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PROBLEMS AND SOLUTIONS. 
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{x (y 3 — z 3 ), y(z 3 —x 3 ), z(x 3 — y 3 )]. The coordinates of the same point may be otherwise 
found by a direct investigation, as follows: Write 
x 2 : y 2 : z 2 = x (y 3 — z 3 ) : у (z 3 — ж 3 ) : z (x 3 — y 3 ); x x : y x : z x — у : z : x. 
If in the equation of the curve we substitute for x, y, z, the values ux x + vx 2 , 
uy x + vy 2 , uz x + vz. 2 , we find 
и [x x 2 x 2 + y 2 y 2 + z x % + 21 {x,y x z x + y 2 z x x x + z 2 x x y x )] 
+ v [x x xf + y x yi + z x z 2 2 + 21 (x x y,z 2 + y x z 2 x 2 + z x x 2 y$\ = 0, 
say uP 4- vQ — 0 ; we may therefore write u = Q, v = — P, and the coordinates of the 
third point are Qx x — Px 2 , Qy x — Py 2 , Qz x — Pz 2 . Now 
Qx x - Px, = y x y 2 (x x y 2 - x 2 y x ) + z x z 2 (x x z 2 - x 2 z x ) + 21 (x x -y 2 z, - x.?y x z x ) 
= yz (z 3 — oc 3 ) \y 2 (z 3 — x 3 ) — zx (y 3 — z 3 )} 
+ zx(x 3 — y 3 ) {yz (X 3 — y 3 ) — X 2 (y 3 — z 3 )} 
+ 21 [y 2 . yz . (z 3 — cc 3 ) (x 3 — y 3 ) — x 2 (y 3 — z 3 ) 2 zx]. 
= (ос?у fi + y 3 z 6 + z 3 x s — 3a?y 3 z 3 ') z 
+ xyz ( x (i + y n + z c ‘ — y 3 z 3 — z 3 x 3 — a?y 3 ) z 
— 21 (x s y 3 + y 6 z 3 + z e P — Sx 3 y 3 z 3 ) z; 
so that we have Qx x — Px 2 — Hz ; and in like manner Qy x — Py 2 = Tlx, Qz x — Pz 2 = IIу; 
and therefore Qx x — Px 2 : Qy x — Qy 2 : Qz x — Pz 2 = z : x : у, which proves the theorem. 
I consider in like manner the following question; viz., if (y, x, z) be joined with 
the tangential of (x, y, z); to find the third point of intersection. We have here 
x 2 : y 2 : z 2 = x(y 3 — z 3 ) : у (z 3 — a?) : z (x 3 — y 3 ); x x : y x : z x = y : x z\ 
and P, Q as before ; and the coordinates of the third point are 
Qxi Px 2 : Qy x Py 2 : Qz x Pz 2 ] 
also 
Qx x — Px 2 — xу (z 3 — x 3 ) [y 2 (z 3 — X s )— ж 2 (y 3 — z 3 )] 
+ Z 2 (X 3 — y 3 ) {yz {x 3 — y 3 ) — zx (y 3 — z 3 )] 
+ 21 {y 3 z (z 3 — x 3 ) (ж 3 — у 3 ) — x 2 {у 3 — z 3 ) 2 zx], 
= x {у 3 {z 3 — x 3 ) 2 — z 3 (x 3 — y 3 ) (y 3 — z 3 )] 
+ у {z 3 (ж 3 — у 3 ) 2 — ж 3 (у 3 — z 3 ) (z 3 — ж 3 )} 
+ 2Iz {у 3 (z 3 — ж 3 ) (ж 3 — у 3 ) — ж 3 (у 3 — z 3 ) 2 ], 
that is Qx x — Рх 2 = (ж + у — 21 z) (x 3 z r ' + у 3 ж 6 + z 3 y e — 3x?y 3 z 3 ); 
similarly Qy x — Py 2 = (ж + у — 2 Iz) (y 3 z 3 + y 6 z 3 + z 3 x 6 — 3oc?y 3 z 3 ); 
also Qz x — Pz 2 ={x + у — 2lz) (ж 6 + у 6 + z 6 — y 3 z 3 — Py 3 ) xyz ; 
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