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157. Proposed by 6. B. M. ZEEE, A. M., Fb. D., Professor of Chemistry and Physics, The Temple College,
Solve the equations
-?— +— y —+^—+— ?L_=i x i y i g i w =1
(1+1 &+* c+r d+X ' a+n^ b+p-^ c+fi d+fi '
a + i' 6+y c-\-v d-\-v ' a-\-p b+p c+p d-\-p
I. Solution by L. E, DICKSON, A. M.. Ph. D., Assistant Professor of Mathematics, University of Chicago,
In view of the given equations, the following equation in t,
a+t ' b+t ' c+t d+t
has the roots X, /*, v, p. Setting f(t)~(t+a)(t+ V) (tf+c)(tf+ri!), we have
*/«(0 + ^(0+*/o(0+«/<*(0=/(0,
where f a (t) denotes the partial derivative of /(<) with respect to a. Hence
is an identity in t. To obtain the value of x, set t=— a. Then
-xfai-a^i-iyia+XXa+^ia + rXa+p),
( a+ A)(a+ A )(a+T')(a + / >)
The value of any other variable, say y, may be obtained similarly, or by
interchanging a with b in the expression for x, as is evident from the symmetry
This method of solution is equally simple for n such equations in n variables.
II. Solution by L. C. WALKEE, A. II., Graduate Student, Leland Stanford University, Cal.
By Art. 586 of Hall and Knight' s Higher Algebra, consider the following
equation in 0,
a + + b+d^~ c+6 d+0 O+0)(H 0)(c+0)(<H-0) '
x, y, z, u being for the present regarded as known quantities.
This equation when cleared of fractions is of the third degree in 0, and is
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.