Challenging Mathematical Problems with Elementary Solutions: Combinatorial analysis and probability theoryHolden-Day, 1964 |
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第 1 到 3 筆結果,共 12 筆
第 40 頁
... equidistant from A , B , C , D with A , B , C on one side of it and D on the other side . By the same reasoning there is exactly one plane equidistant from A , B , C , D with C ( or B or A ) on one side of it and the other three points ...
... equidistant from A , B , C , D with A , B , C on one side of it and D on the other side . By the same reasoning there is exactly one plane equidistant from A , B , C , D with C ( or B or A ) on one side of it and the other three points ...
第 40 頁
... equidistant from A , B , C , D with A , B , C on one side of it and D on the other side . By the same reasoning there is exactly one plane equidistant from A , B , C , D with C ( or B or A ) on one side of it and the other three points ...
... equidistant from A , B , C , D with A , B , C on one side of it and D on the other side . By the same reasoning there is exactly one plane equidistant from A , B , C , D with C ( or B or A ) on one side of it and the other three points ...
第 46 頁
... equidistant from the five points . = 15 spheres and 5. The problem amounts to determining how many points ( the centers of the required spheres ) are equidistant from the four faces of the tetrahedron . The locus of all points equidistant ...
... equidistant from the five points . = 15 spheres and 5. The problem amounts to determining how many points ( the centers of the required spheres ) are equidistant from the four faces of the tetrahedron . The locus of all points equidistant ...
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常見字詞
A₁ A₂ An+m arrangements b₁ B₂ binomial coefficients binomial theorem bishops black squares C₁ chessboard chord circle coefficient color column compute the number Consequently consider corresponding customers denote determine the number diagonals digits dihedral angle divided divisible draw equally likely possible equation equidistant equivalence classes exactly example experiment favorable outcomes follows formula G₁ given Hence inclusion and exclusion intersection k-gons knights length mathematical induction maximum number n-gon number of different number of favorable number of paths number of shortest obtain pairs partition passengers plane polygons positive integers possible outcomes Pr{E probability theory problem 54 prove queens rectangle relatively prime remaining required probability rooks S₁ segment selected at random sequence shortest paths side solution to problem solved sphere square controlled Suppose T₂ total number triangle unfavorable values vertex vertices