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.-. 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* 


+ 2(fg'+fg-ch'-e'h)X,i. 

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- 
delphia, Pa. 

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). 


c from (1) in (3) gives 

2b' ! x(l+eosA)+2l[my-mx-D(l+QOsA')~\+2Dm-m 2 y=0. 

The condition for equal roots of b is 

2ic(l+cos J 4)(2Dm— m 2 y)=\my— mx— D(l+cosJL)] s 

or [Z>(l + cosA)— m(# + y)] 2 =2»» 2 a;#(l— cosA)=4m s ;n/sin 2 £.A. 

.-. D(l+cosA)— m(x-\-y) ±2m,}/ (xy)$,m^A=Q. 

.-. x*+2xy<nsA+y*-2^^-j[x+y±2(xy)smbA-\=0. 

.-. x 2 +2xy+y* -4xysm^A-2^rj[x+y±2 ] /(xy)siniA']=0. 
.-. lx±2^/(xy)smiA+y][xT2 l /(xy)sinlA -+y - 2eQg2 ^ ]=0- 

.-. xT2 v / (xy)siniA + y-^^j=0, 

or a; 8 + 2xyaosA-\-y 2 -\-\_m — 4m(a: + #)cos 2 £.A"|/4cos 4 £.l=0. 
This is the circle. 


144. Proposed by 6. B. M. ZERE. A. M„ Ph. D„ Professor of Chemistry and Physics, The Temple College, 
Philadelphia, Pa. 

Find the volume of the sphere, x % + y 2 + z 2 =2az, (a) within the paraboloid 
z=Ax 2 +By 2 ; (6) within the cone z 2 =Ax 2 +By 2 . 

Solution by the PROPOSER. 

a; 2 +# 2 +z 8 =2az....(l), z=Ax 2 +By 2 ....(2), z 2 =Ax 2 +By i ....(Z). 
From (1), z=a± \/ {a 1 -x s -y 2 }=a ± y {a 2 — r 2 }. 

From v= I I zrdrdO we get 

v^iC" f y (a* -r 2 )d$rdr=% f V'-Ca 8 -*")* ]d0. 

J o J «^ 

(a). From (1) and (2), 

x 2 +y 2 + (Ax 2 +By* ) s =2a( Ax 2 +By 2 ) . 
r s +r 4 (Acos s + Bsin 8 0) 8 =2ar s Gicos 2 0+£sin 2 0).