Advanced book on Mathematics Olympiad

Geometry and Trigonometry 651

A

D

B

C

E

FP

M N

Figure 93

A(EP D)+A(N P DC)+A(BN C)=A(EN F )+A(AMF )+A(MN BA).

AddingA(MN P )to both sides, we obtain

A(EP D)+A(DMC)+A(BN C)=A(EN F )+A(AMF )+A(AP B).

Writing the other two similar relations and then subtracting these relations two by two,

we obtain

A(AMF )=A(DMC), A(AP B)=A(EP D), A(BN C)=A(EN F ).

The equalityA(AMF )=A(DMC)implies thatMF·MA·sin∠AMF=MC·MD·

sin∠CMD, henceMF·MA=MC·MD. Similarly,BN·CN=EN·FNand

AP·BP=DP·EP. If we writeAM=a,AP=α,BN=b,BP=β,CN=c,

CM=γ,DP=d,DM=δ,EP=e,EN=η,FM=f,FN=φ,then

a

δ

=

γ

f

,

b

η

=

φ

c

,

e

β

=

α

d

.

Also, any Latin letter is smaller than the corresponding Greek letter. Hence

a

δ

=

γ

f

>

c

φ

=

η

b

>

e

β

=

α

d

>

a

δ

.

This is a contradiction. The study of the case in whichPis betweenAandMyields a

similar contradiction, sinceMis now betweenDandP, andDcan take the role ofA

above, showing that the three main diagonals must intersect.

(Revista Matematica din Timi ̧soara ̆ (Timi ̧soara Mathematics Gazette))

654.(a) Definef:Z→[ 0 , 1 ),f(x)=x

√

3 −x

√

3 . By the pigeonhole principle, there

exist distinct integersx 1 andx 2 such that|f(x 1 )−f(x 2 )|< 0 .001. Seta=|x 1 −x 2 |.

Then the distance either between(a, a

√

3 )and(a,a

√

3 )or between(a, a

√

3 )and

(a,a

√

3 + 1 )is less than 0.001. Therefore, the points( 0 , 0 ),( 2 a, 0 ),(a, a

√

3 )lie in

different disks and form an equilateral triangle.

(b) Suppose thatP′Q′R′is an equilateral triangle of sidel≤96, whose vertices

P′,Q′,R′lie in disks with centersP , Q, R, respectively. Then