57

BV, CW will all be perpendicular to G H ; and the triangle UVW

will circumscribe the triangle ABC.
Let N, P, Q be the feet of the interior bisectors of the angles
A, B, 0, and N', P', Q' the feet of the exterior bisectors; then the
six straight lines UN, VP, WQ, UN', V P , WQ' pass three and
three through four points which are the points of contact of the
nine-point circle with the inscribed and escribed circles.*

Geometrical Note.
By R. TUCKER, M.A.
If in a triangle ABO, points are taken on the sides such that
BP:CP = CQ: A Q ^ AR : BR = m : w = C F : BP'
= AQ':CQ' = BR':AR'
then the radical axis of the circles PQR, P'Q'R' passes through the
centroid and " S." points of ABC; and if QR, Q'R' cut in 1,
RP, R'P' in 2, PQ, FQ' in 3, then the equation to the circle 123 is
abcSa/3y = mw2aa.2aa{ - mna" + (m? + mn + »2)(62 + c2)}.
FIGURE 20.
The points P, Q, R are given by
(0, nc, mb), (vie, 0, no), (nb, ma, 0),
i.e., P, in trilinear co-ordinates, is (0, nc sinA, mb sinA), etc. ;
and F , Q', R' by
(0, me, nb), (nc, 0, ma), (nib, na, 0).
It is hence evident that the pairs of triangles are concentroidal
with each other and with ABC.
It is also evident that PQ', P'Q are parallel to AB, and so on ;
also that P'Q, PR' intersect on the median through A ; and so on.
The triangle PQR = (m? - mn + w2)A = the triangle P'Q'R'.
The equation to the circle PQR is
(nt2 - mn + n-)abc.^(a^y) = mw2(aa).2(aa. - mno? + m?b' + «V),
and to P'Q'R' is
(TO2 - mn + n2)a6c.2(a/?y) = TOw2(aa).2(aa. - mna? + i!?b2 + mV).

* (20) Rev. \V. A. Whitworth in Mathematical Questions from the Educational

Times, X. 51 (1868).

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 58
The radical axis of these circles is, therefore,
2(aa.6 2 -c 2 ) = 0, hence (1)
The radical axis of either of the circles and of the circumcircle is
of the form PP - JQ + R = 0, where P, Q, R are linear functions of
a, /?, y ; and the envelope of each of these axes is the conic
... (a).
4
The tangents in (a) intersect in the point oa/(a - b~c*) =... = ....
The radical centre of the three circles is
aa.j[al - 6V + mnk(k - 3a2)] =... = ...;
where k = a2 + b2 + c2.
The equations to QR, Q'R' are
- mnaa + nib/3 + m?cy = 0 \ ,,,
-mnaa + m-bji + nicy = 0) ''
and 1, their point of intersection, is on the median through A, and
is given by
aa/(m? + n l ) = b/3/(mn) — cyjtnn.
Similarly 2, 3 are
aajmn = 6/3/(m2 + n2) = cy/mn,
aajnm = bfi/mn = c-y/(m2 + n2).
The above lines (b) envelope the parabola a2a2 = ibcfiy, and so on.
The triangle 123 is readily found to be

The circle 123 has its equation

abc 2(a/?y) = »in2(aa).2{aa. - mna? + (m2 + mn + w2)(i2 + c2) } ... (2)
The radical axis of this circle and the circumcircle can be written

hence it is a straight line parallel to the chord of contact of the

conic (a).
The lines PR, P'Q', ... intersect in 4, 5, 6, given by
aa/(mn -n") = 6/3/m2 = cy/m2, ... ,
showing that these points are also on the medians, as is evident from
the symmetry of the figure.

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 59

The lines PR', P'Q, ... intersect in p, q, r, where p is given by

aa/(m - n) = bji/n = cyjn.
The conic through PP'QQ'RR' has for its equation
mn(aa + bfZ + cy)' = bcl3y + caya + aba(l ... (4),
which, in the figure, is an ellipse, concentric, similar and similarly
situated with the minimum circum-ellipse of ABC.
The polar of A, with regard to (4), is
2amna - (wr + n-)(b/3 + cy) = 0,
therefore it is parallel to BC, and cuts AC in J (say) ; so that
A J = Imn. AC. The triangle formed by the three polars (for
A, B, C) is
= 4(m2 - inn + n2)-A.
The tangents to the conic a t P, P ' are given by
aa(m 2 + n2) + bf$m(m - n) - cyn (m - n) = 0,
aa(m2 + w2) - bfin (m - n) + cym(in — n) = 0,
and intersect, on the median through A, in the point
aa. _ b/3 _ cy
— (m-n) m? + n 2 mr + n?
and the triangle formed by this and the corresponding points equals
the above triangle.
To find the " S." points of PQR, P'Q'R', assume the sides of these
triangles to be p, q, r ; p', q', r'; then
p* = m?c-+
+ nV
nV -- 2mw6ccosA»
2mw6ccosA"» , ,
12 \\ etc. = etc. :
p' = + »V - 2mnbccosA)
2(p2) = (m- - inn. + n2)(a2 + 62 + c2) = K (suppose) = 2(^' 2 ).
The " S." lines through Q, R, respectively, are
(3 y a
nbcr2 ca^mr3 + np*) mabp* mbcq* neap2
me o na nb ma
i.e., - naa-lmr* + np2) + (/tV - rri?p-)bp + mcy(mr + np2) = 0,
maa(mp2 + nq2) - (mp2 + nq*)nbfi - cy(w'y2 - n'p2) = 0 ;
whence we get, for the " S." point of PQR(K,),
aa _ bji _ cy _ 2A
_.

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 60

Similarly, for the " 8." point of P'Q'R'(K2), we have

aa _ b/3 _ cy 2A
nq'^ + mr'2 nr'^ + mp"2 np'2 + mq'* K
The triangle 123 is directly in perspective with ABC, and has
the centroid of the triangles for centre of perspective ; hence we can
readily obtain the co-ordinates of the principal points.
For (1) the " S." point

(2a) the positive " B . " point

aa/[(m? + n*)cW + mnb (c» + a2)] = . . . = . . . ;
(2b) the negative " B." point
aa/[(m? + w> 2 6 2 + wnc2(a2 + ¥)] = . . . = . . . ;
(3) the in-centre
aa/[a(m2 + n?) + (b + c)mn] =... = ... ;
(4) the orthocentre
a/[(m2 + w2)cosBcosC +TOWCOSA]= ... = ... ;
(5) the circumcentre
a/[(m2 + «2)cosA + mwcos(B - C)] =... = ....
It is readily seen that the lines (AP, BQ), (AP', BQ') intersect
on the conic c2y = aba/3, which touches CA, CB at A and B, and
passes through the centroid.
The co-ordinates of the centre are
{£(2csinB), J(2csinA), J(-osinB)} ;
like results hold for the other points of intersection.

[The preceding Note consists of a solution of Questions 11599

and 11670 of the Educational Times, and is published in vol. lviii.
(pp. 119-123) of the "Reprint" from that journal. It is given
here with the editor's kind consent. Part also of Question 11599
was proposed by Prof. Neuberg as Question 787 of Mathesis.
In the number for January 1893, Prof. Neuberg points out that
(a) supra is a conic touching the Brocardians of the Lemoine-line,
where they meet the reciprocal of that line.]

https://doi.org/10.1017/S0013091500031175 Published online by Cambridge University Press