232 
ON THE PROBLEM OF THE IN-AND-CIRCUMSCRIBED TRIANGLE. 
[514 
Case 8. a = c— B = x. 
X = (X - 2)(x — 3) DeFx, %' = FeDx (X -2)(es- 3) (= *), 
g = lx (x — 3) (X — 2) g_DX. 
Case 9. D = F — e = x. By reciprocation of 8. 
No. =2X(X- 3) (x - 2) acB. 
Case 10. a = c = D = x. 
X = B (x — 1) (X — 2) eFx, x — FeX (x—2)B(x — 1), 
g = 2 (x — 1) (Xx — X — x) eBF. 
Case 11. D = F = a—x. By reciprocation of 10. 
No. = 2 (X — 1) (Xx — X — x) ceB. 
Second process : form a = B = D = x. 
X = (X - 2) c (X - 1) eFx, x ' = FeX <> (X-l)(x- 2), 
giving the former result. 
Case 12. c — e — x, a = D = y. 
X = BxY(x-l)Fy, x = FxY ( x ~ 1) % (= X)> 
g = 2x (x — 1 )y YBF. 
Case 13. F = B = x, a = D — y. By reciprocation of 12. 
No. =2X(X-1 )Yyce. 
Case 14. c — e — x, a—B = y. 
X = (Y-2)xD(x-l)Fy, x ~ FxF ( x — ^) Y(y — 2), 
g = 2x (x — 1) (Yy — Y — y) DF. 
Case 15. F = B — x, D = e = y. By reciprocation of 14. 
No. = 2X (X -l)(Yy - Y - y) ac. 
Case 16. c = e = x, D = F= y. 
X = BxY (x— 1) (F— 1) a, Tt = Yx(Y-l)(a>-l )Ba(= x ), 
g = 2x (x — 1) Y (Y — 1) aB. 
Case 17. c = e — x, B = F=y. 
X = D(x- 1) Ya(Y-l)x, %'= Ya(Y-l)xD(x — 1)(= %), 
g = 2x (x— 1) Y (Y — 1) aD. 
But we have here aD as an axis of symmetry, so that each triangle is counted 
twice, or the number of distinct triangles is =^g.