Advanced book on Mathematics Olympiad

344 Methods of Proof

D(O 1 ,r 1 ), D(O 2 ,r 2 ),...,D(On,rn),

in decreasing order of their radii.

Choose the diskD(O 1 ,r 1 ), and then delete all disks that lie entirely inside the disk

of centerO 1 and radius 3r 1. The remaining disks are disjoint fromD(O 1 ,r 1 ). Among

them choose the first in line (i.e., the one with maximal radius), and continue the process

with the remaining circles.

The process ends after finitely many steps. At each step we deleted less than eight

times the area of the chosen circle, so in the end we are left with at least^19 of the initial

area. The chosen circles satisfy the desired conditions.

(M. Pimsner, S. Popa,Probleme de geometrie elementara ̆(Problems in elementary

geometry), Editura Didactica ̧ ̆si Pedagogica, Bucharest, 1979) ̆

54.Given a circle of radiusrcontainingnpoints of integer coordinates, we must prove

thatn< 2 π^3

√

r^2. Becauser>1 and 2π>6 we may assumen≥7.

Label thenlattice points counterclockwiseP 1 ,P 2 ,...,Pn. The (counterclockwise)

arcs



P 1 P 3 ,



P 2 P 4 ,...,



PnP 2 cover the circle twice, so they sum up to 4π. Therefore, one

of them, say



P 1 P 3 , measures at most^4 nπ.

Consider the triangleP 1 P 2 P 3 , which is inscribed in an arc of measure^4 nπ. Because

n≥7, the arc is less than a quarter of the circle. The area ofP 1 P 2 P 3 will be maximized

ifP 1 andP 3 are the endpoints andP 2 is the midpoint of the arc. In that case,

Area(P 1 P 2 P 3 )=

abc

4 r

=

2 rsinπn· 2 rsinπn· 2 rsin^2 nπ

4 r

≤

2 rπn· 2 rπn· 2 r^2 nπ

4 r

=

4 r^2 π^3

n^3

.

And in general, the area ofP 1 P 2 P 3 cannot exceed^4 r

(^2) π 3
n^3. On the other hand, if the
coordinates of the pointsP 1 ,P 2 ,P 3 are, respectively,(x 1 ,y 1 ),(x 2 ,y 2 ), and(x 3 ,y 3 ), then
Area(P 1 P 2 P 3 )=±

1

2

∣∣

∣

∣∣

∣

111

x 1 x 2 x 3

y 1 y 2 y 3

∣∣

∣

∣∣

∣

=

1

2

|x 1 y 2 −x 2 y 1 +x 2 y 3 −x 3 y 2 +x 3 y 1 −x 1 y 3 |.

Because the coordinates are integers, the area cannot be less than^12. We obtain the

inequality^12 ≤^4 r

(^2) π 3
n^3 , which proves that 2π
√ (^3) r (^2) ≥n, as desired.
Remark.The weaker inequalityn(r) < 63

√

πr^2 was given in 1999 at the Iranian Mathe-

matical Olympiad.

55.Order the eight integersa 1 <a 2 <···<a 8 ≤2004. We argue by contradiction.
Assume that for any choice of the integersa, b, c, d, eithera+b+c<d+4or