2. To find the radius p of Taylor's circle of a triangle ABC. [Taylor's circle for the triangle ABC is the circle in Ex. 1 for PQR; hence we have to express r and s of the latter triangle* in terms of the parts of ABC. We easily obtain

also

p2=4R2 (sin2A sin B sin C+cos24 cos2B cos2C)

Pi=4R2(sin A cos B cos2C+cos A sin'B sin3C)

with similar values for p22 and p32.

From these expressions we have the result given in Ex. 1: Σp2=4R2.]

3. The lines B'C', C" A', A′′B', parallels to the sides of ABC, are the chords of contact of the ex-circles of PQR with its sides.†

[Let A'B' meet PR in the point Q'. Then B'B'RQ' is a parallelogram, therefore RQ'=semiperimeter of PQR, etc.]

4. Employing the notation of Art. 35, prove that the lines joining the corresponding vertices of the two triangles PQR and XYZ are concurrent at the circum-centre of the latter.

[Let p and q be the perpendiculars from R on the sides YZ and ZX of the triangle XYZ. Then p/q=RB'sin B/RA'sin A. But RB" RA'=QR/RP=a cos A/b cos B. Substituting and reducing we have p/q=cos A/cos B.

But if Z be joined to the circum-centre of XYZ, the joining line is the locus of a point such that perpendiculars from it on the sides are in this ratio; hence ZR passes through the circum-centre of XYZ. And similarly for the lines PX and QY.]

*The sides of the pedal triangle are equal to a cos A, b cos B, c cos C, or R sin2A, R sin2B, R sin2C; hence its perimeter = 4R sin A sin B sin C ; its s-a=2R sin A cos B cos C, its s-b=2R cos A sin B cos C, etc.; its r=2R cos A cos B cos C; its r1=2R cos A sin B sin C, etc.

+ The polars of the vertices of a triangle with respect to the ex-circles meet the sides in six points which lie on the same circle.-Mathesis, t. 1, p. 190.

‡PX, QY, and RZ are perpendiculars to antiparallels to the sides of XYZ and therefore meet the sides of PQR at right angles.

Hence the circum-centre of XYZ is the orthocentre of the triangle PQR.

5. The Simson lines of the median triangle LMN of a given one ABC with respect to the vertices P, Q, R of the pedal triangle pass through the centre of Taylor's circle.*

[The circum-centre O of ABC is the orthocentre of LM N. Hence RO is bisected by the Simson line XYZ of R. Also CZ-RZ; therefore the line XYZ is parallel to OC. But the centre of Taylor's circle O, is (Art. 47) the middle point of RH,; therefore, etc.] 6. The Simson lines of PQR, whose poles are L, M, N, pass through

*The point on the circum-circle from which perpendiculars or other isoclinals are let fall on the sides of an inscribed triangle is called the Pole of the Simson line.-V. Mathesis, t. 2, p. 106, "Sur la Droite de Simson," par M. Barbarin.

F

[For the perpendicular NZ from N on PQ bisects it (Euc. III. 3); and the perpendiculars NX and NY are equally inclined to AB (Euc. I. 26), hence the line XYZ is a perpendicular to AB through the middle point of PQ; therefore, etc. (Art. 47.)]

7. Prove that the common inclination (0) of the sides of the triangles A'B'C' and A'B'C" to those of ABC is given by the equation tan 0=-tan A tan B tan C.

(Taylor)

8. The intercepts made by Taylor's circle on the sides are a cos A cos (B-C), bcos B cos (C-A), ccos C'cos (A – B).

=

9. The circum-centre of a triangle, its symmedian point, and the orthocentre of its pedal triangle are collinear. (Tucker.)

[The orthocentre of the pedal triangle has been shown to be (Ex. 4) the circum-centre of XYZ, and K is the centre of similitude of ABC and XYZ; therefore, etc.]

10. The circum-centre and the orthocentre of its pedal triangle are equidistant from, and collinear with, the centre of Taylor's circle. (Neuberg.)

[For CH, and ZR are parallel, since both are at right angles to PQ; also RH, is bisected at 01 (Art. 47), therefore, etc., by Art. 37.]

CHAPTER IV.

GENERAL THEORY OF THE MEAN CENTRE OF A SYSTEM OF POINTS.

49. We now proceed to the discussion of the general linear relation connecting the distances of a system of points from a given line.

Let A, B, C, D... be the system of points, AL, BL CL... their distances from any line L, and Σ(a. AL) the algebraic sum

a. AL+b. BL+c. CL+...

where a, b, c ... are given quantities.

By (a.AL) is therefore meant the sum of given multiples of the distances of the system of points from the line; perpendiculars from points on opposite sides of L being taken with opposite signs.

50. Theorem.—For any two lines M and N and systems of points A, B, C... and multiples a, b, c ... having given

to prove that

Σ(a. AM) = 0 and Σ(a. A N) = 0

Za. AL=0,

where L is any line passing through O the intersection of M and N.