514] ON THE PROBLEM OF THE IN-AND-CIRCUMSCRIBED TRIANGLE.
235
Second process. Taking the form
C — I) = e = x, B = F-y\
here
No - = X + X ~ Red ->
X=Yx(X- 2) 0 - 3) Ya, = y,
and
X + X ~ 2F 2 x — 3) (X — 2) a.
There is a first-mode reduction,
aY {2 r + 28 (Z - 4) + 3* (Z - 3)},
viz. this is
aY { Z 2 — Z + 8« — 3£
+ (Z — 4) (« 2 — x + 8Z — 3£)
+ (Z — 3) ( — 9Z + 3£)},
which is
= «7 [X (x 2 — « — 6) — 4« 2 -f 12«};
and a second-mode reduction
= aYX (« —2)(x— 3).
Hence the two together are
= aY[X (2« 2 — 6«) — 4« 2 + 12«}
= 2F« (« — 3) (Z — 2) a,
whence the result is
= 2 (F 2 — F) « (« — 3) (Z — 2) a,
which agrees with that obtained above.
On account of the symmetry we must divide by 2.
Case 30. e = D = F — x, a = c = y. By reciprocation of 29.
No. =2Z (X-3)(x-2)y(y-l)B.
On account of the symmetry we must divide by 2.
Case 31. c — e = D — x, a — B = y.
x = (F- 2)«(Z - 2) («-3) Fy, x ' = Fx ( X - 2) (« - 3) F(y-2),
g=x(x-3)(X-2)F{(Y-2)y+Y(y-2)}
= 2x (x — 3) (Z — 2){yY — y—Y)F.
Case 32. F =B = a — x, D = e = y. By reciprocation of 31.
No. = 2Z(Z- 3)(x-2){yY-y- Y) c.
30—2