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 14 筆
第 46 頁
... angle is the bisector of the dihedral angle , i.e. , the plane which divides it into two congruent dihedral angles . Since any two intersecting planes form two pairs of vertical dihedral angles , the locus of all points equidistant from ...
... angle is the bisector of the dihedral angle , i.e. , the plane which divides it into two congruent dihedral angles . Since any two intersecting planes form two pairs of vertical dihedral angles , the locus of all points equidistant from ...
第 111 頁
... angles of all the polygons into which the n - gon is divided . Since the sum of the angles of a k - gon is ( k - 2 ) 180 ° , the required sum equals • [ rs + 2r1 + 3rs + ··· + ( m − 2 ) rm ] 180 ° . On the other hand , the sum of the ...
... angles of all the polygons into which the n - gon is divided . Since the sum of the angles of a k - gon is ( k - 2 ) 180 ° , the required sum equals • [ rs + 2r1 + 3rs + ··· + ( m − 2 ) rm ] 180 ° . On the other hand , the sum of the ...
第 116 頁
... angles of the triangles is therefore equal to the sum of the angles of the n - gon , which is ( n - 2 ) 180 ° . It follows from this that k 180 ° ( n − 2 ) · 180 ° , or k = n2 . = - Hence the number of triangles into which an n - 116 ...
... angles of the triangles is therefore equal to the sum of the angles of the n - gon , which is ( n - 2 ) 180 ° . It follows from this that k 180 ° ( n − 2 ) · 180 ° , or k = n2 . = - Hence the number of triangles into which an n - 116 ...
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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