Advanced book on Mathematics Olympiad

Methods of Proof 353

When we write this number as a product of two factors, one of the factors is congruent to
1 and the other is congruent to 3 modulo 4. Adding or subtracting a 2 from each factor
produces numbers congruent to 3, respectively, 1 modulo 4. We deduce that what stays
invariant in this process is the parity of the number of numbers on the blackboard that
are congruent to 3 modulo 4. Since initially this number is equal to 1, there will always
be at least one number that is congruent to 3 modulo 4 written on the blackboard. And
this is not the case with the sequence of nines. This proves our claim.
(St. Petersburg City Mathematical Olympiad, 1997)

75.Without loss of generality, we may assume that the length of the hypotenuse is 1 and
those of the legs arepandq. In the process, we obtain homothetic triangles that are in
the ratiopmqnto the original ones, for some nonnegative integersmandn. Let us focus
on the pairs(m, n).
Each time we cut a triangle, we replace the pair(m, n)with the pairs(m+ 1 ,n)
and(m, n+ 1 ). This shows that if to the triangle corresponding to the pair(m, n)we
associate the weight 2 m^1 +n, then the sumIof all the weights is invariant under cuts. The
initial value ofIis 4. If at some stage the triangles were pairwise incongruent, then the
value ofIwould be strictly less than

∑∞

m,n= 0

1

2 m+n

=

∑∞

m= 0

1

2 m

∑∞

n= 0

1

2 n

= 4 ,

a contradiction. Hence a configuration with all triangles of distinct sizes cannot be
achieved.
(Russian Mathematical Olympiad, 1995)

76.First solution: Here the invariant is given; we just have to prove its invariance. We
first examine the simpler case of a cyclic quadrilateralABCDinscribed in a circle of
radiusR. Recall that for a triangleXY Zthe radii of the incircle and the circumcircle are
related by

r= 4 Rsin

X

2

sin

Y

2

sin

Z

2

.

Let∠CAD=α 1 ,∠BAC=α 2 ,∠ABD=β. Then∠DBC=α 1 , and∠ACD=β,
∠BDC=α 2 , and∠ACB=∠ADB= 180 ◦−α 1 −α 2 −β. The independence of the
sum of the inradii in the two possible dissections translates, after dividing by 4R, into
the identity

sin

α 1 +α 2

2

sin

β

2

sin

(

90 ◦−

α 1 +α 2 +β

2

)

+sin

(

90 ◦−

α 1 +α 2

2

)

sin

α 1

2

sin

α 2

2

=sin

α 1 +β 1

2

sin

α 2

2

sin

(

90 ◦−

α 1 +α 2 +β

2

)

+sin

(

90 ◦−

α 1 +β 1

2

)

sin

α 1

2

sin

β

2

.