Advanced book on Mathematics Olympiad

Geometry and Trigonometry 637

Figure 84

parallel cutting plane passing through the origin. Because of the conditiona>b>c,a

circular cross-section cannot lie in thexy-,xz-, oryz-plane. Looking at the intersection

of the ellipsoid with theyz-plane, we see that some diameter of the circular cross-section

is a diameter (segment passing through the center) of the ellipsex=0,y

2

b^2 +

z^2

c^2 =1.

Hence the radius of the circle is at mostb. The same argument for thexy-plane shows

that the radius is at leastb, whencebis a good candidate for the maximum radius.

To show that circular cross-sections of radiusbactually exist, consider the intersection

of the plane(c

√

a^2 −b^2 )x=(a

√

b^2 −c^2 )zwith the ellipsoid. We want to compute the

distance from a point(x 0 ,y 0 ,z 0 )on this intersection to the origin. From the equation of

the plane, we obtain by squaring

x 02 +z^20 =b^2

(

x 02

a^2

+

z^20

c^2

)

.

The equation of the ellipsoid gives

y 02 =b^2

(

1 −

x 02

a^2

−

z^20

c^2

)

.

Adding these two, we obtainx 02 +y 02 +z^20 =1; hence(x 0 ,y 0 ,z 0 )lies on the circle of radius
1 centered at the origin and contained in the plane(c

√

a^2 −b^2 )x+(a

√

b^2 −c^2 )z=0.

This completes the proof.

(31st W.L. Putnam Mathematical Competition, 1970)

630.Without loss of generality, we may assumea<b<c. Fix a point(x 0 ,y 0 ,z 0 ), and

let us examine the equation inλ,

f (λ)=

x 02

a^2 −λ

+

y 02

b^2 −λ

+

z^20

c^2 −λ

− 1 = 0.

For the functionf (λ)we have the following table of signs:

f(−∞)f(a^2 −)

++

∣∣

∣∣f(a

(^2) +) f (b (^2) −)
−+
∣∣
∣∣f(b
(^2) +) f (c (^2) −)
−+
∣∣
∣∣f(c
(^2) +) f (+∞)
−−,