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Geometry Difficulty 5.2 AIME, harder Find the answer

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), (2,2,2)(2,2,-2), (2,2,2)(2,-2,-2), where (a,b,c)(a, b, c) represents (a,b,c)(a, b, c) and its permutations.

The planes z=0z=0, z=±2z= \pm 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)(x, y) of these 3 points can only be at
(0,0)(0,0); (1,1),(1,1),(1,1)(1,1),(-1,-1),(\overline{1,-1});
(0,2),(0,2)(\overline{0,2}),(\overline{0,-2}); (0,1),(0,1)(\overline{0,1}),(\overline{0,-1});
(0,12),0,12)\left.\left(\overline{0, \frac{1}{2}}\right), \overline{0,-\frac{1}{2}}\right);
(1,3),(1,3),(1,3),(1,3)(\overline{1,3}),(\overline{1,-3}),(\overline{-1,3}),(\overline{-1,-3});
(13,13),(13,13),(13,13)\left(\frac{1}{3}, \frac{1}{3}\right),\left(-\frac{1}{3},-\frac{1}{3}\right),\left(\frac{1}{3},-\frac{1}{3}\right).
Therefore, the center of the sphere for 8 points on a common sphere can only be at
(1) (0,0,0)(0,0,0);
(2) (1,1,1),(1,1,1)(1,1,1),(-1,-1,-1),
(1,1,1),(1,1,1)(\overline{1,1,-1}),(\overline{1,-1,-1});
(3) (0,0,2),(0,0,2)(\overline{0,0,2}),(\overline{0,0,-2});
(4) (0,0,1),(0,0,1)(\overline{0,0,1}),(\overline{0,0,-1});
(5) 0,1,1),(0,1,1),(0,1,1)\overline{0,1,1}),(\overline{0,1,-1}),(\overline{0,-1,-1});
(6) (0,0,12),0,0,12)\left.\left(\overline{0,0, \frac{1}{2}}\right), \overline{0,0,-\frac{1}{2}}\right);
(7) (1,1,3),(1,1,3),(1,1,3)(\overline{1,1,3}),(\overline{1,1,-3}),(\overline{1,-1,3}),
(1,1,3),(1,1,3),(1,1,3)(\overline{1,-1,-3}),(\overline{-1,-1,3}),(-1,-1,-3);
(8) (13,13,13),(13,13,13)\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right),\left(-\frac{1}{3},-\frac{1}{3},-\frac{1}{3}\right),
(13,13,13),(13,13,13)\left(\overline{\frac{1}{3}, \frac{1}{3},-\frac{1}{3}}\right),\left(\overline{\frac{1}{3},-\frac{1}{3},-\frac{1}{3}}\right).
Below, we denote the sphere with center XX and radius rr as S(X,r)S(X, r). The number of points among the 27 points on this sphere is S(X,r)|S(X, r)|.
(1) A(0,0,0)A(0,0,0).
S(A,2)=6,S(A,22)=12|S(A, 2)|=6,|S(A, 2 \sqrt{2})|=12,
S(A,23)=8|S(A, 2 \sqrt{3})|=8;
(2) B(1,1,1)B(1,1,1).
S(B,3)=8,S(B,11)=12|S(B, \sqrt{3})|=8,|S(B, \sqrt{11})|=12;
(3) C(0,0,2)C(0,0,2).
S(C,2)=5,S(C,22)=8|S(C, 2)|=5,|S(C, 2 \sqrt{2})|=8,
S(C,23)=S(C,25)|S(C, 2 \sqrt{3})|=|S(C, 2 \sqrt{5})|
=S(C,26)=4=|S(C, 2 \sqrt{6})|=4;
(4) D(0,0,1)D(0,0,1).
S(D,1)=2,S(D,5)=8|S(D, 1)|=2,|S(D, \sqrt{5})|=8,
S(D,3)=9|S(D, 3)|=9,
S(D,13)=S(D,17)=4|S(D, \sqrt{13})|=|S(D, \sqrt{17})|=4;
(5) E(1,1,0)E(1,1,0).
S(E,2)=S(E,10)=4|S(E, \sqrt{2})|=|S(E, \sqrt{10})|=4,
S(E,6)=S(E,14)=8|S(E, \sqrt{6})|=|S(E, \sqrt{14})|=8;
(6) F(0,0,12)F\left(0,0, \frac{1}{2}\right).
s(F,12)=S(F,32)=1,S(F,172)=S(F,332)=4,S(F,52)=5,S(F,412)=8; \begin{array}{l} \left|s\left(F, \frac{1}{2}\right)\right|=\left|S\left(F, \frac{3}{2}\right)\right|=1, \\ \left|S\left(F, \frac{\sqrt{17}}{2}\right)\right|=\left|S\left(F, \frac{\sqrt{33}}{2}\right)\right|=4, \\ \left|S\left(F, \frac{5}{2}\right)\right|=5,\left|S\left(F, \frac{\sqrt{41}}{2}\right)\right|=8 ; \end{array}
(7) G(1,1,3)G(1,1,3).
S(G,3)=4,S(G,11)=8|S(G, \sqrt{3})|=4,|S(G, \sqrt{11})|=8,
S(G,19)=S(G,33)=5 |S(G, \sqrt{19})|=|S(G, 3 \sqrt{3})|=5 \text {; }
(8) H(13,13,13)H\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right).
S(H,33)=1,S(H,3)=3,S(H,513)=S(H,11)=6,S(H,533)=7. \begin{array}{l} \left|S\left(H, \frac{\sqrt{3}}{3}\right)\right|=1,|S(H, \sqrt{3})|=3, \\ \left.S\left(H, \frac{\sqrt{51}}{3}\right)|=| S(H, \sqrt{11}) \right\rvert\,=6, \\ \left.S\left(H, \frac{5 \sqrt{3}}{3}\right) \right\rvert\,=7 . \end{array}

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).  \begin{array}{l} \left(\mathrm{C}_{n}^{8}+1\right)+8\left(1+\mathrm{C}_{12}^{8}\right)+6+6\left(1+\mathrm{C}_{9}^{8}\right)+ \\ \quad 2(1+1)+6+24 \\ =4584 \text { (groups). } \end{array}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.