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Problems
Contests
National and Regional Contests
Brazil Contests
Rio de Janeiro Mathematical Olympiad
2018 Rio de Janeiro Mathematical Olympiad
2018 Rio de Janeiro Mathematical Olympiad
Part of
Rio de Janeiro Mathematical Olympiad
Subcontests
(6)
6
1
Hide problems
Friend squares
Two perfect squares are friends if one is obtained from the other adding the digit
1
1
1
at the left.For instance,
1225
=
3
5
2
1225 = 35^2
1225
=
3
5
2
and
225
=
1
5
2
225 = 15^2
225
=
1
5
2
are friends.Prove that there are infinite pairs of odd perfect squares that are friends.
5
2
Hide problems
Cadence number of a Permutation
Let
n
n
n
be an positive integer and
σ
=
(
a
1
,
…
,
a
n
)
\sigma = (a_1, \dots, a_n)
σ
=
(
a
1
,
…
,
a
n
)
a permutation of
{
1
,
…
,
n
}
\{1, \dots, n\}
{
1
,
…
,
n
}
. The cadence number of
σ
\sigma
σ
is the number of maximal decrescent blocks.For example, if
n
=
6
n = 6
n
=
6
and
σ
=
(
4
,
2
,
1
,
5
,
6
,
3
)
\sigma = (4, 2, 1, 5, 6, 3)
σ
=
(
4
,
2
,
1
,
5
,
6
,
3
)
, then the cadence number of
σ
\sigma
σ
is
3
3
3
, because
σ
\sigma
σ
has
3
3
3
maximal decrescent blocks:
(
4
,
2
,
1
)
(4, 2, 1)
(
4
,
2
,
1
)
,
(
5
)
(5)
(
5
)
and
(
6
,
3
)
(6, 3)
(
6
,
3
)
. Note that
(
4
,
2
)
(4, 2)
(
4
,
2
)
and
(
2
,
1
)
(2, 1)
(
2
,
1
)
are decrescent, but not maximal, because they are already contained in
(
4
,
2
,
1
)
(4, 2, 1)
(
4
,
2
,
1
)
.Compute the sum of the cadence number of every permutation of
{
1
,
…
,
n
}
\{1, \dots, n\}
{
1
,
…
,
n
}
.
Possible values for angle.
Let
Θ
1
\Theta_1
Θ
1
and
Θ
2
\Theta_2
Θ
2
be circumferences with centers
O
1
O_1
O
1
and
O
2
O_2
O
2
, exteriorly tangents. Let
A
A
A
and
B
B
B
be points in
Θ
1
\Theta_1
Θ
1
and
Θ
2
\Theta_2
Θ
2
, respectively, such that
A
B
AB
A
B
is common external tangent to
Θ
1
\Theta_1
Θ
1
and
Θ
2
\Theta_2
Θ
2
. Let
C
C
C
and
D
D
D
be points on the semiplane determined by
A
B
AB
A
B
that does not contain
O
1
O_1
O
1
and
O
2
O_2
O
2
such that
A
B
C
D
ABCD
A
BC
D
is a square. If
O
O
O
is the center of this square, compute the possible values for the angle
∠
O
1
O
O
2
\angle O_1OO_2
∠
O
1
O
O
2
.
4
2
Hide problems
Simple Geometry Problem
Let
A
B
C
ABC
A
BC
be an acute triangle inscribed on the circumference
Γ
\Gamma
Γ
. Let
D
D
D
and
E
E
E
be points on
Γ
\Gamma
Γ
such that
A
D
AD
A
D
is perpendicular to
B
C
BC
BC
and
A
E
AE
A
E
is diameter. Let
F
F
F
be the intersection between
A
E
AE
A
E
and
B
C
BC
BC
. Prove that, if
∠
D
A
C
=
2
∠
D
A
B
\angle DAC = 2 \angle DAB
∠
D
A
C
=
2∠
D
A
B
, then
D
E
=
C
F
DE = CF
D
E
=
CF
.
When this system has a solution?
Find every real values that
a
a
a
can assume such that
{
x
3
+
y
2
+
z
2
=
a
x
2
+
y
3
+
z
2
=
a
x
2
+
y
2
+
z
3
=
a
\begin{cases} x^3 + y^2 + z^2 = a\\ x^2 + y^3 + z^2 = a\\ x^2 + y^2 + z^3 = a \end{cases}
⎩
⎨
⎧
x
3
+
y
2
+
z
2
=
a
x
2
+
y
3
+
z
2
=
a
x
2
+
y
2
+
z
3
=
a
has a solution with
x
,
y
,
z
x, y, z
x
,
y
,
z
distinct real numbers.
3
2
Hide problems
Number of good functions
Let
n
n
n
be a positive integer. A function
f
:
{
1
,
2
,
…
,
2
n
}
→
{
1
,
2
,
3
,
4
,
5
}
f : \{1, 2, \dots, 2n\} \to \{1, 2, 3, 4, 5\}
f
:
{
1
,
2
,
…
,
2
n
}
→
{
1
,
2
,
3
,
4
,
5
}
is good if
f
(
j
+
2
)
f(j+2)
f
(
j
+
2
)
and
f
(
j
)
f(j)
f
(
j
)
have the same parity for every
j
=
1
,
2
,
…
,
2
n
−
2
j = 1, 2, \dots, 2n-2
j
=
1
,
2
,
…
,
2
n
−
2
. Prove that the number of good functions is a perfect square.
k-good functions are k-powers
Let
n
n
n
and
k
k
k
be positive integers. A function
f
:
{
1
,
2
,
3
,
4
,
…
,
k
n
−
1
,
k
n
}
→
{
1
,
⋯
,
5
}
f : \{1, 2, 3, 4, \dots , kn - 1, kn\} \to \{1, \cdots , 5\}
f
:
{
1
,
2
,
3
,
4
,
…
,
kn
−
1
,
kn
}
→
{
1
,
⋯
,
5
}
is good if
f
(
j
+
k
)
−
f
(
j
)
f(j + k) - f(j)
f
(
j
+
k
)
−
f
(
j
)
is multiple of
k
k
k
for every
j
=
1
,
2.
⋯
,
k
n
−
k
j = 1, 2. \cdots , kn - k
j
=
1
,
2.
⋯
,
kn
−
k
.(a) Prove that, if
k
=
2
k = 2
k
=
2
, then the number of good functions is a perfect square for every positive integer
n
n
n
. (b) Prove that, if
k
=
3
k = 3
k
=
3
, then the number of good functions is a perfect cube for every positive integer
n
n
n
.
2
2
Hide problems
Fake geometry problem
Let
A
B
C
ABC
A
BC
be an equilateral triangle with side 3. A circle
C
1
C_1
C
1
is tangent to
A
B
AB
A
B
and
A
C
AC
A
C
. A circle
C
2
C_2
C
2
, with a radius smaller than the radius of
C
1
C_1
C
1
, is tangent to
A
B
AB
A
B
and
A
C
AC
A
C
as well as externally tangent to
C
1
C_1
C
1
. Successively, for
n
n
n
positive integer, the circle
C
n
+
1
C_{n+1}
C
n
+
1
, with a radius smaller than the radius of
C
n
C_n
C
n
, is tangent to
A
B
AB
A
B
and
A
C
AC
A
C
and is externally tangent to
C
n
C_n
C
n
. Determine the possible values for the radius of
C
1
C_1
C
1
such that 4 circles from this sequence, but not 5, are contained on the interior of the triangle
A
B
C
ABC
A
BC
.
111's in this strange sequence?
Let
(
a
n
)
(a_n)
(
a
n
)
be a sequence of integers, with
a
1
=
1
a_1 = 1
a
1
=
1
and for evert integer
n
≥
1
n \ge 1
n
≥
1
,
a
2
n
=
a
n
+
1
a_{2n} = a_n + 1
a
2
n
=
a
n
+
1
and
a
2
n
+
1
=
10
a
n
a_{2n+1} = 10a_n
a
2
n
+
1
=
10
a
n
. How many times
111
111
111
appears on this sequence?
1
2
Hide problems
On the sum of factorials of digits.
A natural number is a factorion if it is the sum of the factorials of each of its decimal digits. For example,
145
145
145
is a factorion because
145
=
1
!
+
4
!
+
5
!
145 = 1! + 4! + 5!
145
=
1
!
+
4
!
+
5
!
. Find every 3-digit number which is a factorion.
Fake geometry problem
Let
A
B
C
ABC
A
BC
be a triangle and
k
<
1
k < 1
k
<
1
a positive real number. Let
A
1
A_1
A
1
,
B
1
B_1
B
1
,
C
1
C_1
C
1
be points on the sides
B
C
BC
BC
,
A
C
AC
A
C
,
A
B
AB
A
B
such that
A
1
B
B
C
=
B
1
C
A
C
=
C
1
A
A
B
=
k
.
\frac{A_1B}{BC} = \frac{B_1C}{AC} = \frac{C_1A}{AB} = k.
BC
A
1
B
=
A
C
B
1
C
=
A
B
C
1
A
=
k
.
(a) Compute, in terms of
k
k
k
, the ratio between the areas of the triangles
A
1
B
1
C
1
A_1B_1C_1
A
1
B
1
C
1
and
A
B
C
ABC
A
BC
.(b) Generally, for each
n
≥
1
n \ge 1
n
≥
1
, the triangle
A
n
+
1
B
n
+
1
C
n
+
1
A_{n+1}B_{n+1}C_{n+1}
A
n
+
1
B
n
+
1
C
n
+
1
is built such that
A
n
+
1
A_{n+1}
A
n
+
1
,
B
n
+
1
B_{n+1}
B
n
+
1
,
C
n
+
1
C_{n+1}
C
n
+
1
are points on the sides
B
n
C
n
B_nC_n
B
n
C
n
,
A
n
C
n
A_nC_n
A
n
C
n
e
A
n
B
n
A_nB_n
A
n
B
n
satisfying
A
n
+
1
B
n
B
n
C
n
=
B
n
+
1
C
n
A
n
C
n
=
C
n
+
1
A
n
A
n
B
n
=
k
.
\frac{A_{n+1}B_n}{B_nC_n} = \frac{B_{n+1}C_n}{A_nC_n} = \frac{C_{n+1}A_n}{A_nB_n} = k.
B
n
C
n
A
n
+
1
B
n
=
A
n
C
n
B
n
+
1
C
n
=
A
n
B
n
C
n
+
1
A
n
=
k
.
Compute the values of
k
k
k
such that the sum of the areas of every triangle
A
n
B
n
C
n
A_nB_nC_n
A
n
B
n
C
n
, for
n
=
1
,
2
,
3
,
…
n = 1, 2, 3, \dots
n
=
1
,
2
,
3
,
…
is equal to
1
3
\dfrac{1}{3}
3
1
of the area of
A
B
C
ABC
A
BC
.