Volume 42, 1909 This text is also available in PDF
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– 327 – Art. XLI.—Some Theorems relating to Sub-Polar Triangles.

[Read before the Philosophical Institute of Canterbury, 1st December, 1909.]

§1. Let the straight lines joining the vertices of the triangle ABC to the points O1, O2 meet the opposite sides BC, CA, AB, in D1D2, E1E2, F1F2, respectively; then the triangles D1E1F1, D2E2F2 are termed sub-polar triangles, the points O1, O2 being their respective poles.

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The vertices of any two sub-polar triangles lie on a conic, for, taking the co-ordinates of the points O1, O2 to be (Α1 β1 γ1), (Α2 β2 γ2) respectively, it may be at once verified that the conic Α21Α2 + β21β2 + γ21γ2 - βγ(1/β1γ2 + β2γ1) - γΑ(1/γ1Α2 + 1/γ2Α1) - Αβ(1/Α1β2 + 1/Α2β1) = 0 . . . . . (i) passes through the vertices of D1E1F1 and D2E2F2.

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If the points O1, O2 be regarded as being determined by the intersection of the line L ≡ lΑ + mβ + nγ = o with the conic S ≡ Α0βTγ + βT0 γΑ + γ0 Αβ = 0, then the conic (i) takes the form lΑ20 + mβ20 + nγ20 + βγ/β0γ0 (mβ0 + nγ0 - lΑ0) + γΑ/γ0Α0 (nγ0 + lΑ0 - mβ0) + Αβ/Α0β0 (lΑ0 + mβ0 - nγ0) = 0 . . . . . (ii)

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Let the line L = 0 pass through the fixed point O′(Α′β′γ′); then, eliminating l between equation (ii) and the relation lΑ′ + mβ′ + nγ′ = 0, the conic (i) takes the form mS1 + nS2 = 0, when S1 ≡ β′Α20 - Α′β20 - (Α0β′ + Α′β0)γ/γ0(Α/Α0 - β/β0) - Αβ/Α0β00β′ - Α′β0) = 0, S2 ≡ γ′Α20 - Α′γ20 + (γ0Α′ + γ′Α0)β/β0(γ/γ0 - Α/Α0) - γΑ/γ0Α0(γ′Α0 - γ0Α′) = 0. Hence we derive the theorem,—

If the poles of two sub-polar triangles be determined by the intersection of a variable line passing through a fixed point with a fixed conic circumscribing the triangle of reference, the conic circumscribing the two sub-polar triangles passes through four fixed points.

It will be seen on inspection that one of the points of intersection of the two conics S′, S″ is the point (Α0β0γ0)—the pole of the conic So.

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A particular case arises if we suppose the variable line L to pass through the point (Αoβoγo). Conic (ii) reduces to lΑ20 + mβ20 + nγ20 - 2lΑ0βγ/β0γ0 - 2mβ0γΑ/γ0Α0 - 2nγ0Αβ/Α0β0 = 0 . . . . (iii) subject to the relation lΑo + mβo + nγo = o. By eliminating l we obtain mβ0(Α/Α0 - β/β0)(Α/Α0 + β/β0 + 2γ/γ0) - nγ0(γ/γ0 - Α/Α0)(Α/Α0 + 2β/β0 + γ/γ0) = 0, which shows that all conics of this family pass through the four fixed points (Αoβoγo), (-3Αoβoγo), (Αo - 3βoγo), (Αoβo - 3γo).

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If the poles of two sub-polar triangles be the circular points at infinity, conic (i) takes the form Α2 + β2 + γ2 + 2βγ cos A + 2γΑ cos B + 2Αβ cos C = o.

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If the two points O1, O2 be the extremities of a diameter of the circle ABC, then conic (i) takes the form λΑ2/sin A cos B + μ β2/sin B cos B + ν γ2/sin C cos C + βγ/sin B sin C(μ tan B + ν tan C - γ tan A) + γΑ/sin C sin A(ν tan C + λ tan A - μ tan B) + Αβ/sin A sin B(λ tan A + μ tan B - ν tan C) = 0, where λ + μ + ν = o.

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Since the equation of any diameter of the circle ABC is γΑ/cos A + μβ/cos B + νγ/cos C = 0 if the diameter pass through the symmedian point λ tan A + μ tan B + ν tan C = 0, and the above equation reduces to sin (B-C)/sin A Α2 + sin (C-A)/sin B β2 + sin (A-B)/sin C γ2 - 2 sin A sin (B-C)/sin B sin C βγ - 2 sin B sin (C-A)/sin C sin A γΑ - 2 sin C sin (A-B)/sin A sin B Αβ = 0, a conic which passes through the symmedian point of the triangle ABC.

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If O1, O2 be the extremities of a diameter of the Steiner ellipse 1/aΑ + 1/bβ + 1/cγ = 0, then the conic (i) reduces to λ(a2Α2 - 2bcβγ) + μ(b2β2 - 2caγΑ) + ν(c2γ2 - 2abΑβ) = 0, where λ + μ + ν = 0.

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§ 2. The condition that conic (i) should be a rectangular hyperbola is 1/Α1Α2 + 1/β1β2 + 1/γ1γ2 + cos A(1/β1γ2 + 1/β2γ1) + cos B(1/γ1Α2 + 1/γ2Α1) + cos C(1/Α1β2 + 1/Α2β1) = 0.

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Hence, if O11β1γ1) be fixed, the locus of O2 will be the circum-conic 1/Α(1/Α1 + cos C/β1 +cos B/γ1) + 1/β(cos C/Α1 + 1/β1 +cos A/γ1) + 1/γ(cos B/Α1 + cos A/β1 + 1/γ1) = 0 . . . . (iv)

If O1 be the centroid of the triangle ABC, then the conic (iv) reduces to the circle ABC.

– 329 – Hence the theorem,—

A rectangular hyperbola can be drawn through the middle points of the sides of a triangle and the points in which the sides of the triangle are cut by lines joining any point on the circum-circle of the triangle to the vertices of the triangle.

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If O1 be the orthocentre of the triangle ABC, then conic (iv) reduces to cos (B-C)/Α + cos (C-A)/β + cos (A-B)/γ = 0.

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§ 3. If the circle described about the triangle D1E1F1 meet the sides BC, CA, AB of the triangle ABC again in the respective points D1E1F1, then the lines AD1, BE1, CF1 are concurrent. Let the co-ordinates of the point of concurrence O1 be (Α1β1γ1). The vertices of the triangles D1E1F1, D1E1F1 lie on the circle Α21Α1 + β21β1 + γ21γ1 + βγ(1/β1γ1 + 1/β1γ1) - γΑ(1/γ1Α1 + 1/γ1Α1 - Αβ(1/Α1β1 + 1/1/Α1β1) = 0 . . . . . (v)

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From the conditions for a circle, we at once obtain 1/Α1 : 1/β1 : 1/γ1 = a21Α21 + Α21β21 - β21γ21) - 2Α1β1γ1 cos A(aΑ + bβ1 + cγ1) : b21β21 + β21γ21 - γ21Α21) - 2Α1β1γ1 cos A(aΑ + bβ1 + cγ1) : c21γ21 + γ21Α21 - Α21β21) - 2Α1β1γ1 cos A(aΑ + bβ1 + cγ1).

It may be at once verified that if Α1 : β1 : γ1 = 1/a : 1/b : 1/c, then Α1 : β1 : γ1 = sec A : sec B : sec C, and the circle in this case is the nine-point circle of the triangle ABC.

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If aΑ1 + bβ1 + cγ1 = 0, then κ(1/bβ1 + 1/cγ1) = Α12, κ(1/cγ1 + 1/aΑ1) = β12, κ(1/aΑ1 + 1/bβ1) = γ12, hence the locus of the point Α1β1γ1 is the quartic curve a√(1/bβ + 1/cγ) + b√(1/cγ + 1/bβ) + √(1/aΑ + 1/bβ) = 0 . . . . (vi)

This result may be stated as follows:—

If lines drawn through the vertices of the triangle ABC parallel to a given line L meet the sides of that triangle in D1E1F1, and the circle through D1E1F1 intersect the sides again in D1E1F1, then the lines AD1 BE1, CF1 are concurrent in the point O1, and as L turns about a fixed point in the plane ABC, the locus of O1 will be the above curve (vi).

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This quartic curve is the isogonal transformation of √a(bγ + cβ) + √b(cΑ + aγ) + √c(aβ + bΑ)=0 a conic inscribed in the triangle formed by drawing tangents to the circle ABC at the vertices of the triangle of reference.

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The lines joining the points of contact of the conic with the sides of the circumscribing triangle to the opposite vertices of that triangle are concurrent in the point [1/b2 + 1/c2 - 1/a2, 1/c2 + 1/a2 - 1/b2, 1/a2 + 1/b2 - 1/c2].

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If the point O11β1γ1) lie on the circle ABC, then the locus of O11β1γ1) is a quartic curve whose isogonal transformation is the conic Α2 cot A + β2 cot B + γ2 cot C - 2 (βγ sin A + γΑ sin B + Αβ sin C) = 0

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If the point O1 lie on the Steiner ellipse 1/aΑ + 1/bβ + 1/cγ = 0, then the locus of O1 is the quartic curve whose isogonal transformation is the conic abc2 + β2 + γ2) - (a2 + b2 + c2)(aβγ + bγΑ + cΑβ) = 0.

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§4. The remainder of the paper is concerned with the cases arising when the poles of the two sub-polar triangles are isogonally conjugate with respect to the triangle ABC. Let a point O′(Α0β0γ0) and its isogonal conjugate O′(1/Α0 1/β0 1/γ0) be taken, and let their sub-polar triangles be DEF, D′E′F′. Also let p ≡ β00 - γ00, q ≡ γ00 - Α00, r ≡ Α00 - β00 p′ ≡ β00 + γ00, q′ ≡ γ00 + Α00, r′ ≡ Α00 + β00

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The equation of the line Oo′ is L ≡ ΑΑ002 - γ02) + ββ002 - Α02) + γγ002 - β02) = 0. which, after dividing out by Αoβoγo, reduces to L ≡ pΑ + qβ + rγ = 0.

The sub-polar triangles whose poles are isogonal conjugates are self-conjugate with respect to the conic which is the isogonal transformation of the line joining their poles.

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The co-ordinates of the points D, D′ are respectively (O βoγo) (O 1/β0 1/γ0), and the equations of EF and E′F′ are -Α/Α0 + β/β0 +γ/γ0 = 0 -ΑΑ0 + ββ0 + γγ0 = 0.

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The equations of the polars of D, D′ with respect to the conic S0pβγ + qγΑ + rΑβ = 0 are p(βγ0 + γβ0) + Α(qγ0 + rβ0) = 0, p(ββ0 + γγ0) + Α(qγ0 + rβ0) = 0, which at once reduce to the equations found for EF, E′F′, and so prove the theorem.

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The vertices of the triangles DEF, D′E′F′ lie on the conic S ≡ Α22 + γ2 - p′βγ - q′γΑ - r′Αβ = 0, and the sides of the two triangles touch the conic Σ ≡ Α2/pp′ + β2/qq′ + γ2/rr′ = 0.

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The polars of the points O, O′ with respect to the triangle ABC are Α/Α0 + β/β0 + γ/γ0 = 0, ΑΑ0 + ββ0 + γγ0 = 0, and these lines meet in the point Ω whose co-ordinates are (p q r).

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Let the sides EF, FD, DF, E′F′, F′D′, D′E′ touch Σ in the points P, Q, R, P′, Q′, R′ respectively: the equations of PP′, QQ′, RR′, are -Α/p′ + β/q′ + γ/r′ = 0, Α/p′ - β/q′ + γ/r′ = 0, Α/p′ + β/q′ - γ/r′ = 0, Hence the triangle formed by PP′, QQ′, RR′ is the sub-polar triangle of the point Ω′ (p′q′r′).

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The conic S is the harmonic conic of Σ1 ≡ √Α/Α0 + √β/β0 + √γ/γ0 = 0 Σ2 ≡ √Α0Α + √β0β + √γ0γ = 0. The co-ordinates of intersection of the two conic Σ1, Σ2 are* (p′-2, q′-2, r′-2) (p′-2, q′+2, r′+2) (p′+2, q′-2, r′+2) (p′+2, q′+2, r′-2). The lines joining these points are of the form Α(q′-γ′) - (p′-2)(β-γ)=0, Α(q′+γ′) - (p′+2)(β+γ)=0, and it is easily shown that the co-ordinates of the intersections of the joining lines not lying on Σ1 and Σ2 are (-p q r) (p - q r) (p q - r). These are the points in which the corresponding sides of the triangles DEF, D′E′F′ meet. Hence the theorem,—

The intersections of corresponding sides of two sub-polar triangles whose poles are isogonal conjugates determine the vertices of the diagonal triangle of the quadrangle formed by the intersections of the two conics inscribed in the triangle of reference which are the envelopes of the polars with respect to that triangle of points lying on the polars of the poles of the two sub-polar triangles.

[Footnote] * “Messenger of Mathematics,” No. 451, November, 1908, p. 117.

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The sides of the triangles PQR, P′Q′R′ pass through the vertices of the triangle ABC, the equations of QR and Q′R′ being respectively ββ0/qq′ + γγ0, β/β0qq′ + γ/γ0rr′ = 0.

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The tringles PQR, P′Q′R′ are self-conjugate with respect to the following inscribed conie :- √Α/p′ + √β/q′ + √γ/r′ = 0 Any point on the conic S0 ≡ Α0βγ + β0γΑ + γ0Αβ = 0 is expressed by the co-ordinates [-κΑ0, κβ0 + γ0, κ(κβ0 + γ0)], where κ is a variable parameter.

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The Equations of the sides EF, FD, DE of the tringle DEF are respectively κβ0(Α/Α0 + β/β0) + γ0(γ/γ0 + Α/Α0) = 0, κβ0(Α/Α0 + β/β0) + γ0(γ/γ0 - Α/Α0) = 0, κβ0(Α/Α0 - β/β0) + γ0(γ/γ0 + Α/Α0) = 0. Hence EF passes Through the point of intersection of the tangents to S0 at B and C; FD passes through the point in which the line joining B to the pole of the conic meets the tangent at C; DE passes through the point in which the line joining C to the pole of the conic meets the tangent at B.

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The sides of the tringle D′E′F′ are respectively κ2β0γ + κ(Α0Α + β0β + γ0γ) + βγ0 = 0, κ2β0γ - κ(Α0Α + β0β - γ0γ) - βγ0 = 0, κ2β0γ + κ(Α0Α - β0β + γ0γ) - βγ0 = 0, and these lines envelop respectively the conics (Α0Α + β0β + γ0γ)2 - 4β0γ0βγ = 0, (Α0Α + β0β - γ0γ)2 + 4β0γ0βγ = 0, (Α0Α - β0β + γ0γ)2 + 4β0γ0βγ = 0 Hence the Theorem,—

If the pole of a sub-polar triangle move on a conic circumscribing the triangle of reference, the sides of the sub-polar triangle will pass through three fixed points, while if the pole move on a straight line the sides of the sub-polar triangle will envelop three fixed conics.