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The only possibility is that C is the smallest side in the first triangle, and D the largest one in the second, and then K(C,A,B)=(A,B,D), so we get A=CK, B=AK, D=BK, and so D=CK³. As sides are integers, K must be a rational = M/N, and CM³/N³ must be an integer. The sides C, A, B, D are then C, CM/N, CM²/N², CM³/N³. Obviously N³ must divide C, so we can write C=EN³ and the sides are now EN³, EMN², EM²N, EM³. We have D-C=20141, so E(M³-N³)=20141. Factoring, E(M-N)(M²+MN+N²)=11x1831, and the only way this works is with E=1, M-N=11 and M²+MN+N²=1831 (so M=30 and N=19) and the triangles are (6859, 10830, 17100) and (10830, 17100, 27000). |
