MathDB
Problems
Contests
National and Regional Contests
Costa Rica Contests
Costa Rica - Final Round
2006 Costa Rica - Final Round
2006 Costa Rica - Final Round
Part of
Costa Rica - Final Round
Subcontests
(3)
3
1
Hide problems
Prove the existence of this triangle.
Let
A
B
C
ABC
A
BC
be a triangle. Let
P
,
Q
,
R
P, Q, R
P
,
Q
,
R
be the midpoints of
B
C
,
C
A
,
A
B
BC, CA, AB
BC
,
C
A
,
A
B
respectively. Let
U
,
V
,
W
U, V, W
U
,
V
,
W
be the midpoints of
Q
R
,
R
P
,
P
Q
QR, RP, PQ
QR
,
RP
,
PQ
respectively. Let
x
=
A
U
,
y
=
B
V
,
z
=
C
W
x=AU, y=BV, z=CW
x
=
A
U
,
y
=
B
V
,
z
=
C
W
. Prove that there exist a triangle with sides
x
,
y
,
z
x, y, z
x
,
y
,
z
.
1
2
Hide problems
Find SUM 1/(i_1 i_2 ... i_r)
Consider the set
S
=
{
1
,
2
,
.
.
.
,
n
}
S=\{1,2,...,n\}
S
=
{
1
,
2
,
...
,
n
}
. For every
k
∈
S
k\in S
k
∈
S
, define
S
k
=
{
X
⊆
S
,
k
∉
X
,
X
≠
∅
}
S_{k}=\{X \subseteq S, \ k \notin X, X\neq \emptyset\}
S
k
=
{
X
⊆
S
,
k
∈
/
X
,
X
=
∅
}
. Determine the value of the sum
S
k
∗
=
∑
{
i
1
,
i
2
,
.
.
.
,
i
r
}
∈
S
k
1
i
1
⋅
i
2
⋅
.
.
.
⋅
i
r
S_{k}^{*}=\sum_{\{i_{1},i_{2},...,i_{r}\}\in S_{k}}\frac{1}{i_{1}\cdot i_{2}\cdot...\cdot i_{r}}
S
k
∗
=
{
i
1
,
i
2
,
...
,
i
r
}
∈
S
k
∑
i
1
⋅
i
2
⋅
...
⋅
i
r
1
in fact, this problem was taken from an austrian-polish
A functional relation. - compute f(2004).
Let
f
f
f
be a function that satisfies :
f
(
x
)
+
2
f
(
x
+
2001
2
x
−
1
)
=
4014
−
x
.
\displaystyle f(x)+2f\left(\frac{x+\frac{2001}2}{x-1}\right) = 4014-x.
f
(
x
)
+
2
f
(
x
−
1
x
+
2
2001
)
=
4014
−
x
.
Find
f
(
2004
)
f(2004)
f
(
2004
)
.
2
2
Hide problems
inequality m13
If
a
a
a
,
b
b
b
,
c
c
c
are the sidelengths of a triangle, then prove that \frac {3\left(a^4 \plus{} b^4 \plus{} c^4\right)}{\left(a^2 \plus{} b^2 \plus{} c^2\right)^2} \plus{} \frac {bc \plus{} ca \plus{} ab}{a^2 \plus{} b^2 \plus{} c^2}\geq 2.
The binomial coefficient congruent to [n/p] mod p.
Let
n
n
n
be a positive integer, and let
p
p
p
be a prime, such that
n
>
p
n>p
n
>
p
. Prove that :
(
n
p
)
≡
⌊
n
p
⌋
(
m
o
d
p
)
.
\displaystyle \binom np \equiv \left\lfloor\frac{n}{p}\right\rfloor \ \pmod p.
(
p
n
)
≡
⌊
p
n
⌋
(
mod
p
)
.