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Problems
Contests
National and Regional Contests
Indonesia Contests
Indonesia Regional
2022 Indonesia Regional
2022 Indonesia Regional
Part of
Indonesia Regional
Subcontests
(6)
5
1
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As Inefficient As Possible
Numbers
1
1
1
to
22
22
22
are written on a board. A "move" is a procedure of picking two numbers
a
,
b
a,b
a
,
b
on the board such that
b
≥
a
+
2
b \geq a+2
b
≥
a
+
2
, then erasing
a
a
a
and
b
b
b
to be replaced with
a
+
1
a+1
a
+
1
and
b
−
1
b-1
b
−
1
. Determine the maximum possible number of moves that can be done on the board.
4
1
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Elementary Geometry for Mathematical Olympiads
Suppose
A
B
C
ABC
A
BC
is a triangle with circumcenter
O
O
O
. Point
D
D
D
is the reflection of
A
A
A
with respect to
B
C
BC
BC
. Suppose
ℓ
\ell
ℓ
is the line which is parallel to
B
C
BC
BC
and passes through
O
O
O
. The line through
B
B
B
and parallel to
C
D
CD
C
D
meets
ℓ
\ell
ℓ
at
B
1
B_1
B
1
. Lines
C
B
1
CB_1
C
B
1
and
B
D
BD
B
D
intersect at point
B
2
B_2
B
2
. The line through
C
C
C
parallel to
B
D
BD
B
D
and
ℓ
\ell
ℓ
meet at
C
1
C_1
C
1
. Finally,
B
C
1
BC_1
B
C
1
and
C
D
CD
C
D
intersects at point
C
2
C_2
C
2
. Prove that points
A
,
B
2
,
C
2
,
D
A, B_2, C_2, D
A
,
B
2
,
C
2
,
D
lie on a circle.
3
1
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Not This Type of Inequality
It is known that
x
x
x
and
y
y
y
are reals satisfying
5
x
2
+
4
x
y
+
11
y
2
=
3.
5x^2 + 4xy + 11y^2 = 3.
5
x
2
+
4
x
y
+
11
y
2
=
3.
Without using calculus (differentials/integrals), determine the maximum value of
x
y
−
2
x
+
5
y
xy - 2x + 5y
x
y
−
2
x
+
5
y
.
2
1
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Perfect Square, or otherwise
(a) Determine a natural number
n
n
n
such that
n
(
n
+
2022
)
+
2
n(n+2022)+2
n
(
n
+
2022
)
+
2
is a perfect square. In case you didn't realize,
n
=
1
n=1
n
=
1
works lol (b) Determine all natural numbers
a
a
a
such that for every natural number
n
n
n
, the number
n
(
n
+
a
)
+
2
n(n+a)+2
n
(
n
+
a
)
+
2
is never a perfect square.
1
1
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How many subsets belong to either set?
Let
A
A
A
and
B
B
B
be sets such that there are exactly
144
144
144
sets which are subsets of either
A
A
A
or
B
B
B
. Determine the number of elements
A
∪
B
A \cup B
A
∪
B
has.
1
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Indonesian Regional MO 2022 - Part 1*
The test this year was held on Monday, 22 August 2022 on 09.10-11.40 (GMT+7) for the essay section and was held on 12.05-13.30 (GMT+7) for the short answers section, which was to be done in an hour using the Moodle Learning Management System. Each problem in this section has a weight of 2 points, with 0 points for incorrect or unanswered problems, whereas in the essay section each problem has a weight of 7 points. As with last year, calculators, protractors and set squares are prohibited. (Of course abacus is also prohibited, but who uses abacus?)Anyway here are the problems.Part 1: Speed Round (60 minutes) The problem was presented in no particular order, although here I will order it in a rough order of increasing difficulty.Problem 1. The number of positive integer solutions
(
m
,
n
)
(m,n)
(
m
,
n
)
to the equation
m
n
=
1
7
324
m^n = 17^{324}
m
n
=
1
7
324
is
…
.
\ldots.
…
.
[url=https://artofproblemsolving.com/community/c6h2909996_7_digits_numbers]Problem 2. Consider the increasing sequence of all 7-digit numbers consisting of all of the following digits:
1
,
2
,
3
,
4
,
5
,
6
,
7
1,2,3,4,5,6,7
1
,
2
,
3
,
4
,
5
,
6
,
7
. The 2024th term of the sequence is
…
.
\ldots.
…
.
Problem 3. Suppose
(
a
,
b
)
(a,b)
(
a
,
b
)
is a positive integer solution of the equation
a
+
15
b
=
a
15
b
.
\sqrt{a + \frac{15}{b}} = a \sqrt{ \frac{15}{b}}.
a
+
b
15
=
a
b
15
.
The sum of all possible values of
b
b
b
is
…
.
\ldots.
…
.
Problem 4. It is known that
A
B
C
D
ABCD
A
BC
D
is a trapezoid such that
A
B
AB
A
B
is parallel to
C
D
CD
C
D
, with the length of
A
B
=
6
AB = 6
A
B
=
6
and
C
D
=
7
CD = 7
C
D
=
7
. It is known that points
P
P
P
and
Q
Q
Q
are on
A
D
AD
A
D
and
B
C
BC
BC
respectively such that
P
Q
PQ
PQ
is parallel to
A
B
AB
A
B
. If the perimeter of trapezoid
A
B
Q
P
ABQP
A
BQP
is the same as the perimeter of
P
Q
C
D
PQCD
PQC
D
and
A
D
+
B
C
=
10
AD + BC = 10
A
D
+
BC
=
10
, then the length of
20
P
Q
20PQ
20
PQ
is
…
.
\ldots.
…
.
Problem 5. Suppose
A
B
C
ABC
A
BC
is a triangle with side lengths
A
B
=
16
AB = 16
A
B
=
16
,
A
C
=
23
AC=23
A
C
=
23
, and
∠
B
A
C
=
3
0
∘
\angle{BAC} = 30^{\circ}
∠
B
A
C
=
3
0
∘
. The maximum possible area of a rectangle whose one of its sides lies on the line
B
C
BC
BC
, and the two other vertices each lie on
A
B
AB
A
B
and
A
C
AC
A
C
is
…
.
\ldots.
…
.
Problem 6. Suppose
a
,
b
,
c
a,b,c
a
,
b
,
c
are natural numbers such that
a
+
2
b
+
3
c
=
73
a+2b+3c=73
a
+
2
b
+
3
c
=
73
. The minimum possible value of
a
2
+
b
2
+
c
2
a^2+b^2+c^2
a
2
+
b
2
+
c
2
is
…
.
\ldots.
…
.
Problem 7. An equilateral triangle with a side length of
21
21
21
is partitioned into
2
1
2
21^2
2
1
2
unit equilateral triangles, and the sides of the small equilateral triangles are all parallel to the original large triangle. The number of paralellograms which are made up of the unit equilateral triangles is
21
k
21k
21
k
. Then the value of
k
=
…
.
k = \ldots.
k
=
…
.
Problem 8. Define the sequence
{
a
n
}
\{a_n\}
{
a
n
}
with
a
1
>
3
a_1 > 3
a
1
>
3
, and for all
n
≥
1
n \geq 1
n
≥
1
, the following condition:
2
a
n
+
1
=
a
n
(
−
1
+
4
a
n
−
3
)
2a_{n+1} = a_n(-1 + \sqrt{4a_n - 3})
2
a
n
+
1
=
a
n
(
−
1
+
4
a
n
−
3
)
is satisfied. If
∣
a
1
−
a
2022
∣
=
2023
\vert a_1 - a_{2022} \vert = 2023
∣
a
1
−
a
2022
∣
=
2023
, then the value of
∑
i
=
1
2021
a
i
+
1
3
a
i
2
+
a
i
a
i
+
1
+
a
i
+
1
2
=
…
.
\sum_{i=1}^{2021} \frac{a_{i+1}^3}{a_i^2 + a_i a_{i+1} + a_{i+1}^2} = \ldots.
i
=
1
∑
2021
a
i
2
+
a
i
a
i
+
1
+
a
i
+
1
2
a
i
+
1
3
=
…
.
Problem 9. Suppose
P
(
x
)
P(x)
P
(
x
)
is an integer polynomial such that
P
(
6
)
P
(
38
)
P
(
57
)
+
19
P(6)P(38)P(57) + 19
P
(
6
)
P
(
38
)
P
(
57
)
+
19
is divisible by
114
114
114
. If
P
(
−
13
)
=
479
P(-13) = 479
P
(
−
13
)
=
479
, and
P
(
0
)
≥
0
P(0) \geq 0
P
(
0
)
≥
0
, then the minimum value of
P
(
0
)
P(0)
P
(
0
)
is
…
.
\ldots.
…
.
Problem 10. The number of nonempty subsets of
S
=
{
1
,
2
,
…
,
21
}
S = \{1,2, \ldots, 21\}
S
=
{
1
,
2
,
…
,
21
}
such that the sum of its elements is divisble by 4 is
2
k
−
m
2^k - m
2
k
−
m
; where
k
,
m
∈
Z
k, m \in \mathbb{Z}
k
,
m
∈
Z
and
0
≤
m
<
2022
0 \leq m < 2022
0
≤
m
<
2022
. The value of
10
k
+
m
10k + m
10
k
+
m
is
…
.
\ldots.
…
.