642 Geometry and Trigonometry
An=(
1
n,
1
n− 1·c 1 ,...,1
3
·cn− 3 ,1
2
·cn− 2 ,−cn− 1)
,
An+ 1 =(
1
n,
1
n− 1·c 1 ,...,1
3
·cn− 3 ,1
2
·cn− 2 ,cn− 1)
,
where
ck=√(
1 +
1
n)(
1 −
1
n−k+ 1)
,k= 1 , 2 ,...,n− 1.One computes that the distance between any two points is
√
2√
1 +
1
n>
√
2 ,
and the problem is solved.
(8th International Mathematics Competition for University Students, 2001)
636.View the ring as the body obtained by revolving about thex-axis the surface that
lies between the graphs off, g :[−h/ 2 ,h/ 2 ]→R,f(x)=
√
√ R^2 −x^2 ,g(x) =
R^2 −h^2 /4. HereRdenotes the radius of the sphere. Using the washer method we find
that the volume of the ring is
π∫h/ 2−h/ 2(
√
R^2 −x^2 )^2 −(√
R^2 −h^2 / 4 )^2 dx=π∫h/ 2−h/ 2(h^2 / 4 −x^2 )dx=h^3 π
12,
which does not depend onR.
637.Let the inscribed sphere have radiusRand centerO. For each big face of the
polyhedron, project the sphere onto the face to obtain a diskD. Then connectDwith
Oto form a cone. Because the interiors of the cones are pairwise disjoint, the cones
intersect the sphere in several nonoverlapping regions. Each circular region is a slice
of the sphere, of widthR( 1 −^12
√
2 ). Recall the lemma used in the solution to the first
problem from the introduction. We apply it to the particular case in which one of the
planes is tangent to the sphere to find that the area of a slice is 2πR^2 ( 1 −^12
√
2 ), and this
is greater than^17 of the sphere’s surface. Thus each circular region takes up more than^17
of the total surface area of the sphere. So there can be at most six big faces.
(Russian Mathematical Olympiad, 1999)
638.Keep the line of projection fixed, for example thex-axis, and rotate the segments in
AandBsimultaneously.
Now, given a segment with one endpoint at the origin, the length of its projection
onto thez-axis isr|cosφ|, where(r,θ,φ)are the spherical coordinates of the second
endpoint, i.e.,ris the length of the segment,φis the angle it makes with the semiaxis