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Volume 50, 1918
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– 317 –

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 (αo, βo, γo), is

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U ≡ a;αoX + bβoY + cγoZ − 2RSo − 2δ ρ2 = 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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R2 (aαo + bβo + cγo) − 2RSo − 2 δ d2 = o;

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i.e., 2RSo = 2 δ (R2–d2): hence U may be written

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oX + bβoY + cγoZ = 2 δ (R2 + ρ2 − d2) (ii)

If the circle U cut the circle ABC at the angle θ, then R2 + p2d2 = 2Rp cos θ whence

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U ≡ aαoX + bβoY + cγoZ − abc ρ cos θ = o (iii)

The equations of the circles of radius ρ and centre (αoβo γo) touching the circle ABC internally and externally are respectively

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oX + bβoY + cγoZ − abc ρ = o

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oX + bβoY + cγoZ + abc ρ = o

the trilinear co-ordinates of the point of contact being

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R/R ± ρ [ao ± ρ cos A, βo ± ρ cos B, γo ± 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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oX + bβoY + cγoZ. = (aαo + bβo + cγo) (R2 −d2);

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i.e., aαo (X + d2 − R2) + bβo (Y + d2 − R2) + cγo (Z + d2 − R2) = o.

Let × = ρ12, Y = ρ22, Z = ρ 82, and let the radii HA, HB, HC subtend at O the angles λ, μ, ν, respectively; we then have

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oρ1 cos λ + bβoρ2 cos μ + cγoρ3 cos ν = o (iv)

– 318 –

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

m1 cos λ + m2 cos μ + m3 cos ν = o

am1 cos λ + bm2 cos μ + cm3 cos ν = o,

where m1, m2, and m3 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 aρ1 = bρ2 = cρ3,

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a √Xoαoα + b √Yoβoβ + c √Zo γoγ = 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 √Xoαoα + b √Yoβoβ + c √Zo γoγ = o.

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Comparing this with the equation of the maximum inscribed ellipse, √ aα + √bβ √γ = o, we have aXoαo = bYoβo = cZoγo, 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 U1 be the result of substituting in U the co-ordinates (X1, Y1, Z1) for (X, Y, Z). Suppose a circle of radius p1 concentric with U to pass through the point (X1Y1Z1), then

U1 = aαoX1 + bγoY1 − 2RSo − 2δ ρ2

O = aαoX1 + bβoY1 − 2RSo − 2δ ρ1 2

∴ U1 = 2δ (ρ1 2 − ρ2) = 2δ t1 2,

where t1 is the length of the tangent from (X1Y1Z1) to the circle U = O.

Let t1, t2, and t3 be the lengths of the tangents to the circle U from A, B, C respectively: then

2 δ t21 = bβoc2 + cγob2 − 2 δ (R2 − d2 + ρ2);

i.e., 2 δ (t12 + R2 − d2 + ρ2) = bβoc2 + cγob2.

We have Xo = t12 + ρ2, whence, if OH subtend at A, B, and C the angles θ, φ ψ respectively,

abc ρ1 cos θ = bβoc2 + cγob2

abc ρ2 cos φ = aαoc2 + cγoa2

abc ρ1 cos ψ = αob2 + bβoa2

2 δ = aαo + bβo + cγo.

– 319 –

Eliminating aαooo we have

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2Rρ1 cos θ o c2 b2
2Rρ2 cos φ c2 o a2
2Rρ3 cos ψ b2 a2 = o, o
1 1 1 1

which reduces to the relation

o cos A θ + bρ2 cos B cos φ + cρ3 cos C cosψ = 2 δ.

From the equations

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αρ1 cos θ/bc = bβo/b + γo/c, bρ2 cosφca = γo/o + αo/a, cρ3 cos ψ/ab = αo/a + βo/b,

we see that if O lie on the trilinear polar of the symmedian point BC2 AO cosθ + CA2BO cosφ + AB2CO cosψ = o.

Solving for αo, βo, γo 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 to be the length of the tangent to any circle from the middle point of BC, then

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2 δ t22 = aαoc2 + cγoa2−2 (RSo + δρ2)

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2 δ t23 = aαob2 + bβoa2−2 (RSo + δρ2)

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2 δ t2o = aαom21 + a2/4(bβo + cγo)−2(RSo + δρ2)

where m1 is the median drawn through A.

Hence 2δ(t22 + t23−2t2o

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= aαo(b2 + c2−2m21) + a2/2(bβo + cγo)

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= a2/2 δ;

i.e. t22 + t23 = 2t2o + a2/2,

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 t22 = κ (c2 tan A + a2 tan C − 2δ),

where κ = 4R2cos A cos B cos C.

which reduces to t22 = ca cos B. Also t32 = ab cos C, hence t32 + t32 = a2.

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 BC2.

If U = o pass through the point (X1Y1Z1), then the locus of O (αoβoγo) is the circle

V ≡ aαX1 + bβY1 + cγZ1 − 2RS −2δρ2 = o.

– 320 –

If t1, t2, t3 be the lengths of the tangents to the circle V from A, B, C respectively, then X1 = t12 + p2, Y1 = t22 + p2, Z = t32 + p2, and the equation of the circle takes the form

V ≡ (t21aα + t22bβ + 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

at1 ± bt2± ct3 = 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 t22 = BO. BE and t32 = CO. CF. Also OE/OF = OB/OC, and therefore BE/CF = OB/OC, hence t22: t32 = OB2: OC2,

i.e., t1: t2: t3 = OA: OB: OC.

By Ptolemy's theorem BC . OA = AC . OB + AB . OC;

i.e., at1 = bt2 + ct3.

If t be the length of the tangent to the circle V = o from any point P (αoβoγo), then

2 δ t2 = aαot12 + bβot22 + cγot82 − 2RSo.

For the circle BPC, t = t2 = t3 = o.

Consider now the circles BPC, CPA, APB. We have

ot21 = bβot22 = cγot23 = 2RSo

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Hence aαo + bβo − cγo/2RSo = 1/t21 + 1/t22 + 1/t32

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i.e. 1/t21 + 1/t22 + 1/t23 = 1/R2−d2

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 at1 = bt2 = ct3: if G be the centroid of the triangle t12 = t22 = t32 = ⅓ (a2)

If the circle V reduce to a point P1 whose tripolar co-ordinates are (X1Y1Z1) we have

X1aα + Y1bβ + Z1o = 2RS.

For a point-circle at P2 (X2Y2Z2)

X2aα + Y2bβ + Z2cγ = 2RS.

Hence the radical axis of the pair of circles is

(X1 − X2) aα + (Y1 −Y2) bβ + (Z1 − Z2) cγ = o.

– 321 –

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If P l, P2 be inverse points determined by the equations X/l = Y/m = Z/n, then X1 − X2/l = Y1 − Y2/m = Z1 − Z2/n = κ1 − κ2, where κ1 and κ2 are the roots of the equation κ2 [Σ (a2l2) −2 Σ (bc cos Amn]

− 2κ abc [(a cos Al)] + a2b2c2 = 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 a2X = b2Y = c2Z. Hence these points are equidistant from the line α/a + β/b + γ/c = o, which is therefore perpendicular to the Brocard diameter Σ [a2 (b2 – c2) 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/a2 = Y/b2 = Z/c2: they therefore lie on Euler's line, Σ[(b2 − c2) X] = o, and are equidistant from the line a3α + b3β + c3γ = o.

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Let two pairs of inverse points be determined by the equations X/l1 = Y/m1 = Z/n1 and X/l2 = Y/m2 = Z/n2; the centre of the circle through them will be determined by the equations

l1aα + m1bβ + n1cγ = o

l2aα + m2bβ + n2cγ = o,

whence aα : bβ : cγ = m1n2 −m2n1 : n1l2 = n2l1 : l1m2 − l2m1, and the equation of the circle will be

(m1n2 −m2n1) × + (n1l2 − n2l1) Y + (l1m2 − l2m1) Z = o.

The two pairs of points will lie respectively on the diameters

(m1 − n1) × + (n1 − l1) Y + (l1 − m1) Z = o

(m2 − n2) × + (n2 − l2) Y + (l2 − m2) 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 − h2 = 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), (X1, Y1, Z1) be the tripolar co-ordinates of P and Q respectively, then

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X/X1 = Y/Y1 = Z/Z1 = HP2/R2,

where H is the circum-centre.

Hence HP2 (lX1 + mY1 + nZ1) = h2R2.

– 322 –

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 = 4RHP2 sin A sin B sin C. Hence

Hence

HP2 (a cos AX1 + b cos BY1 + c cos CZ1 − 2 δ R) = 2 δ R3.

Eliminating HP2, 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/h2

or h2 (S + 2 δ R) = 2 δ R (h2 + U);

i.e. h2S − 2 δ RU = o.

The equation of the circle which is the inverse of the line lX + mY + nZ – h2 = 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.