Advanced book on Mathematics Olympiad

Geometry and Trigonometry 641

Note thatd(X,Y) >

√

2 if and only if

d^2 (X, Y )=

∑n

k= 1

x^2 k+

∑n

k= 1

y^2 k− 2

∑n

k= 1

xkyk> 2.

Therefore,d(X,Y) >

√

2 implies

∑n
k= 1 xkyk<0.
Now letA 1 ,A 2 ,...,Amnbe points satisfying the condition from the hypothesis, with
mnmaximal. Using the symmetry of the sphere we may assume thatA 1 =(− 1 , 0 ,..., 0 ).
LetAi=(x 1 ,x 2 ,...,xn)andAj=(y 1 ,y 2 ,...,yn),i, j≥2. Becaused(A 1 ,Ai)and
d(A 1 ,Aj)are both greater than

√

2, the above observation shows thatx 1 andy 1 are
positive.
The conditiond(Ai,Aj)>

√

2 implies

∑n
k= 1 xkyk<0, and sincex^1 y^1 is positive, it
follows that

∑n

k= 2

xkyk< 0.

This shows that if we normalize the lastn−1 coordinates of the pointsAiby

x′k=

xk

√∑

n− 1

k= 1 x

2

k

,k= 1 , 2 ,...,n− 1 ,

we obtain the coordinates of pointBiinSn−^2 , and the pointsB 2 ,B 3 ,...,Bnsatisfy the
condition from the statement of the problem for the unit sphere inRn−^1.
It follows thatmn≤ 1 +mn− 1 , andm 1 =2 impliesmn≤n+1. The example of
then-dimensional regular simplex inscribed in the unit sphere shows thatmn=n+1.
To determine explicitly the coordinates of the vertices, we use the additional information
that the distance from the center of the sphere to a hyperface of then-dimensional simplex
is^1 nand then find inductively

A 1 =(− 1 , 0 , 0 , 0 ,..., 0 , 0 ),

A 2 =

(

1

n

,−c 1 , 0 , 0 ,..., 0 , 0

)

,

A 3 =

(

1

n

,

1

n− 1

·c 1 ,−c 2 , 0 ,..., 0 , 0

)

,

A 4 =

(

1

n

,

1

n− 1

·c 1 ,

1

n− 2

·c 2 ,c 3 ,..., 0 , 0

)

,

···

An− 1 =

(

1

n

,

1

n− 1

·c 1 ,...,

1

3

·cn− 3 ,−cn− 2 , 0

)

,