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Problems
Contests
International Contests
Cono Sur Olympiad
1992 Cono Sur Olympiad
1992 Cono Sur Olympiad
Part of
Cono Sur Olympiad
Subcontests
(3)
3
2
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Find all values
Consider the set
S
S
S
of
100
100
100
numbers:
1
;
1
2
;
1
3
;
.
.
.
;
1
100
1; \frac{1}{2}; \frac{1}{3}; ... ; \frac{1}{100}
1
;
2
1
;
3
1
;
...
;
100
1
. Any two numbers,
a
a
a
and
b
b
b
, are eliminated in
S
S
S
, and the number
a
+
b
+
a
b
a+b+ab
a
+
b
+
ab
is added. Now, there are
99
99
99
numbers on
S
S
S
. After doing this operation
99
99
99
times, there's only
1
1
1
number on
S
S
S
. What values can this number take?
Find conditions
Consider a
m
∗
n
m*n
m
∗
n
board. On each box there's a non-negative integrer number assigned. An operation consists on choosing any two boxes with
1
1
1
side in common, and add to this
2
2
2
numbers the same integrer number (it can be negative), so that both results are non-negatives. What conditions must be satisfied initially on the assignment of the boxes, in order to have, after some operations, the number
0
0
0
on every box?.
2
2
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Find two points
Let
P
P
P
be a point outside the circle
C
C
C
. Find two points
Q
Q
Q
and
R
R
R
on the circle, such that
P
,
Q
P,Q
P
,
Q
and
R
R
R
are collinear and
Q
Q
Q
is the midpopint of the segmenet
P
R
PR
PR
. (Discuss the number of solutions).
Proof in a triangle
In a
△
A
B
C
\triangle {ABC}
△
A
BC
, consider a point
E
E
E
in
B
C
BC
BC
such that
A
E
⊥
B
C
AE \perp BC
A
E
⊥
BC
. Prove that
A
E
=
b
c
2
r
AE=\frac{bc}{2r}
A
E
=
2
r
b
c
, where
r
r
r
is the radio of the circle circumscripte,
b
=
A
C
b=AC
b
=
A
C
and
c
=
A
B
c=AB
c
=
A
B
.
1
2
Hide problems
Find n
Find a positive integrer number
n
n
n
such that, if yor put a number
2
2
2
on the left and a number
1
1
1
on the right, the new number is equal to
33
n
33n
33
n
.
Proof about solutions
Prove that there aren't any positive integrer numbers
x
,
y
,
z
x,y,z
x
,
y
,
z
such that
x
2
+
y
2
=
3
z
2
x^2+y^2=3z^2
x
2
+
y
2
=
3
z
2
.