Advanced book on Mathematics Olympiad

4.1 Geometry 227

cosφ 1 dx−cosφ 1 d 1 −(x− 1 )sinφ 1 dφ 1 +sinφ 1 dy−sinφ 1 dη 1

+(y−η 1 )cosφ 1 dφ 1 = 0.

This expression can be simplified if we note thatdηd 11 is the slope of the tangent, namely

tan(φ 1 +π 2 ). Then cosφ 1 d 1 +sinφ 1 dη 1 =0, so

cosφ 1 dx+sinφ 1 dy=[(x− 1 )sinφ 1 −(y−η 1 )cosφ 1 ]dφ 1.

And now a little Euclidean geometry. Consider the right triangleO 1 A 1 P with legs
parallel to the axes. The altitude fromO 1 determines onA 1 Ptwo segments of lengths
(x− 1 )sinφ 1 and−(y−η 1 )cosφ 1 (you can see by examining the picture that the signs
are right). This allows us to further transform the identity obtained above into

cosφ 1 dx+sinφ 1 dy=t 1 dφ 1.

The same argument shows that

cosφ 2 dx+sinφ 2 dy=t 2 dφ 2.

The Jacobian of the transformation is therefore the absolute value of

1

t 1 t 2

(cosφ 1 sinφ 2 −sinφ 1 cosφ 2 )=

1

t 1 t 2

sin(φ 1 −φ 2 ).

Andφ 1 −φ 2 is, up to a sign, the supplement ofα. We obtain

2 π^2 =

1

2

∫ 2 π

0

∫ 2 π

0

dφ 1 dφ 2 =

∫∫

P/∈D

sinα

t 1 t 2

dxdy.

The theorem is proved. 

636.A ring of heighthis obtained by digging a cylindrical hole through the center of a

sphere. Prove that the volume of the ring depends only onhand not on the radius

of the sphere.

637.A polyhedron is circumscribed about a sphere. We call a facebigif the projection

of the sphere onto the plane of the face lies entirely within the face. Show that there

are at most six big faces.

638.LetAandBbe two finite sets of segments in three-dimensional space such that the

sum of the lengths of the segments inAis larger than the sum of the lengths of the

segments inB. Prove that there is a line in space with the property that the sum of

the lengths of the projections of the segments inAonto that line is greater than the

sum of the lengths of the projections of the segments inB.