or A'B" = 2p cosC, with similar expressions for BC" and CA".

Hence

The segments intercepted by the circle on the sides of ABC are proportional to the cosines of the opposite angles.* It is from this property the circle derives its name.

40. The middle point M of A"B′ is on the median through C to the opposite side c; hence the perpendicular through K to this side passes through M, or as has been shown otherwise (Art. 30, Ex. 4). If a perpendicular be drawn through K to the base meeting it in N and the median in M, MK = NK, from which it follows that the lines joining the middle points of the sides to the middle points of the corresponding perpendiculars meet at the symmedian point (Hain).

41. The sides of the triangles A'B'C' and A"B"C" are perpendicular to the corresponding sides of ABC. The cosine circle may therefore be obtained by rotating the two inscribed triangles in opposite directions until 0 = 90°. (Art. 39.)

The ratio of similitude of A'B'C' and ABC = tano.

42. II. Triplicate Ratio Circle.-Let the parallel in figure of Art. 35 pass through K.

Then L, M, N are the middle points of AK, BK, and CK, since AA'A"K, etc., etc., are parallelograms; and the centre O of the corresponding Tucker circle bisects OK.

The sides of A'B'C' are inclined to those of ABC at an

angle = w. For consider the angles in the equal segments A'A", B'B", C'C", and it is obvious (Euc. III. 21) that A'B'A" A'C'A"=B'C'B" = B'A'B" = C'A'C" =C'B'C".

=

* See Mathesis, t. i., p. 185 :—

"Sur le centre des Médianes Antiparallèles," Neuberg (1881).

Hence K is the negative Brocard point of A'B'C'. Similarly it is the positive Brocard point of A"B"C". It follows generally that the locus of the negative Brocard point of A'B'C' is a line passing through K.

43. The ratio of similitude of A'B'C' and ABC is sin w/sin 2w since w; hence

=

44. The intercepts B'C", C'A", A'B" made by the circle on the sides are thus determined:-The triangles A'KB" and ABC are similar, therefore A'B" /c=ratio of 2cA 2A

altitudes

= a2+b2+c2/ с a2+b2+c2 ;

hence A'B" =

၉မိ
a2+b2+c2°

(1)

with similar expressions for B'C" and C'A". The general property of the circle may be thus stated:-Parallels through the symmedian point meet the non-corresponding sides in six points which lie on a circle; and the intercepts made on each side are in the ratios a3: b3: c3. From the latter property the circle takes its name. For the sake of brevity it is often written "T.R." Circle.*

45. III. Taylor's Circle.-Let the antiparallels A'A", B'B", C'C", which, it will be remembered, are always parallel to the sides of the pedal triangle (PQR) of ABC, pass through the middle points a, B, y of the sides of PQR.

Consider the segments into which A'A" is divided by B and y. We have By=QR, yA′′ = {PQ (Euc. I. 5), and

* An account of the circle will be found in Mathesis in the article by Neuberg already referred to (Art. 39). See also Nouvelles Annales, 1873, p. 264.

for the same reason BA'=RP; therefore A'A" is equal to the semiperimeter of PQR

=(a cos A+b cos B+c cos C)=2R sin A sin B sin C. Hence generally

=

=

A'A" B'B" C'C"=2R sin A sin B sin C...... (1) Again, since B"aC' is an isosceles triangle, the perpendicular to the chord B"C" of Tucker's circle at the middle point bisects the vertical angle a and passes through the in-centre of aẞy. Similarly for the chords C"A' and A"B'. Hence

The centre of the circle coincides with the in-centre of the median triangle (aßy) of PQR.

Many properties of this circle are proved in Neuberg's article in Mathesis, t. 1, p. 185, but it was described independently in England by Mr. H. M. Taylor, and now bears his name. (Proc. Lond. Math. Society, vol. xv. p. 122.)

46. Since aQ= aR = aB" = aC, the circle on QR as diameter passes through B" and C and RB"Q= RCQ = 90°; or B and C are the projections of Q and R on the sides AB and AC; hence

The six projections of the vertices of the pedal triangle on the sides of ABC lie on Taylor's circle.

47. The triangle B'aC" is isosceles, therefore O1a the bisector of its vertical angle a is at right angles to BC; hence generally

The lines O1a, O1ẞ, O1y are perpendiculars to the sides of ABC.

Let H, denote the orthocentre of CPQ; then QH, and O1a are parallel; similarly PH, and 0,ẞ are parallel; hence the triangles PQH, and aßО, are similar, their ratio of similitude being = 1⁄2, or HR is bisected at O̟1.

Similarly PH, and QH, are each bisected at 01; and therefore the triangles H1HH, and PQR are equal in all respects.

48. Theorem.-Taylor's circle of the triangle ABC is the common orthogonal circle of the ex-circles of PQR. In the triangle AA'A" we have by rule of sines AA" = A'A" sinC/sinA = 2R sinB sin2C (Art. 45 (1)), AC ARCOSA = b cos2A;

also

=

multiplying these results and reducing

AA". AC4R2 sin2B sin2C cos2A,

but AQ=ccosA; substituting we obtain

AA". ACAQ2 sin2B,*

or the square of the perpendicular from A on QR. Hence the tangent from A an ex-centre of PQR to Taylor's circle

* Otherwise from the right-angled triangle A A′′P and ACP we have AA"=b sin3C; and from the triangles ACR and AC'R,

AC=b cos2A; therefore AA". AC' = b2 sin2C cos2A.

is equal to the radius of the ex-circle; similarly for the ex-centres B and C; therefore, etc.*

EXAMPLES.

1. To find the value of the radius p of a circle cutting the excircles of a triangle PQR orthogonally.

[In figure of Art. 45 p2=01A'2. But if a perpendicular be drawn from 0 to ẞy it is equal to the radius of the in-circle of the triangle aßy or half the radius (r) of P.QR; and the distance of its foot from A' is equal to the semiperimeter of aẞy-i.e., s of PQR.

Similarly for the radii P1, P2, P3 of the circles cutting two escribed and the inscribed of PQR orthogonally we obtain

and by adding these results we have, on reducing,

the sum of the squares of the radii of the four circles cutting orthogonally the inscribed and escribed circles of any triangle taken in threes is equal to the square of the diameter of the circum-circle.

* In the triangle PQR since perpendiculars PA' and QB" are let fall from the extremities of the base PQ on the external bisector AB of the vertical angle R, by a well-known property yA'=yB′′=¿ sum of sides. But the distance of the middle point of any side from the points of contact of the ex-circles which touch it externally=sum of sides. Hence if a circle be described with y as centre and yA'=yB" as radius, it cuts the ex-circles of PQR whose centres are at A and B orthogonally. It follows that the locus of the centre of a circle cutting these two orthogonally is the line yO̟1, since it is perpendicular to the line of centres ; similarly a0, and ẞO1 are the loci for the centres of circles orthogonal to the remaining pairs of ex-circles, whose centres are at B and C, C and A respectively.

19

Therefore O, is the centre and O1A'=01B" etc., the radius of the common orthogonal circle, i.e., Taylor's circle.