MathDB
Problems
Contests
National and Regional Contests
Costa Rica Contests
Costa Rica - Final Round
2010 Costa Rica - Final Round
2010 Costa Rica - Final Round
Part of
Costa Rica - Final Round
Subcontests
(6)
3
1
Hide problems
Sequence of numbers
Christian Reiher and Reid Barton want to open a security box, they already managed to discover the algorithm to generate the key codes and they obtained the following information:
i
)
i)
i
)
In the screen of the box will appear a sequence of
n
+
1
n+1
n
+
1
numbers,
C
0
=
(
a
0
,
1
,
a
0
,
2
,
.
.
.
,
a
0
,
n
+
1
)
C_0 = (a_{0,1},a_{0,2},...,a_{0,n+1})
C
0
=
(
a
0
,
1
,
a
0
,
2
,
...
,
a
0
,
n
+
1
)
i
i
)
ii)
ii
)
If the code
K
=
(
k
1
,
k
2
,
.
.
.
,
k
n
)
K = (k_1,k_2,...,k_n)
K
=
(
k
1
,
k
2
,
...
,
k
n
)
opens the security box then the following must happen:a) A sequence
C
i
=
(
a
i
,
1
,
a
i
,
2
,
.
.
.
,
a
i
,
n
+
1
)
C_i = (a_{i,1},a_{i,2},...,a_{i,n+1})
C
i
=
(
a
i
,
1
,
a
i
,
2
,
...
,
a
i
,
n
+
1
)
will be asigned to each
k
i
k_i
k
i
defined as follows:
a
i
,
1
=
1
a_{i,1} = 1
a
i
,
1
=
1
and
a
i
,
j
=
a
i
−
1
,
j
−
k
i
a
i
,
j
−
1
a_{i,j} = a_{i-1,j}-k_ia_{i,j-1}
a
i
,
j
=
a
i
−
1
,
j
−
k
i
a
i
,
j
−
1
, for
i
,
j
≥
1
i,j \ge 1
i
,
j
≥
1
b) The sequence
(
C
n
)
(C_n)
(
C
n
)
asigned to
k
n
k_n
k
n
satisfies that
S
n
=
∑
i
=
1
n
+
1
∣
a
i
∣
S_n = \sum_{i=1}^{n+1}|a_i|
S
n
=
∑
i
=
1
n
+
1
∣
a
i
∣
has its least possible value, considering all possible sequences
K
K
K
.The sequence
C
0
C_0
C
0
that appears in the screen is the following:
a
0
,
1
=
1
a_{0,1} = 1
a
0
,
1
=
1
and
a
0
,
i
a_0,i
a
0
,
i
is the sum of the products of the elements of each of the subsets with
i
−
1
i-1
i
−
1
elements of the set
A
=
A =
A
=
{
1
,
2
,
3
,
.
.
.
,
n
1,2,3,...,n
1
,
2
,
3
,
...
,
n
},
i
≥
2
i\ge 2
i
≥
2
, such that
a
0
,
n
+
1
=
n
!
a_{0, n+1} = n!
a
0
,
n
+
1
=
n
!
Find a sequence
K
=
(
k
1
,
k
2
,
.
.
.
,
k
n
)
K = (k_1,k_2,...,k_n)
K
=
(
k
1
,
k
2
,
...
,
k
n
)
that satisfies the conditions of the problem and show that there exists at least
n
!
n!
n
!
of them.
6
1
Hide problems
Least element of 2010-set
Let
F
F
F
be the family of all sets of positive integers with
2010
2010
2010
elements that satisfy the following condition: The difference between any two of its elements is never the same as the difference of any other two of its elements. Let
f
f
f
be a function defined from
F
F
F
to the positive integers such that
f
(
K
)
f(K)
f
(
K
)
is the biggest element of
K
∈
F
K \in F
K
∈
F
. Determine the least value of
f
(
K
)
f(K)
f
(
K
)
.
4
1
Hide problems
Two variable diophantine equation
Find all integer solutions
(
a
,
b
)
(a,b)
(
a
,
b
)
of the equation
(
a
+
b
+
3
)
2
+
2
a
b
=
3
a
b
(
a
+
2
)
(
b
+
2
)
(a+b+3)^2 + 2ab = 3ab(a+2)(b+2)
(
a
+
b
+
3
)
2
+
2
ab
=
3
ab
(
a
+
2
)
(
b
+
2
)
5
1
Hide problems
Trisected angle
Let
C
1
C_1
C
1
be a circle with center
O
O
O
and let
B
B
B
and
C
C
C
be points in
C
1
C_1
C
1
such that
B
O
C
BOC
BOC
is an equilateral triangle. Let
D
D
D
be the midpoint of the minor arc
B
C
BC
BC
of
C
1
C_1
C
1
. Let
C
2
C_2
C
2
be the circle with center
C
C
C
that passes through
B
B
B
and
O
O
O
. Let
E
E
E
be the second intersection of
C
1
C_1
C
1
and
C
2
C_2
C
2
. The parallel to
D
E
DE
D
E
through
B
B
B
intersects
C
1
C_1
C
1
for second time in
A
A
A
. Let
C
3
C_3
C
3
be the circumcircle of triangle
A
O
C
AOC
A
OC
. The second intersection of
C
2
C_2
C
2
and
C
3
C_3
C
3
is
F
F
F
. Show that
B
E
BE
BE
and
B
F
BF
BF
trisect the angle
∠
A
B
C
\angle ABC
∠
A
BC
.
1
1
Hide problems
Triangles with equal area
Consider points
D
,
E
D,E
D
,
E
and
F
F
F
on sides
B
C
,
A
C
BC,AC
BC
,
A
C
and
A
B
AB
A
B
, respectively, of a triangle
A
B
C
ABC
A
BC
, such that
A
D
,
B
E
AD, BE
A
D
,
BE
and
C
F
CF
CF
concurr at a point
G
G
G
. The parallel through
G
G
G
to
B
C
BC
BC
cuts
D
F
DF
D
F
and
D
E
DE
D
E
at
H
H
H
and
I
I
I
, respectively. Show that triangles
A
H
G
AHG
A
H
G
and
A
I
G
AIG
A
I
G
have the same areas.
2
1
Hide problems
Divisibility with recurrence relation
Consider the sequence
x
n
>
0
x_n>0
x
n
>
0
defined with the following recurrence relation:
x
1
=
0
x_1 = 0
x
1
=
0
and for
n
>
1
n>1
n
>
1
(
n
+
1
)
2
x
n
+
1
2
+
(
2
n
+
4
)
(
n
+
1
)
x
n
+
1
+
2
n
+
1
+
2
2
n
−
2
=
9
n
2
x
n
2
+
36
n
x
n
+
32.
(n+1)^2x_{n+1}^2 + (2^n+4)(n+1)x_{n+1}+ 2^{n+1}+2^{2n-2} = 9n^2x_n^2+36nx_n+32.
(
n
+
1
)
2
x
n
+
1
2
+
(
2
n
+
4
)
(
n
+
1
)
x
n
+
1
+
2
n
+
1
+
2
2
n
−
2
=
9
n
2
x
n
2
+
36
n
x
n
+
32.
Show that if
n
n
n
is a prime number larger or equal to
5
5
5
, then
x
n
x_n
x
n
is an integer.