[Through O draw OAA' parallel and PA' perpendicular to p1. The projection of p1 on OP-projection of AA'; but A, B, C, ... and

A', B', C', ... are the vertices of regular polygons, whose mean centres are both on OP. Therefore the sum of the projections of pi.. OP=0.]

53. Theorem. For any line L to prove that Za. AL=2(a)OL.

Draw M through O parallel to L.

...

on

Multiplying these equations respectively by a, b, c, ... and adding, we have

(a. AL)=(a. AM)+2(a)OL;

but Σ(a.AM)=0 since M passes through the mean centre; therefore, etc.

This property enables us to find the mean centre. For by taking a line L in an arbitrary position and calculating Σ(a. AL)/2(a) we have for the locus of O a line parallel to Lat this distance from it. Again, take a line in another position and construct the locus of O as before. The intersection of these loci is the point required.

COR. 1. If Ea. AL is a constant, the line L touches, or envelopes, a circle concentric with 0.

COR. 2. If the multiples are all equal ΣAL=n. OL, where n denotes the number of points in the system. COR. 3. For systems of points and multiples and their

the mean centre O of all the points and their corresponding multiples is the mean centre of 01, 02, ... On for the multiples (a,), Σ(α2), ... Σ(an).

[For since Σα11L=Σ(α ̧)OL‚ Σã‚¿AL=Σ(ɑ„)OL, etc., on adding these equations

Hence the mean centre of a system of points can be found as follows:-Find the mean centre 0, of two of the points A and B; next find the mean centre of O1 and C for multiples a + b, c. Denote this by 02, and find the mean centre of O, and D for multiples a+b+c, d, and so on. When the entire system has thus been exhausted the last mean centre found is that of the system.

EXAMPLES.

1. The sum of the distances of the vertices of a triangle from any line is equal to three times the distance of its centroid from the line.

2. Draw a tangent to a circle such that Za. AL may be a maximum, minimum, or have any given value.

[The extremities of the diameter passing through the mean centre are obviously the points of contact in the extreme cases. The general case reduces to draw a common tangent to two circles.] 3. If Z touches the in-circle Za. AL-2A when the multiples are equal to the sides of the triangle.

3a. For the ex-circle to the side c the equation becomes

aAL+bBL-cCL=2A.

4. The projection of the mean centre on any line is the mean centre of the projections of the system of points on the line.

[Let the projections be denoted by O', A', B', C'... and Z the line 00'. Then A'O' AL, B'O'=BL, etc. Hence

Za. A'O' Za. AL=0; therefore, etc.]

5. If 0, 01, 02, 03 denote the in- and ex-centres of a triangle, (s-a)OL+(s-b) O2L+ (s−c)O2L=s. OL.*

[For O is the mean centre of the remaining points for multiples s-a, s-b, s-c (Art. 52, Ex. 3), and since

(s-a)=s; therefore, etc.]

6. Let three similar triangles BCA', CAB' and ABC' be described on the sides of ABC in the same aspect; to prove that the mean centres of the triangles ABC and A'B'C' coincide (Brocard).

Complete the
The triangles

[Let X be the middle point of BC and Z' of A'B'. parallelogram BA'CP. Join AX, C'Z', Z'X and PB'. BPC and B'CA are similar, therefore CP/CB=B'C/AC (Euc. VI. 4), or by alternation B'C/CP-AC/BC; also the angles B'CP and ACB are equal, therefore the triangles BPC and ABC are similar (Euc. VI. 6); hence CB'|B'P=CA/AB; alternately CB'|CA=PB'|AB; but CB/CA=C'A/AB (hyp.); therefore PB'/AB=C'A/AB from which PB'=AC'.

Again LPB'C=LBAC, to these add the equals ACB' and BAC" respectively; therefore PB' and AC' are parallel. But Z'X is parallel and equal to half of PB'; therefore it is parallel and equal to half of AC'. Hence the medians AX and C'Z' trisect each other.* Otherwise thus:†-Let another triangle ABC" be described below the base AB symmetrically equal to ABC'. It is easy to see that

*For another proof see Milne's Companion to the Weekly Problem Papers, Art. 123.

+ Educational Times. Reprint. Vol. liv., p. 102.

=

the triangles ABA' and CBC" are equal in area; similarly ABB' and CAC" are equal. By addition we have ABA' + ABB' ABC+ ABC" or ABA'+ABB' - ABC' ABC, i.e. the algebraic sum of the perpendiculars on AB from A', B', C'=the perpendicular from Con AB. Similar results are obtained for the sides BC and CA; therefore, etc. Syamadas Mukhopadhyay.]

7. If two points A and B be displaced to new positions A' and B', their mean centre M for any multiples is displaced to M' found by the following construction :

Through M draw lines MP and MQ equal and parallel to AA' and BB' respectively. Join PQ and divide it in M' such that PM'/QM' AM|BM.

[For since AA'PM and BB'QM are parallelograms, A'P=AM and B'Q=BM; therefore by similar triangles PA'M' and QB'M',

A'P A'M' PM

:. therefore, etc.]

8. If three points A, B and C be displaced to new positions A', B' and C', their mean centre M is displaced to M' found by the following construction :—

Through M draw lines MP, MQ and MR equal and parallel to the displacements AA', BB' and CC' respectively; M' is the mean centre of P, Q, R.

[For let X denote the mean centre of A and B, X' which is found by Ex. 7 of A' and B'. Draw MO equal and parallel to XX. Join OX', RC' and X'C'.

It is evident by parallels that O is the mean centre of P and Q; also MX-OX' and MC=RC'; therefore in the similar triangles