian point of the triangle ABC, then at_{1} = bt_{2} = ct_{3}: if G be the centroid of the triangle t_{1}^{2} = t_{2}^{2} = t_{3}^{2} = ⅓ (a^{2})
If the circle V reduce to a point P_{1} whose tripolar co-ordinates are (X_{1}Y_{1}Z_{1}) we have
X_{1}aα + Y_{1}bβ + Z_{1}cγ_{o} = 2RS.
For a point-circle at P_{2} (X_{2}Y_{2}Z_{2})
X_{2}aα + Y_{2}bβ + Z_{2}cγ = 2RS.
Hence the radical axis of the pair of circles is
(X_{1} − X_{2}) aα + (Y_{1} −Y_{2}) bβ + (Z_{1} − Z_{2}) cγ = o.
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If P _{l}, P_{2} be inverse points determined by the equations X/l = Y/m = Z/n, then X_{1} − X_{2}/l = Y_{1} − Y_{2}/m = Z_{1} − Z_{2}/n = κ_{1} − κ_{2}, where κ_{1} and κ_{2} are the roots of the equation κ^{2} [Σ (a^{2}l^{2}) −2 Σ (bc cos Amn]
− 2κ abc [(a cos Al)] + a^{2}b^{2}c^{2} = o.
Hence the equation of the line bisecting perpendicularly the above pair of inverse points is
laα + mbβ + ncγ = o.
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The two points whose pedal triangles are equilateral are determined by the equations a^{2}X = b^{2}Y = c^{2}Z. Hence these points are equidistant from the line α/a + β/b + γ/c = o, which is therefore perpendicular to the Brocard diameter Σ [a^{2} (b^{2} – c^{2}) X] = o, on which the points lie.
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The two points whose distances from A, B, C are proportional respectively to a, b, c are given by the equations X/a^{2} = Y/b^{2} = Z/c^{2}: they therefore lie on Euler's line, Σ[(b^{2} − c^{2}) X] = o, and are equidistant from the line a^{3}α + b^{3}β + c^{3}γ = o.
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Let two pairs of inverse points be determined by the equations X/l_{1} = Y/m_{1} = Z/n_{1} and X/l_{2} = Y/m_{2} = Z/n_{2}; the centre of the circle through them will be determined by the equations
l_{1}aα + m_{1}bβ + n_{1}cγ = o
l_{2}aα + m_{2}bβ + n_{2}cγ = o,
whence aα : bβ : cγ = m_{1}n_{2} −m_{2}n_{1} : n_{1}l_{2} = n_{2}l_{1} : l_{1}m_{2} − l_{2}m_{1}, and the equation of the circle will be
(m_{1}n_{2} −m_{2}n_{1}) × + (n_{1}l_{2} − n_{2}l_{1}) Y + (l_{1}m_{2} − l_{2}m_{1}) Z = o.
The two pairs of points will lie respectively on the diameters
(m_{1} − n_{1}) × + (n_{1} − l_{1}) Y + (l_{1} − m_{1}) Z = o
(m_{2} − n_{2}) × + (n_{2} − l_{2}) Y + (l_{2} − m_{2}) Z = o
We now proceed to find the equation in tripolar co-ordinates of the inverse of the circle U ≡ o with respect to S, the circum-circle of the triangle ABC.
U ≡ lX + mY + nZ − h^{2} = o
S ≡ a cos AX + b cos BY + c cos CZ − abc = o.
Let P be any point on U and Q its inverse. Let (X, Y, Z), (X^{1}, Y^{1}, Z^{1}) be the tripolar co-ordinates of P and Q respectively, then
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X/X^{1} = Y/Y^{1} = Z/Z^{1} = HP^{2}/R^{2},
where H is the circum-centre.
Hence HP^{2} (lX^{1} + mY^{1} + nZ^{1}) = h^{2}R^{2}.
The equation of the circle of radius HP concentric with the circum-circle is
a cos AX + b cos BY + c cos CZ − 2 δ R = 4RHP^{2} sin A sin B sin C. Hence
Hence
HP^{2} (a cos AX^{1} + b cos BY^{1} + c cos CZ^{1} − 2 δ R) = 2 δ R^{3}.
Eliminating HP^{2}, the equation of the inverse of U becomes
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a cos AX + b cos BY + c cos CZ − 2 δ R/lX + mY + nZ = 2 δ R/h^{2}
or h^{2} (S + 2 δ R) = 2 δ R (h^{2} + U);
i.e. h^{2}S − 2 δ RU = o.
The equation of the circle which is the inverse of the line lX + mY + nZ – h^{2} = o takes the same form as the above, subject to the condition l + m + n = o.
The equation of the circle which is the inverse with respect to the circle ABC of the line pα + qβ + rγ = o is
aX {p cos 2A + q cos (A −B) + r cos (C − A)}
+ bY {p cos (A − B) + q cos 2B + r cos (B − C)}
+ cZ {p cos (C − A) + q cos (B − C) + r cos 2C}
= abc (p cos A + q cos B + r cos C).
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The equation of the circle which is the inverse with respect to the circle ABC of the line α/a + β/b γ/c = o is
a cos (A − ω) × + b cos (B − ω) Y + c cos (C − ω) Z = abc cos ω, where ω is the Brocard angle of the triangle ABC.