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Problems
Contests
International Contests
Cono Sur Olympiad
2001 Cono Sur Olympiad
2001 Cono Sur Olympiad
Part of
Cono Sur Olympiad
Subcontests
(3)
3
2
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Three acute inscribed triangles determine a nonobtuse one
Three acute triangles are inscribed in the same circle with their vertices being nine distinct points. Show that one can choose a vertex from each triangle so that the three chosen points determine a triangle each of whose angles is at most
9
0
∘
90^\circ
9
0
∘
.
How long before a function with g(n+1)=g(n)+-1 reaches 2001
A function
g
g
g
defined for all positive integers
n
n
n
satisfies [*]
g
(
1
)
=
1
g(1) = 1
g
(
1
)
=
1
; [*]for all
n
≥
1
n\ge 1
n
≥
1
, either
g
(
n
+
1
)
=
g
(
n
)
+
1
g(n+1)=g(n)+1
g
(
n
+
1
)
=
g
(
n
)
+
1
or
g
(
n
+
1
)
=
g
(
n
)
−
1
g(n+1)=g(n)-1
g
(
n
+
1
)
=
g
(
n
)
−
1
; [*]for all
n
≥
1
n\ge 1
n
≥
1
,
g
(
3
n
)
=
g
(
n
)
g(3n) = g(n)
g
(
3
n
)
=
g
(
n
)
; and [*]
g
(
k
)
=
2001
g(k)=2001
g
(
k
)
=
2001
for some positive integer
k
k
k
. Find, with proof, the smallest possible value of
k
k
k
.
1
2
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Distinct row & column sums in 2000x2000 array of 0s, 1s, -1s
Each entry in a
2000
×
2000
2000\times 2000
2000
×
2000
array is
0
0
0
,
1
1
1
, or
−
1
-1
−
1
. Show that it's possible for all
4000
4000
4000
row sums and column sums to be distinct.
Distance between two points in a polygon in a square
A polygon of area
S
S
S
is contained inside a square of side length
a
a
a
. Show that there are two points of the polygon that are a distance of at least
S
/
a
S/a
S
/
a
apart.
2
2
Hide problems
Increasing sequence of positive integers w/ a_{3n+1}=2a_n+1
A sequence
a
1
,
a
2
,
…
a_1,a_2,\ldots
a
1
,
a
2
,
…
of positive integers satisfies the following properties.[*]
a
1
=
1
a_1 = 1
a
1
=
1
[*]
a
3
n
+
1
=
2
a
n
+
1
a_{3n+1} = 2a_n + 1
a
3
n
+
1
=
2
a
n
+
1
[*]
a
n
+
1
≥
a
n
a_{n+1}\ge a_n
a
n
+
1
≥
a
n
[*]
a
2001
=
200
a_{2001} = 200
a
2001
=
200
Find the value of
a
1000
a_{1000}
a
1000
.Note. In the original statement of the problem, there was an extra condition:[*]every positive integer appears at least once in the sequence.However, with this extra condition, there is no solution, i.e., no such sequence exists. (Try to prove it.) The problem as written above does have a solution.
Solve 2001*(sum of the digits of m) = m
Find all positive integers
m
m
m
for which
2001
⋅
S
(
m
)
=
m
2001\cdot S (m) = m
2001
⋅
S
(
m
)
=
m
where
S
(
m
)
S(m)
S
(
m
)
denotes the sum of the digits of
m
m
m
.