Advanced book on Mathematics Olympiad

636 Geometry and Trigonometry

A glimpse at these formulas suggests the following computation:

x^2 +y^2 −

1

3

z^2

=t^2 cos^2 θ+sin^2 θ+ 2 tsinθcosθ+cos^2 θ+t^2 sin^2 θ− 2 tcosθsinθ−t^2

=t^2 (cos^2 θ+sin^2 θ)+cos^2 θ+sin^2 θ−t^2 = 1.

The locus is therefore a hyperboloid of one sheet,x^2 +y^2 −^13 z^3 =1.

Remark.The fact that the hyperboloid of one sheet is a ruled surface makes it easy to

build. It is a more resilient structure than the cylinder. This is why the cooling towers of

power plants are built as hyperboloids of one sheet.

627.The equation of the plane tangent to the hyperboloid at a pointM(x 0 ,y 0 ,z 0 )is

x 0 x

a^2

+

y 0 y

b^2

−

z 0 z

c^2

= 1.

This plane coincides with the one from the statement if and only if

x 0

a^2

1

a

=

y 0

b^2

1

b

=

z 0

c^2

1

c

.

We deduce that the point of contact has coordinates(a,b,c), and therefore the given

plane is indeed tangent to the hyperboloid.

628.The area of the ellipse given by the equation

x^2

A^2

+

y^2

B^2

=R^2

isπABR^2. The section perpendicular to thex-axis is the ellipse

y^2

b^2

+

z^2

c^2

= 1 −

x 02

a^2

in the planex =x 0. HenceSx =πbc( 1 −x

02

a^2 ). Similarly,Sy =πac(^1 −

y^20

b^2 )and

Sx=πab( 1 −z

(^20)
c^2 ). We thus have
aSx+bSy+cSz=πabc

(

3 −

x^20

a^2

+

y 02

b^2

+

z 02

c^2

)

= 2 πabc,

which, of course, is independent ofM.

629.Figure 84 describes a generic ellipsoid. Since parallel cross-sections of the ellipsoid

are always similar ellipses, any circular cross-section can be increased in size by taking a