Advanced book on Mathematics Olympiad

Geometry and Trigonometry 643

Oz, andθis the oriented angle that its projection onto thexy-plane makes withOx.If
we average the lengths of the projections onto thex-axis of the segment over all possible
rotations, we obtain

1

4 π

∫π

0

∫ 2 π

0

r|cosφ|sinφdθdφ=

r

2

.

Denote byaandbthe sums of the lengths of the segments inAandB, respectively.
Then the average of the sum of the lengths of the projections of segments inAisr 2 a, and
the average of the same sum forBisr 2 b. The second is smaller, proving that there exists
a direction such that the sum of the lengths of the projections of the segments fromA
onto that direction is larger that the corresponding sum forB.

639.This is just a two-dimensional version of the previous problem. If we integrate
the length of the projection of a segment onto a line over all directions of the line, we
obtain twice the length of the segment. Doing this for the sides of a convex polygon, we
obtain the perimeter (since the projection is double covered by the polygon). Because
the projection of the inner polygon is always smaller than the projection of the outer, the
same inequality will hold after integration. Hence the conclusion.

640.Fori= 1 , 2 ,...,n, letaibe the lengths of the segments and letφibe the angles
they make with the positivex-axis (0≤φi≤π). The length of the projection ofaionto
some line that makes an angleφwith thex-axis isfi(φ)=ai|cos(φ−φi)|; denote by
f(φ)the sum of these lengths. The integral mean offover the interval[ 0 ,π]is

1

π

∫π

0

f(φ)dφ=

1

π

∑n

i= 1

∫π

0

fi(φ)dφ

=

1

π

∑n

i= 1

ai

∫π

0

|cos(φ−φi)|dφ=

2

π

∑n

i= 1

ai=

2

π

.

Here we used the fact that|cosx|is periodic with periodπ. Since the integral mean off
isπ^2 and sincefis continuous, by the intermediate value property there exists an angle
φfor whichf(φ)=^2 π. This completes the proof.

641.The law of cosines in triangleAP Bgives

AP^2 =x^2 +c^2 − 2 xccosB

and

x^2 =c^2 +AP^2 =x^2 +c^2 − 2 xccosB− 2 c

√

x^2 +c^2 − 2 xccosBcost,

whence