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Problems
Contests
National and Regional Contests
Brazil Contests
Brazil Team Selection Test
2019 Brazil Team Selection Test
2019 Brazil Team Selection Test
Part of
Brazil Team Selection Test
Subcontests
(4)
4
2
Hide problems
Path in chessboard without a square
Consider a checkered board
2
m
×
2
n
2m \times 2n
2
m
×
2
n
,
m
,
n
∈
Z
>
0
m, n \in \mathbb{Z}_{>0}
m
,
n
∈
Z
>
0
. A stone is placed on one of the unit squares on the board, this square is different from the upper right square and from the lower left square. A snail goes from the bottom left square and wants to get to the top right square, walking from one square to other adjacent, one square at a time (two squares are adjacent if they share an edge). Determine all the squares the stone can be in so that the snail can complete its path by visiting each square exactly one time, except the square with the stone, which the snail does not visit.
Element can be written as sum of two squares
Let
p
≥
7
p \geq 7
p
≥
7
be a prime number and
S
=
{
j
p
+
1
:
1
≤
j
≤
p
−
5
2
}
.
S = \bigg\{jp+1 : 1 \leq j \leq \frac{p-5}{2}\bigg\}.
S
=
{
j
p
+
1
:
1
≤
j
≤
2
p
−
5
}
.
Prove that at least one element of
S
S
S
can be written as
x
2
+
y
2
x^2+y^2
x
2
+
y
2
, where
x
,
y
x, y
x
,
y
are integers.
1
1
Hide problems
Many 60 degrees
Let
A
B
C
ABC
A
BC
be an acute triangle, with
∠
A
>
6
0
∘
\angle A > 60^\circ
∠
A
>
6
0
∘
, and let
H
H
H
be it's orthocenter. Let
M
M
M
and
N
N
N
be points on
A
B
AB
A
B
and
A
C
AC
A
C
, respectively, such that
∠
H
M
B
=
∠
H
N
C
=
6
0
∘
\angle HMB = \angle HNC = 60^\circ
∠
H
MB
=
∠
H
NC
=
6
0
∘
. Also, let
O
O
O
be the circuncenter of
H
M
N
HMN
H
MN
and
D
D
D
be a point on the semiplane determined by
B
C
BC
BC
that contains
A
A
A
in such a way that
D
B
C
DBC
D
BC
is equilateral. Prove that
H
H
H
,
O
O
O
and
D
D
D
are collinear.
3
2
Hide problems
Coloring and counting intersection points of circles in general position
Let
n
≥
2
n \geq 2
n
≥
2
be an integer. There are
n
n
n
distinct circles in general position, that is, any two of them meet in two distinct points and there are no three of them meeting at one point. Those circles divide the plane in limited regions by circular edges, that meet at vertices (note that each circle have exactly
2
n
−
2
2n-2
2
n
−
2
vertices). For each circle, temporarily color its vertices alternately black and white (note that, doing this, each vertex is colored twice, one for each circle passing through it). If the two temporarily colouring of a vertex coincide, this vertex is definitely colored with this common color; otherwise, it will be colored with gray. Show that if a circle has more than
n
−
2
+
n
−
2
n-2 + \sqrt{n-2}
n
−
2
+
n
−
2
gray points, all vertices of some region are grey.Observation: In this problem, a region cannot contain vertices or circular edges on its interior. Also, the outer region of all circles also counts as a region.
Inequality with numbers with sum 1
Let
n
≥
2
n \geq 2
n
≥
2
be an integer and
x
1
,
x
2
,
…
,
x
n
x_1, x_2, \ldots, x_n
x
1
,
x
2
,
…
,
x
n
be positive real numbers such that
∑
i
=
1
n
x
i
=
1
\sum_{i=1}^nx_i=1
∑
i
=
1
n
x
i
=
1
. Show that
(
∑
i
=
1
n
1
1
−
x
i
)
(
∑
1
≤
i
<
j
≤
n
x
i
x
j
)
≤
n
2
.
\bigg(\sum_{i=1}^n\frac{1}{1-x_i}\bigg)\bigg(\sum_{1 \leq i < j \leq n}x_ix_j\bigg) \leq \frac{n}{2}.
(
i
=
1
∑
n
1
−
x
i
1
)
(
1
≤
i
<
j
≤
n
∑
x
i
x
j
)
≤
2
n
.
2
2
Hide problems
Prove that 3 tangents are concurrent
Let
A
B
C
ABC
A
BC
be a triangle, and
A
1
A_1
A
1
,
B
1
B_1
B
1
,
C
1
C_1
C
1
points on the sides
B
C
BC
BC
,
C
A
CA
C
A
,
A
B
AB
A
B
, respectively, such that the triangle
A
1
B
1
C
1
A_1B_1C_1
A
1
B
1
C
1
is equilateral. Let
I
1
I_1
I
1
and
ω
1
\omega_1
ω
1
be the incenter and the incircle of
A
B
1
C
1
AB_1C_1
A
B
1
C
1
. Define
I
2
I_2
I
2
,
ω
2
\omega_2
ω
2
and
I
3
I_3
I
3
,
ω
3
\omega_3
ω
3
similarly, with respect to the triangles
B
A
1
C
1
BA_1C_1
B
A
1
C
1
and
C
A
1
B
1
CA_1B_1
C
A
1
B
1
, respectively. Let
l
1
≠
B
C
l_1 \neq BC
l
1
=
BC
be the external tangent line to
ω
2
\omega_2
ω
2
and
ω
3
\omega_3
ω
3
. Define
l
2
l_2
l
2
and
l
3
l_3
l
3
similarly, with respect to the pairs
ω
1
\omega_1
ω
1
,
ω
3
\omega_3
ω
3
and
ω
1
\omega_1
ω
1
,
ω
2
\omega_2
ω
2
.Knowing that
A
1
I
2
=
A
1
I
3
A_1I_2 = A_1I_3
A
1
I
2
=
A
1
I
3
, show that the lines
l
1
l_1
l
1
,
l
2
l_2
l
2
,
l
3
l_3
l
3
are concurrent.
Always possible to choose some friends
We say that a distribution of students lined upen in collumns is
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
i
t
a
l
i
c
′
>
b
a
c
a
n
a
<
/
s
p
a
n
>
<span class='latex-italic'>bacana</span>
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
i
t
a
l
i
c
′
>
ba
c
ana
<
/
s
p
an
>
when there are no two friends in the same column. We know that all contestants in a math olympiad can be arranged in a
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
i
t
a
l
i
c
′
>
b
a
c
a
n
a
<
/
s
p
a
n
>
<span class='latex-italic'>bacana</span>
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
i
t
a
l
i
c
′
>
ba
c
ana
<
/
s
p
an
>
configuration with
n
n
n
columns, and that this is impossible with
n
−
1
n-1
n
−
1
columns. Show that we can choose competitors
M
1
,
M
2
,
⋯
,
M
n
M_1, M_2, \cdots, M_n
M
1
,
M
2
,
⋯
,
M
n
in such a way that
M
i
M_i
M
i
is on the
i
i
i
-th column, for each
i
=
1
,
2
,
3
,
…
,
n
i = 1, 2, 3, \ldots, n
i
=
1
,
2
,
3
,
…
,
n
and
M
i
M_i
M
i
is a friend of
M
i
+
1
M_{i+1}
M
i
+
1
for each
i
=
1
,
2
,
…
,
n
−
1
i = 1, 2, \ldots, n - 1
i
=
1
,
2
,
…
,
n
−
1
.