Full text of "185"
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satisfied by the four values $=X, $=/*, d=v, 8=p, in virtue of the given equa-
tions ; hence it must be an identity.
To find the value of x, multiply up by a + 6, and then put a +0=0; thus
x= (a +A)(o+ Ji .)(a+y)(o+/.)
By symmetry, we have
v= <fi+W+t>-)<b + v)(b+p)
y (b-c)(b-d)(b-d) '
= _ (<>-M)(<>+/')(<>+*0(g+/»)
and U ~ (d-a)(d-b)(d-c) '
Similarly solved by Q. B. M. ZERR.
186. Proposed by W. J. GBEENSTBEET, M. A., Editor of The Mathematical Gazette, Stroud. England.
Given the tangential equations to two conies S, S' , find the tangential co-ordinates
of the join of the poles of two given parallel lines with respect to S. Deduce the tangent-
ial equation of the center of S, and find that of the intersection of S and S' .
Solution by G. B. M. ZEEE, A.M., Ph.D., Professor of Chemistry and Physics, The Temple College, Philadel-
Let bc-f*=A, ca-g*=B, ab-h*=C, gh-af=F, Jif-bg^G, fg-ch=R.
Then S=AX* +.B/* 2 + <M +2F/j.v+2GvX+2HX/i.
Similarly, 8 , =A'X*+B' l i* + C'v* + 2F»v+2G'vX+2H'Xv.
Let Aa+./i/3+vy and Xa-\-iJ.p+vy-\-m be the two given parallel lines; p, q, t
and p', q', t' their poles with respect to S. Then for the first line
ap + hq + gt=X, hp + bq +ft=v, gp +fq + ct=v.
Solving these equations for p, q, t,
p=(AX+Hfi+Gv)/ A , q=(HX+B,x+Fv)/A,
t=(GX+Fv+Cv)/A, where A--=abc+2fgh-af*-bg*-cJi 2 .
For the second line,
ap'+hq'+gf+m=X, hp'+bq'+ft'+m=p., gp'+fq' + d'+m=v.
.-. p'=lAX+H r x+Gv-m(A+E+Q)y a .
q'=[m+Bfi+Fv-m(B:+B+F)y a .
t'=^lGl+Fn+Cv-m(G+ F+C)]/ a .
(P> Q> 0> (.P't Q'> Q are the tangential co-ordinates of the join of the poles.
Let A', B', C be the angles of the triangle of reference. The center is the
pole of the line at infinity asinA'+/?sinB'-t-j'sinO'=0. The tangential co-ordin-
ates of the center are obtained by substituting sinA', sinJB', sinO' for A, p., v in
p, q, t and are
8, =(As,inA'+HsinB'+ GsmC')/A ,
8 2 =(SsmA'+BsmB'+FsmG')/ A ,
8 3 =( <?sinA' + FsinB' + CsinC")/ a .
.-. The tangential equation of the center is XS 1 -\-p8 i +y8 i =0.
Write a + M' for a, b-t lib' for 6, c-\-kc' for c,f+lcf for/, g-\-bg' for g, A-f-
hh' for h in AX*+BpL* + Cr*+2Ffiv + 2GrX + 2mft=0.
Then the tangential equation of the four points of intersection of 8 and 8" is
S+lc$-\-lc s S'=0 where 1c is undetermined, and
$=(bc'+b'c-2ff)W+(ca'+c'a-2gg') ! x i +(ab'+a'b-2M')r*
The condition for equal roots for Zr is $ 2 =4##', which is the equation of
the four points of intersection.
186. Proposed by J. E. HOT; Professor of Mathematics, Coronal Institute. San Marcos, Texas.
If. two sides of a triangle and its in-circle be given in position, tne envelope of its
circumcircle is a circle (Mannheim). [From Casey's Sequel to Euclid.]
Solution by 6. B. M. ZEEE, A. M, Ph. D., Professor of Chemistry and Physics in The Temple College, Phila-
Let vertex A be origin, sides b, c the axes. Then x i -\-2xyoosA+y s — bx—
cy=0 is the equation to the circumcircle. Let this equation be written
Since the sides b, c and the inscribed circle are fixed in position, the tan-
gents from A to the in-circle are constant.
.•. b-\-c— a=& constant=m....(2).
a=l/(6 8 -(-c 8 — 26ccosJ.). This in (2) gives after reduction,
m 8 +2fc(l+cosA)-2m(6+c)=:0....(3).