233 
514] ON THE PROBLEM OF THE IN-AND-CIRCUMSCRIBED TRIANGLE. 
Case 18. a = D — x, c = B = y. 
X = 7(y - 2) XeFx, x = FeXy (F-2) a? (= %), 
g = 2xX (Fy — Y — y) eF. 
Case 19. c = F — x, e = B = y. 
X = YxDyXa, x = XyDxYa (= X )> 
g = 2xyXYaD. 
Case 20. c = I) = x, e = F=y. 
x = Bx (X — 2)y (Y — 2) a, % '= F(y - 2) X (x - 2) Ba, 
g = {xy (X - 2) (Y- 2) + XY(x- 2) (y- 2)j aB 
= 2 [xyXY - xy (X + Y) - XY(x + y) + 2xy + 2XY} aB. 
Case 21. c— B = x, e = F = y. 
X = X (x - 2) By (F - 2) a, %' = F(y — 2) Dx (X - 2) a, 
g = {X (F- 2) y (* - 2) + F (X - 2) a (y - 2)} aD 
= 2 [xyXY- xy (X + F) - XY {x + y) + 2 xY + 2yX} aD. 
Case 22. a = D = x, c = F = y, e = B — z. 
X = ZyXzYx, x = YzXyZat (= %), 
g = 2 xyzXYZ. 
Case 23. a = B = x, c = D = y, e = F = z. 
X = (X-2)y(Y-2)z(Z-2)x, x ' = % (* - 2) F(y-2) X{x-2), 
g = xyz{X — 2)(F — 2)(F— 2) + XF£(# — 2) (y — 2) (2 — 2) 
= 2 [xyzXFF — xyz (FF + ZX + XY) — XYZ(yz + zx + xy) 
+ 2xyz {X + F+ Z) -f 2XYZ (x + y + z) — 4icy£ — 4>XYZj. 
Case 24. a = D = x, c = B = y, e — F— z. 
X= F(y — 2) Xz(Z— 2) x, x' = F(s-2)Xy(F-2)dj, 
g = aX {F(F- 2) ^ (y — 2) + F(F— 2)y (z— 2)} 
= 2a?X [yzYZ — yz (Y + Z)— YZ(y + z)+ 2yZ + 2zYj. 
Case 25. a — c = x, D — F = y, e = B = z. 
X = Z(x-l)Yz(Y-l)x, x= 7z(Y-l)xZ(x-l)(=x), 
g = 2x (x — 1) Y(Y — l)zZ. 
But we have here eB as an axis of symmetry, so that each triangle is counted 
twice, or the number of distinct triangles is = Jg. 
C. VIII. 
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