THE MODERN GEOMETRY OF THE TRIANGLE
BY
WILLIAM GALLATLY, M.A.
SECOND EDITION
LONDON : FBANCIS HODGSON, 89 EARRINGDON STREET, E.G.
PRICE HALF A CROWN NET
THE MODERN GEOMETRY OF THE TRIANGLE.
BY
WILLIAM GALLATLY, M.A.
SECOND EDITION. *'< •
LONDON : FKANCIS HODGSON, 89 EARRINGDON STREET, E.G.
(J2,
PREFACE.
IN this little treatise on the Geometry of the Triangle are presented some of the more important researches on the subject which have been undertaken during the last thirty years. The author ventures to express not- merely his hope, but his con- fident expectation, that these novel and interesting theorems — some British, but the greater part derived from French and German sources — will widen the outlook of our mathematical instructors and lend new vigour to their teaching.
The book includes some articles contributed by the present writer to the Educational Times Reprint, to whose editor he would offer his sincere thanks for the great encouragement which he has derived from such recognition. He is also most grateful to Sir George Greenhill, Prof. A. C. Dixon, Mr. V. R. Aiyar, Mr. W. F. Beard, Mr. R. F. Davis, and Mr. E. P. Rouse for permission to use the theorems due to them.
W. G.
SYNOPSIS OF CHAPTEES.
[The numbers refer to Sections.]
CHAPTER I. — DIRECTION ANGLES. (1-13.)
Relations between 616-263 and pgr : lines at rt. angles, condition: re- lation between pqr : distance between (a#y) and (a'&'y') : perpr. on line : Properties of Quad. : Centre Circle.
CHAPTER II.— MEDIAL AND TRIPOLAR COORDINATES. (14-21.)
Medial formulae : Feuerbach Point : mid-point line of quad. : N.P. Circle : Tripolar eqn. to circumdiameter : straight line (general) : circle : to find points with given trip. c. : trip. c. of Limiting Points.
CHAPTER III. — PORISTIC TRIANGLES. (22-34.)
Poristic condition : R.A. of 0(R) and I(r) : points S\ S%<rHiGi poristically fixed : circular loci of FGHO'I'NM : properties of Nagel Point N : Gergonne Point M : poristic formulae : locus of If found.
CHAPTER IV. — SIMSON LINES. (35-55.)
Draw S.L. in given direction : S.L. bisects TH: OAT — crl : perps. from A, B, C on S.L. : eqn. to S.L. : perpr. TU, length and direction :* the point .ZVfor quad. : pairs of S.L. : the point a : circle (kkf) : eqn. to TOT' ; ABC and A'B'C' n.c. of o> : the centre Ot of (kk1) : tricusp Hypocycloid : Cubic in sin<£ or cos4> : Greenhill and Dixon's Theorem.
CHAPTER V.— PEDAL TRIANGLES. (56-69.)
Pedal triangle def similar to LMN : BSC = A + \ : sin \ a an : Limiting Points : II in terms of xyz : Radical Axes (group) : Feuerbach 's Theorem : U, II, OS'2, and (afry) in terms of \p.v : Artzt's Parabola : So for Sj, inverse of S : S.L. of T, T' are axes of similitude for def, d\e\f\.
VI MODERN GEOMETRY.
CHAPTER VI. — THE OBTHOPOLE. (70-80.)
Concurrence of pp', qq', rr' : 5 = IE cos el cos 02cos 03 ; S found geo- metrically : general n.c. of S : point <r lies on N.P. circle : ABC and A'B'C' n.c. of ff : a- coincides with « : three S.L. through a point, reciprocal relations: Lemoyne's Theorem: pedal circles through <r(o>): constant b.c. of S for def: Harmonic system of lines, a common to the 4 circles. (See Appendix I and II.)
CHAPTER VII. — ANTIPEDAL TRIANGLES. (81-87.)
Angular c. : Orthologic and Antipedal triangles: nc. of S' : V = $M i n.c. of centre of similitude of U and V : similar properties of S, and S\ : Vl-V'l = 4 A ; S' and /S", are called Twin Points.
CHAPTER VIII. — ORTHOGONAL PROJECTION OF A TRIANGLE. 88-100.)
Shape and size of projection of ABC on plane passing through fixed axis in plane of ABC : ABC projected into triangle with given angles : projec- tion of ABC on planes at constant inclination : 2tf''2 cot A — 2 A (1 + cos2 ») . . . : cquilat. triangle and Brocard Angle : Antipedal triangles and projection : calculation of a', b' , c' : Pedal triangles and projection: general theory for any triangle XYZ : Schoute Circles.
CHAPTER IX. — COUNTER POINTS. (101-116.)
a, II', U', 31', q2 in terms of A/XI/ : nn' — &R-q-: equation to minor axis : SA . S'A = AB , Ac : A'B'C' b.c. of <TO : p* = £ A/iV : Ratio Bl : Cl : pedal circle 88' cutsN.P. circle at o>, «": Aiyar's Theorem, OS. OS' = 2R. OV0 : M'Cay's Cubic : Counterpoint conies, direction of asymptotes and axes : For TOT ', conic is R.H. : centre, asymptotes, (semi-axis)2 : similarly for S' and S\ : join of Twin Points ^ S\ bisected at « ; circle centre S radius SS\ passes through I'm'ri.
CHAPTER X.— LEMOINE GEOMETRY. (117-129.)
K found : n.c. and b.c. of K: K centroid of def: equation to OK: II : a2 + #2 + y'2 and u- + v* + w2 each a min. at K : Artzt's Parabola: triangle T^T-.T-j, : Lemoine Point of Ii/a-^ : list of "(«-«)" points : AK bisects chords parallel to T.2T3: Harmonic quad. : A'l passes through K : locus of centres of rect. inscribed in ABC : cosflj or a(62 — c2) : tripolar equation to OK ; Apollonian Circles : Lemoine Axis : Harmonic quad, inverts into square.
SYNOPSIS OF CHAPTERS. VU
CHAPTER XI. — LEMOINE-BROCARD GEOMETRY. (130-152.)
Forms for cot a> : n.c. and b.c. of n and ft' : equation to rift' : o> not greater than 30° : useful formulae : Neuberg Circles : Steiner Angles : Pedal triangle of ft : Triangles XBC, TCA, ZAB (see also Appendix III) : First Brocard Triangle PQR : centre of perspective D : eqn. to axis of perspective : G double point of ABC, PQR : OK bisects tin' at right angles : On = eR, OK = eR sec <a : Steiner and Tarry Points : figure KPROQ similar to 'S.ACTB : G mean centre of points yzx, zxy, xyz : apply to PQR ; G centroid of Dftft' : D lies on 20T: OD = e-R : Isodynamic Points 5 and St : Isogonic points 3' and 8^ : Circum-Ellipse and Steiner Ellipse.
CHAPTER XII. — PIVOT POINTS. TUCKER CIRCLES. (153-166.)
Pairs of homothetic triangles, inscribed and circumscribed to ABC : family of circles touch conic : Tucker Circles : list of formulae : Radical Axis : First Lemoine Circle : Pedal circle of ftft' : Second Lemoine Circle : Taylor Circle : trip. c. of Limiting Points for Taylor Circle are as cot A, cotB, coi C.
CHAPTER I.
INTRODUCTION: DIRECTION ANGLES.
1. IN this work the following conventions are observed: — the Circumcentre of ABC, the triangle of reference, will be denoted by 0 ; the Orthocentre by J3"; the feet of the perpendiculars from A, B, G on BG, CA, AB respectively by Hlt J3"2, Hs (the triangle HJIJi.& being called the Orthocentric Triangle) ; the lengths of AH^ BH^ GHS by /?l? &2, 7?,3 ; the Centroid or Centre of Gravity by G ; the in- and ex-centres by I, J,, J2, 73 ; the points of contact of the circle I with the sides of ABG by X, Y, Z ; the corresponding points for the circle II being X^ Y^ Zr
The Circiimcentre of JjJjJg is /,• which lies on 01} also OJ — OJ, and the Circumradius of ^IjJj = 2E.
2 MODERN GEOMETRY.
The mid-points of BC, CA, AB are A', B', C" ; the triangle- A'B'C' being called the Medial Triangle. Its Circumcircle is the Nine-Point Circle, with Centre 0' and radius = \E '. • the- Orthocentre of A'B'C' is 0, the Centroid is G, while the in-cen Ire- is denoted by 1'.
The lines drawn through A, J5, C parallel to BC, CA, AB form the Anti-Medial Triangle A1B1G1.
Its Circumcentre is JT, its Centroid is 6r, its Medial or Nine- Point Circle is ABG ; and its Nine-Point Centre is 0.
The letters "n.c." stand for normal or triUnear coordinates- denoted by a/?y.
The letters " b.c." stand for barycentric or areal or triangular coordinates, denoted by xyz ; also x = aa, &c., so that
2. Let _p, g, r be the lengths of the perpendiculars from A, B, C on a straight line TT'.
Let 01? <9.2, 08 be the Direction Angles of TT' ; that is, the angles which the sides of ABC make with TT', these angles being measured from TT' as axis, and in the same sense.
By projecting the sides of ABC along and perpendicular to.
TT', we obtain 5 . a cos 0t = 0, 2. a sin ^ = 0.
The diagram shows that a sin 6l = q — r, &c., so that S.op sin^ = 2.^(3 — r) = 0.
' a,
r
Two sets of direction angles should be particularly noted. For 01, cos 0t = A'X/OI = \ (b-c)/OI oc (6-c). For OG^if,
cos 0, = A'HJOH = R sin(5-0)/0//oc (/r-r)/^.
DIRECTION ANGLES. 3
To express cos Ol in terms of p, g, r. Since CHl:HlB = b cos C . c cos 5,
.'. g.6 cos C + r.c cos B = (b cos (7 + c cosB)Hld
= a(p — Hld') = ap — a.AJIl cos 0t ; .*. 2A.cos 0j — ap— bq cos 0 — crcosB; When Hl falls outside dd', the right-hand signs are changed.
3. To determine the condition that la. -\-mj3-\-ny = 0, and I'a + m'ft + n'y = 0 may be at right angles.
Let OH 0.2, #3 and <£a, <£2, <£3 be the direction angles of the two lines, so that Ol = <^ =b \TT.
Let p, q, r, p', 5', r' be the perpendiculars on the lines from A, 5, (7, so that Z cc op, Z' oc op'.
Xow ap' sin <£j+ ... = p' (%' — r') + ••• '— 0.
And 2 A . sin fa = 2A . sin (^ ± |ir) = 2 A cos 0,
= ap — bq cos (7 — cr cos 5. .;. ap'(ap—bq cos G—cr cos 5) + ... = 0, or ZZ' -f mm' + ww' — (mnr + m'w) cos J. — . . . = 0,
which is the required condition.
4. To determine TT, the length of the perpendicular on TT' from a point P, whose b.c. are («, y, «), in terms of p, q, r.
Since P is the centre for masses at A, B, C proportional to
so that TT is determined when p, q, r are known. Note that the ratios only of #, ?/, z are needed.
5. To determine TT, when T21' is Za + m/8 + wy = 0.
Put Z2+...— 2mwcos A— ... = D2.
Now Z oc ap ^ k.ap.
Also 2 (a2/ — gr . 26c cos ^1) = 4A2,
and % + y + z = aa+bft+cy = 2A.
Hence TT = (Za + m/3 + ny)/D.
A form of little use, as it is almost always difficult to evaluate D.
4 MODERN GEOMETRY.
6. A straight line TT' is determined when any two of the three perpendiculars p, q, r are given absolutely. It follows that there must be some independent relation between them.
From elementary Cartesian Geometry we have 2 A = Ap.qr + Bq.rp+Cr.pq
= p.a cos Ol + q.b cos #2 4- r . c cos 03
/. 4A2 = op.2A cos #! + ... = ap (ap—bq cos C — cr cos P) -f ... = S (ay-26tfcos-4.gr) = 2{ay-(-a2 + Z>2 + c2)grj. This is the relation sought. When TT' passes through A, p = 0, so that &2g2 + cV2 — 2bc cos A.qr = 4A2.
7. The points P, P' lying on TT' have absolute n.c. («, /?, y) and (a//3', y'). It is required to determine d, the length of PP', in terms of these coordinates.
We have 2 { o-p2 - ( - a2 + 62 + c2) gr } = 4A2. But a2 = 2 A (cot B + cot C),
and — a2 + 62-i-c2 = 4A cot A.
Hence (q— r)2 cot ^4 + (r— p)2 cot B+ (p — r/)2 cot 0 = 2A.
Now ^7— r = a sin $a — a. (a — a!)/d.
Hence
.da = (a-a')2 sin 2A + (P-fi')2 sin 27?+ (y-y')2 sin 2(7.
8. To prove thatv when TT' is a circumdiameter,
op/JB = b cos 03+ c cos #2 .................. (G)f
From (2) we can express the right side in terms of p, q, r.
Then apply the condition a cos A .p-}~ ... = 0, and the result follows.
Hence prove that —
for Of, p = R/OI.(b-c)(8-a)/a ;
for OF, p = R/OH. (62-c2) cos A/a.
The equation to TOT', which is p.aa.+ ... = 0, now takes the form (fr cos Os + c cos 02) a+ ••• = 0.
For 01, (6-c)(*-a).tt+... = 0
[more useful that (cos B — cos C) a + ... = 0]. For OGH, (&*-cs)cos4.a+... = 0.
t For theorems and proofs marked (G) the present writer is responsible.
DIRECTION ANGLES. 5
9. The results just obtained are useful in investigating some algebraic relations of a quadrilateral, which has BC, CA, AB for three of its sides : the fourth side being PQR.
The circum circles of the four triangles AQR, BRP, CPQ, ABC, have a common point (call it If) so that the parabola which touches the four sides of the quadrilateral will have M for focus ; while the four orthocentres lie on the directrix, which is also the common radical axis of the circles (AP)(BQ)(CR).
The four circumcentres 0, Olt 02, 0.A are known to lie on a circle (call it the Centre Circle) which also passes through If.
Let p be the radius of this circle and pv p2, p& the radii of AQR, BRP, GPQ.
Since the angles at Q, R are #2, 03,
2pl sin 0S = AQ (in circle AQR) — p/sin 02 ; /. Pl = p sin Ql . l/(2 sin ^ sin 02 sin 03) = m . p sin 6V In circle AM QR, AM = 2Pl sin ARM, ;md in BMRP, BM = 2p2 sin (B RM or ARM) ; .-. AM : BM : CM — pl : pz : pa = p sin #a : q sin 6.2 : r sin 03 ; therefore the n.c. of M are as
l/p sin OH l/q sin 0S, 1/r sin 08,
or a/p(q—r),&c. or a/(l/q — l/r). (R.F.Davis)
1-Iote that in the circle AMQR,
QR = 2pl sin A cr, ap sin Or
6 MODERN GEOMETRY.
10. The join 00, of the centres ABCM, ARQM is perpendi- cular to the common chord MA,
Similarly 002 is perpendicular to the chord MB. .-. L 0,00, = AMB = ACB (in ABCM) = C;
.'. 0,0,0,= C, ...; and the triangle 0,0,03 is similar to ABC. But, from
MA : MB : MO = Pl : p, : Pt = MO, : MO, : MO,. Hence M is the double point of the similar triangles ABO, 0,0,0,.
Since 0,0 is perpendicular to MB, and 0,0, to MR, .'. Z 00,0, or OMO, = BMR = BPR (circle JBMPR) = 0,.
Hence the chords 00,, 00,, 00, in the Centre Circle subtend angles 01} 0,, 03, so that
00! : 002 : 003 = sin^ : sin03 : sin 08.
11. To determine the equation of the 4-orthocentre line. The equation to PQR being p. aa + q.bfi -\-r.cy = 0, the point P is determined by a = 0. q.bp + r.cy = 0.
The perpendicular from P on An proves to be
cos A . aa/r -f (c/p — a cos B/r) ft + cos A . cy/p = 0,
and the perpendicular from R on AC is
cos A . aa/q + cos A . bft/p + (b/p—a cos C/q) y = 0.
By subtraction their point of intersection is found to lie on
or p sin.0, cos A .aa+ ... = 0,
or cos.4/a'.aa-f ... = 0,
or p(q—r} cos^.a+ ... = 0. (M is a'/3V)
From the symmetry of this equation the line clearly passes
through the other three orthocentres and is therefore the
directrix of the parabola.
For example, let PQR be xla?-\-... = 0, which will prove to be the Lemoine Axis (128).*
Here p cc I/a2 : so that the focus of the parabola, known as Kiepert's Parabola, has n.c. a/(62 — c2) &c., the directrix being (62— c2) cos A.a+... = 0, which is OGH.
The mid-points of diagonals lie on
( — 1/p + l/g-f l/r)aj+... = 0, or cotA.x + ... = 0, the well known Radical Axis of the circles ABC, A'B'C', &c.
* The bracketed numbers refer to sections.
DIRECTION ANGLES. 7
12. To determine p, the radius of the Centre circle. (G)
M0l = pl = m.p sinflj, so M02 = m.q sin02, and O^MO^ = 0,0.60.2 — C;
sin2 Oz—2pq sin 6l sin #2 cos G
—pY _ (q—r}(p-r} a2+62-c2 J ab ~ab~~
But since Z Oa0302 = 0, O^ = 2p sin (7,
.'. p = mB.k^/abc.
To determine the length of AP, one of the diagonals of the quadrilateral.
The distance d between (a/?y) and (a'fl'y') is given by
^2.A/B2 = (a- a')2 sin 2A+.... (6)
Now, for P, a = 0; ^.fc^ + c.ry — 0;
... £=(_r)/(3-r).2A/6, y = gr/(g-r).2A/c. Also for ^1, a' = 2A/a, )8' = 0, / = 0 ;
+ <fl(q—ry. 4A2/c2 . sin 20. Hence .4P2 (g-r)2 = q*b2 + r'2c*
13. Let o>15 w.2, a)5 be the centres of the diameter circles, or mid-points of AP, BQ, CR. To determine the length of ci^to,. Forcu,, a1==A/a, ft = (_r)/(gr-r) . A/6, 7l = gr/(g-r).A/c. For <o2, a2 = r/(r—p) . A/a, /32 = A/6, yt = ( -p)/(r—p) . A/c. Now, a^V.A/E2 = (aa — a2)2 sin
/. 2 . w, V/ A - _J^ — , . cot J. + <y8 X9 . cot (p-rr (^-^)2
,
Cp—
Hence, since cot .4 — (— ^+
D = a sin $a I 6 sin ^2 I c sin 03
CHAPTER II.
MEDIAL AND TRIPOLAR COORDINATES.
14. Medial Coordinates. — If A', J5', C' are the rnid-points of BC, CA, AB, the triangle A'B'G' is called the Medial Triangle of ABO ; its circumcircle A'B'G' is the Nine-Point Circle whose centre 0' bisects OH.
For every point P in ABO there is a homologous point P' in A'B'G', such that P' lies on GP, and GP = 2. OP'.
Let a/?y, a'fi'y' be the n.c. of a point P referred to ABC, A'B'C' respectively.
A diagram shows that a -fa' = ±hr
.'. aa + aa' = i.a^ = A = 4. area of A'B'C'
= oa
/. aa = b(3' + cy', &c.,
so that 2aa' = — aa + bft + cy,
or, in b.c.,
a; = y + s', 2^' =— % + y + z.
For example, the ^4'5'C" b.c. of the Feuerbach Point F being a/(b — c), 6/(c— «), c/(a — 6), to determine the J.£0 b.c. of this point.
Here #' oc a/(b — c), &c. ;
.-. a; = ^ + ^ oc b/(c—a) + c/(a—b) oc (6 — c)(s — a)/(c — a)(a— b)
ac (b — c)*(s-a).
If the J.'5'C' b.c. are to be deduced from the ABC b.c., then oc _(6-c)2(s-a) + (c-a)2( oc a(a — ?>)(« — c) oc a/(6 — c).
15. If the ABC equation to a straight line is lx + my + nz = 0, the A'B'C' equation is
l(y' + z') + ... = 0, or (w + w)aj'+... = 0.
If the A'B'G' equation to a straight line is I'x' + m'y' + n'z' = 0, the ABC equation is
+ ... = 0, or ( — Z' + mf + w')a' + »- = °- 8
MEDIAL AND TRIPOLAR COORDINATES. 9
Example. — The well known Radical Axis whose ABC equation is cotA.x-\- ... = 0 becomes
(cot B + cot C)x' + . . . = 0 or aV + b*y' + cV - 0, when referred to A'B'C'.
If Hlt H,, H3 are the feet of perpendiculars, the ABC equation to H.2HS is
— cot^.a- + cot.B.?/ + cot0.z = 0.
Therefore the A'B'C' equation is
(cot B + cot C) x+ (cot 6'— cot 4)y + (— cot ^ + cot B)z = 0, reducing to a~x-\- (a8 — c2) y+ (a2— fr2) «.
Returning to the quadrilateral discussed in section (13) we see the b.c. of o^ given by a, = A/a, &c.
Hence the ABC equation to the mid-point line of the quadrilateral is ( — I/p + 1/q + 1/r) x + . . . =0. The 4'£'C' equation therefore is
x'/P + y'/q + z'/r — 0, or aqr.a+ ... = 0. The perpendicular on this from J/ is therefore given by
TTj = aqr.h-i/D, where D2 = 2 { a2gsrs -_p2gr ( - a2 -f 62 + c2) } = k\
16. To determine the A'B'C' equation to a circumdiameter TOT', whose direction angles are 01} 0^ Os.
Letjp', 7', r' be the perpendiculars from A', B', C' on TOT'. A diagram shows that
p' = OA' cos 6l = R cos J. cos 6r Hence the required equation is
cos A cos 6l . x' -H . . . =0. Example.— For OI? cos^ = %(b — c)/OL Hence the A'B'C' equation to 01 is
(b — c) cos ^1. #'+... = 0.
The J..B(7 equation to the circumcircle ABC is a/a + ... = 0, or a2/#+... = 0. Therefore its A'B'C' equation is
a2/0/ + ,') + ... = 0, or aV + 6y + cV + (a2 + 62 + c2) (y V + z'x' 4- «-''?/') = °-
Referred to A'B'C', the equation of the Nine- Point or Medial Circle is a2#'+ ... = 0.
10 MODERN GEOMETRY.
Referred to A BC, this becomes
reducing to the well known form
a cos A.d:+ ... — a3— ... = 0.
17. Tripolar Coordinates. —
The tripolar coordinates of a point are its distances, or ratios of distances, from A, J5, C.
To determine the tripolar equation to a circumdiameter whose direction angles are 0T, 02, 03.
Let P be any point on the line, x, y, z, the projections of OP on the sides.
Then, if rjt r2, rs be the tripolar coordinates of P,
r22— r32 = a.2x = a. 2. OP cos 0t, and (rf^ftrMvf-rflrn+\r*-rf)rf = °5
.'. a cos 0! . rj2 4- 6 cos 02 . r22 4- c cos 03 . r32 = 0, and a 0080!+... = 0.
So that an equation of the form
lr* + mrf + nrs* = 0, where I + m + n = 0,
represents a circumdiameter.
Note that, for every point P or (r15 r.2, rs) on the line, the ratios r22 — r32 : r32 — ra2 : r/2— r,a are constant ; being, in fact, equal to the ratios a cos 0j : b cos $2 : c cos 03.
The tripolar equations to 01, OGH should be noted. (i) The projection of 01 on BG — J (& — c) ;
.'. cos 0j cc (b—c) ; and the equation to 01 is
o(fe-c)r12+... =0. (ii) The projection of OH on BG
= (M~Ca)/2a; and the equation to OH is
18. To find the equation to a straight line with direction angles Olt 02, 03, and at a distance d from 0. (Gr)
Transferring to Cartesian coordinates, we see that the equa- tion differs only by a constant from that of the parallel circumdiameter.
MEDIAL AND TRIPOLAR COORDINATES.
11
It must therefore be of the form
a cos 6l . rf + . . . = k.
Let A', B', C' be the mid-points of the sides, and let A'O meet the line in A".
A
Then if (p1? p2, p3) be the coordinates of A",
k = a cos Ol . pf 4- 6 cos 0., . p22 4- c cos 03 . p22 [p2 = p3] (plz — p./)
= acos^.Sc.C'lf
= a(OD/OA")2c.OA"sinB
= cZ . 2ac sin />'
Hence the required equation is
a cos
. . . = 4<iA.
19. When (/4-m + w) is not zero.
To prove that if Q -be the mean centre of masses Z, m, « at ^t, 5, 0 ; or if (Z, m, w) are the b.c. of Q, then, for any point P whose tripolar coordinates are (rl5 r2, r3),
Zr12 + mr22-f nrs2 = Z.^!Q2 + m. J5Q2 + w. CQ2+ (l + m + n) PQ*. Take any rectangular axes at Q, and let (c^Og), (^i?>2), (^2)1
be the Cartesian coordinates of .4, -B, (7, P. Then ZTY* = I (al-x)~^-l(at,-yY
= LAQ^l.PQi-2x.lal-2y.lar But, since Q is the mean centre for masses Z, wi, TO, -i -f w^ = 0 ; laz + m62 -H wcs = 0 I
12 MODERN GEOMETRY.
.'. Irf + mrf+nr* = I . AQ2-}- m . BQ2 + n . CQ2 + (Z + m + w) PQ2. This, being true for any point P, is true for 0 ;
n) OQ2;
The power II of the point 'Q for the circle ABU is K2-OQ2 or OQ2 — JR2, according as Q, lies within or without the circle. If P describes a circle of radius />(= PQ) round Q, an internal point, then the tripolar equation to this circle is
If the circle cuts the circle ABG orthogonally, then OQ»«JP+rt
so that the circle becomes
Zr/2 + mr/ + nr32 = 0. (R. F. Davis) .
Examples. —
(A) The circumcircle :
Here p = JK, QO = 0, Z oc sin 2^1 ;
(B) The inscribed circle:
p = r, QO8 = I02 = E2-2Sr, Z oc a ; / . arT2 + fer22 + crs2 = 2 A (r -f 2E) .
(C) The Nine- Point circle :
p = ±R- Q& = JOB2 = ±R--2R cos ^l cos B cos (7. Since the n.c. of the Nine-Point centre are cos (B — (7)...., the b.c. are sin A cos (B — C1),..., so that
Z oc sin 2£ + sin 2(7; /. 2 (sin 2B+ sin 2(7) ras = 4A (1 4- 2 cos ^4 cos 5 cos C).
2O. To determine the point or points whose tripolar coordi- nates are in the given ratios p : q : r.
Divide BG, CA, AB internally at P, Q, R and externally at P', Q', R', so that
BP : CP = q : r = BP' : GP',
GQ:AQ = r:p=GQr: AQ',
AR : BR = p : q = AR' : BR'.
Let Wj, w2, <o3 be the centres of the circles described on PP', QQ', RR' as diameters, and let the circles (PP), (QQ) intersect at T, T'.
MEDIAL AND TRI POLAR COORDINATES.
13
Then since GPBP' is harmonic, we have, for every point T or T' on the circle PP'
a'
BT : CT = BP : CP = q : r - BPr : GP' = BT' : CT',
and TP, TP' bisect the angles at T, while T'P, T'P' bisect those at T'.
So for every point T or T' on the circle (QQr) GT : AT ^ CQ : AQ = r : p, Ac.
Hence at T, T' the points of intersection of the circles (PP'), (QQf) AT: JIT : CT = p: q:r = AT' : BT' : CT'.
The symmetry of the result shows that T and T' lie also on the circle (BE').
Hence there are two points whose tripolar coordinates are as p : q : r, and these points are common to the three circles
(QQ')> (RR').
Since (GPBP'} is harmonic,
.-. WlP2 = («>,£. WlO.
Hence the circle (PP1), and similarly the circles (QQ'), '), cut the circle ABG orthogonally, so that the tangents from 0 to these circles are each equal to R.
It follows that —
(a) 0 lies on TT', the common chord or Radical Axis of
the three circles (PP';, (W), (RR1).
(b) OT.OT' = E\ so that T, T are inverse points in circle
ABC.
(c) The circle ABG cuts orthogonally every circle through
T, T' including the circle on TT' as diameter, so that wT2 = Ow2 — B2, where w is the mid-point of TT'.
14 MODERN GEOMETRY.
(d) The centres o^, o>2, w3 lie on the line through w, bisect- ing TT' at right angles.
21. The tripolar coordinates of Limiting Points. (Gr>
Since OT . OT = _R2, the circle ABC belongs to the coaxal system which has T, T' for Limiting Points, and therefore O^CD^ for Radical Axis ; so that, if 7ra, 7r.2, 7r3 are the perpen- diculars from A, J5, G on a>1(o2o>3, we have by coaxal theory
2.0T.7TJ =~AT* or ^ocp2. Hence the equation to the Radical Axis w^Wg is
pzx-\-q2yi-T'z = 0. And conversely, if the Radical Axis be
Xx + /XT/ + vz = 0, then the tripolar coordinates of T or T' are VA, v'/x, \/v.
Examples. —
(1) For the coaxal system to which the circle ABC and the in-circle XYZ belong, the Radical Axis is
(s — a)-x+ ... = 0.
Hence the tripolar coordinates of the limiting points lying on 01 are as (s — a), (« — &), (s—c).
(2) For the circles ABC, 1^1* the Radical Axis is
a + /? + y = 0 or x/a + . . . = 0 ; .-. p : q : r = I/ Va : I/ Vb : 1/v/c, the limiting points lying on 01.
(3) The circle ABC and the Antimedial circle A1B1C1 (1) have Radical ALxis a?x + bzy + (?z = 0 ;
/. p : q : r = a : b : c, the limiting points lying on OGH.
(4) The circles ABC, A'B'C', Polar Circle, &c., have for their common Radical Axis
cot A . x + cot B . y -f cot C . z = 0 ; .*. p : q : r = A/cot J. : v/cot 5 : \/cot 0, the limiting points lying on OGH.
CHAPTER III.
PORISTIC TRIANGLES.
22. LET I be any point on the fixed diameter DD' of the circle 0(R). With centre D and radius DI cut the circle at e, and e, ; let 6,6, cut DD' at e. Let 01 = d, and le = r.
Then De + el+IO = R, and J>e = De*/2R = DP/2R • .'. (R—d)*/2B + r + d = E ;
, or OI2 = d» = R--2Br.
(Greenhill)
L'
An infinite number of triangles can be inscribed in the circle 0(R) and described about the circle /(r), provided
On 0(JJ) take any point A, and draw tangents AB, AC to /(r). Let BI, CI meet 0(Ej in 5', C" respectively.
16 MODERN GEOMETRY.
Then, since BI.IB' = &-OI* = 2Br, and #/ = r/sin 0 ;
.-. B'l = 2E sin 0 = 7?M So C'l = C'A.
Hence B'C1 bisects AT &t right angles, so that L B'C' A = £'C"I or £'C"(7 ;
.'. BC touches I(r).
It follows that by taking a series of points J along DTJ and calculating r from r = (JS3— • d~)/2R, we have an infinite number of circles /(r) ; each of which, combined with 0(7?), gives a poristic system of triangles.
23. The Radical Axis of 0(R) and I(r).
Let .L, U be the Limiting Points of the two circles, and let El .E/2, the Radical Axis, cut 01 in 1J7.
Bisect 01 in &.
Then W- R3 = -E/^2 = EP-i*t
by ordinary coaxal theory ;
= (2#2-2fir-r*)/2d,
and J5I = »- Jd = (2Rr-v*)/2d.
Also ^i2 = ^i2-r2 =
24. We now proceed to discuss some points which remain unchanged in a system of poristically variable triangles ABC.
(a) The inner and outer centres of similitude (8l and S2)
of the circles ABC, XYZ. (6) The centre of similitude (<r) of the homothetic triangles
XYZ and 1,1 Jy
(c) The orthocentre (Ht) of the triangle XYZ.
(d) The Weill Point (^), the centroid of XYZ.
25. (a) To determine the distances of Sl and $2 from the Radical Axis.
Since 01 is divided at Slt so that
03, : 13, = E : r.
PORISTIC TRIANGLES.
17
So that the circle known theorem.
is coaxal with 0(R) and I(r), a well
26. To show that <r, the centre of similitude of the homothetic triangles XYZ, 1^1^ is poristically fixed, and to determine its distance from the Radical Axis.
Since the circumcentre of XYZ is J, while the circumcentre of /jJa/s is /, lying on 01, and such that OJ = 01;
.'. a- also lies on 01 \ and al/aJ = ratio of circumradii = r/2R = constant ;
.*. <r is a fixed point. Again a-I/IJ = r/(2R — r), from above ;
18 MODERN GEOMETRY.
And EI=r(2R-r)/2d.
.-. Ea . El = EL\ (Greenhill)
so that the circle Jo- belongs to the coaxal system. Note that the homothetic triangles XYZ, J^/g slide on fixed circles, the joins XIlt Y72, ZI^ passing through the fixed point IT.
It will he convenient here to determine the n.c. of the point a. From figure p..., drawing cro-j perpendicular to 7, (7, and noting that X and 7| are homologous points in XYZ, ^ZjJg, we have a-X/a-I-i = ratio of circumradii of the triangles
= r/2R ; .-. ao-j/JjX, = a-X/I.X = r/(2R-r).
0-0-, = a = r/(2R-r).rl; .'. a : ft : y = TI : r, : r, = l/(s-a) : l/(s-b) : l/(*-c).
Note also that, since 7 and J are the circumcentres of XYZ,
vI/IJ = r/(2R - r) , and IJ=2.0I=2d-,
.'. o-I= 2dr/(2R—r).
27. To prove that 23",, the orthocentre of XYZ, is poristically fixed, and to determine its distance from the Radical Axis. Since H{ and I are the orthocentres of XYZ, j^/s/j,
/. jffi lies on o-J, that is, on 01;
.'. crHJal = r/2R, a fixed ratio ;
/. Hi is a fixed point. Again, since o-J = 2dr/(2R— r), (26)
Note also that
7T r 7?»7 7?7T
PORTSTIC TRIANGLES.
19
To determine the n.c. of JET,-.
The orthocentre JT, of XYZ is the centre of masses tan X, tan Y, tan Z placed at -V, Y, Z.
/. a oc tanY. X Y sin Z+ tan Z. ZX sin Y, oc cot ±B cos2 |6Y + cot J(7 cos2 £J5, /(*-a), Ac.
28. TAe JF^ZZ
When an infinite number of n-gons can be inscribed in one fixed circle, and described about another fixed circle, the mean centre of the points of contact X, Y, Z, . . . with the inner circle is a fixed point, which may be called the Weill Point of the polygon (M'Clelland, p. 96). For a triangle the Weill Point is Gh the centroid of XYZ.
In the triangle XYZ, since / is the circumcentre and /T, is the orthocentre,
/. Gf is a fixed point on 01, and ^,-H, = 2.GJ.
To determine the n.c. of 6r,.
Since Gt is the mean centre of masses 1, J, I at JT, Y, Z.
aoc
Yoc cos*
cos
29. We now proceed to discuss the loci of some well known points related to ABC, which are poristically variable : (a) The Feuerbach Point F, (6) The Centroid G,
(c) The Orthocentre H,
(d) and (e) 0' the circumcentre, and I' the in-centre of
the Medial Triangle A'B'G'.
20 MODERN GEOMETRY.
Draw Hh, Gg, I'k parallel to O'l:
(a) The point F moves alonsr the in-circle, (6) G describes a circle, for OG = f . 00' ;
/. 00 = f.OI; thus # is fixed, and
6ty - |.0'I= i(fi— 2r) = constant,
(c) H describes a circle, for
OH = 2. 00'; /. Oh = 2. 01, so that & is fixed, and
Hh = 2.0'I = R—2r = constant,
(d) 0' obviously describes a circle, centre I, radius
(4B-f),
(e) I' describes a circle.
For since I, I' are homologous points in the triangles A BO, A'B'C', whose double point is 6r,
/. IGT is a straight line, and GI-2. GF, /. Ik = f . 10, so that fc is a fixed point, and &/' = f . Gg = %(R-2r) = /O'.
3O. (/) The Nagel Point. — This point also belongs to the series whose circular loci may be found by inspection.
Let XIx be the diameter of the in-circle which is perpendi- cular to BC, and let the ex-circles /15 /2, /3 touch BC in X^ OA in Y2, AB in Zz respectively.
Then BX, = s—c, CXl = s—b,
so that the equation to AXl is
y 1(8-1) = zl(s-c),
and thus AX^ BT^ CZS concur at a point N whose b.c.'s are as (s— a), (s— 6), (s—c).
This point is called the Nagel Point of ABC. If the absolute n.c. of N are a/2y, then
aa/0 — a) = ... = 2A/s;
a = hl. (s — a)/s. Draw NP, NQ, NR perpendicular to AH,, BK2, CHS, then
AP = h, — a = h^a/s = 2r.
.'. the perpendiculars from N on B-f!^ 0}A^ AVB^ the sides of the anti-medial triangle A^B-^G-^ (...) are each = 2r. Hence N is the in-centre of the triangle A1B1C1.
PORISTIC TRIANGLES.
The two triangles ABC, A1B1C1 have G for their common centroid and centre of similitude, / and N for their in-centres, 0 and H for circumcentres, 0' and 0 for Nine-Point centres.
The corresponding joins are parallel, and in the ratio 1:2. Thus ON = 2 . 0'l = E-2r,
and ON, O'l are parallel.
.*. N describes a circle, centre 0, radius = B— 2r.
31. An additional note on this interesting point may here be interpolated.
Since AB, AC are common tangents to the circles / and /15 and Ix, ZjX, are parallel radii of the circles, drawn in the same direction,
/. AxXl or AxNXl is a straight line. And since AP = 2r = Xx,
.'. PX is parallel to AxNXl ;
/. PXX^ is a parallelogram, and PN — XX^ = b — c. Again XI = Ix and XA' = A'Xl •
/. A' I is parallel to AN and PX. Let /P meet BC in T.
22 MODERN GEOMETRY.
Then TH, :TX = TP:TI=TX: TA',
= TH..TA'.
.-. T lies on the Radical Axis (common tangent) of the in-circle and Nine-Point circle.
32. The Gergonne Point.
This is another point whose poristic locus is a circle.
Since BX = s— b and GX = s—c,
the barycentric equation to AX is y(s— b) = z(s — c), so that AX, BY, CZ concur at a point whose b.c. are I/ (s — a), ....
This point, the Centre of Perspective for the triangles ABC and XYZ, is called the Gergonne Point, and will be denoted byM.
To determine the absolute b.c. of JSf,
a? 2A 2A3 2A'2
The join of the Gergonne Point M and the Nagel Point N passes through jSTt. (G.)
Proceeding as usual, the join proves to be a(b — c)(s— a).a;4-... — 0, which is satisfied by the n.c. of Ht, which are (&-f c)/(s — «)
The join of JH and 6r passes through <r. (G.)
For this join is (b— c)(s— a) .aa+ ... — 0, which is satisfied by the n.c. of (r, which are l/(s — a) ....
33. In the poristics of a triangle the following formulae are often required.
(1) A = ra. (2) abc — 4AR =
Put s— a = $], &c. : so that a = sa
(3) sl8& = AV* = r2*-.
(4) l/^ + 1/^+l/*, = ifi+v* r3)/A
(5) Va+ ...
(6) aX -»•."
= s . r (4E -h r) + 3r25 = 4r (E 4-
PORISTIC TRIANGLES. 23
34. * The poristic locus of the Gergorme Point If is a circle coaxal with 0 (R) and I(r). (Greenhill)
Let TTj, 7r2, TT, be the perpendiculars from A, B, C on the Radical Axis E^E^.
The power of A for the circle J(r) = (s— a)".
But by coaxal theory this power is also equal to 2ir^d ; .-, ^ = (s-a)2/2d, &c.
But, if TT be the perpendicular on the Radical Axis from any point whose b.c. are (x, y, z), then
(X + y + Z) 7T = 7TrC -f 7T2y -f 7T3Z.
Hence, for If, whose b.c. are as 1/s,, &c.,
and, finally,
Again, the power II for the circle ABC of a point whose b.c. are (x, y, z) is given by
ncu'iiz -f- o'zx -f- c'xu //?A\
= / . .. , ..v2 •
.Hence, for M, whose b.c. are II = (a>sl+...)sl*
Of. the power of M for the circle ABO varies as the distance of M from the Radical Axis. Hence M describes a circle coaxal with 0(K) and f(r).
[f m is the centre of this circle, then
II = 2 . Om . TT, by coaxal theory ;
4(E + r) ,
/. Om — ;^^ ' .d. 4R+r
It may be shown that the radius of the M circle = r^ ~ , but the proof is long.
* The original proof belongs to Elliptic Functions. For the proof here given the present writer is responsible.
CHAPTER IV.
THE SIMSON LINE.
35. FROM any point Ton the circumcircle ABC, draw perpen- diculars TX, TY, TZ to the sides BC, CM, AB. Produce TX to meet the circle in t.
Then, since BZTX is cyclic,
L BZX = BTX or BTt = BAt • .'. ZX is parallel to At. So XY is parallel to At.
Thus XYZ is a straight line, parallel to At. It is called the Simson Line of T, and T is called its Pole.
24
THE SIMSON LI1STE. 25
To draw a Simson Line in a given direction At.
Draw the chord tT perpendicular to BC, meeting BC in X. A line through X parallel to At is the Simson Line required, T being its pole.
Let AH19 AO meet the circle ABC again in a, a'; it is required to determine the Simson Lines of A, a!, a.
(a) For A, the point X coincides with Hlt while Y and Z coincide with A ; therefore AHl is the Simson Line of A .
(b) Since a'BA, a'GA are right angles, it follows that BC is the Simson Line of a'.
(c) Drawing ay, az perpendicular to AC, AB. it is at once seen that yz, the Simson Line of a, passes through Hlt and that it is parallel to the tangent at A.
36. To prove that XYZ bisects TH (E orthocentre) . If Q be the orthocentre of TBC,
TQ = ZRcosA — AH, and QX = Xt — Ax,
since At, XYZ are parallel ; .-. Hx=TX;
.'. HxTX is a parallelogram. /. XYZ bisects TH, say at h. It follows that h lies on the Nine-Point Circle. TOT' being a circumdiameter, prove that, when the Simson Line of T passes through T', it also passes through G.
(W. F. Beard)
37. Let o-,, o-2, 0-3 be the direction angles of XYZ, taking BXZ = ov
To prove that the base angles of the triangles OAT, OBT, OCT are equal to the acute angles which XYZ makes with the sides of ABC; i.e., to o^, o-2, <r3, or their supplements, as the case may be.
L OAT (or OTA) = fr—%.AOT
= fr—AtT = \K- YXT = BXZ = ov A relation of fundamental importance. To determine QX or Xt,
Xt — Bt.Ct/2R,
and Bt — 2R sin BA t = 2R sin BZX — 2R sin <r3 ;
.'. QX = Xt = 2R sin <r2 sin crs.
26 MODERN GEOMETRY.
38. To determine the n.c. of T, the direction angles of XYZ being o-j, o-2, o-3.
TA = 212. cos OTJ. = 2# cos o-, ; /. a . 2R — TB . TC = 4<R- cos <r.2 cos cr3 ;
a = 2R cos o-., cos cr3 ; so that the n.c. are as sec <rl : sec<r2 : seco-3.
To determine the segments YZ, ZX, XY. In the circle AYTZ, with AT as diameter,
YZ = AT sin A = 2R cos <rl . sin yl = a cos ar
39. To determine p, q, r, the lengths of the perpendiculars from A, B, C on XYZ, the Simsori Line of T.
From (37) Jft = 2R sin o-, sin o-8,
and / ^lf X = ZXT - JTT - o-j ;
.-. p = JC^ . gin AtX = 2R cos <rl sin (r2 sin o-3 cc cot o^!. The equation to XYZ is therefore
cot <T! . a; + cot <r2 . y -t- cot 0-3.2; = 0.
To determine TT, the length of the perpendicular TU 011 the Sim son Line of T.
TT = TU = TX sin TXZ = 2R . cos o-2 cos <rs cos o-^ Or, since 27^1 — 27t cos o-15
4O. Let a parabola be drawn touching the sides of ABG and having T for focus. The Simson Line XYZ is evidently the vertex-tangent and U is the vertex.
Since XYZ bisects TH, the directrix is a line parallel to XYZ and passing through H.
Kiepert's Parabola. — Let T be the pole, found as in (35), of the Simson Line parallel to the Euler Line OGH. Let the direction angles of OGH be 015 02, 08, then
cos Ol oc (62 — c2)/a, &c. ; so that the n.c. of T are a/(62 — c2)....
The directrix will be OGH, or (6s — c2) cos A.a+....
THE SIMSOX LINE. 27
41. Denote the vectorial angles OTA, OTB, OTC, OTU by ^n ^j' ^3< ^ respectively, so that <£x, <£2, <£3 are equal to <rlt a2, OT3 or their supplements.
To prove that S = fa + tfr.-, — <£3. Since TtT' = JTT ; .-. arc Ct = BT' •
.'. L tTC = BTT' = <£, ;
.-. T'Tt = T'TC-tTC - c£3-4>2 ;
... ITT-S = TYZ = ZXT+ T'TX = (^r-^) + 4>3-4>2 5
/. 8 = ^4-<£|— <£8.
Making the usual convention that <£s is to be negative when (7 falls on the side of TOT' opposite to A and B, we may write
42. Consider the quadrilateral formed by the sides of ABC and a straight line PQR.
The circles ABC, AQR, BRP, CPQ have the common point M", and their centres 0, 0,, 02, 03 lie on a circle called the Centre Circle, passing through M. (9)
Let 015 02, 03 be the direction angles of PQE.
If pl be the circumradius of AQR, the perdendiculars from Oj on AQ, AR are /o1 cos E, /cx cos Q.
So that, if a, /?, y are the n.c. of 0I5
ft :y = cos jB : cos Q = sec 0.2 : sec 03
Hence A015 BO^ CO.A meet at a point N whose coordinates are (sec015 sec 02, sec03).
And, since -^- + -^-7-+ -^ = o cos ^+ ... = 0 ; (2) sec ^ sec 0, sec 08
/. JV lies on the circle ABC, and is the pole of the Simson Line parallel to PQR.
Again L 0,NO, = -rr—ANB
(BNO% being a straight line) = TT — 0 ; and since 002, 00., are perpendicular to MA, MB,
.-. 0,00, = AMB = 0;
so that N lies on the Centre Circle, and is therefore the second point in which the Centre Circle intersects the circle ABC.
28
MODERN GEOMETRY.
43. We will now deal with pairs of Simson Lines. T and T' being any points on the circle ABC, it is required to determine 8, the point of intersection of their Simson Lines.
Let
0J0203 be the direction angles of the
Simson Lines of T, T', and of the chord TT'. Then, if a, /?, y are n.c. of 8,
a = SX sin <rl — XX' sin o^ sin cr//sin (o^ — <r/)
= TT' cos 6l . sin o^ sin o-//sin (o^ — o-/) .
But OJT = o-,, OAT' = o-/, (37)
so that TAT' --- o^ — <r/;
.'. TT' = 2Esin(o-1-o-/),
and a = 2J2 cos ^ sin cr, sin o-/. (G.)
The equation to the chord which joins the points (aj/S^) and (a2/32y2) on the circle ABC is
aa/ajttg -f . . . = 0.
Hence the equation to TT' joining the points T, T' whose n.c. are (seco-,, ...)(sec<r/, ...) is
cos o-! cos <r/ . aa + . . . = 0. And the tangent at T is
cos2 o-j . aa + . . . = 0.
44. To prove that the Simson Lines of T and T', the ex- tremities of a circumdiameter, intersect at right angles on the Medial Circle.
THE SIMSON LINE.
29
Let these Simson Lines Xh, X'h' intersect at o>.
Produce TX, T'X' to meet the circle in t, £', so that TtT't' is a rectangle, and tt' a diameter.
The Simson Lines of Tl\ T' are parallel to At, At' (35), and are therefore at right angles.
Again, since h, // are mid-points of HT, HT\ (36)
But h, h' lie on the Medial Circle ; therefore they are the ends of a Medial diameter.
Also huh' = JTT, as shown above.
Therefore w lies on the Medial Circle. Since A'X - A'X', and X<*X' = JTT,
/. A'u = A'X or A'X' = E cos 0t, where 6^ 02, O.d are the direction angles of TOT'.
30 MODERN GEOMETRY.
45. Let TOT' cut the Simson Lines wh, wh' in k, /,•' ; and cut O'to in 01? where 0' is the Nine-Point centre.
Since Hh = hT and Hh' = h'T',
.'. fcfc' is parallel to TOT' or fcfc' : And since hh1 is parallel to kk', and 07i = 07i',
.-. 0,k = OJc ; also kuk' = ±TT :
Hence kk' is a diameter of the circle described with Oj as centre and Oaw as radius.
And since O'OjOj is a straight line, this circle touches the Nine-Point circle at oj.
46. Let or/, (r./, o-./ be the direction angles of X'h'u, the Simson Line of T.
Then, since the Simson Lines of T and T1 are at right angles,
.'. cr/ = O^i^TT,
so that the n.c. of T' are 2Bcos (^TT— o-.2) cos (^TT— <r3), &c., or 2/t sin o-2 sin or3, &c. ; which are as cosec^ : coseccr2 : cosec <r3. Or, since T't is parallel to EC,
.-. T'X' — tX = 2R sin cr., sin cr3. Therefore the equation to the diameter TOT' is cos o-j sin o-a . a; + . . . = 0.
47. Consider the figures ATOT', uXA'X'.
Since TAT' and XwX' are right angles, while 0, A' are the mid-points of TOT', XA'X', and
the angle OTA or (MT = A'Xu or 4'wX ; (37)
it follows that
the figures ATOT' , uXA'X' are similar, a fact to be very carefully noted.
48. Through o> draw Pwp perpendicular to BC. Then from similar triangles AOp, wA'p',
OpjR = A'p'/A'u = Ap'/R cos 0, ; /. A'p = Op cos ^! = perpendicular from p on 0^4' ;
.*. Pwp' passes through p.
Hence, if Ap, Bq, Or be perpendiculars to TOT', and pp', qq\ rr' be perpendiculars to BO, CA, AB, then pp', qq', rr' are concurrent at that point o> on the Nine-Point circle, where the Simson Lines of T, T' intersect.
It follows that o> is the Orthopole of the diameter TOT' (see Chap. VI).
THE SIMSON L1XE.
31
49. Since A'u = R cos (915 (44)
and A'<t) = R sin A'P<a, in the Nine-Point Circle,
.-. ^4'Pw = i-Tr — #! = angle Op// ;
.-. ^L'P is parallel to TOT'.
Hence, to find <u when 27O.T" is given, draw the chord A'P in the Nine-Point circle, and then draw the chord P<o perpendicular to BG.
A'
SO. The A'B'C' n.c. of <o.
From (35) the Simson Line of o> in the Nine-Point circle is parallel to A'P, and therefore to TOT'.
Hence the direction angles of this Simson Line are #1? 0«, Qy
It follows from (38) that the J/£'C" n.c. of o> are
a' = 2.JEcos^2cos^, &c., which are as sec 015 sec ^2, sec 03.
The ^BCn.c. of w.
From the similar triangles wA'p', AOp,
a = up = A'u.Ap/A 0 = p cos 6r
Hence the absolute n.c. of w are p cos 015 ^ cos $2, 9* cos ^8. To determine a in terms of p, q, r.
32 MODERN GEOMETRY.
Let p', q', r' be the perpendiculars on TOT' from A', B', C'. Then since A' is mid-point of 50,
also p' = A'O.cos 6l = R cos A cos 0l ;
The formulae of (14) and (50) supply us with a very simple proof of (8).
For aa = bp + rf. (14)
.*. a .p cos #! = b . R cos 0S cos 6l + c.R cos ^ cos #., ; = b cos 03 + c cos 02.
51. To illustrate the use of these formulae, take the case when TOT' passes through J, the in-centre.
Here cos0a = %(b-c)/d, p = R/d.(b-c)(*-a)/a \d=OI]. So that 4o; = R(c-a)(a-b)/d* oc l/(&-c), and aa = ^E/d2. (6-c)2(5-a) a (6-c)2(.s-a).
Hence w is the Feuerbach Point.
52. To determine the A'B'G' n.c. of 01? the centre of the circle in (45).
Taking A'B'G' as triangle of reference, the equation to TOT' is cos ^.cos Ol.aajr ... = 0. (16)
The equation to the diameter of the Nine-Point circle passing through oo, whose n.c. are (sec 01S ...) is
sin Ol cos 6l . aa' + . . . = 0. (46)
Hence at Olt where these lines intersect, aa'.cos Ol oc cos B sin 03 — cos C sin $2
a cos B (p — q)/siu G — cos C (r — p}/ sin B ; /. a'cosflj oc p (sin 2 J5 + sin 2 (7) —5. sin 25— r sin 2 6'. Bat since TOT' is a diameter of ABC,
p sin 2 J. + ^ sin 27? + r sin 20 = 0. Hence a <x p sec 0X, &c.
To determine the radius Ojw (= p) of this circle.
The perpendicular TT from o> on the A'B'C' Simson Line of <u
= 2 . JE . cos 0t cos 0, cos 6y (39)
The angle &' which this makes with the diameter wO^'
= ^ + ^ + 0,. (41).
THE SIMSON LIXE. 33
The perpendicular from oo on TOT'
= 2 perpendicular on Simson Line = 2?r ; /. p = Oj<o = 2?r sec 8' = 2.R cos ^ cos 02 cos 03 . sec (0X + 02 + 08) .
The parabola, which touches the sides of A'B'G' and has o> for focus, has the Simson Line of o> for its vertex tangent, and for its directrix the diameter TOT', which is parallel to the Simson Line and passes through the orthocentre 0 of A'B'G'.
53. The envelope of the Simson Line is a Tricusp Hypocy- cloid .
Since /.wA'H1 = AOT,
.-. «>0'Hl = 2.AOT. But Medial Radius = JB ;
/. arc -E^w = arc AT = 2 arc dh, by similar figures Hdh, HAT.
Now take arc A'L = ^ arc A'H^
and draw Medial diameter LO'l.
Then, since A'd is also a Medial diameter,
arc A'L = arc Id ;
/. arc H^L = 2 arc A'L — 2 arc Id. Also arc H^ = 2 arc JW ;
.'. arc LCD = 2 arc Z&.
Therefore Xo>/> touches* a Tricusp Hypocycloid, having 0' for centre, the Medial Circle for inscribed circle, and LOl for one axis.
54. Let DOD' be the circumdiameter through I, the in-centre. Let <£j, </>.2, 03 denote the vectorial angles D'DA, D'DB, D'DG. Draw the chord CIP.
Then P is the mid-point of the arc AB ; and
Also IPD' or GPD' = GDD' = 03 ;
/. IPD = ^r-03. Let r/R = in ; then
I + m_lljf__ny IP _ sin IPD' sin P DP' l—m ~ ID " IP ' ID ~~ sinDD'P ' sin IPD
sin 0.< sin -| (^ + <ft.2)
cos -| (0j + </>2) cos 03
D
34
MODERN GEOMETRY.
.'. tan
I —
. cot </>,
Put <k + 02— 0s = ^ 5 sin S EEB ,S', sin 03 = Then # = sin (<k + <£2) cos <£8 — cos (<k -f 02) si . 2tan^(<k + <M , 1 - tan
- l + tan'i^ + i.) ' C°S - /. fif/s. (l-m)2sm2
= 2 (1 — m2) cos2 <^)3— (1 — m)2 sin2
cos
And finally,
8/s. f or .s-3-m^.
= (3 —
^ = 0.
B D
From (42) the angle S is the vectorial angle of the perpendi- cular from _D on the Simson Line of -D, and this Simson Line passes through the Feuerbach Point F, and through /, one of the fixed points, where the circle I(r) cuts the axis DIOD' (45). The cubic then gives the values of <k, <£2> 0»i *ne vectorial
THE SIMSON LINE.
35
angles of DA, DB, DC ; and thus the triangle ABC is deter- mined for any given position of F.
Again, if we put cos 8 = (7, cos <£3 = c, we shall obtain
(Greenhill).
55. In a poristic system of triangles ABC, the Simson line of a fixed point 8 on 0 (-R), passes through a fixed point.
Let SSv S8Z be tangents to I(r) ; then 8^ is also a tangent.
Draw DE touching I (r) and parallel to /SyS'.,.
Draw 8T perpendicular to DE.
Then shall the Simson Line of 8 pass through the fixed point T.
Let DE be cut by AB. AC in y, /?.
Let AC, AB cut S, £2 in F, G ; let AS» ASl cut D.2? in H, K.
Now ip, IE, Iy, ID are perpendicular respectively to IF, T8^ 1G, ISl ; and hence
Thus
or
ftD.pH yD.yH
Ao-nin, L SDH = SS& = SAS, or Sf/lfi".
.*. HDAH is cyclic, and similarly SEAK.
36 MODERN GEOMETRY.
The relation (i) shows that the powers of fi and y for these circles are in the same ratio ; and hence that ft, y lie on a circle coaxal with these circles, and therefore passing through A and 8.
That is, ASj3y is cyclic.
Now the four circles circumscribing the four triangles formed by AB, AC, BC, fiy have a common point, which must be /S, since this point lies on the circles ABC, Afiy.
Hence o^o^cr, and T, the feet of the perpendiculars from S on the lines jRO, GA, AB, /3y, are collinear.
In other words, the Simson Line o^o-gOg of S for the triangle ABG passes through T, a fixed point, being the foot of the perpendicular from S on the fixed line fty.
(Grreenhill and Dixon)
The properties given in (34), (54), (55) present themselves in the Cubic Transformation of the Elliptic Functions.
CHAPTER V.
PEDAL TRIANGLES.
56. FROM a point S within the triangle ABC draw Sd, Se, Sf perpendiculars to BC, CA, AB respectively, so that def is the Pedal Triangle of 8 with respect to ABC.
Let angle d = A, e = /*, / = v.
Produce AS, BS, CS to meet the circle ABC again in L, If, N.
From the cyclic quadrilateral SdCe,
L Sde = SCe or NCA = NLA, so Sdf = ML A ;
d or A = MLN, &c.
Thus LMN is similar to the pedal triangle of S with regard to ABC.
Similarly ABC is similar to the pedal triangle of S with regard to LMN.
.'. BSC=BMC+SCM
= BMC + NCM or NLM = A + \.
To determine a point 8 whose pedal triangle has given angles A, /x, v, describe inner arcs on BG, CA, AB respectively, containing angles A-}- A, J5 + /x, 0 + v, any two of these arcs intersect at the point S required.
37
38 MODERN GEOMETRY.
This construction also gives S as the pole of inversion when ABC is inverted into a triangle LMN with given angles X, /x, v.
57. If rx, r2, ?*3 represent SA, SB, 8C, the tripolar coordinates of 8, and if p be the circumradius of def, then 2p sin A = e/ = SA . sin e J/
(in circle SeAf, diameter AS) = rl sin JL ;
.*. MN : NFj : LM = ef : fd : de = sin A : sin /x : sin v
= ari\ : ferj : ci\.
Limiting Points. (G""0
In section (21) let TA, TB, TC meet the circle ABC again in IMN.
Then from (57) MN oc sinX or arr
In (a), TA&S— a; .'. MNcta(s—a).
In {&), 77J. a 1/ya; /. M"JV oc a/v/a a A/a.
In (c), TAaza-, /. If JV oc a2.
In (a1), T^l oc v/cot^.; .'. JfJVocsin^. -s/cot J. oc \/sin 2J..
Let Jx, I3 be the in -centres of LMN in (a) and (c) ; H^ H4 the orthocentres in (b) and (rf).
It is very easy to prove that
01, = 01, OI3 = OH, OH, = 01, OiT4 =
58. In (21), since
p : q : r oc sin A/a : sin fji/b : sin i//c, the Radical Axis becomes
sin2 A/a2 . x + sin2 /x/62 . ?/ 4- sin2 vjc~ . z = 0.
To determine J, the pole of this Radical Axis for the circle ABC.
The polar of a'ft'y' is (by1 + c)8') a + . . . = 0. /. sin2A/« oc ~by -\-cfi } .'. sin2A a ac/3' •+- aZ>y' ;
.• . 6ca' oc — sin2 A -f sin2 /x + sin2 v oc — e/2 +/^2 + de* a cos A .fd . de oc cos A . sin ^ sin v oc cot A ;
.*. a! oc a cot A; so that the n.c. of /are a cot A, 6 cot/x, c coti/. (Dr. J. Schick)
PEDAL TRIANGLES. 39
59. To show that Z7, the area of def, is proportional to II. the power of 8 for the circle ABC.
2U = de . df. sin X = r2r3 . sin B sin C sin X, and 2 A = 4 R- sin A sin B sin C.
Now r3.&M"=I
and in the triangle SMC,
r3 or 86' = $flf.si Hence Z7/A = Jll/fi3 = J (B2- OS')/Rr.
When U is constant, OS is constant, and S describes a circle, centre 0.
Let a, ft, y be the n.c. of S.
Then fty sin .1 + . . . = 2(AeS/+...)
Putting 4A2 = (aa + &^ -f cy )2,
we h ave afiy + . . . = (R2 — 0 S°)/abc, . (aa + 6^-4- cy) 2 as the locus (a concentric circle) of S9 when OS is constant.
When 08 = R, U vanishes, as the pedal triangle becomes™a Simson Line, and
GO. To determine the power II of a point S in terms of #, y, z, the b.c. of S.
Since aa/x = bft/y — cy/z —
.'. a = 2A/ (a: -\-y + -z). x/a, &c. ;
C 2R
Note that only the ratios x : y : z are required.
Examples. — (i) For I, x oc a ;
40 MODERN GEOMETRY.
(ii) For #, x oc tan ^4 ;
a1 tan B tan 0 + • . •
(tan A + tan 1? + tan (7)J = 8 7T2. cos ^4. cos B cos C.
(iii) For 6r, a; = y = z ;
61. When 8 lies on a known circle for which the powers of J., J5, G are simple expressions, the power II of S for the circle ABC usually takes a simpler form than that given by the above general formula.
Let d be the distance between the centres of the two circles Q and ABC ; tftftf the known powers of J., #, G for the circle Q ; TrTTjTToTr.j, the perpendiculars from $, J., 5, (7 on the Radical Axis of the two circles.
From (7), we have ;r = "
But, from coaxal theory,
t* = 2d . 7r1? &c., while II =
Equating this to the expression for II, found in (60 j we have a-yz + tfzx + rfxy = (tiX + t*y+t£z)(p+y+t} as the locus of 8 : i.e. the circle Q.
As an example, let us find II for the point w (50).
This point lies on the Nine-Point Circle, for which the power of A = t* = \ be cos ^4, &c.
Also for w, x = aa = p (q + r} tan A, (50)
and x + y + z = 2A.
/. H= [Jfoccos-4._p(2 + r) tan -4 -f.. . J/2A
62. The Radical Axis of the circles Q and ^5(7 is
t?x+tfy + t'fz = 0, for TTj oc t^.
Examples. —
(i) When Q is In-circle : t? = (s—af. (ii) When Q is Circle J^Jj : ^2 = fee.
PEDAL TRIANGLES. 41
(iv) When Q is Anti-medial Circle (1) : t* = a\ (v) When Q is Nine-Point Circle : ^2 — ^ be cos A oc cot -4.
(vi) When Q is circle T^./l1.^ circumscribed to the triangle formed by the tangents at A, B, C : t* — E2 tan B tan C oc cot A
(vii) When Q is the circle ((91?) : t? = 2E cos A.%h,ac cot A. (viii) When Q is Polar Circle :
t? = AH*+4<B? cos A cos B cos 0 a cot A
So that circles (v) (vi) (vii) (viii) have the same Radical Axis cot A.x + cot B . y + cot C . z = 0.
63. Feuerbach's Theorem.
To determine the Radical Axis of the Nine-Point circle and In-circle :
Take A'B'C' as triangle of reference,
The power of A' for the In-circle = t* = A'X* = J(6— c)2 ; /. the Radical Axis is (6— c)2#'-|- ... = 0.
But this is the tangent to the circle A'B'C' at the point whose n.c. are l/(6 — c)....
Hence the two circles touch, and !/(& — c)... are the n.c. of the point of contact.
64. To express II in terms of X, //,, v.
We have L BSG = A + A ; (Fig., p. 37)
/. aa = 2.&BSC = BS.SO.aintA + X) = BS . SM . sin (A + X) sin A/sin X = n {sin2 A cot X + i sin 2^1} ;
. 2A/H or (aa + &/34-cy)/LI = S.sinM cot X + i^.sin2^. Multiply each side by 452. Then
87?2A/I1 or 2E . abc/tt = a? cot X + 62 cot /x, + c2 cot v + 4A
= 2E . abc or 8E2A, n = 2R . abc/M ; giving II in terms of X, /x, v.
* The expression "a2cotA+..." was first used, I believe, by Dr. J. Schick, Professor in the University of Munich. The relations he deals with are different from those treated in this work.
42 MODERN GEOMETRY.
Since IV
cr cotX-K..+4A
65. Observe also that the area of def
= U= J. A/E2.n = 2A2/M ;
so that, if p be the radius of the pedal circle def, 2p~. sin X sin /UL sin v — U = 2A2/M ;
and now all the elements of def are found in terms of X, /UL, v. From above, aa = fl . sin (A + X) sin A/sin. X.
(ibc sin (A +X)
Hence a = — . : — • .
M sin X
66. To illustrate one of these results, take $, the focus of Artzt's Parabola, which touches AB at B and J.Cf at C (see Figure, p. 88). (G-.)
It is known that L SAC = SB A, SAB = SO A ;
/. 5^% = BAS + ABS = BAS + CAS = A. So GSK^ = A.
.'. BSA = IT— A — #+ G = 06VI, and 58(7 = 2,1.
So that, if X, /x, v are the pedal angles of $,
Putting 4A.cot^L = — a2
we obtain 2A.M = a2(2
= 4crra12, where ml ^
Then p~ sin ^4. sin G sin J? — A2/lf ;
and finally p — J&c/m,.
Then $^4 . sin J. = e/ = 2p sin X ;
/. SA = %. bc/m, ; so 85 = | . c^jm^ SG =
The n.c. of S can be at once obtained from
a = abc/M. sin ( A + A.) /sin X, &c.
67. If, with respect to the circle ABC, the point Sl be the inverse point of S, lying on OS produced ; then OS.OSl = -B2,
PEDAL TRIANGLES .
43
so that S, £>i may be taken as the limiting points of a coaxal system, to which the circle ABC belongs.
If OSS, cuts this circle in T,. T', then AT, AT' bisect the angles between AS and AS19 &c., so that
S,A : SA = ^T : ST = S,B : SB = S,C : SC. Let d^f, be the pedal triangle of Sl ; then in the cyclic quad- rilateral SlelAfl : BI/I = SiA . sin A ;
•"• ei/i :/i^i : ^iei = #i-4. sin J. : S^B.sinB : S-^C . sin C = SA .sin A : SB .sin B : SO .sin C - ef : fd : de.
Hence the pedal triangles of the inverse points S and Si are inversely similar : so that
di = X, el = ^ fi = v.
To prove that the triangle .L1.M1.W1 is similar to d^fi, the pedal triangle of Sr
In the cyclic quadrilateral #1d1ce],
So
.'. eidji or X = M^N^ &c.
Similarly ABC is similar to the pedal triangle of Sl with respect to the triangle LjJtfjJVj.
To determine &\ first find 8 (56), and then obtain Sl as the inverse point of S.
44 MODERN GEOMETRY.
We now have a second pole ($T) from which ABC can be inverted into a triangle J^M^ with given angles X, p., v.
68. To express IIj, the power of Sl for the circle ABC, in terms of X, /u,, v.
l_ B8,C = BM1G—M1CS1 (M.CN,)
= HM.C-M.L.N, [cyclic quad. = A—\.
So A&\B = C-v.
Also A8,C = A81B + B81C = (A-\) + (C-v)
= n-B. |> > B]
.-. oo! = 2.^B81C = BSl.C81.sinB8lC = BSl . 8^ . sin (A - X) sin A/sin X = Ilj.sin (A — X) sin J./sinX. So cy-t = Hi . sin ( C—v) sin (7/sin v,
and 6ft = n: . sin (B—p) sin 7?/sin /*,
giving y8x a negative value [/A > B], as the figure indicates. Proceeding as before, we find
n^ = ZR.abc or 852A, where M,= a2 cot X+ ...— 4A, giving IIj in terms of i¥15 and therefore of X, u, i^. Since H = 02-E2
Jtt2 ' a2cotX+...-4A'
[or from
Hence 0/S2/^2 = ^/M, and 0<S\W = OT E—08 VM—
_
8T' ~ JB+ 08 ~ VM+ VMl ' The area of A^A/I = #1 = SA2/!/!. Hence l/U—l/Ul = 4!/^.
If ^ be the radius of the pedal circle d^f^
2pi* . sin X sin /x, sin v = ^ = 2A2/M1. And now all the elements of dlelfl are found in terms of
/A, V.
From above, ac^ = IIj . sin ( A — X) sin A/sin X.
a6c sin (A — X)
Hence a, = -— . - \ . } .
M sin X
PEDAL TRIANGLES.
45
69. In section (47) it was proved that the figures AT SOT' and uXdA'X' are similar, and just as S is homologous to d, so is Sl homologous to dr
Now AT bisects the angle SASl ;
.*. toX bisects the angle dn)dr It follows that
<ad : wdi = Xd : Xd, = TS : TS» &c., so that <*>d : we : w/ = <adl : o^ : <ofr
Hence <o is the double point of the similar figures ; and wX, <uX', the Simson Lines of T, T\ are the axes of similitude for the inversely similar triangles def and ^i/i- (Neuberg)
CHAPTER VI.
THE ORTfiOPOLE.*
*7O. LET jo, g, r be the lengths of the perpendiculars Ap, Bq, Cr from A, 7?, C on any straight line TT', whose direction angles are 0M 02, 0,.
Draw p$j perpendicular to BC, qS2 perpendicular to OA, rS& perpendicular to AB.
These lines shall be concurrent.
For B8l = BD+p sin^, CSl = GD—p sinfl,, and i.apsinWj^O; (2)
/. KBSf—GSfi = 5(£'/)2-OD2)4-2.2^sin^ = 0.
Hence $!_p, *S2g, /93r are concurrent.
The point of concurrence, denoted by <S', is called (by Professor J. Neuberg) the Orthopole of TT.
The Orthopole theorems are nearly all due to Professor J. ISeuberg.
46
THE ORTHOPOLE. 47
Let 9 be the length of the perpendicular 88' on TT1.
Now, for this one particular case, take 0a, Or 03 to represent the acute angles which the sides of ABO make with TT'.
We have pq = c cos 0S.
Also pSS' = Ol :
since 8p, 88' are perpendicular to BC, TT' ; also #£$' = 03 and ^$2 = C ;
since $p, $g are perpendicular to 50, (7 A
.-. Sp = c cos 03.sin (90°— 02)/sin G = 2R cos 02 cos 6.. and 9 = Sp cos 0X = 2 J^ . cos 0, cos 02 cos 03.
As T2" moves parallel to itself, the figure SqS'pr remains unchanged in shape arid size.
71. To determine the Orthopole geometrically.
Let Ap meet the circle ABG again in R.
Draw the chord RR'R" perpendicular to BG.
Let BA, OA meet TT' in r', r".
Then BR = 2B.sin (BAR or r'Ap) = 2R cos 08, so OS = 2E cos 02.
/. RR = BR . OB/2B = 2E cos 02 cos <93 = ^8p.
Hence 8 is found by drawing R'S, pS parallels to Ap, RR'.
As TT' moves parallel to itself, 8 slides along R'8 perpendi- cular to TT'.
Also R'8' 8, being parallel to AR, is the Simson Line of R".
72. To determine the ABC n.c. of S, (G.)
a. = 8Sl = projection of J9gr$ on fi^^ = q cos 0j + 85 cosp8q = q cos 0j + 2R cos 6*! cos $3 . cos 0. .'. asectfj — q + 2R cosC (cr — ap cos B — bq cos J.)/2A,
from (2).
Multiply first term on right side by sin A sin B sin 0 and the other terms by 2A/4R2. Then
! . A/2/t2 = sin ^4 sin 0.^ sin B — cos A cos G .q sin 5
+ cos C.r sin G— cos jB cos O.jp sin A — cos J5.(/ sin J9 + cos0.r sin 6'— cos^cos O.
MODERN GEOM ETRY .
48
Also 2A.COS0J = ap — cos C .bq — i
*.-. a = R/2&2.(ap—bq cos C — cr cos B)
x (cos 73 cos C.ap — cos B . bq — cos C .cr). Multiplying the two factors, and using the form
2 . ay -S.gr. 2bc cos A = 4A2, (5)
we obtain
2Aa = abc cos B cos C+aqr — rpb cos C — pq.c cos B. When TT', whose equation is ap.a+ ... = 0, passes through the circumcentre 0, then ap . cos A + . . . =0. In which case
a = JK/2A2 . 2 A cos Ol . (cos B cos C.ap + ap. cos ^4) = p cos 0r
73. Let ^Oi' be the circumdiameter parallel to TT'.
Then cr, the orthopole of tt\ lies on the Ts"ine-Point circle.
Let H be the orthocentre of ABC.
Since ATT = irR, .'. J^'o-j = o^-Hp
Again 5"E' = 2?. J2' + 2.0A' = tnr + AH;
* It was Mr. T. Bhimasena Rao who first pointed out that the expression for each coordinate involves two linear factors in p, q, r. His method is different from that given above.
THE ORTHOPOLE. 49
But ATT' = RR' (equal triangles AmerRarR'] = o-rr. And Tr'H, = 2.7TO-,;
Hence o- is the mid-point of /TZ?", and therefore lies on the Nine-Point circle.
74. To find a0£0y0, the ABO n.C. of cr. (G.)
Bisect Tr'jETj in m.
Then CTTT = JV and TT<TI == TT'W.
.*. a0 = o-^ = J.m = ATT COS m^TT — jo COS 0j, &C., as found in preceding; article.
To find a0'£0'y0', the J/£'C' n.c. of cr. (G.)
Since cr^ is = and parallel to Am,
.'. Acr is = and parallel to o^m. Also TT^Hjm is a rectangle ;
AH^cr is a symmetrical trapezium.
Hence the circle (^Lw), passing through Hl and TT, passes also through or.
.'. <rA' = R sin o-Jfj J/ = U cos (9T ;
/. a/ = orB' .irC'/R — R COS ^2 COS #3, &C.,
so that cr is (sec^, sec<92, sec ^3) referred to A' JVC'.
It follows that the point cu of sections (44-50) coincides with the Orthopole <r of a circumdiameter TOT'.
75. The Simson Lines of the extremities of any chord TT' of the circle ABC pass through $, the orthopole of TT' .
For, since L TpR = 90° = TR1R,
.'. TR^R is cyclic ; .'. pR^X or pR^ = TRp or TRA = T1\A ;
.'. Rrp is parallel to ATV But XEa = 72'JS = Sp ;
.'. XA^ is parallel to R\p, and therefore to T^. Hence XS is the Simson Line of T. So for T'.
50
MODERN GEOMETRY.
Thus we have three Simson Lines SX, SXr, 8R', all passing through 8, their poles being T, T', R" respectively.
They are the three tangents drawn from 8 to the tricusp- hypocycloidal envelope of the Simson Lines, so that each of the points T, T\ R" has similar relations to the other two ; for example :
(«) Just as R'S is perpendicular to TT'. so XS is perpen- dicular to T'R", and X'S to TR" ; or each Simson Line is per- pendicular to the join of the other two poles.
(6) As jp, the foot of the perpendicular from A on TTf1 lies on the perpendicular to BC from S : so also do pl and p3, the feet of perpendiculars from A on TR", T'R".
Hence ppip$ is the Simson Line of A in the triangle TT'R".
Thus (S. Narayanan) the Simson Lines of A, B, C for th triangle TT'R" as well as the Simson Lines of T, T\ R" for AhO all pass through 8.
THE ORTHOPOLE.
51
76. Let TT' cut BG in £7, and let the circle (AU)9 passing through Hl and p, cut Sp in p'.
Then Sp . Sp' = twice product of perpendiculars from 0 and S on TT'. (G.)
In the circle (A U ) the chords pp\ H^A are parallel ;
.'. the trapezium Ap'pH^ is symmetrical. But the trapezium Aa"jrHl is also symmetrical ; /. p'a- = pir — d and p-rr = S<r ;
Sp' = 2 So- cos crSp' = 2d cos Qr But Sp — 2R cos 0.2 cos 0;i = d sec 6^ ;
Note that 2^8 is therefore the power of S for the circle (AU).
Similarly, if TT' cut CA in 7, ,15 in TF, the power of S = 2c/3 for each of the circles (£7), (017).
Therefore in the quadrilateral formed by 730, CA, AB, TT', the orthopole ( S) of TT' lies on the 4-orthocentre line, or common Radical Axis of the three diameter circles.
77. Lemoyne's Theorem.
If P be any point on the straight line TT', whose orthopole
is S, then the power of S' with regard to XT/, the pedal circle of P, is constant.
52 MODERN GEOMETRY.
Draw pfi, py parallel to RB, Ti'O, so that tlie figures Aftpy, ABRC are honiothetic.
Draw AX' parallel to BC, XPX' perpendicular to PC. Let YZ cut /3y in x.
Since Zi lies on the circle A BC, therefore^ lies on the circle Afty. Lpyx or pyfi = pA3 or pAZ i = pYZ or pYx
(circle AYpZPX', diameter AP) ; Yxpy is cyclic ; /. pxy - pYy = 180°-pr,4 = jpX'4 ;
/. pxX' is a straight line. Let Xx cut pS in 8. Then Sp/XX' = px/xX'
— ratio of perpendiculars from p and J. on fiy = „ „ „ It and ^L on BG
(from similar figures Aftpy, ABRC) = SB'/ XX' ; /. 8p = RR' = pS;
.'. 8 coincides with $.
Observe that, since U^pp A is symmetrical,
.*. Xpp'X' is symmetrical, and therefore cyclic. Denote the circles XYZ, XX'pp, AYpZPX' by L, M, N respectively.
Let L and M intersect again in x. Xow the common chord of L and M is A'a''.
If and N is A~>.
,, ,, „ L and JV is YZ.
These common chords are concurrent. But X'p, YZ pass through x ;
.'. Xx' passes through &, and therefore through S. Finally, since Xpx'p'X' is cyclic,
/. SX.Sx' = Sp.Sp' = 2d&. But X, x' are two points on the circle XYZ. Hence the power of $ for the circle XYZ = 2<iS. ISTote that the circles (AU), (#P), (GW) are the pedal circles of
ur, F, w.
78. When TT' is a circumdiameter ^Of', then t7 vanishes, and then the pedal triangles all pass through the orthopole <r of tOt'.
THE ORTHOPOLE. 53
Since in this case AT = TrR, the dimensions of the homothetic figures A^iry, ABRC are as 1 : 2, so that fty becomes B'G'. Hence Xo- and YZ intersect on B'C'.
79. Returning to the general case, the diagram shows that if x, y, z are the b.c.'s of 8 with regard to the triangle XYZ, then
8 SP
AXYZ~xX XX'
_ 27? cos #2 cos 0^ >
2R sin B sin 0
.'. x : y : z = sec 0, sin A : sec 02 sin B : sec #3 sin C. Thus the Orthopole has constant b.c. for every one of the pedal triangles XYZ. (Appendix I.)
8O. In section (20) we determined the inverse points T, T' whose tripolar coordinates are p, q, r by dividing BG at P, P' ; CA at Q, (/ ; AB at 1?, J?/ ; so that BP : PC = q : r, &c., and describing circles on PP', ^(/, _BK'.
Conversely we may begin by taking two inverse points T and T', whose tripolar coordinates are p, q, r. Then the internal and external bisectors of BTC, (BT'G) meet the sides in the points P, P', &c. ; for
J3P : PC = BT : TC = q : r, &c.
It is known that the triads of points P'^'JB', P'QR, Q'RP, E'PQ lie 011 four straight lines, the equation of P'Q'R' being
px + qy + rz = 0, while that of P'QR is — px-}-qy + rz = 0, &c.
To prove that the point common to the four circumcircles of the four triangles formed by these four straight lines is w, the Orthopole of OT1*.
Describe a parabola touching P'Q'R' and the sides of the Medial Triangle A'B'C', and take A B'C' as triangle of reference.
From (15) the A' B'G' equation of P'Q'R' is
(g + r) x' 4- (r +p) y'+(p + q) z' = 0.
Then from (9) the n.c. of the focus are as , ^-, &c.,
where [g + r = p'], ..., or as a^y3,
Changing +p into — jo makes no change in the focus, therefore the parabola also touches P'QR, and similarly Q'RP and R'PQ
54
MODERN GEOMETRY.
Now let 0,, 02, 03 be the direction angles of OTT\ and draw Tm perpendicular to BG.
Then -
oc
—GT* oc 2a. J/w oc 2a.OITcos01;
a /)
-= - 5 oc sec 0,. 2-r
Hence the orthopole w, whose n.c. are (sec 015 ...) is the focus of the parabola : the Simson Line of w (in A'B'G') being the vertex tangent, and OTT' the Directrix. And, the circum- circles of the triangles formed by any three of the four tangents P'Q'R', ..., pass through the focus o>.
On the fixed Directrix TOT' may now be taken an infinite number of inverse pairs (TT*). For each pair we have a set of four harmonic lines touching the parabola, the circumcircles of the four triangles passing through <o, and the four orthocentres lying on TOT'.
In addition to each set of four harmonic tangents, there are also the three tangents BfG'1 G'A\ A'B'. The student may develop this hint. (G.)
Remember also that the pedal circles of all points T or T' also pass through o>. (Appendix II}.
CHAPTER VII.
ANTIPEDAL TRIANGLES.
81. If S be any point within the triangle ABC, the angles BSC, CSA, ASB are called the Angular Coordinates of ' 8 -, they are denoted by .X", Y, Z.
If A, /A, v are the angles of the pedal triangle of S, then
(65)
So that, if a, /?, y are the n.c. of 8,
aa = Il.sin (A+\) sinA/siu A = n . sin X sin A/sin (X—A).
82. Orthologic Triangles. N
Draw any straight lines MN, NL, LM perpendicular to 8 A, SB, SO, so that the perpendiculars from J, 5, C on the sides of LMN are concurrent (at /S). Then shall the perpendiculars from LMN on the sides of ABC be concurrent.
Let AS, MN intersect at Z, &c.
55
56 MODERN GEOMETRY.
Then SM2- SN2 = MP—NP = AM*- AN2 •
8V—SM* = CL*-CM*',
.-. (BL2- CL2) + (CM '-^Jtf 2) 4- (AN'-BN*) = 0. Hence the perpendiculars from L, If, JV on £0, (7-4, AB are concurrent, say at R.
Triangles ABC, LMN which are thus related, are said to be mutually Orthologic.
To determine the trigonometrical relation between 8 and R. Since SB, SC are perpendicular to LN, LM,
.-. /_ BSC = X = TT-L ;
so that aa = Il.sin (TT — L) sin ^4 /sin (?r — L — J.), from (65)
or x = II .sin i sin^4/sin (lv + ^1).
Now let »', ^/', 2' be the b.c. of R referred to LMN, and let II' be the power of R for the circle LMN. Reasoning as before, we have
x' = n'.sin (TT — A) sin _L/sin (TT — A — L)
= IT. sin ^4 sin _L/sin (J. + L) ; .-. x/x' = y/yf = z/zf - area ABC/LMN
= n/n'.
Hence the b.c. of & with reference to ABC are as the b.c. of R with reference to LMN ; and the proportion is that of the areas of the triangles, or of the powers of S and R for their respective circumcircles.
83. Antipedal Triangles.
Let def be the pedal triangle of 8.
Then, since dS is perpendicular to BC, eS perpendicular to CM, fS perpendicular to AB are concurrent at S, the triangles ABC, def are orthologic, so that the perpendiculars from A on ef, from B on fd, from 0 on de, are concurrent — say at /S".
Through A, B, C draw perpendiculars to S'A, S'B, S'G, form- ing the triangle D'E'F' — the Antipedal Triangle of S'.
The sides E'F', ef (being each perpendicular to S'A) are parallel ; and therefore def, the pedal triangle of S, and I)' E'F', the antipedal triangle of >$", are homothetic.
It follows that
jy = d = >, E' = e = ,M, F' =f= v.
ANTJPEDAL TRIANGLES.
57
Also BS'C = TT— D' = TT— A, OS' A = TT— n, AS'B - -R—V.
Again, since ABC and def are orthologic, the b.c. of 8' for
5(7 are as the b.c. of S for def.
Therefore, taking a/3y, afly' as the ii.c. of S, /S", respectively,
k.BS'C _ area of ABC _ A^ .
^
area of de/ " U"
aa =
But, from (65),
_ afrc sin (.£?+/*.) . _ a^c sin ( G + v) .
: ~' " " '
sn /A
sn r
and
« n , «?^c sin ( B + fi) sin ( C +
so that, finally, «.' — — — . - * - r-^-2 — :
J M sm /x sin v
But
abc
sin A
a'2?>2c2 sin (^ +X) sin (g + /*) sin
- • - — • — M* sin A sin fj. sin v
= ftp = -yy', from symmetry. Thus 8, S' are the foci of a conic inscribed in- ABO.
(65)
In Germany these points are called " Gegenpunkte." In this work the name " Counter Points " will be used.
58
MODERN GEOMETRY.
84. To determine the area (V) of D'E'F', the antipedal triangle of S\ in terms of X, /A, v. (G-.)
Henceforth the areas of the pedal triangles of $, Sf, &c., will be denoted by U, Z7', &c., antipedal triangles by F, F', &c.
Let 01? 02, 03 be the circumcentres of the circles S'BD'C, ffCE'A, S'AF'B ; S'D', S'E', S'F' being diameters.
Then 0,Af = i.acotD' = i.
.-. . AO.BC = J. a2 cot A. Now Oj is the mid-point of S'D' ; .-. area S'BD'C = 2 . BO^S' = 2 (J . a2 cot X + £' .-. V = S'BD'C + S'CE'A+ S'AF'B
= \ (a2 cot X -j- 62 cot /x + c2 cot i/ + 4A)
But Z7 = 2A2/M. (65)
/. C7F' = A2 ;
or, the area of ABC is a geometric mean between the area of the pedal triangle of any point, and the homothetic antipedal triangle.
D'
This is a particular case of the more general theorem given in (158).
ANTIPEDAL TRIANGLES. 59
85. To determine the n.c. (X v, w) of T, the centre of similitude
of the homothetic triangles def and D'E'F'. (Gr.)
The ratio of corresponding lengths Td, TD' in def and D'E'F'
= VU : VV = N/J7F : V = A : V = U : A. Therefore, since e£, D' are homologous points, M/JW = Td/D'd = U/(±—U).
Now the perpendicular from d' on AC = J31 ' -\-a! cos 0,
„ ^L7? = y'+a' cos#;
.-, d'C = (P + a! cos 0)/sin 6', d'5 = (/ + a; cos £)/sin £ ; .-. D'D" = d'd" = d'B.d'C/d'S' ;
where m = Z7/(A—
Example. — The orthocentric triangle, or pedal triangle of jff, is homothetic to T^T^ (Fig., p. 89), which is formed by the tangents at A, 5, 0, and is therefore the antipedal triangle of 0.
Here a = R cos J., &c., and the formula gives
0 7-j sin2 ^1 cos J3 cos (7 ,, ,
u - 2R.- — - — -ccsm^tan/1.
1 — 2 cos J. cos ±? cos U
Since H and 0 are homologous points, being the in-centres of the two triangles, the point T lies on OH.
86. Let S1 be the inverse of 8 for the circle ABC, as in (67), and let dlelfl be the pedal triangle of Sv
Then, since d^S^ 6^,/jSj, perpendiculars to BC, CA, AB, are concurrent at Slt therefore the triangles ABC and dlejl are orthologic.
It follows that the perpendiculars from J., 2?, G on e^,, /^p ^G! respectively meet at a point — call it $/•
Through A, B, G draw perpendiculars to S^A, S^B, S^O forming the triangle D/J^j'J?/, the antipedal triangle of /S/.
The sides E^F-{ and ejlt being each perpendicular to *S/A, are parallel, so that d^f^ the pedal triangle of $19 and D^E^F^ the antipedal triangle of $/, are homothetic.
Hence D/ = di = X, -
60
MODERN GEOMETRY.
We have now four triangles, viz. : the pedal triangles of >V, ,S',, and the antipedal triangles of S\ $,', each of which has angles X, /x, v.
Let c^ftyj and a/ft'y/ be the n.c. of £4 and $/.
From (82) the b.c. of #/ in ABC are as the b.c. of 8l in d^i/i-
i = ... = - ; (Z7, = area of d^)
.'. aa,' =
But, from (68), ft =
so for yl ; and I Hence, finally,
.
J/1 sin /x,
a, a, —
/ abc sin (B — /x) sin ((.7 — v)
MI " sin fji. sin v
, _ «262c2 sin (.-I— X) sin (5— /x) sin(0— v)
sin A sin /< sin v = ft/3/ = y^/, from symmetry.
Hence Sl and /S'/ are a second pair of Counter Points, being the foci of a conic touching the sides of ABC.
ANTIPEDAL TRIANGLES .
61
87. To determine the area (F/) of the antipodal triangle ^'EiFi in terms of A, /tx, v. (Gr.)
Let O/, 02', 08' be the circumcentres of the circles S^BD^C, \'CEiA, Si'AFiB} SI'DI, SiJB^, SI'FI being diameters, and >/, O./, 03' lying on A'O, B'O, C'O respectively.
Then
0,'A' = ±a cot D/ = Ja cot A ;
O/ is the mid-point of Sl'Dl ' .
Therefore, if (a/yS/y/) ^e ^ne n-c- pendicular from _D/ on J5C',
i be the^per-
So
Adding, \ve have on the left side
62 MODERN GEOMETRY.
Hence F/ = | (a2 cot A + &'- cot /A + c2 cot v— 4 A)
- P*r Now area (Z7J of ^e^ is 2^/Ml ; (68)
,-, JW = A2.
N"ote also that difference of antipedal triangles of *S", S}' = F'-F/ = ^M-^M, = 4A.
The points 8' and S^ having similar antipedal triangles, are called "Twin Points."
Of the four points 8, S\ 8^, 8^ :
(a) 8 and 6\ are Inverse Points, with similar Pedal triangles.
(6) 8' and $/ are Twin Points, with similar Antipedal triangles.
(c) (SSf) and ($!$/) are pairs of Counter Points, the Pedal triangle of either point of a pair being homothetic to the Antipedal triangle of its companion point.
CHAPTER VIII.
THE ORTHOGONAL PROJECTION OF A TRIANGLE.
88. * AL being a given axis in the plane of the triangle ABC, it is required to determine the shape and size of the orthogonal projection of ABC on a plane X, passing through AL and inclined at an angle 6 to the plane ABG.
Draw AM perpendicular to AL.
In AM take a point T, such that AT/AM = cos 0. Draw BUb, CVc perpendicular to AL, cutting LT in U and V.
Then, since Ub/Bb = Vc/Cc = AT/ AM = cos 6, AUV represents, in shape and size, the orthogonal projection of ABG on the plane X.
S1
* In writing Sections (88-90) I have drawn on Professor J. Neuberg's " Projections et Contre-projections d?un Triangle fixe," with the author's permission. To avoid overloading the chapter, the Counter-projection theorems are omitted.
63
64 MODERN GEOMETRY.
On either side of the common base BG a series of triangles is described similar to the orthogonal projections of ABC on a series of planes X passing through the common axis AL.
It is required to determine the locus of the vertices of these triangles.
Draw At 8 perpendicular to LT, cutting the circle ALA^I in S. The triangle SBC will be similar to A UV, and therefore to the projection of ABC on the plane X.
Draw SNS' perpendicular to BC.
Then L ALT = tAT or SAM
= SLM.
And A, t, S, N are right angles.
Therefore the figures ALtT, SLNM are similar.
And LU : LB = LV : LC = LT : LM.
Therefore the figures ALUVtT and SLBONM are similar.
Therefore the triangle SBC is similar to AUV.
Hence the vertices of all triangles SBC, described on BC, and having SBC = AUV, SCB = A VU, BSC = UA V, lie on the circle ALAr
The angles of projection range from
(9 = 0 to 0 = -J7T.
When 6 = 0, cos 0 = 1, so that T coincides with Af, S with AV and S' with A.
As 6 increases from 0 to ^TT, T travels from M to A, S from A l to M, and S' from A to M.
Hence the locus of S is the arc AMAl on the side of AA' remote from the axis AL.
89. When a series of variable axes AL19 AL2, ... are taken, the point AH being the image of A in BC, remains unchanged.
Hence each position of the axis AL gives rise to a circle ALA-^M passing through the two fixed points A and A^ so that this family of circles is coaxal.
Let DA or DAl = £, and let DR = h, where R is the centre of the circle ALAr
Then, with D as origin and DA as 7/-axis, the equation of the circle is x* + y* — 7r = 2hx,
h being the variable of the coaxal system.
We may now deal with the problem : To determine the plane X on which the orthogonal projection of ABC has given angles X, /x, v.
ORTHOGONAL PROJECTION OF A TRIANGLE.
65
Construct the triangle SBC, so that SBC = /*, SCB = v, and hence B8C = A.
Let the circle ASAl cut BC in L and M".
Then, if AL and $ are on opposite sides of AA{, AL is the required axis.
Draw LtT perpendicular to SA.
Then the required inclination 0 of the plane X to the plane ABC is given by
cos 6 = AT/ AM.
9O. The triangle ABC being projected on a series of planes making a constant angle a with the plane of ABC, and the triangles SBC being drawn, as before, similar to the successive projections, it is required to determine the locus of S.
Draw AL parallel to the line of intersection of the planes ; then the original projection is equal and similar to the pro- jection on a parallel plane through AL.
Determine S as before, taking T such that
AT/ AM = cos a.
Join 8L, SM, cutting AAl in H and H'. Then H is the orthocentre of LH'M, so that DH.DH' = DL.DM=k-.
66 MODERN GEOMETRY.
AT tsLvAL
Again, cos a —
AM t&nALM tan SLM HD
" tan ^ L.I/' A^D* .'. HD = k cos a = constant, H'D = k sec a = constant, so that H and Hf are fixed points.
Hence, since HSH' is a right angle, the point ,S describes a circle on HH' as diameter.
The equation to the circle HSH' is
x" + (y—k cos a) (y — It sec a) = 0.
For another series of planes inclined at a constant angle a', the points H and H' would be changed to Bl and If/, where
DH, = k cos a', DH,' = k sec a', and DH..DH,' = Jr.
The series of circles are therefore coaxal, having A and A} for limiting points, and therefore cutting orthogonally the former series of circles, which pass through A and Ar
91. A triangle ABG, with sides a, 6, c and area A, is projected orthogonally into a triangle A'B'C', with sides «', &', c', angles A, yu, ?•, and area A' ; the angle of projection being 0, so that
A' = A cos 0. To prove 2 . a!* cot J = 2A (1 4- cos2 0) ;
^.ar cot A. = 2A (sec 04- cos 0). Let /^, hfr h% be the heights of the points A, 7?, C above the
plane A'B'O. '
Then //,—/?.„ = vV-a'2 ;
... v/V2-a"2± v/62-Z/2± vV-c'2 = 0, the signs of the surds depending on the relative heights of
A, n, c.
This leads to
-(62 + c2-a2)2 or 16A-+16A'2
No \v b~ + c2— a2 = 4A . cot J ;
fe'2-fc'2-a'2 = 4A'.cotX = 4A. cos 0 cot A. Hence 2 . a'2 cot 4 = 2A (1 + cos2 6) ;
2. a2 cot A - 2 A (sec 0 + cos 0).
ORTHOQONAL PROJECTION OF A TRIANGLE. 67
The Equilateral Triangle and the Brocard Angle. When ABC is equilateral, we have
(a'2 + 6'2 + c'2) . I/ A/3 = 2A (1 + cos2 0) = 2A' (sec (9 + cos 6) . But, if u/ be the Brocard Angle of A'B'C',
cot «' = (a'2 f fe'2+ c'2)/4A' ; (131)
.-. cot a/ - -v/3/2 . (cos 0 + sec 0) .
The Brocard Angle therefore depends solely on the angle (#) of projection. It follows that all equilateral coplanar triangles project into triangles having the same Brocard Angle.
92. Antipedal Triangles and Projection. (Gr.)
Let I'm'n'j Z/w/n/ be the sides ; F', F/ the areas ; and A, /*, v the angles of D'E'F', JD/JW. the antipedal triangles of 8', £/• From (84),
2F' = 2.a2cotA + 4A = 2A (sec (9 + cos 0) +4A ;
/. F'/A = ( v/sec<9+ vxcos <9)2. So F//A = ( v/s^c~£- v/c^sl)2 ;
'- ^F/ = 2. %/ A cos 6> = 2. A/ A' ;
and yF'+A/F/^ 2. x/A sec^ = 2. v/A, ;
where Ax is the area of the counter-projection of ABC : that is,
the triangle whose projection (for 0) is ABC.
Xow the triangles F', F/, with the projection and counter- projection, are all similar, having angles X, /x, v ; so that their corresponding sides are as the square roots of their areas ; /. Z'-Z/ = 2.a'; Z' + Z/ = 2.a1.
Hence, if two antipedal triangles be drawn having the same angles as the projection, the sides of the projection are half the difference of the corresponding sides of the antipedal triangles.
This theorem, of which the above is a new proof, is due to Lhuillier and Neuberg.
93. The triangle ABC is projected orthogonally on to a plane inclined at an angle 6 to the plane of ABC.
If U, the line of intersection of the planes, make direction angles uv u^, u^ with the sides of ABC, it is required to deter- mine the lengths a', &', c' of the sides of the projection A'B'G' in terms of u^ u.2, %, and 0.
Let cos 6 = k.
68 MODERN GEOMETRY.
Draw perpendiculars Bb, Cc to the line U, and take
so that B'C' (= a') is equal to the projection of BC. Now the projection of B'C' on Bb
= B'b-C'c = k(Bb-Cc~) = A-.asin^; and the projection of B'C' on U = a cos u^ ;
.', a!" = B'C'- = a2 (cos2 u^ + k~ sin2 ?tt) = a2 (1-(1- A;2) sin2 u,} = a2 (1 — Binst*, siri'2^).
The dimensions of the projection depending, of qourse, only on the direction, not on the position, of U '.
94. Pedal Triangles and Projection. (G.)
To prove that, as the plane of projection revolves round Z7,
the projections of ABC are similar to the pedal triangles of
points lying on the circumdiameter TOT', where T is the pole
of the Simson Line parallel to U.
Let XYZ, X'Y'Z' be the Simson Lines of T7, T'.
Then XYZ, being parallel to ?7, makes angles ult u^ u$ with the sides of ABC, and therefore
L OTA or OAT = ur (37)
ORTHOGONAL PROJECTION OF A TRIANGLE. 69
In TOT' take any point P, and let OP = kr. R. Let u, v, w be the sides of the pedal triangle of P. Then, in the triangle GAP, AP2 = tf + k'*E2
So that 4tt* = 4 . AP2 Bin* A = a2 (1 + k')z j 1 - 4k> 2 sin2 u, I .
But a'- = rt2 {1 — (l-/v2) sinX}. (93)
a'2 1
Hence — , = — - — — ,
4w* (! + &')
and a' : b' : c' = u : v : if '
(so that the projection on the plane 0 is similar to the pedal triangle of P), provided that
so that P, and its inverse point P', are given by TP/T'P = TP'/T'P' = cos 0.
From (36) the point Tis found by drawing chord At parallel to £7", and the chord tXT perpendicular to BC.
95. Next, let the plane of projection be inclined at a constant angle a to the plane of ABC. (Gr.)
T j.i • it 1 — k 1 — COS a 21
In this case, k' — — - = - - - - = tan2 ia ;
1 + COSa
.'. OP — R tan2 Ja, OP' = B cot2 ia.
If, therefore, circles be drawn with centre 0 and radii tair^a.jB and cot2|a.JK. any point P on one circle, and its inverse point P' on the other circle, will have their pedal triangles similar to the projections of ABC.
Another Proof. — Let ABC be projected on the plane a into A'B'G', whose angles are A, /x, v. Let P (or its inverse P') be a point whose pedal triangle has angles A, /x, v.
70 MODERN GEOMETRY.
Then OP--IB' = "'cotA + fr cotg + r cotK-jA
or cot \ 4- o cot /x + c cot i> 4- 4A
_ 2 A (sec a+ cos a) — 4 A _ (1— cos a)' ~ 2A(seca + cosa)-f-4A ~~ (1 + cosa)2'
/. OP IE = tan2 Ja ; so OP'/R = cot2 ia, as before.
96. A triangle XYZ is projected orthogonally on a series of planes inclined to its own plane at a given angle 0. A point is taken such that its pedal triangle with respect to ABC is similar to one of these projections. To determine the locus of the point.
Draw inner arcs BQC, CQA, AQB containing angles A+X, B + Y , C -{- Z : a point Q being thus determined whose pedal triangle has angles X, Y, Z.
Let AQ, BQ, CQ (or AQ', BQ', CQ') meet the circle ABC again in yyz (or x'y'z') : then it is known that
the angle x or x' = X, y or y' = F, z or z' = Z. (56) Take Q as inversion centre, and let
(inversion-radius)2 or fc2 = power of Q for ABC
= E'-OQ* = OQ.OQ-OQ? = QO.QQ'; so that in this system Q' is inverse to 0.
Also tf = AQ.Qx = BQ.Qy = CQ.Qz;
so that #i/;z inverts into ABC,
97. Lemma.
If with any centre JT and any radius p, any four points I), jB, .F7, G are inverted into D', E', F', G', then the pedal triangle of G with respect to DEF is similar to the pedal triangle of G' with respect to D'E'F'. For
D'G'/DG = KD'/KG, and E'F'/EF = KE'/KF ; /. (D'G'.E'F')/(DG.EF) = (KD'.KE')/(KG.KF)
= (KD'.KE'.KF')/(p\KG) = (E'G'.F'D')/(EG.FD) = (F'G'.D'E')/(FG.DE)9 by symmetry, so that the theorem is proved.
ORTHOGONAL PROJECTION OF A TRIANGLE.
71
In Section (95) substitute the triangle xyz for ABC. Then the circles with common centre 0 and radii Rt2, R/t*, are the loci of points whose pedal triangles with respect to xyz are similar to the projections of XYZ on planes inclined to the plane of X YZ at an angle 6.
N"ow invert, with centre Q, radius k. Then xyz inverts into ABC, 0 into Q', and the two concentric circles (0, Rt2) and (0, R/t*) into circles of the coaxal system which has Q and Q' for its limiting points.
Let R be a point on either concentric circle, inverting into $, a point on one or other of the coaxal circles. Then, by the Lemma, the pedal triangle of 8 with respect to ABC is similar to the pedal triangle of R with respect to xyz ; that is, to one of the projections of XYZ.
Hence the required locus consists of these two coaxal circles.
98. Let the circum diameter OQQ' be cut by the concentric circles in LL' , MM' : and by the coaxal circles in II', mm', : so that L and Z, ..., are inverse points in the (Q, k) system.
The centres o>, w' and the radii p, p' of the circles II', mm' will now be determined in terms of X, Y, Z, t,
Ql = W/QL = W .-. 01= OQ+Ql= 01' = [_R(R + OQ
p/R = i (01- 01'} IU = k*t-/(OQ2-RW),
72 MODERN GEOMETRY.
And, writing l/t2 for f',
Om = [R(OQ-Rt*)y(R-OQ.t*), Om' =
Hence OLOm = R- — 01' . Om',
so that the circles II', mm', which are inverse to LL', MM' in the (Q, -K") system, are mutually inverse (as are LL', MM') in the (0, J?) system. Since the pedal triangle of Q has angles X, Y, Z, it follows, from Section (64), that
QQzi-p* - ^ cot x + fr2 cot r+ c2 cot Z- 4A a* cot X+b* cot Y+ c2 cot Z+ 4A ' so that now /a, p', Ow, Oo>' are expressed in terms of X, Y, Z, t.
When OQ = -Bf2, or when Q lies on the inner concentric circle LZ/, then Oo> and p are infinite.
Hence the (Q, &) inverse of the circle (0, OQ) is DJS'D', the Radical Axis of the coaxal system, bisecting QQ' at right angles.
And if N be the pole of DED' for the circle ABC, it is easily proved that the (Q, k) inverse of the circle (0, OQ') is the circle ON.
99. A case of great beauty and interest presents itself when the triangle XYZ is equilateral. For then Q, Q', having equilateral pedal triangles, are the Isodynamic points 8, S,, lying on 0-ST, the Radical Axis of the coaxal system is now the Lemoine axis ; and N, the pole of the Lemoine axis for the circle ABC, is the Lemoine point K.
It follows that in the (8, fc) system, the Lemoine axis is the inverse of the circle (0, 08) ; while the Brocard circle is the inverse of (0,
Suppose the equilateral triangle XYZ to be projected into a triangle with angles A/xv, sides Zwm, area A', and Brocard angle <£. Then, from Section (91),
.-. ;4 = (cot <£— v/3)/cot <£+ V3).
Therefore <£ is constant, as 0 is constant.
Hence the remarkable property of this coaxal system of Schoute circles, viz., the pedal triangles of all points on the circumference of any one circle have the same Brocard angle.
ORTHOGONAL PROJECTION OF A TRIANGLE. 73
1OO. To prove that the Brocard angle <£ is equal to the acute angle (less than 30°) between wO and OO. (G-.)
From Section (98),
Otf/R* = (a2 cot 60°+ ... -4A)/(a2 cot 60°+ ... + 4A)
= (cotw — v/3)/(cotu> + A/3).
And t* = (cot<£— v/3)/(cot<£ + A/3) (Section 99). Therefore, from Section (98),
= (cotw+ \/3)/(cot a> — In our calculations, Section (98), we took Rt* < OQ, so that to lies to the right of Q, and therefore of Q' (here 8a). Let OOo> = 180— x.
It is known that
008 = 30°, OQ^ = 150° ; (ISO/.)
= sin x/sin (x — o>) .sin (o> + 30°)/sin 30° = (cot co -|- v/3)/(cotw— cot^) ;
/. <^> = x-
Many years ago Prof. Dr. P. H. Schoute proved that the locus of a point whose pedal triangle had a constant Brocard angle is a circle of the coaxal system whose limiting points were 8, 8r His proof, I believe, was analytical.
Bead sections 99, 100 after Chapter XI.
CHAPTER IX.
COUNTER POINTS.
1O1. WE now proceed to a further examination of the two pairs of Counter Points ($, S') and (Sv $/), where the pednl triangle of 8 is homothetic to the antipodal triangle of S1, while the pedal triangle of 8' is homothetic to the antipedal triangle of $, and so for Sl and $/.
Let afiy, a'P'y' be the n.c. of 8, 8'.
It has been shown that
W sin (A + X) sin (B + /*) sin ( C + v M* sin A sin /x sin v
74
= yy =
. (83)
COUNTER POINTS. 75
It follows that S and 8' are the foci of a conic inscribed in ABC ; that the two pedal triangles def, d'e'f have the same circumcircle, the Auxiliary Circle of the conic ; that the major axis = 2p, the diameter of this circle, the centre lying midway between S and 8f, while the semi-minor axis q is given by q- = aa', &c
The points S, S' are often called i; Isogonal Conjugates," because, by a well known property of conies,
L SCB = S'CA, SB A = tf'BG, 8 A C = 8'AB,
the pairs (SA, S'A), &c., being equally inclined to the corre- sponding sides.
The line 8' A is then said to be isogonal to SA, S'B to MB, S'C to SC', so that S', the counter point of S, lies on any line isogonal to SA or SB or SG.
Let afty be any point Z/ on Z?0, and a'/S'y' its counter point.
Then ftft1 = yy' = aa' = 0. (a = 0)
But /?, y are not zero ;
.-. /3' = y = 0;
i.e. the counter point of L is A.
If S be on the circle ABC, draw chord ST perpendicular to BC, and draw diameters S8l and T'L\.
Obviously AT^ is isogonal to AS; also AT^ is parallel to the Simson Line of 8^ and therefore perpendicular to the Simson Line of 8.
Hence the counter point of 8 is at infinity, being common to the three parallels ATlt BT%, CT3 which are perpendicular to the Simson Line of S.
1O2. To determine the relations between X, /x, i/, the angles of def, and A', //, i/, the angles of d'e'f.
It is known that tBSC = A + \; so BS'C = A+X. (56) But BS'C = -rr-BD'C = TT-X ; so B8C = «—V.
And ^4 +X - 5/SC = TT— X' ;
.-. X + V = 7T — A.
.'. sinX' = sin(^. + X), sin X = sin (J. + X') ;
76 MODERN GEOMETRY.
abc sin X' , abc sin // sin v'
also a = — . -T—, - ; a' = — - . -^ —. - ; (83)
M sin A M sin /u, sin v
• 2 a262c2 2 sin X' sin u! sin v'
ana a- = aa = m . , where m = — : — - — r— — — : - .
M sin X sin /x sin v
Let n' be the power of S' ; U1 the area of d'e'/' ;
M' = a2cotX'+...+4A.
Then UM = 272 . afcc = U'M'. (64)
And 2/.sinA sin//. sinv = U = 2A'-/3f ; (65)
2/.sinX' sin/x' sinv' = 17' = "2
.-. m- = u'/u = M/M' = n'/n ;
.-. Uf = m*U = m\2tf/M-, and I/' = M/m?, U' = maH.
Since aa = n. sin ^ +A) sln^ = n.B1'pX/Bin^ ; sin X sin X
and, similarly,
; _ ri, sin(^.4-Xf) sin A _ ^,, sin X sin A sin X' sin X'
a result due to Professor Genese.
1O3. The equation to the minor axis of the conic which has 8, Sf for foci. (H. M. Taylor)
Let 7r15 7r2, 7r3 be the perpendiculars on the minor axis from -4, J3, 0, so that the equation is
•jr^a . a -\- 7r.2b . j3 -r TT^C . y = 0. A diagram shows that
C A 2 <" j 2 O v1 ^
\SJBL — O./1 =^.00.^;
also 8 A sin A = ef — 2p sin X ; (57)
.*. Trj.sin2^. oc (sin'2X — sin'^X') a sin (X — X') sin A
Hence the equation to the minor axis is
a.sin(X— X')+y3.siii(/<— /x') + y.sin(v— v') = 0. The proof here given is by the present writer.
COUNTER POINTS.
77
1O4. Lemma. — The tangent be to the conic being drawn parallel to BC, to prove
SA.S'A = AC.Ab = AB.Ac.
Let a, /?, y, 8 be the angles subtended at & by the tangents from 5, (7, c, b.
Then
- = y;
and, regarding the parallel tangents as drawn from a point }D at infinity, so that SD is parallel to B(7,
Z DSl = DSn = i(2/3-h2y) = /. DSC= DSl-CSl= (/? + y)- /. S'CJL = SCI = DSC = y = Also >S'^(7= SAb.
Hence the triangles ASb, ACS' are similar;
.*. AS : Ab = AC : AS' ; /. SA.S'A = AC.Ab = AB.Ac.
The theorem is due to Mr. E. P. Rouse, the demonstration to Mr. B. F. Davis.
78 MODERN GEOMETRY.
1O5. The coordinates of the centre cr0 of the conic. (G.)
Let a0/80y0, an7?0'y0' be the n.c. of o-0, the centre of the conic, referred to ABO and to the mid-point triangle A'B'C' respect- ively.
Let &i be the perpendicular from A on B G. Then a0 -j- a0' = J^ ;
/. 4E.o0' = 2R(h, — 2a0) = 2Experp. from A on be = ZRxAbsiuB or 2Rx4csinCf = AC. Ab or AB.Ac = 8 A . S'A (from Rouse's Theorem).
But 8 A .sin A = ef — 2p . sin X ; (57)
/. 4 71 . a0' = 4p2. sin X sin X'/sin2 J. ;
aa(' = "2p2. sin X sin X'/sin yl, giving the absolute A'B'C' b.c. of cr0 ; and since
aa0' + fy80' + cy0' = A, , where N = 3,. sin X sin X'/sin ^4.
1O6. The conic touching BG at Z, to prove
Bl : Cl = sin /u, sin /x,'/sin J5 : sin v sin v'/sin (7.
Project the conic (ellipse) into a circle of radius q, the angle of projection being 6 = cos"1 q/p.
Let the centre cr0 be projected into o-a, and ^4.5(7 into A^B^J^. Then A. Bla-lCl = A. Bo-^C X q/p ; .'. q.a^ = ac^Xq/p ;
/. a, cc aa0 ; /. 5Z : C? = J?^! : C% = «j— &! : *?,— ca
= aa0 — 6/30 -f cy0 : aa0 -(- ?>^0 — cy0
= feft' : cy0' (14)
= sin /x. sin /x'/sin B : sin v sin v'/sin 6'.
The theorem is by Mr. H. M. Taylor; the proof is by the present writer.
COUNTER POINTS.
79
1O7. Let w, a/ be the orthopoles of the circumdiameters passing through 8 and S' ; i.e. the points on the Nine-Point circle whose Nine-Point circle Simson Lines are parallel to the diameters OS and 08' (49).
The pedal circle of every point on the diameter through 8 passes through o> ; therefore def, the pedal circle of 8, passes through w ; similarly the circle d'e'f passes through a/. (78)
Therefore defd'e'f, the common pedal circle of 8 and $', passes through <o and <*/.
And therefore ww' is the Radical Axis of the two circles. (Gr.)
0)
s'/L
When OS' falls on OS, a/ coincides with w, and the pedal circle touches the Nine-Point circle at w.
108. Aiyars Theorem.
If 0' be the Nine-Point centre
Then OS . OS' = 2 R . 0'<r0.
Let OJA and %$&' be the direction angles of OS, OS'.
The power of A' for the pedal circle = A'd.A'd'
= OS cos 0,. OS' cos 0/.
But since GMO' is the Radical Axis of the two circles, the power of A' for pedal circle
= perpendicular from A' on wo/ x 2 . 0'<r0. And this perpendicular from A'
= A'w.A'u'/R = E cos e,.R cos 6,' ; (44;
/. OS.OS' = 2R.O'<rQ.
(V. Ramaswami Aiyar.)
80 MODERN GEOMETRY.
Xote that the A'B'C' equation to o>o/ is
COS Ol COS 6-[ . aa + . . - = 0 :
for, from (50), w and o>' have n.c. (sec^... sec ft/...).
1O9. M'Cay's Cubic. Tor this occasion take (Imn) as the n.c. ABC coordinate? of S ; so that (1/7, 1/m, 1/n) are the n.c. of ,S".
The equation to &$' then is
l(m~-n-)a+... = 0. When /8/S>' passes through 0, we have
I (m2 — rr) cos A+ ... =0. So that £ and 8' He on M'Cay's Cubic,
And conversely, if any diameter TOT' cut this cubic at S, 8' (the third point is 0), then ;S, /S' are counter points, and their pedal circle touches the Nine-Point circle at w, the orthopole of TOT'.
We have already met with this circle in Section (45) : its centre is On O'Q-p* is a straight line, so that now Aiyar's Theorem may be written 08. OS' = 2R(R—p), and if the circle cut TSS'OOfl' in k, k', then w&, wk' are the Simson Lines of T, T'.
110. Counterpoint Conies. (G.)
The point P moving along a given line TT', it is required to determine the locus of its counter point Q.
Let TT cut the sides of ABC in L, M, N.
Then the Q locus passes through J., B, C the counter point of £, I/, A7. (101)
Draw the chord Aa parallel to TT', and at parallel to BC.
The counter point of t (101) is the point at oo on Aa, and therefore on TT'. But this point being on TT', its counter point must be on the locus of Q.
Hence t is a point where the Q locus cuts the circle ABC ; also (101) t is the pole of the Simson Line perpendicular to TT.
Denote perpendiculars from Q, the counter point of P, on AC, AB, Bt, Ct by ft y, £', /.
COUNTER POINTS. 81
sin (1A(" — siu PA /; -
Then
y sin QA'Ji sin
H _smQCB ainQBt _smPCA sin QBt
Q<J .sin Q.Ct ~ sin ^5(7 ' sin QCt ~ ninPBA ' sin QCt'
Let Z?P, (7P cut Aa in ?w, and w.
Then QBt = QBC+CKt = PBA+ BCa - FBA + BAm
= Bma = BPT. So QCt = QCB+BCt = PCA + CBa = PCA + nAC
= Pna = OPT.
^ q
~
yy' PC' PA' sin CPT PC . sin OPT ~ r '
where p, g, r are perpendiculars from. A, B, C on T27'.
Hence the locus is a come passing through ^4, J5, C, ^.
If the sides of the quadrilateral were BA, BO, tA, tC, we should have •yy'/aa' = r/p, .....
111. To determine the directions of the asymptotes and axes.
Draw Tx, T'x' parallel to BC.
The points at infinity of the Q conic are the counter points of T, Tf ; and therefore lie on Ax, Ax', which are isogonal to AT, AT', so that Ax, Ax' are parallel to the asymptotes.
But the arc tx = aT = AT'.
Hence tT, tT' are parallel to Ax1, Ax, and therefore to the asymptotes.
G
82
MODERN GEOMETRY.
112. To determine the Counter Point Conic of a circum- diameter TOT'.
From (111), the Asymptotes are at right angles, since they are parallel to tT, tT' ; therefore the conic is ^a Rectangular Hyperbola. .
If TOT' be la + mp + ny=Q or p.aa+... = 0, the conic is Z/a-f... = 0.
To find the centre, we have
my + nfi cc a, &c. ; » a a Z( — al -\-brn -\-cti).
But I oc op, and, from (8), ap/R = b cos 08 + c cos ^2 ; whence a ex. p cos #j, &c.
Hence, from (50), the centre is w, the Orthopole of TOT'.
T'
113. To determine the Asymptotes. (G.)
Let wZ, wX' be the Simson Lines of T, T'. From Section (41), if ul9 «,„ i(?,, At ai-e the perpendiculars from A, 5, 0, .7? on ojj^ ; arid w2, r2, w^ hz those on coX2 ; then
nl = 2E . cos <rl sin cr2 sin cr3, ^2 = 2/i'.cos (^TT— o-J... = 2B . sin o-j cos o-2 cos cr3 ;
COUNTER POINTS. 83
.'. 2 . u^ = R*. sin 2o-l sin 2<r2 sin 2<r3
= 2 . t^ro = 2 . iv }w*.
Also Aj = perp. from -ff on o>Xj
= perp. from T on w.X'j = 2R . cos o-j cos o-2 cos <rs. So h, = 2R.cos(i7r-(T1)...
= 2 R . sin o-j sin o-., sin cr3 ; /. 2.w1«2 = ... 27^,.
Hence w^, wXg are the Asymptotes of the Rectangular Hy- perbola and the square of the semi-axis
= -B2. sin 2^ sin 2cr2 sin 2<r3. Since OT.4 or OAT = o-, ; (37)
/. |? = E.sin^OT = E.sin2o-. Hence the square of semi-axis = pqr/R.
114. Produce H<o'to cut the circle ABC in t'. Then, since oj lies on the Nine-Point Circle,
flu> = wt'.
But H is on the Eectangular Hyperbola, and w is the centre. Therefore t' lies on the Rectangular Hyperbola. Therefore t' coincides with t, the fourth point where the Rectangular Hyperbola cuts the circle ABC.
Let ttj, &, yx be the n.c. of S^ inverse to 8; and a/, /?/, y/ the n.c. of $/, the Twin Point of 6'.
It has been shown that
_ abc sin(J.— A.) , _ abc sin (B — p) sin (C — v)
If sin A. JVJT ' sin /x sin v
(68) and (86)
So that /S»/ is the counter point of $,, and therefore lies on the Rectangular Hyperbola, which is the counter point conic of TOT'.
84
MODERN GEOMETRY.
115. To prove that o> is the mid-point of S'SJ.
We have shown that the pedal circles of all points on TOT' pass through the orthopole w.
Therefore the pedal circles of (SSf) and (fi^/S/) pass through w. Also, from Section (69), w is the Centre of Similitude of def, the pedal circle of S, and of d^ej^ the pedal circle of $,. Hence, if p, p1 are the circumradii of these circles,
tad <ad}
P Pi ' d, d' being homologous points in the similar triangles def, d^fr
S,/-
Draw to& perpendicular to BG. Then, since CD is on the circle defd'e'f,
2p
and similarly, w/c = ud^ . wdl'/2pl ;
so that the projection of w on BG is midway between the pro- jections of 8' and Si on BG.
So for OA, AB.
Therefore o> is the mid-point of S' and /S'/.
Hence, as the inverse points S and Sl travel along TOT' in contrary directions from T, their counter points S' and $/ travel along the Rectangular Hyperbola which passes through -4, B^ G and has o> for its centre. $', $/ are always at the extremities of a diameter of the Rectangular Hyperbola, and the difference between the areas of their antipedal triangles is always 4A.
COUNTER POINTS. 85
116. Let I', m', n' be the images of 8' in EG, CA, AB. (Gr.) Then, because cr0 is the centre of the pedal circle of ($$'),
.-. <r0S=:o-0£'; .'. SI' = 2.<r0d' = 2p. Therefore a circle, centre 8, radius 2p, passes through I'm'n'. Again, since S'<rQ = ar08 and 8'<o = oi/8/,
/. 88^ = 2.o-0<u = 2p, since the circle def passes through <u.
Hence a circle described with centre S and radius 2p passes through I'm'n' and $/.
CHAPTER X.
LEMOINE GEOMETRY.
117. The Lemoine Point. — On the sides of the triangle ABC construct squares externally, and complete the diagram as given, the three outer sides of the squares meeting in A,, B,, C,.
The perpendiculars from A, on AC, AB being b and equal to c, the equation to AAl is (3/b = y/c.
Hence AA,, BB,, CCl meet at the point whose n.c. are (a, 6, c),and whose b.c. are therefore (a2, 62, c2).
This point is called the Lemoine or Grebe or Symmedian Point and will be denoted by K.
AK, BK, CK are called the Symmedians of A, B, C.
The absolute values of the n.c. are given by a = ha, &c.,
2A where k =
Produce AK to meet BC in Kl ; then, BK, : CK, = &AKB : AAKC
= ratio of b.c. z and y
o , ,>
= c . o" ;
so that the segments BK,, CK, are as the squares of adjacent sides.
LEMOINE GEOMETRY. 87
118. K is the centroid of its pedal triangle def. For &eKf= ^.Ke.KfsinA
cc be sin A ;
.-. A eKf = AfKd = A dKe. So that K is the centroid of def.
Since the n.c. of 0 are as (cos A, cos -B, cos 0), while those of ET are as (sin A9 sin 5, sin C)t therefore the equation to OK is sin(£— C).a+sm(C-4).j8 + sin(4— 5).y, or (b*-<?)/a*.x+;.. = 0.
The Power of JT for the circle ABC. Using the form of Section (60), we have
119. If a, /?, y are the n.c. of any point, then a2 + /?2-fy2 is a minimum at K. Since a2+Z)2 + c2
= 4A2+... ;
therefore (a3 + &2 + c2)(a2 + /S2 + y2) is always greater than 4A2 except when a/a = (3/b •-= y/c
or when the point coincides with K.
In this case, therefore, a24-/82-fy2 has its minimum value,
4A2 which is — 2 — 7j 7t •
The sides of the pedal triangle of a point 8 are u, v, «•. To show that u*-\-v*+v& is a minimum when 8 coincides with the Lemoine Point K.
From (57), u = r,.sin A ; (SA = r,)
But, since a2, 62, c2 are the b.c. of K, we have, from (19),
a\ r,2 + Z>2' r22 + c2. r32 = (a2 + 62 + c2) CfiTtf2 - KO* + E2) ; and the right-hand expression is a minimum when KS = 0, or when 8 coincides with K.
The minimum value of u2
(118) and (131)
88
MODERN GEOMETRY.
12O. Let figures Yand Z, directly similar, be described exter- nally on AC, AB. so that A in Y is homologous to B in Z, and 0 in Y to JL in Z.
Then, if 8 be the double point of Y and Z, the triangles SAO, SBA are similar, having
L SEA = SAC ; SAB = SGA ; ASB = 4SC, so that $ is the focus of a parabola known as Artzt's First Parabola, touching AB, AO at B, C.
Let Pv Pz be the perpendiculars from S on AC, AB.
Then, from similar triangles SB A, SAC,
so that S lies on the J.-Symmedian.
Again, / BSK, = SAB + SB A
= SAB + SAC
.;. BSC = 2.BSK, = 2A = BOG; so that S lies on the circle BOG.
Also, OSKi = OSB+A = OGB + A (in circle B08C)
= 90°;
/. 8A = 5A-
Hence the double point 8 of the two directly similar figures on AB, AC may be found by drawing the Symmedian chord AAl through K and bisecting it at 8.
LEMOINE GEOMETRY.
89
121. Let rl\'l\'l\ be the Tangent Triangle, formed by drawing tangents to the circumcircle at A, B, C.
Then AT,, BT2, CTS, are concurrent at K.
For, if q, r be the perpendiculars from T^ on AC, AB,
q__ _ T.GsinACT,, _sinff_ b
r
*inC c '
Hence AT^ passes through K ; so also do BT.2, (7T3. Note that Tl is the point of intersection of the tangents at
B and C, whose equations are y/c + a/a = 0, a/a-f 6//S = 0, so
that the n.c. of '1\ are ( — a, 6, c).
122. The Lemoine Point A of I^A,.
In the figure of the preceding section, it will be seen that the Lemoine Point K of the inscribed triangle ABC, is the Gergonne Point (32) of rI\T2Ts,
Therefore the point M, which is the Gergonne Point of ABC, is the Lemoine Point of XYZ.
But XYZ and I-J^ are homothetic, the centre of similitude being <r. (26)
Therefore the Lemoine Point — call it A — of I^/a lies on crJIf, and <rM : crA = linear ratio of XYZ, IJJs =r:2B.
90 MODERN GEOMETRY.
The point A has n.c. (s — a), ... ; for the perpendiculars from the point {(s— a), (s — &), (s— c)} on the sides of 1^1^ are found to be proportional to the sides of this triangle.
Note the following list of " (s — a) " points : —
(1) Nagel Point: b.c. are (s-a), (5-6), (s— c). (30)
(2) Grergonne Point : b.c. are 1/0— a), &c. (32)
(3) Lemoine Point A of I-J^: n.c. are (.<? — a), —
(4) Centre of Sim. o- of XTZJJJ.,: n.c. are 1/0 -a),.... (26)
123. To prove that JJT bisects all chords of the triangle ABC, which are parallel to the tangent at A, or perpendicular to OA, or parallel to the side HZHS of the orthocentric triangle H^H^Hy
Let BAK = B, OAK = <£.
Then sin 0/sin <£ = y/ft = c/b
= sin 0/sin B = sin 5^T2/sin CA'1\.
Therefore AK and T^AI\ are harmonic conjugates with respect to AB and AC.
It follows that J..K bisects all chords which are parallel to T2Tx or JTo-Hg, or are perpendicular to OA.
124. The Harmonic Quadrilateral. — The angles of the har- monic pencil at A are seen to be B, <£, 6 or (7, 0, <f> (Fig., p. 89).
The same angles are found, in the same order, at B and 0 and.4r
Hence the pencils at B, (7, A and Al are harmonic; ABA}C being called, on this account, a Harmonic Quadrilateral.
In the triangle ABAl the tangent at B is harmonically con- jugate to BK^ so that BKl is the B-symmedian for this triangle.
Similarly, A^K^ is the Jlj-symmedian in £A1C1 ; and CK^ the C-symmedian in AGAr
To prove that rectangle AB.A-^G =• rectangle AC .A^B. sin 0/sin <t> = sin BC A/sin CBAl = A^/A.C.
.-. AB.A.G = AC.A1B.
If x. //, z, u are the perpendiculars from Kl on AB, AC, AJ3, A£, then x/AB = y/AC = z/A,B.
For, since AI^ is the J.-symmedian of ABC,
.-. x/AB = y/AC. And, since 5^ is the .B-symmedian of
x/AB = z/A,B. And, since AlKl is the J-^symmedian of
LEMOINE GEOMETRY.
91
125. The Lemoine Point is the point of intersection of lines joining the mid-points of the sides of ABC to the mid-points of corresponding perpendiculars.
A
Let A', B', C' be the mid-points of BC, CA, AB, and let A'K meet AD in I.
Then, since BK and the tangent BTl at B form a harmonic pencil with BA, BC, therefore the range (AKK^T-^ is harmonic : therefore the pencil A'(AWco ) is harmonic, so that AD is bisected at I.
A rectangle XYVU being inscribed in ABC with the side XY on BC, to find the locus of its centre P.
The diagram shows that the mid-point of UV lies on A'A, and that P lies on A'L
The Lemoine Point K, common to A'l, B'm, C'n is the com- mon centre of three inscribed rectangles, standing on BC, CA, AB respectively.
126. To determine the direction angles 619 02, 03, which OK makes with the sides of ABC. From the diagram,
OK cos 0l = A'd = A'D—.
HJ
Now and
A'D =
fr2-c2 2a
2A
(117)
92 MODERN GEOMETRY.
.-. COS0! = m.a(b~ — c2), where 1/m = .-. COS0! : cos08 : cos 0& = a(b"—c2) : 6 (c2— a2) : c(a2— Z>2). Hence the tripolar equation to OK is
a1^8— Ci)r1i-ffe2(cs-os)r8*-f6l(aa--6»)rtl = 0. (17)
127. !TAe Apollonian, Circles. — Let the several pairs of bisectors of the angles A, B, C meet BG in a, a! ; OJ. in ft £' ; J.B in y, y'.
The circles described on aa', /?/?, yy' as diameters are called the Apollonian Circles.
Let the tangent at A to the circle ABC meet BG in Lr Then angle L,aA = aBA + BAa. = L^a = L,AC -\-GAa. = so that /^a = L^.
And, since aAa' = 90°,
.', L,A = LX ;
so that Z/j is the centre of the Apollonian Circle (aa'), passing through A and orthogonal to the circle ABG.
Since AL^ is a tangent, the polar of Ll passes through A ; and, since (BKfiL^ is harmonic, the polar of Lt passes through Kr
It follows that AKK^Ti is the polar of L, ; so that L,A, is the other tangent from Lv
Hence the common chords AA^ BB^ GGl of the circle ABG and the Apollonian Circles intersect at K, which is therefore equipotential for the four circles.
Note that OLl bisects AKK1A1 at right angles, and therefore passes through the point IS of Section (120).
128. The Lemoine Axis.
Since the polars of L^ L>2, Ls pass through K, therefore L^L.^ lie on the polar of K.
The equation to the tangent at A is ft/b+y/c = 0, &c. Hence L^L.^LS is
a/a + /3/6-fy/c = 0, or x/ar + y/tf + z/c* — 0. This is called the Lemoine Axis.
129. A Harmonic Quadrilateral such as ABA^ can be in- verted into a square.
LEMOINE GEOMETRY.
93
Let the circle described with centre 'L\ and radius cut the Apollonian Circle Ll in E and E' .
or
Then, since the tangents OA, OB, 00, OA{ to the two circles are equal, 0 lies on their common chord EE', and
OE.OE' = S2; so that E, E' are inverse points for the circle ABC.
Let AE, BE, CE, A}E meet the circle again in LMNL'.
Then, taking E as pole and VII as radius of inversion (where n = power of E = RZ—OE*), we have
n n
LM= AB.
LN = AC.
EA.EB' ' EA.EC
But BE : EC = BA : AC,
since E is on the Apollonian Circle aAa! ; /. LM = LN, Ac. Hence LMNL' is a square. Another square may be obtained by taking E' as pole.
In the above figure, E is on the Apollonian Circle aa';
/. r2 : r3 = BE : CE = Ba : aC = c : b ;
.•. brz = cr& ; .*. de = df or /x = v. Also A + \ = BEC = TT-^.BT.C = .-. \ — ?r so that x = v =
CHAPTER XI
LEMOINE-BROCARD GEOMETRY.
13O. The Brocard Points.— On the sides BC, CA, AB let triads of circles be described whose external segments contain the angles
O) A,£,C;
(b) C,A, B-
(c) B, C, A ; the cyclic order (ABC) being preserved.
The first triad of circles intersect at the orthocentre H. Let the second triad intersect afc 17, and the third at O'. O and O' are the Brocard Points of ABC.
Since the external segment of BQC contains the angle C, this circle touches CA at C.
Similarly, CttA touches AB at A, and A£lB touches BC at B.
(Memorize the order of angles for O by the word " CAB."}
In like manner. BQ!G touches AB at B, Cti'A touches BC at C, ASl'B touches CA at A.
Again, since J.O/> touches BC at B, the angle 056' = VAB.
So QCA = SIBC.
Similarly, Q'AC = Q'CB = &BA.
Denote each of the equal angles £IBC, QCA, SlAB by w, and each of the equal angles 0,'CB, &c., by <•/.
94
LEMOiNE-BROCARD GEOMETRY. 95
131B To determine CD and to'. In the triangle AtoB,
toB =: c sin w/sin B = 2R sin to.c/6. So to A = 2B sin to . 6/a, 12 6Y = 2 JB sin co . a/c ;
sin (01 -to) _ sin CLIO _ 120 _ a2 . sin to ~~ sin 12 0.4 ~ toA~ be '
... cot to = £±^±£1 - cot A + cot 5 + cot 0 4A
_ sin2 J, + sin2 B + sin2 C. _ 1 4- cos A cos j? cos C
2 sin ^. sin B sin C sin ^1 sin B sin 0
The same expressions are found for a/ ;
a> ^^ o>.
The angle <o, which is equal to each of the six angles QA B, , QCA, &BA, Q'CB, Q'AC, is called the Brocard Angle of ABC.
Since oV = o>, it follows that
to' A = 2E sin w . c/a, Q'C = 2 Z? sin co . fe/c, O'5 = 2.R sin w . a/6. Observe that the n.c. of K may now be written
2A
a = — — — - - . a = Ja tan to = R sin A tan to, &c. (117) CL" -\~ b -\~ c"
132. To determine the n.c. and b.c. of to and fi'. From the diagram,
a = toB sin to — 212 sill2 to . c/6.
So fi = 2R sin2 to . a/c, y = 2E sin2 to . 6/a.
And for O',
a' = 2B sin2 to . 6/c, ft' = 2R sin2 to . c/a, 7' = 2JB sin2 to . a/6,
.*. aa' = y8y8' = yy' : so that O, 12' are Counter Points. The b.c. of O are given by
x:y:z = l/b- : 1/c2 : I/a2 ; and for to', x' : y' : z' = 1/c2 : I/a2 : 1/61
The line toto' is then found to be
(a4-62c2).^/a2 + (64-cV).?//62+(c4-a262)^/c2 = 0.
The power U of to = /V+";2 . ^ , °?'f ^
22 22
.-. on = on'.
96 MODERN GEOMETRY.
133. The Brocard Angle is never greater than 30°. For cot <o = cot A -\- cot B + cot (7,
and cot V cot ('+... = 1 ;
.*. cot- co = cot2J.+ ... + 2; /. (cotl?-cot(7)2+... = 2(cotM + ...)-2 = 2 (cot2 (o-3).
Hence cot to is never less than A/3, and therefore w is never greater than 30°.
134, Some useful formulae.
(a) cosec2 CD = 1 4- cot2 <o = 1 4- (a2 4- &2 + c2)2/16A2
s -, , . „ a4 and 1 — 4 sin- cu =
This expression will be denoted by e2.
(6) cos CD == — - — — .
. 0 2A(a2 + 62 + c2)
(c) sin 2to — — ^ — .2 a .
(d) COS l>oi = _- A±^
(e) cotL^^j. a
4 A(
(/) Since sin (A — w)/sinw = as/66;
/. sin(J.— w) : sin(l?— a>) : sin (0— w) = a3 : Is : c8, .and sin(J. — w) sin (/?— <o) siii(6'— to) = sin3 to.
(gr) sin (^.4 -j-w)/sin o> = sin A (cot to -f cot A)
Note that, when 6 — c, sin (^4. + w) = 2 sin to. (/&) cos (^4 + <o)/sin J. sin to = cot A cot to — 1
/. cos(yl+(o) = sin A sinto/8A2.(fc4 + c4— a2/;2 — <rc2) a sin (A—E) sin 7> + sin ( J.— C) sin f '.
LE:-IOINE-BROCARD GEOMETRY.
97
135. JTeuberg Circles.
The base BG of a triangle ABC being fixed, to determine the locus of the vertex A, when the Brocard Angle of the triangle ABC is constant.
Bisect BG in I) : draw DA^A.2 perpendicular, and A M parallel to BG.
Then
= 4. area of ABC 4,a.DM
.'. AD*— a. DM .coto) + 3/4.a2 = 0. Take DN = \ a . cot o>, so that L BND = GND = u>. Then NA* = AD* + ND2-2.DN .DM
-3/4.o8 = 1/4. a2 (cot'w-3) = constant.
Hence the locus of -4 is a circle, called a Neuberg Circle, centre N, and radius p = \a
136. Let Sj&C', #_E7'(7 be equilateral triangles on opposiie nicies of the common base BC, so that DE = |a. V3.
Let the Neuberg Circle cut DE in A» A.2. Then DA,. DA, = DJV2-P2 = 3/4. a2 = DE2.
And thus, for different values of <o, the Neuberg Circles form
H
98 MODERN GEOMETRY.
a coaxal family, with E and E' for Limiting Points, and EG for Radical Axis.
Let GAl cut the circle in Tv Then, since E is a limiting point,
A GA^.CT^ = CE* = CB\ so that the triangles GBA19 GT^B are similar ;
.-. B'l\ : A,B = BG : A,G. But A,B = A,C, /. BT, = BG = BE ;
/. BTl is a tangent at Tr Similarly, if BA% cuts the circle at T^ ;
then CT» is a tangent at T2.
137. The Steiner Angles.
From the similar triangles BA,G, T^BC,
L rl\BG = BAf.
Also, from the cyclic quadrilateral BT^ND^ with right angles at T, and Z>, /_ BT^D = BND or w ;
BG = u + BA.C = Al + «>;
n o) _ sin r,D(7 _ ^^ _ BG „
sino> ~*inBrl\D~ BD ~ BD
o- -i i Similarly,
sin w Thus J.15 A% are the values of x obtained from
sin (x + w) = 2 sin o>. This gives cot2^ — 2 cot J# . cot w + 3 = 0;
whence cot^^ = cot w— -v/cot'2 w— 3,
cot^A, = cot 00+ v/cot'2w— 3 ; as is obvious from the diagram.
For A.D/DB - DJNT/DJ5 - NAJDB ;
-j = cot to— p/^a = cot GJ— \/cot2a> — 3.
A denoted by S19 /S'2.
So cotJA2 = cot w -f \/cot2 w — 3.
The angles %AV \A^ will be called the Steiner Angles, and
LE.MOINE-BROCARD GEOMETRY. 99
138. Either Brocard Point ft or ft' supplies some interesting illustrations of the properties of pedal triangles. For ft, A + \ = BQO = 180-0 = A + B;
.'. \ = B, so /x = 0, v = A. And for ft', \' = 0, // — A, v' = B. So that the pedal triangles of the Brocard Points are similar to ABO.
To determine p, the circumradius of the pedal triangle of ft.
ef = 2p sin A. — 2p sin B. But ef=QARiuA = 2R sin to . b/a . sin A (131)
p = J\ sin to.
Hence, the triangles def, ABO have their linear ratio equal to sin <o : 1.
U — A sin2w. Also II = 4_B2/A . U ;
.-. E2— Oft2 = n = 4JR2 sin2 to, /. Oft2 = E2(l— 4 sin2o>) = e'E*. (134 a)
139. Lemma I.
Let XBC, YCA, ZAB be isosceles triangles, described all inwards or all outwards, and having a common base angle 6. To prove that AX, BY, CZ are concurrent. Let (a^yO, (a2/32y2), (a3&y3) be the n.c. of X, Y, Z. Then 04 = ^ a tan 0 ;
ft = JO7sin(C'-6>) = i. a sec 0. sin (C-0) ;
yx = ia sec 0. sin (B — 6).
A
C
So that fll : ft : yj = sin0 : sin (0—0) : sin (.# — 0) ; a2 : ft : y.2 = sin (0—0) : sin 0 : sin (A— 0), a» • ft : 7;i == sin (^ — 0) : sin (^ — 0^ : si11 0-
100 MODERN GEOMETRY.
The equation to AX is ft/fa = y/y1? or ft. sin (7>>— 0> = 7 sin (C-0), &c.
Hence AX, J3Y, 0^ concur at a point 8, the centre of Perspective for triangles ABC, XYZ, whose n.c. are as I/sin (A- 0), I/sin (7?-0), I/sin (0-0).
The point 8 obviously lies on Kiepert's Hyperbola, the Counter Point conic of OK, for the K.H. equation is
sin (B— G) . 1/a-f- ... =0. (Appendix TIL a)
14O. Lemma II.
The centroid (£') of XYZ coincides with O. For if (a, 0, 7) be the n.c of G'.
3. (la = ttj + Oo + a..
= \ sec0 [a sin 0-1-6 sin (0 — 0) +c.sin (7?— 0) } oc 26 sin (7 oc I/a ; &c.
6r' coincides with G. (See Appendix 771.6)
141. Illustrations, — (A) In the diagram of Neuberg's Circle (p. 97), change N into Nv and on BC, CA, AT? describe the isosceles triangles N^BC, N.2CA, N^AB (all inwards), with the common base angle (^-TT — <o), so that N1N^N9 are the centres of the three Neuberg Circles, corresponding to 7? (7, CA, AB.
Then since sin (A — 0} becomes cos (A + <*>), the lines AiV,, BN.2, CNS concur at a point whose n.c. are as sec (A + w)..., that is, at the Tarry Point. (143)
For a second illustration take the triangles PBC, RAB having the common base angle w measured inwards.
The triangle PQR, called the First Brocard Triangle, has G for centroid from (140), while AP, BQ, CR meet at a point D, the Centre of Perspective for PQR, ABC, its n.c. being as I/sin (A-cu) Ac., or I/a3, 1/63, 1/c8, the b.c. being as I/a2, 1/6% 1/c2.
For the n.c. of /',
aj : ^ : yl = sin a> : sin (C — co) : sin (7^— co) ;
= 1 : cr/ab : b*/ac, BO that the b.c. of P are as a2, c2, 62.
LEMOINE-BROCARD GEOMETRY.
101
Similarly those of Q are c'2, ?r, cr, and those of R are ?/2, a2, c2. The equation of QR is found to be
(a4— &3e2)0+(c4— a262)2/+(6*— cV)s = 0. This meets J56' at a pointy, for which
(c4-o86s)y+(fe*-caa2)« = 0, or, ?//(64-cV) + s/(c'-aW) = 0.
Hence the Axis of Perspective _p<?r of the triangles PQR, ABC is z/O4 — 6V2) + . . . =0. (4p_pewdi'a' 7IZ". c)
142. O, O', A'.
Some of the relations between O, Q', and K, will now be
investigated.
A
B
(a) Since PA' = 5.4'. tan PjBA' = Ja.tan w = KM,
/. AP is parallel to BC,
so £Y^, 7CK are parallel to OA, AB
respectively.
Since OPK = 90° = OQK = ORE,
it follows that P, Q, E lie on the Brocard Circle (OA').
(6) Since the angles PPG', QBG are each = o>,
/. PP passes through O.
Similarly OQ, AE pass through fl, while C'P, AQ, BR pass through O'.
102 MODERN GEOMETRY.
(c) Since KQ, KR, are parallel to AC, AB,
.; QPR = QKR = A. So RQP = B, PRQ=C;
and thus PQR is inversely similar to ABC, the triangles having the common centroid G as their double point.
(d) Since
POE = VAB + ttBA = u+(B— o>) =B = PQR, .'. O (and similarly O') lies on the Brocard Circle.
(e) L VOK = &PK = OJ5(7 = o> ; so Q'OK = <o.
OK bisects flO' at right angles (at Z). From Section (138).
00 = R (1-4 sin2 oo)4 = eR ;
OK (diameter of Brocard Circle) = eR secw.
DO' = 2 . 012 sin o> = 2eR sin to.
143. The Steiner and Tarry Points.
The ABC Steiner Point denoted by 2 is the pole of the ABC Simson Line which is parallel to OK.
To determine 2 geometrically, draw Aa- parallel to OK, and crS perpendicular to BC. (35)
If 0j, 0.2, #3 are the direction angles of OK, it has been proved that cos 6l oc a (62— c2). (126)
The n. c. of 5 are 2 R cos 02 cos 03, &c., which are as
sectf], &c., or as 5 r ; and the b.c. are as l/(62 — c2).
a (o — c )
The point diametrically opposite to S on the circle J^-BO is called the Tarry Point, and is denoted by T ; therefore the Simson Lines of 2 and T are at right angles. Hence the n.c. of Tare 2 R sin 02 sin 08, &c. (46)
Now 03T sin ^ = J^M- OA1 (KM perp. to 56')
= R sin A tan o> — IZ cos -4 oc cos(J.-f to).
Hence the n.c. of Tare as sec(J. + oo), sec(5+(o), sec((7 + u>). In (42) let PQR be the Lemoine Axis ; then N is the pole of the Simson Line parallel to this axis, and therefore per- pendicular to OK.
Hence N is the Tarry Point.
LEMOINE-BROCARD GEOMETRY.
103
144. PQR being the First Brocard Triangle, inversely similar to ABC, to prove that the figure KPROQ is inversely similar to 2ACTB.
Since ZSOT is a diameter, 3o-T is a right angle, so that a-T is parallel to BC, and arc B<r = CT.
Now, since KR is parallel to AB (142 a.), and KO to A<r (as above), .'. L OKR = <rAB = T2C, from the equal arcs.
Similarly, OKQ = T2B ;
also ABC, PQR are inversely similar.
Hence 0 and T are homologous points in these two triangles.
Therefore K and 2 are homologous.
Hence the figures KPROQ, ^AGTB are inversely similar.
Since L 2BA = KQP (similar figures)
= JKTJKP, and AB is parallel to KR,
.'. 2B is parallel to EP, &c. Hence TA is perpendicular to QE, T5 to EP, TO to PQ.
145. Lemma.
The points L, If, JV have b.c. proportional to yzx, zxy, xyz, arranged in cyclic order ; to prove that 6?', the mean centre of LMN, coincides with G.
Let (0,^70, OAh)* (oa^sys)* (aW) be tne absolute n.c. of L, 3f, N and 6r' respectively. Then, for L,
So and
/. a, = 2 A/ (aj + y + z).y/a. 04 = 2A/(o; + y + «).«/«, a8 = 2 A/ (a; + y + 2) . a?/a ; /. 3a' = ai + a24-a3 = 2A/o, Ac. Hence G' coincides with G.
104
MODERN GEOMETRY.
146. The b.c. of P, Q, It being aVfe8, cW, 7/W, in cyclic order, it follows that G is the centroid of PQR, as already proved, and is therefore the double point of the inversely similar triangles ABC, PQR.
Let L be the circumcentre of PQJR, and therefore of the Brocard Circle (OK) ; then L in PQR is homologous to 0 in ABC. Therefore the axes of similitude of the two inversely similar triangles bisect the angles between GO and GL.
Again, the b.c. of 1), O, 12' are I/a2, 1/62, 1/c2; 1/6', 1/c2, l/aj; 1/c2, I/a2, 1/62, in cyclic order.
Hence G is the centroid of DOQ'.
Bisect DO' in Z, then (9 lies on DZ, and £Z : GD = 1 : i>.
But OG:GH=l:2. (H orthocentre of ABO)
.'. D.S" is parallel to OK, and Dfi" =2.0Z = 2.eP cos to.
Let LG meet DIf in H'.
Since i is the circumcentre, and G the centroid of PQR, .'. LGH' is the Euler Line of PQR.
But LG : GH' = ZG : GD = 1 : 2 ;
/. J^ is the orthocentre of PQE.
LEMOINE-BROCARD GEOMETRY. 105
Again, HH' — 2 . OL = OK,
.'. H'HKO is a parallelogram and E'K bisects OH at the Nine-Point centre.
147. Since G is the double point of the inversely similar figures 2ACTB, KPROQ, therefore the points 6?, 0, T in the former figure are homologous to G, L, 0 in the latter.
Hence angle OGT = LGO, so that G, L, T are collinear. Again, if p(= ±eR secco) be the radius of the Brocard Circle, then GO : GL = R : p (by similar figures).
So GT:GO=R:P;
.-. GO'=GL.GT- and GT2 : GO- = P- : p- = GT : <!L.
148. To prove that D lies on the circurndiameter SOT. From (146) H'D : ZL = H'G : GL = 2 : 1 ;
and ZL = OZ—p = eR cos w—^eR sec w = p cos 2w ;
/. H'D = 2.ZL = 2f> cos 2o>. Again, H'T:LT= GT+2.GL: GT-GL; and, from above, GT : GL = R- : p~ ;
.-. H'TiLT = jB2 + 'V : J^-p2
= 2 cos 2w : 1 = H'D : LO (or p), /. D lies on SOT.
To determine OD.
DT: OT= H'T: LT = 2cos2w:l, .-. OD:B = 2cos2o>-l:l;
/. OD = e-K Note also that
OD. 02 = e2E2 = Ofi2 or OO'2.
So that OO, OO' are tangents to the circles respectively.
106 MODERN GEOMETRY.
149. The Isodynamic Points.
These are the pair of inverse points 8 and 81? whose pedal triangles are equilateral ; so that
A = ^ = v = 60°. In this case, M = a? cot X + . . . + 4A (64)
= 4 A (cot o> cot 60° + 1).
So for ST, M, — 4A (cot & cob 60°— 1). (68)
The Powers 11(11^ are given by
H (BO = 8EsA/Jf (1Q = 2EV(cotoi cot60°=bl). Areas of pedal triangles = 2A*/M (-3^)
= -|A/(cot(ocot600±l).
abc Absolute n.c. = a (al} =
oc sin (A ±60°).
The circamradii of the pedal triangles are given by 2p* (2p*) . sin 60° sin 60° sin 60° = area of pedal triangles
= iA/(cot<ocot60°±l).
Let (piP2p3), (p/paW) be the tripolar coordinates of 8, 8a. Then pl sinA = ef = p sin 60° = constant.
•'• Pi-P-2'-Pz= 1> •' 1/& : !/c = />/ : Ps' ^ Pa'- The tripolar equation to OK is
a«(6*_cs)rii_,_.i> = 0, for cos^ oc a(62-c2); (126) and this is satisfied by )\ oc I/a.
Hence 8, 8! lie on OK.
1 50. Consider the coaxal system which has 8 and 8t for its limiting points.
From (149), since p : q : r — I/a : l/b : 1/c, therefore the Radical Axis of the system becomes
which is the Lemoine Axis Jj^L^Jj^. Let this Radical Axis cut OK in X ; then, since K is the pole of L-JJ^LZ for the circle ABC, .'. \0 . ^K = square of tangent from X to ABO
= X82 or XSj2, since ABO belongs to the coaxal system.
Therefore the Brocard Circle (OK) belongs to this system, and therefore is coaxal with ABC.
LEMOINE-BROCAKIJ GEOMETRY. 107
151. The Isogonic Points.
These are the Counter Points of 8 and Bl ; they are therefore denoted by 8' and £/.
Their antipedal triangles are equilateral, having
areas %M (MJ = 2A (cot w cot 60°d=l). Their n.c. are
,, !} ale sin(Bd=60°) sinf fJ=b60°)
~ M(Ml)'' sin 60° sin 60° oc I/Bin (4 ±60°).
Hence, from (139), if equilateral triangles XBC, YCA, ZAB be described inwards on BC, CA, AB, then XA, YB, ZC concur at 8' ; for the outward system, the point of concurrence is S/.
These points lie on Kiepert's Hyperbola, whose equation is sin (5— C)/a+... —0.
152. The Circum-ellipse.
Let l/a + m//3 + n/y — 0, be the ellipse, with axes 2p, 2q.
Let a/?y, a//?V be the n.c. of the centre O, referred to ABC and A'B'C', so that 2aa' = — oa + fyS + cy.
Project the ellipse into a circle, centre co, the radius of the circle being therefore g, while the angle of projection 6 is cos"1 q/p.
Let LMN, with angles X/xv, be the triangle into which ABC is projected.
/. A.lfo>J\r= A.J3OC'xcos0; /. g2 sin 2X = aa x g/p ;
.-. aa =: pg' sin2X;
so that the ABC b.c. of O are as sin 2X, sin 2/t, sin 2p. Also | (aa + 6/3 + cy) = A ;
.*. 2pg sin X sin /x sin i/ — A. Again 2«a' = — aa -\- b/3 + cy ;
aa' = 2pg sin X cos /x cos v ; so that the A'B'G' b.c. of O are as tan A, tan/x, tanv.
Since A = 2pq sin A sin ti sin v ;
and — aa+6/2-fcy = 4>pq sin A cos/x cos i/ ;
and aa = pq sin 2A ;
it follows that
(aa-f bfi— cy)
= 2/?y . sin2 2A sin2 2p sin2 2v ;
108 MODERN GEOMETRY.
aa — b/3+cy) (aa+bfi-cy)
giving the locus of the centre, when the area of the ellipse (irpq) is constant.
Let PQB he the triangle formed by tangents to the Ellipse at A, B, Cf, then AP, BQ, GR have a common point— call it T — whose b.c. are as «Z, fern, en.
Let PQR be projected into pqr, whose sides touch the circle LMN.
* Then the projection of T is evidently the Lemoine Point of LMN (fig., p. 89), and therefore its b.c. are as
HN*, NL2, LM- or as sin2 A, sin2tt, sin'2v;
.'. al : bm : en = sin2 A : sin2 //, : sin2 v. Hence the Ellipse is
sin2A./a./3y + ... = 0; and its Counter Point Locus is
sin2 A/a. a -|- ... =0.
This is the Radical Axis of the coaxal system, whose Limit- ing Points have A/U.I/ for the angles of their pedal triangles. To calculate the axes of the ellipse in terms of A, /A, v. From (91), a2 cot A + b" cot ^ -f c2 cot v •=. 2A (sec 6 + cos 6)
= 2A (q/p+p/q) = 2A (p* + q*)/l>q. And from (152), A = 2pq sin A sin/x sinv.
J/
..1 . . . .
sin A sin //. sin v sin A sin /x sin v
So (P~g)8 = i. • x ^ •
sin A siti /x sin v
A well known example is the Steiner Ellipse, whose centre is 6r, so that LMN is equilateral, and
A = /u, = v — ^TT. It will be found that
p~ + cf = § (a2 -4- fe2 4- c3) = f . A cot w ; _pg = 4A/3 \^3 ; p/q — (cot w-f \/cot2w— 3)/ \/3 = cot ;ST V3 ; ^/p = (cotw— v/cot2a>— 3)/v/3 = cot jS2 \/3 ; where ^, /S2 are the Steiner Angles.
The Counter Point Locus of the Steiner Ellipse is
a/a+0/6+?/c s= 0, which is the Lemoine Axis.
* The point T may be called the Sub- Lemoine Point of the conic.
CHAPTER XII.
PIVOT POINTS. TUCKER CIRCLES.
153. Let DEF be any triangle inscribed in ABC, and let its angles be A, /x, \>.
The circles AE F, BFD, ODE meet in a point— call it S.
Let def be the pedal triangle of S, 17 its area, andp its circumradius.
In the circle SDBF, angle SDF = SBF or SBf = Sdf.
So L SDE = Sde •
.*. d = X, so e = /LI, /=:»/; and thus the triangle D/*/.?1 is similar to the pedal triangle of 8.
The point $ can be found, as in (56) by drawing inner arcs (A + X)... on BC,CA,AB.
Again, / dSf — ir—B — DSF (circle tiD'BF).
Hence L dSD = fSF = eSE.
Denote each of these angles by 0.
Then SD = a sec (9, SE = ft sec 0, SF = y sec b, where (a/3y) are the n.c. of 8.
Hence /S is the double point for any pair of the family of similar triangles DEF, including def, so that it may be fitly named the " Pivot Point " (Drehpunkt) of these triangles, which rotate about it, with their vertices on the sides of ABC, chang- ing their size but not their shape.
109
110
M ODERN GEOMETRY .
The linear dimensions of DEF, def are as sec 6 : 1 ; so that, if M, m are homologous points in these triangles ; then
MSm = 0, M S = mS sec 0 ;
and the locus of M for different triangles DEF is a line through m perpendicular to 8m.
An important case is that of the centres of the triangles DEF.
These lie on a line through <TO, the centre of the circle def, perpendicular to >Scr0 ; and, if a- is the centre of DE F, then <rSo-0 = 6.
154. All the elements of DEF may now be determined absolutely in terms of X/xv and 0.
For
M sin A
(65)
where M = a~ cot X-f fe2 cot /x. + c2 cot v + 4d ;
and 2p2 . sin X sin /* sin i/ = Z7 ;
so that p is known.
Hence SD = a sec 0, circumradius of DEF = p = EF = 2p sin X ; area of D#^ = U sec2 6^ = 2&-/M . sec- ^.
155. The circle DEF cutting the sides of ABC again in D'E'F', let X', //, v' be the angles of the family of triangles D'E'F', and S' their Pivot Point.
In the triangle AF'E,
180°— A — FI'E + F'EE' = FDF + F'D'E' = So 18G°-B
PIVOT POINTS. TUCKER CIRCLES.
HI
It follows that S' is the Counter Point of $, and that the angles of the family D'E'F' are 180°— A-X, ... . (102)
The triangles def, d'e'f therefore have the same circumcentre cr0 and the same circumradius p0.
To find or', the centre of D'E'F', we draw a perpendicular to »SV0$' through o-0, and take
L o-'SV0 = & ; where 6' = d'S'D'.
But a-' coincides with a-, either being the centre of the circle DD'EE'FF'.
Hence 0' = 0.
156. To prove that the circle DD'EE'FF' touches the conic which is inscribed in ABC, and has $, S' for foci.
Let TT be a point where the circle So-ti' meets the conic. Then since arc So- = S'ar ;
L STrS' is bisected by TTO-. Therefore CTTT is normal to the conic at TT. Now, in the cyclic quadrilateral o-iSV/S",
O-TT . SS' = S'(T . STT -j- Str . S'ir = S& (Sir + S'TT) ;
=: So-.2p : for 2p = major axis of conic. But SS'/S<r = '2 . S<rJS<r = 2 cos 0 ;
<T7r = p sec 6 =• p.
Also cr is the centre of the circle DD' ... . Hence this circle touches the conic at r.
112
MODERN GEOMETRY.
157. Triangles circumscribed about ABO.
Through A, B, C draw perpendiculars to S'A, S'B, S'O, forming pqr, the Antipedal Triangle of S'.
This triangle (o, S' being Counter Points) is known to be homothetic to def, the pedal triangle of S, and therefore to have angles X, /A, v.
Obviously Sp, 8q, Sr are diameters of the circles BS'Cp, &c.
Through A draw QR parallel to EF, and therefore making an angle 0 with qr.
Let QC, RB meet at P.
Since AQC = AqC — /A, and ARB = v ;
so that P lies on the circle BS'C'p. R
Hence, as the vertices of DEF slide along the sides of AB( ', the sides of PQR, homothetic to DEF, rotate about A, B, C, and its vertices slide on fixed circles.
Since S'q is a diameter of S'qQC,
S'Q= S'q cos 0, ....
Therefore 8' is the double point of the family of triangles ', including jp^r.
The linear dimensions of the similar triangles PQR, pqr are as cos 6 : 1 ; so that, if 2V, n be homologous points in the two triangles, N describes a circle on S'n as diameter.
PIVOT POINTS. TUCKER CIRCLES.
113
158. To determine the elements of PQR. From (84) V = area of pqr = %M ;
so that, if p' be the circumradius of pqr,
2p'2. sin \ sin /x sin v =• ^M. Then, for PQR, circumradius p' — p' cos 0.
Area of PQR = V cos20 = i Afcos2 0. But area of DBF - 2A2/M . sec2 0.
Hence, the area of ABC is a geometric mean between the areas of any triangle DEF inscribed in ABC, and the area of the triangle PQR which is homothetic to DEF, and whose sides pass through A, B, C.
159. Tucker Circles.
An interesting series of circles present themselves, when for " Pivot Points " we take the Brocard Points O and £1'. The circle DD'EE'FF', is then called a Tucker Circle, from K. Tucker, who was the first thoroughly to investigate its properties.
Since def, d'e'f are now the pedal triangles of 12, O' ; /. D = d = B, E = e= C, F = f = A-,
Denote by Z (corresponding to o-0) the centre of the common pedal circle of O, O', and by Z' (corresponding to a-) the common circumcentre of DEF, D'E'F'.
The line of centres ZZ', bisecting Ofi' at right angles, falls on Oil; also Z'VZ = 0 = Dtid = EVe = Ftlf.
And since SlOZ = u = VAF = O5D = Q(IE7,
it follows that the figures VOZ'Z, QAFf, VBDd, VCEe are similar.
114
Let
Then
MODERN GEOMETRY .
' = 0' = ACIF = BttD = CQE.
Now the radius of the pedal circle of fii)' = JP — 7i* sin w, (138) .'. circumradius of DD' ... = p = p sec 0 = Tt.siii to/sin (o> 4-0').
The quadrilateral BDSIF being cyclic, EFT) = B&D — 6' : therefore the arcs DF', FE', ED' subtend each an angle 6' at the circumference, and are therefore equal.
Hence the chord E'D is parallel to AB, F'E to BC, D'F to AC : and a circle with centre Z' and radius p cos 0' will touch the three equal chords.
In the cyclic quadrilateral E!EDF, AE'F = EDF=B; so that the equal chords are anti-parallel to the corresponding sides of ABC.
L TJE'D' = F'E'D'- F'E'D = A-0'. F'E'E = F'E'D' + D'E'E — A + 6'. Chord EF' parallel to BO = 2p sin F'E'E = '2,> sin (4 + 6') ;
chord DD' cut from EC = 2/> sin DE'D' = 2P sin (A- 6'). And if afiy be the n.c. of the centre Z1 ',
a = p cos^DZ'D' = p cos DE'D' = p cos (A-&).
PIVOT POINTS. TUCKER CIRCLES. H5
16O. The following list of formulae will be found useful.
(a) Radius of circle DD' ... = p — E sin co/sin (w + fl').
(b) Radius of circle touching equal chords = p' = p cos $'.
(c) N.c. of centre Z' ; a = p cos (A — 6').
(d) Length of equal anti-parallel chords = 2p sin &'.
(e) Chord DD' cut from PC = 2p sin (A — &). (/) Chord #/<" parallel to EG — 2p sin (A + (>').
((/) If d! and dl are points on 0.K" such that 00d = 30°, OOdj = 150°;
then a x cos (A—3Q) a sin (4 + 60°),
and aa oc cos (^t — 150) a- sin (J.— 60°).
Hence d and dl coincide with the Isodynamic Points 8 and 8r
161. The Radical Axis of the Tucker circle (parameter #'), and the circle ABC.
If ^2, £22, I.* are the powers of A, B, C for the Tucker circle, then the required Radical Axis is tl*x + ... = 0. (62)
Now,
'sin A 'sin (w-f 6')'
o sin 5 . /,, = 2p . — — - . sin 6. Kin A
And since EF' is parallel to BO,
.
sin A
.'. ^2 = AF. AF' = 4p2. sin ^. sin S sin C sin 0'. sin (4 -f ^)/sin8 ^1, so that the Radical Axis is
sin^+0'^ + .-. = 0.
116
MODERN GEOMETK Y .
162. The properties of four Tucker Circles, whose centres are certain standard points on OK, will now be discussed.
(A) The First Lemoine Circle, or Triplicate Ratio Circle. This has its centre at L, the mid-point of OK, or the centre of the Brocard Circle.
so that & = <o.
Then, (a) p = R sin w/siii 2w = ^Ii* sec w.
(]}) p = p COS to = ^R.
(c) a = ^Ksec 00. cos (A— w).
(d) Anti-parallel chord = R sec o> . sin <o = Z? tan oo.
(e) Chord DD' cut from 5C = E sec o> sin (A— o>). So that
: FF' = sin (A— o>) : sin(S-w) : sin (6y— o>)
Hence the name "Triplicate Ratio Circle."
(/) Chord EF' parallel to BC = R sec <o.sin
These chords pass through K. For since chord ED' is anti- parallel to AB, ED'C = A,
.'. perp. from E on BC = ED' sin A = li tan o> . sin A, from (^) = perp. from K on BC.
PIVOT POINTS. TUCKER CIRCLES H7
163. (B) The Pedal Circle of OQ'. The centre being Z, 6' = %TT — w.
(a) p = R sin oj.
(fo) p' = 72 sin2 (o.
' c) a = R sin a) . sin (A -f- a>) := .R sin2 a> . £ 6/
(d) Anti-parallel chord = R sin 2w.
(a) Chord cut from BC = 2R sin w cos
(/) Chord parallel to BO = 2R sin w cos (J.— <o).
164. (C) The Second Lemoine Circle, or Cosine Circle.
A
The centre of this circle is -K", so that & = O&K = -Jir. (a) p = R tan w. (6) P' - 0.
(c) a = R tan <o sin ^1. (131)
(d) Anti-parallel chords E'F, F'D, D'E each equal
the diameter 2R tan w; so that they each pass through the centre K, as is also obvious from (6).
(e) DD' = 2R tan co cos A ;
.'. J>D' : EE' : FF' = cos A : cos 5 : cos (7. Hence the name "Cosine Circle."
(/) Chord parallel to BC = 2R tan <o cos A
= chord cut from BC.
118
MODERN GEOMETRY.
165. The Taylor Circle.
Let H!, H.2, Hs be the feet of the perpendiculars from A, B, C on the opposite sides.
Draw H^ perpendicular to AB, H.2D to BC, HSE to CA.
Let A&F = <£.
Then AF — AH, sin B = 211 sin2 B sin C
and AQ, = 2R sin o> sin 5/sin A ;
sin to
Also Hence
cot o> =
AF sin A sin 5 sin C 1 + cos J. cos B cos (7
(131)
(131)
sin A sin j6> sin (7 tan <£ = — tan A tan 5 tan 0. Similarly it may be shown that
tan 5OD or tan C£lE = — tan A tan B tan 0 ; BO that AVF = BOD = CUE.
Next, draw HJE' perpendicular to CA, HZF' perpendicular to A 13, jff3D' perpendicular to BC. Then it may be shown that
AQ'E' = Btl'F' = Ctt'D' = </>.
The six triangles A&F, ... being all similar, it follows that D&EE'FF' lie on a Tucker circle, called the Taylor circle, after Mr. H. M. Taylor.
A
B O
The angle <£ is called the Tajdor Angle.
PIVOT POINTS. TUCKER CIRCLES. 119
Since <£ is less than TT, we have
D sin <£ = + sin A sin B sin (7, D cos <£ = —cos J. cos B cos (7, where D2 = cos2 JL cos2 B cos2 C- -f sin2 A sin2 J5 sin2 (7. A diagram shows that, T being the centre of this circle, OT : TK — —tan <£ : tan o> = tan A tan B tan C : tan w.
Note the equal anti-parallel chords DP', FE', ED' ; also the chord E'D, parallel to AB, F'E to BC, D' F to AC.
\
0 K T
166. The list of formulae is now —
(a) Radius of T-circle = R . sma>
. / . .
sin (w + «/>)
(fe) p' .= J?_D cos </> := 7? cos ^4 cos 5 cos 0.
(c) 'a = RDcos(A—<j>}
= R (cos2 A cos 5 cos 0— sin2 A sin 5 sin C).
(d) Anti-parallel chord E'F or F'D or D'E
= 2RD sin </> = 2.B sin A sin 7? sin 0.
(e) Chord cut from BO = 2ED sin (.1 — <£)
= R sin 2.4 cos (B—C).
(/) Chord ^'£7 parallel to BC = R sin 2.4 cos A ; the other chords being D'F and E'D.
To determine the Radical Axis of the circle ABC and the Taylor Circle.
sin ( A -f <£) = sin A . cos </> + cos A sin </>
= D ( — sin A . cos .4 cos B cos 0
+ cos A . sin ^4 sin B sin ( ') oc sin A cos2 A.
Hence, from (61), the Radical Axis is cot2 A. x + ... — 0.
So that the tripolar coordinates of the Limiting Points of these two circles are as cot A : cotB : cot C. (21)
APPENDIX I.
Let LMN..., L'M'N'... be two systems of n points.
Place equal masses p, p at L, L' ; gr, q at M, If', &c.
To determine the condition that the two systems shall have the same mass-centre.
Project LMN... and the mass-centre 011 any axis, and let lmn...x be the distances of these projections from a given point 0 on the axis.
Then
—m')+r(n—n') + ... = 0.
and so for any number of axes.
But I— I', &c., are the projections of LL', MM', NN'....
Therefore the required condition is that a closed polygon may be formed, whose sides are parallel and proportional to p.LL', &c.
In the case of a triangle
p.LL' <x Bin (MM1, NN').
Now, in the case under discussion, take a second point P' on TT, and let its pedal triangle be d'e'f.
Here LL' = dd' = PP . cos ^,
and angle (MM', NN') is (ee', //') or A ;
.'. p cos 6, oc sin A, or £> oc sin A sec 01B
Hence all pedal triangles def of points P on TTf have the same mass-centre for the constant masses sin A sec 6V &c., placed at the angular points d, e, /.
120
APPENDIX II.
To determine the second points in which the four circles out the Nine-point Circle.
Let the circle PQR cut the sides of ABC again in P19 Q19 Rr
Then AQ.AQ^ = AR.AR,, &c. ;
.-. AQ.AQl.BR.BRl.CP.CP, = AR.AR..BP.BP,. GQ. CQr But, by Ceva's Theorem, since AP. BQ, GR are concurrent,
BP.CQ.AR = CP.AQ.BR- .: OP, .AQl.BR, = BP, . CQ, . AR,. Therefore AP^ BQ}, GRl are concurrent. Again, since CQ : QA = r : /;,
and AR : RB = p : q,
.'. AQ=p/(r+p).b; AR=p/(p+q).c. So AQi=pJ(ri+pJ.1>', ARl=pJ(p}-{-ql}.c.
But AQ.AQ, = AR.AR^ &c. ;
62
121
122 MODERN GEOMETRY.
.'. p, a -
g + r r -\-p So ql oc H --- h ; r, <x -M -- ;
: q.-r, a 2&
Tn (80) it was shown that the circle 1JQR cuts the Nine Point Circle at a point w, whose b.c. are as a2/(q2—r*), —
Similarly the circle P1Q1B1 (the same circle) cuts the Nine- Point Circle at a point <o', where the b.c. of <o' are given by
a'J ^ AT
' a 2 '
So, if the circle PQ,' R' cuts the Nine-Point Circle again at o^ the b.c. of o>1 are given by
1
writing — p for
APPENDIX III.
(a) To determine the area of X YZ. Since
A'O = E cos A , A'X = %a tan 0 = E sin 4 tan 6.
.'. OX= B/cos 0 . cos ( A + 0} . .'. Z.&retiYOZ = OY.OZ.sinA
= E2/cos2 0 . cos (B + (9) cos (0 + 0) sin A. /. 2.&XYZ = '2(YOZ + ZOX+XOY) = ... ; and by some easy reduction we obtain,
&XYZ = AM cos2 0. [2 sin oj— sin (20 + o>) }/sin o>. When 0 is equal to either Steiner Angle, (137)
then sin (20 + w) = 2 sin w,
and the triangle XYZ vanishes, so that XYZ is a straight line.
But this triangle always has G for its centroid.
Hence, in this particular case, XYZ passes through G.
(b) Instead of the base angles being equal, suppose that
z XE( ' - YCA = ZAB = 0, BOX = CAY = ABZ = 0, and BXG — CYA - AZB = X-
Then Oj = a. sin 0 sin <£/sin ^,
a2 = b sin 0.sin(C'— 0)/sinx, as = c sin 0.sin (B— ^>)/siii x ; 3a = a1 + a2 + u3 = A,, &c.
Hence 6r is the eentroid of XYZ.
123
124 MODERN GEOMETRY.
(c) Let YZ, ZX, XY meet BG, GA, AB in x, y, z respect- ively.
Then, since AX, BY, GZ&re concurrent, xyz is a straight line, being the axis of perspective of the triangles ABC, XYZ.
To show that the envelope of xyz is Kiepert's Parabola.
The equation of yz is
(Ay.-&y,) «+ (w-y3a.2) 0+ (oA-a,&) y = 0.
Now y2a3 — y3a2 oc sin (J. — 0) sin (B — 0) — sin 0 sin 0 ;
cc sin J. sin B — sin C sin 26.
So a3&"~*a«A <* s^n ^ sin J. — sin 5 sin 2^.
Therefore at # we have
/3/(sin C' sin ^4 — sin 5 sin 20) +y/(sin A sin 5— sin G sin 20) = 0, so that xyz is
aa/(sin J. sin 5 sin (7 — sin2 A sin 20) + . . . =0.
Writing this as px + qy + rz = 0, we know, from (9), that this line touches the parabola, the n.c. of whose focus are a/(l/3-l/r), Ac.
Here 1/g — 1/r oc (sin2 J5— sin2 0) sin 20.
Hence the focus has n.c. a/(fc2— c2) &c., and the directrix is (&2-c2)cos.4-a+... =0.
Hence the envelope of xyz is Kiepert's Parabola, having for focus the point whose Simson Line is parallel to OGH, and OGH for directrix.
INDEX.
[The numbers refer to Sections,
Aiyar, V. R., 108, 109. Angular Coords., 81. Apollonian Circles, 127, 129. Artzt's Parabola, 66, 120. Axis of Perspective : ABC and PQR, 141. „ ,, XYZ, App. Ill (c).
Beard, W. F., 36.
Brocard Circle, 142, 144, 150.
Brocard Angle and Equil. T., 91.
Centres of Similitude :
XYZ and Ii/2I3, 26.
def and D'E'F', 85.
HiHiHs and T^T^ 85. Circles :
Apollonian, 127, 129.
Brocard, 142, 144, 150.
Centre, 9, 12, 42.
Lemoine (First), 162.
(Second), 164.
Neuberg, 135.
Nine Point, 16, 45, 61, 107.
Taylor, 165.
Pedal C. of nn', 163.
Centre Olf 44. Cubic Transformation of Elliptic
Functions, 34, 54, 55.
Davis, R. F., 9, 19, 104. Dixon, A. C., 55.
Feuerbach Point, 14, 51. Theorem, 63.
Gegenpunkte , 83. Genese, Prof. R. W., 102. Gergonne Point (ABC), 32, 34, ,, (TiTtTj, 122. Greenhill, Sir George, 22, 26, 34, 54, 55.
Harmonic system of lines, 80. Harmonic Quad., 124, 129.
Isodynamic Points, 99, 149, 160 (g). Isogonal Conjugates, 101. Isogonic Points, 151.
Kiepert's Parabola, 11, 40, App.
Ill (c). Kiepert's Hyperbola, 139, 151.
Lemoine Point of 1,12/3, i22- Lemoyne's Theorem, 77. Lhuillier, 92. Limiting Points, 21, 57. Lines, harmonic system of, 80.
M'Cay's Cubic, 109.
Nagel Point, 30, 31.
Narayanan, S., 75.
Neuberg, Prof. J., 69, 70, 88, 92,
135. Nine Point Circle, 16, 45,. 61, 107.
125
126
MODERN GEOMETRY.
Orthologic Triangles, 82. Orthopole is » for TOT', 74.
Parabola : Artzt, 66, 120.
Kiepert, 11, 40. Points :
SiSrfHiGi (poristically fixed), 24-28.
FGHO'TNM(on poristic circles), 29-34.
SS'SiSj1 (relations), 87.
Feuerbach, 14, 51.
Gergonne, 32, 34, 122.
Tsodynamic, 99, 149, 160 (g).
Isogonic, 151.
Lemoine Point of IiI2^, 122.
Limiting, 21, 57.
Midpoint of S'S/, 115.
Nagel, 30, 31.
Pivot, 153.
Steiner, 143, 144.
Sub-Lemoine, 152.
Tarry, 141, 143, 144.
Twin Points, 87.
Point D, 141, 146, 148.
Pivot Points, 153. Poristic formulae, 33.
Quadrilateral, Harmonic, 124, 129.
Radical Axis :
ABC and Tucker Circle, 161. ,, ,, Taylor ,, 166.
Groups of R.A.'s, 21, 62. Rao, T. Bhimasena, 72. Rouse, E. P., 104.
Schick, Dr. J., 58, 64. Schoute, Dr. P. H., 100. Steiner : Angles, 137.
Ellipse, 152.
Point, 143, 144.
Tarry Point, 141, 143, 144. Taylor, H. M., 103, 106.
,, Circle, 165. Tricusp. hyp., 52. Twin Points, 87.
Weill Point GM 28.
London : Printed by C. F. Hodgson & Son, 2 Newton Street, Kingsway, W.C.
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