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National and Regional Contests
Norway Contests
Niels Henrik Abels Math Contest (Norwegian Math Olympiad) Final Round
2021 Abels Math Contest (Norwegian MO) Final
2021 Abels Math Contest (Norwegian MO) Final
Part of
Niels Henrik Abels Math Contest (Norwegian Math Olympiad) Final Round
Subcontests
(8)
4b
1
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ED=CD or FD=CD, if BE=BF=(|CD^2 -BD^2|)/BC
The tangent at
C
C
C
to the circumcircle of triangle
A
B
C
ABC
A
BC
intersects the line through
A
A
A
and
B
B
B
in a point
D
D
D
. Two distinct points
E
E
E
and
F
F
F
on the line through
B
B
B
and
C
C
C
satisfy
∣
B
E
∣
=
∣
B
F
∣
=
∣
∣
C
D
∣
2
−
∣
B
D
∣
2
∣
∣
B
C
∣
|BE| = |BF | =\frac{||CD|^2 - |BD|^2|}{|BC|}
∣
BE
∣
=
∣
BF
∣
=
∣
BC
∣
∣∣
C
D
∣
2
−
∣
B
D
∣
2
∣
. Show that either
∣
E
D
∣
=
∣
C
D
∣
|ED| = |CD|
∣
E
D
∣
=
∣
C
D
∣
or
∣
F
D
∣
=
∣
C
D
∣
|FD| = |CD|
∣
F
D
∣
=
∣
C
D
∣
.
4a
1
Hide problems
(BAC)^2+(CAD)^2 +(DAB)^2=(BCD)^2 for tetrahedron, <BAC=<CAD=<DAB=90^o
A tetrahedron
A
B
C
D
ABCD
A
BC
D
satisfies
∠
B
A
C
=
∠
C
A
D
=
∠
D
A
B
=
9
0
o
\angle BAC=\angle CAD=\angle DAB=90^o
∠
B
A
C
=
∠
C
A
D
=
∠
D
A
B
=
9
0
o
. Show that the areas of its faces satisfy the equation
a
r
e
a
(
B
A
C
)
2
+
a
r
e
a
(
C
A
D
)
2
+
a
r
e
a
(
D
A
B
)
2
=
a
r
e
a
(
B
C
D
)
2
area(BAC)^2 + area(CAD)^2 + area(DAB)^2 = area(BCD)^2
a
re
a
(
B
A
C
)
2
+
a
re
a
(
C
A
D
)
2
+
a
re
a
(
D
A
B
)
2
=
a
re
a
(
BC
D
)
2
. .
3b
1
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exists a synchronous set containing 2021 different natural numbers.
We say that a set
S
S
S
of natural numbers is synchronous provided that the digits of
a
2
a^2
a
2
are the same (in occurence and numbers, if differently ordered) for all numbers
a
a
a
in
S
S
S
. For example,
{
13
,
14
,
31
}
\{13, 14, 31\}
{
13
,
14
,
31
}
is synchronous, since we find
{
1
3
2
,
1
4
2
,
3
1
2
}
=
{
169
,
196
,
961
}
\{13^2, 14^2, 31^2\} = \{169, 196, 961\}
{
1
3
2
,
1
4
2
,
3
1
2
}
=
{
169
,
196
,
961
}
. But
{
119
,
121
}
\{119, 121\}
{
119
,
121
}
is not synchronous, for even though
11
9
2
=
14161
119^2 = 14161
11
9
2
=
14161
and
12
1
2
=
14641
121^2 = 14641
12
1
2
=
14641
have the same digits, they occur in different numbers. Show that there exists a synchronous set containing
2021
2021
2021
different natural numbers.
3a
1
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3^m + 3^n + k is perfect square for 0<=k<=9
For which integers
0
≤
k
≤
9
0 \le k \le 9
0
≤
k
≤
9
do there exist positive integers
m
m
m
and
n
n
n
so that the number
3
m
+
3
n
+
k
3^m + 3^n + k
3
m
+
3
n
+
k
is a perfect square?
2a
1
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for every n >=3 exist n different positive integers x_i with sum 1\x_i =1
Show that for all
n
≥
3
n\ge 3
n
≥
3
there are
n
n
n
different positive integers
x
1
,
x
2
,
.
.
.
,
x
n
x_1,x_2, ...,x_n
x
1
,
x
2
,
...
,
x
n
such that
1
x
1
+
1
x
2
+
.
.
.
+
1
x
n
=
1.
\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}= 1.
x
1
1
+
x
2
1
+
...
+
x
n
1
=
1.
1b
1
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max no of chickens, an index card for each chicken, 10 boxes, 2021 cards
Pål has more chickens than he can manage to keep track of. Therefore, he keeps an index card for each chicken. He keeps the cards in ten boxes, each of which has room for
2021
2021
2021
cards. Unfortunately, Pål is quite disorganized, so he may lose some of his boxes. Therefore, he makes several copies of each card and distributes them among different boxes, so that even if he can only find seven boxes, no matter which seven, these seven boxes taken together will contain at least one card for each of his chickens. What is the largest number of chickens Pål can keep track of using this system?
1a
1
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no of tidy 3n-tables
A
3
n
3n
3
n
-table is a table with three rows and
n
n
n
columns containing all the numbers
1
,
2
,
…
,
3
n
1, 2, …, 3n
1
,
2
,
…
,
3
n
. Such a table is called tidy if the
n
n
n
numbers in the first row appear in ascending order from left to right, and the three numbers in each column appear in ascending order from top to bottom. How many tidy
3
n
3n
3
n
-tables exist?
2b
1
Hide problems
Ineq from Abel competition 2020-2021
If
a
1
,
⋯
,
a
n
a_1,\cdots,a_n
a
1
,
⋯
,
a
n
and
b
1
,
⋯
,
b
n
b_1,\cdots,b_n
b
1
,
⋯
,
b
n
are real numbers satisfying
a
1
2
+
⋯
+
a
n
2
≤
1
a_1^2+\cdots+a_n^2 \le 1
a
1
2
+
⋯
+
a
n
2
≤
1
and
b
1
2
+
⋯
+
b
n
2
≤
1
b_1^2+\cdots+b_n^2 \le 1
b
1
2
+
⋯
+
b
n
2
≤
1
, show that:
(
1
−
(
a
1
2
+
⋯
+
a
n
2
)
)
(
1
−
(
b
1
2
+
⋯
+
b
n
2
)
)
≤
(
1
−
(
a
1
b
1
+
⋯
+
a
n
b
n
)
)
2
(1-(a_1^2+\cdots+a_n^2))(1-(b_1^2+\cdots+b_n^2)) \le (1-(a_1b_1+\cdots+a_nb_n))^2
(
1
−
(
a
1
2
+
⋯
+
a
n
2
))
(
1
−
(
b
1
2
+
⋯
+
b
n
2
))
≤
(
1
−
(
a
1
b
1
+
⋯
+
a
n
b
n
)
)
2