Advanced book on Mathematics Olympiad

224 4 Geometry and Trigonometry

sphere isR>0. The fixed point, which we callP, becomes the origin. The endpoints
of each chord have only one nonzero coordinate, and in the appropriate ordering, thekth
coordinates of the endpointsXkandYkof thekth chord are the nonzero numbersxkand
yk,k= 1 , 2 ,...,n. The center of the sphere is then

O=

(

x 1 +y 1

2

,

x 2 +y 2

2

,...,

xn+yn

2

)

.

The conditions that the pointsXkandYklie on the sphere can be written as

(

xk−

xk+yk

2

) 2

+

∑

j   =k

(

xj+yj

2

) 2

=R^2 ,

(

yk−

xk+yk

2

) 2

+

∑

j   =k

(

xj+yj

2

) 2

=R^2 ,

withk= 1 , 2 ,...,n. This implies

(

xk−yk

2

) 2

=R^2 −

∑

j   =k

(

xj+yj

2

) 2

,k= 1 , 2 ,...,n.

The term on the left is one-fourth of the square of the length ofXkYk. Multiplying by 4
and summing up all these relations, we obtain

∑n

k= 1

‖XkYk‖^2 = 4 nR^2 − 4

∑n

k= 1

∑

j   =k

(

xj+yj

2

) 2

= 4 nR^2 − 4 (n− 1 )

∑n

k= 1

(

xk+yk

2

) 2

= 4 nR^2 − 4 (n− 1 )‖PO‖^2.

Hence the conclusion. 

632.Letnbe a positive integer. Prove that if the vertices of a( 2 n+ 1 )-dimensional cube
have integer coordinates, then the length of the edge of the cube is an integer.

633.For a positive integerndenote byτthe permutation cycle(n,..., 2 , 1 ). Consider
the locus of points inRndefined by the equation
∑

σ

sign(σ )xσ( 1 )xτ(σ( 2 ))···xτn− (^1) (σ (n))= 0 ,
where the sum is over all possible permutations of{ 1 , 2 ,...,n}. Prove that this
locus contains a plane.