6. Among the 8 vertices, 12 midpoints of edges, 6 centers of faces, and 1 center of the cube, these 27 points, the number of groups of 8 points that lie on a common sphere is:
Pick one
Solution
6.B.
Place the cube in a spatial rectangular coordinate system, with the center of the cube as the origin, and the 8 vertices are (2,2,2), (−2,−2,−2), (2,2,−2), (2,−2,−2), where (a,b,c) represents (a,b,c) and its permutations.
The planes z=0, z=±2 divide the 27 points into three layers. For each group of 8 points on a common sphere, there are at most 4 points on any layer (since there are no 5 points on a circle among the 9 points on each layer), and at least 3 points on a certain layer. The circumcenter (x,y) of these 3 points can only be at (0,0); (1,1),(−1,−1),(1,−1); (0,2),(0,−2); (0,1),(0,−1); (0,21),0,−21); (1,3),(1,−3),(−1,3),(−1,−3); (31,31),(−31,−31),(31,−31). Therefore, the center of the sphere for 8 points on a common sphere can only be at (1) (0,0,0); (2) (1,1,1),(−1,−1,−1), (1,1,−1),(1,−1,−1); (3) (0,0,2),(0,0,−2); (4) (0,0,1),(0,0,−1); (5) 0,1,1),(0,1,−1),(0,−1,−1); (6) (0,0,21),0,0,−21); (7) (1,1,3),(1,1,−3),(1,−1,3), (1,−1,−3),(−1,−1,3),(−1,−1,−3); (8) (31,31,31),(−31,−31,−31), (31,31,−31),(31,−31,−31). Below, we denote the sphere with center X and radius r as S(X,r). The number of points among the 27 points on this sphere is ∣S(X,r)∣. (1) A(0,0,0). ∣S(A,2)∣=6,∣S(A,22)∣=12, ∣S(A,23)∣=8; (2) B(1,1,1). ∣S(B,3)∣=8,∣S(B,11)∣=12; (3) C(0,0,2). ∣S(C,2)∣=5,∣S(C,22)∣=8, ∣S(C,23)∣=∣S(C,25)∣ =∣S(C,26)∣=4; (4) D(0,0,1). ∣S(D,1)∣=2,∣S(D,5)∣=8, ∣S(D,3)∣=9, ∣S(D,13)∣=∣S(D,17)∣=4; (5) E(1,1,0). ∣S(E,2)∣=∣S(E,10)∣=4, ∣S(E,6)∣=∣S(E,14)∣=8; (6) F(0,0,21). s(F,21)=S(F,23)=1,S(F,217)=S(F,233)=4,S(F,25)=5,S(F,241)=8; (7) G(1,1,3). ∣S(G,3)∣=4,∣S(G,11)∣=8, ∣S(G,19)∣=∣S(G,33)∣=5; (8) H(31,31,31). S(H,33)=1,∣S(H,3)∣=3,S(H,351)∣=∣S(H,11)=6,S(H,353)=7.
In summary, the number of groups of 8 points on a common sphere is (Cn8+1)+8(1+C128)+6+6(1+C98)+2(1+1)+6+24=4584 (groups).
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