MathDB
Problems
Contests
National and Regional Contests
Costa Rica Contests
Costa Rica - Final Round
2011 Costa Rica - Final Round
2011 Costa Rica - Final Round
Part of
Costa Rica - Final Round
Subcontests
(6)
3
1
Hide problems
winning strategy for pirate game, archipelago Barrantes -n
The archipelago Barrantes -
n
n
n
is a group of islands connected by bridges as follows: there are a main island (Humberto), in the first step I place an island below Humberto and one above from Humberto and I connect these 2 islands to Humberto. I put
2
2
2
islands to the left of these
2
2
2
new islands and I connect them with a bridge to the island that they have on their right. In the second step I take the last
2
2
2
islands and I apply the same process that I applied to Humberto. In the third step I apply the same process to the
4
4
4
new islands. We repeat this step n times we reflect the archipelago that we have on a vertical line to the right of Humberto. We connect Humberto with his reflection and so we have the archipelago Barrantes -
n
n
n
. However, the archipelago Barrantes -
n
n
n
exists on a small planet cylindrical, so that the islands to the left of the archipelago are in fact the islands that are connected to the islands on the right. The figure shows the Barrantes archipelago -
2
2
2
, The islands at the edges are still numbered to show how the archipelago connects around the cylindrical world, the island numbered
1
1
1
on the left is the same as the island numbered
1
1
1
on the right. https://cdn.artofproblemsolving.com/attachments/e/c/803d95ce742c2739729fdb4d74af59d4d0652f.png One day two bands of pirates arrive at the archipelago Barrantes -
n
n
n
: The pirates Black Beard and the Straw Hat Pirates. Blackbeard proposes a game to Straw Hat: The first player conquers an island, the next player must conquer an island connected to the island that was conquered in the previous turn (clearly not conquered on a previous shift). The one who cannot conquer any island in his turn loses. Straw Hat decides to give the first turn to Blackbeard. Prove that Straw Hat has a winning strategy for every
n
n
n
.
4
1
Hide problems
( \sum y_k \sqrt{p_k} )^2 <= \sum k/ n p_k
Let
p
1
,
p
2
,
.
.
.
,
p
n
p_1, p_2, ..., p_n
p
1
,
p
2
,
...
,
p
n
be positive real numbers, such that
p
1
+
p
2
+
.
.
.
+
p
n
=
1
p_1 + p_2 +... + p_n = 1
p
1
+
p
2
+
...
+
p
n
=
1
. Let
x
∈
[
0
,
1
]
x \in [0,1]
x
∈
[
0
,
1
]
and let
y
1
,
y
2
,
.
.
.
,
y
n
y_1, y_2, ..., y_n
y
1
,
y
2
,
...
,
y
n
be such that
y
1
2
+
y
2
2
+
.
.
.
+
y
n
2
=
x
y^2_1 + y^2_2 +...+ y^2_n= x
y
1
2
+
y
2
2
+
...
+
y
n
2
=
x
. Prove that
(
∑
n
x
≤
k
≤
n
y
k
p
k
)
2
≤
∑
k
=
1
n
k
n
p
k
\left( \sum_{nx\le k \le n }y_k \sqrt{p_k} \right)^2 \le \sum_{k=1}^{n}\frac{k}{n} p_k
(
n
x
≤
k
≤
n
∑
y
k
p
k
)
2
≤
k
=
1
∑
n
n
k
p
k
2
1
Hide problems
Find biggest n
Find the biggest positive integer
n
n
n
such that
n
n
n
is
167
167
167
times the amount of it's positive divisors.
5
1
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Similar to chicken mcnugget theorem
Given positive integers
a
,
b
,
c
a,b,c
a
,
b
,
c
which are pairwise relatively prime, show that
2
a
b
c
−
a
b
−
b
c
−
a
c
2abc-ab-bc-ac
2
ab
c
−
ab
−
b
c
−
a
c
is the biggest number that can't be expressed in the form
x
b
c
+
y
c
a
+
z
a
b
xbc+yca+zab
x
b
c
+
yc
a
+
z
ab
with
x
,
y
,
z
x,y,z
x
,
y
,
z
being natural numbers.
6
1
Hide problems
Tangents of incircles
Let
A
B
C
ABC
A
BC
be a triangle. The incircle of
A
B
C
ABC
A
BC
touches
B
C
,
A
C
,
A
B
BC,AC,AB
BC
,
A
C
,
A
B
at
D
,
E
,
F
D,E,F
D
,
E
,
F
, respectively. Each pair of the incircles of triangles
A
E
F
,
B
D
F
,
C
E
D
AEF, BDF,CED
A
EF
,
B
D
F
,
CE
D
has two pair of common external tangents, one of them being one of the sides of
A
B
C
ABC
A
BC
. Show that the other three tangents divide triangle
D
E
F
DEF
D
EF
into three triangles and three parallelograms.
1
1
Hide problems
Orthocenter is incenter of triangle
Let
A
B
C
ABC
A
BC
be a triangle with orthocenter
H
H
H
. Let
P
,
Q
,
R
P,Q,R
P
,
Q
,
R
be the reflections of
H
H
H
with respect to sides
B
C
,
A
C
,
A
B
BC,AC,AB
BC
,
A
C
,
A
B
, respectively. Show that
H
H
H
is incenter of
P
Q
R
PQR
PQR
.