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Norway Contests
Niels Henrik Abels Math Contest (Norwegian Math Olympiad) Final Round
2023 Abelkonkurransen Finale
2023 Abelkonkurransen Finale
Part of
Niels Henrik Abels Math Contest (Norwegian Math Olympiad) Final Round
Subcontests
(8)
4b
1
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Functional equation f(f(x)+y) = f(y) + x on positive reals
Find all functions
f
:
R
+
→
R
+
f: \mathbb R^{+} \to \mathbb R^{+}
f
:
R
+
→
R
+
satisfying \begin{align*} f(f(x)+y) = f(y) + x, \qquad \text{for all } x,y \in \mathbb R^{+}. \end{align*} Note that
R
+
\mathbb R^{+}
R
+
is the set of all positive real numbers.
4a
1
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Sharp inequality [cyclic sum (a^2+b^2-c^2)/(ab) > 2] in triangle
Assuming
a
,
b
,
c
a,b,c
a
,
b
,
c
are the side-lengths of a triangle, show that \begin{align*} \frac{a^2+b^2-c^2}{ab} + \frac{b^2+c^2-a^2}{bc} + \frac{c^2+a^2-b^2}{ca} > 2. \end{align*} Also show that the inequality does not necessarily hold if you replace
2
2
2
(on the right-hand side) by a bigger by a bigger number.
3b
1
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a^6 + 1 | b^{11} - 2023b^3 + 40b and a^4 - 1 | b^{10} - 2023b^2 - 41
Find all integers
a
a
a
and
b
b
b
satisfying \begin{align*} a^6 + 1 & \mid b^{11} - 2023b^3 + 40b, \qquad \text{and}\\ a^4 - 1 & \mid b^{10} - 2023b^2 - 41. \end{align*}
3a
1
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Solving 2^a + 5^b + 1 = n! in non-negative integers
Find all non-negative integers
n
n
n
,
a
a
a
, and
b
b
b
satisfying
2
a
+
5
b
+
1
=
n
!
.
2^a + 5^b + 1 = n!.
2
a
+
5
b
+
1
=
n
!
.
2b
1
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Arne and Berit play a game on a blackboard
Arne and Berit are playing a game. They have chosen positive integers
m
m
m
and
n
n
n
with
n
≥
4
n\geq 4
n
≥
4
and
m
≤
2
n
+
1
m \leq 2n + 1
m
≤
2
n
+
1
. Arne begins by choosing a number from the set
{
1
,
2
,
…
,
n
}
\{1, 2, \dots , n \}
{
1
,
2
,
…
,
n
}
, and writes it on a blackboard. Then Berit picks another number from the same set, and writes it on the board. They continue alternating turns, always choosing numbers that are not already on the blackboard. When the sum of all the numbers on the board exceeds or equals
m
m
m
, the game is over, and whoever wrote the last number has won. For which combinations of
m
m
m
and
n
n
n
does Arne have a winning strategy?
2a
1
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Covering a triangular grid with rhombi
The sides of an equilateral triangle with sides of length
n
n
n
have been divided into equal parts, each of length
1
1
1
, and lines have been drawn through the points of division parallel to the sides of the triangle, thus dividing the large triangle into many small triangles. Nils has a pile of rhombic tiles, each of side
1
1
1
and angles
6
0
∘
60^\circ
6
0
∘
and
12
0
∘
120^\circ
12
0
∘
, and wants to tile most of the triangle using these, so that each tile covers two small triangles with no overlap. In the picture, three tiles are placed somewhat arbitrarily as an illustration. How many tiles can Nils fit inside the triangle? [asy] /* original code by fedja: https://artofproblemsolving.com/community/c68h207503p1220868 modified by Klaus-Anton: https://artofproblemsolving.com/community/c2083h3267391_draw_me_a_grid_of_regular_triangles */ size(5cm); int n=6; pair A=(1,0), B=dir(60); path P=A--B--(0,0)--cycle; path Pp=A--shift(A)*B--B--cycle;/* label("
A
A
A
",A,S); label("
B
B
B
",B,dir(120)); label("
(
0
,
0
)
(0,0)
(
0
,
0
)
",(0,0),dir(210));fill(shift(2*A-1+2*B-1)*P,yellow+white); fill(shift(2*A-1+2*B-0)*P,yellow+white); fill(shift(2*A-1+2*B+1)*P,yellow+white); fill(shift(2*A-1+2*B+2)*P,yellow+white);fill(shift(1*A-1+1*B)*P,blue+white); fill(shift(2*A-1+1*B)*P,blue+white); fill(shift(3*A-1+1*B)*P,blue+white); fill(shift(4*A-1+1*B)*P,blue+white); fill(shift(5*A-1+1*B)*P,blue+white);fill(shift(0*A+0*B)*P,green+white); fill(shift(0*A+1+0*B)*P,green+white); fill(shift(0*A+2+0*B)*P,green+white); fill(shift(0*A+3+0*B)*P,green+white); fill(shift(0*A+4+0*B)*P,green+white); fill(shift(0*A+5+0*B)*P,green+white);fill(shift(2*A-1+3*B-1)*P,magenta+white); fill(shift(3*A-1+3*B-1)*P,magenta+white); fill(shift(4*A-1+3*B-1)*P,magenta+white);fill(shift(5*A+5*B-5)*P,heavyred+white);fill(shift(4*A+4*B-4)*P,palered+white); fill(shift(4*A+4*B-3)*P,palered+white); fill(shift(0*A+0*B)*Pp,gray); fill(shift(0*A+1+0*B)*Pp,gray); fill(shift(0*A+2+0*B)*Pp,gray); fill(shift(0*A+3+0*B)*Pp,gray); fill(shift(0*A+4+0*B)*Pp,gray);fill(shift(1*A+1*B-1)*Pp,lightgray); fill(shift(1*A+1*B-0)*Pp,lightgray); fill(shift(1*A+1*B+1)*Pp,lightgray); fill(shift(1*A+1*B+2)*Pp,lightgray);fill(shift(2*A+2*B-2)*Pp,red); fill(shift(2*A+2*B-1)*Pp,red); fill(shift(2*A+2*B-0)*Pp,red);fill(shift(3*A+3*B-2)*Pp,blue); fill(shift(3*A+3*B-3)*Pp,blue);fill(shift(4*A+4*B-4)*Pp,cyan);fill(shift(0*A+1+0*B)*Pp,gray); fill(shift(0*A+2+0*B)*Pp,gray); fill(shift(0*A+3+0*B)*Pp,gray); fill(shift(0*A+4+0*B)*Pp,gray); */fill(Pp, rgb(244, 215, 158)); fill(shift(dir(60))*P, rgb(244, 215, 158));fill(shift(1.5,(-sqrt(3)/2))*shift(2*dir(60))*Pp, rgb(244, 215, 158)); fill(shift(1.5,(-sqrt(3)/2))*shift(2*dir(60))*P, rgb(244, 215, 158));fill(shift(-.5,(-sqrt(3)/2))*shift(4*dir(60))*Pp, rgb(244, 215, 158)); fill(shift(.5,(-sqrt(3)/2))*shift(4*dir(60))*P, rgb(244, 215, 158));for(int i=0;i
shipout(bbox(2mm,Fill(white))); [/asy]
1b
1
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The centroid of ABC lies on ME [2023 Abel, Problem 1b]
In the triangle
A
B
C
ABC
A
BC
, points
D
D
D
and
E
E
E
lie on the side
B
C
BC
BC
, with
C
E
=
B
D
CE = BD
CE
=
B
D
. Also,
M
M
M
is the midpoint of
A
D
AD
A
D
. Show that the centroid of
A
B
C
ABC
A
BC
lies on
M
E
ME
ME
.
1a
1
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Show that X,Y, Z are the midpoints of sides of ABC [2023 Abel, 1a]
In the triangle
A
B
C
ABC
A
BC
,
X
X
X
lies on the side
B
C
BC
BC
,
Y
Y
Y
on the side
C
A
CA
C
A
, and
Z
Z
Z
on the side
A
B
AB
A
B
with
Y
X
∥
A
B
,
Z
Y
∥
B
C
YX \| AB, ZY \| BC
Y
X
∥
A
B
,
Z
Y
∥
BC
, and
X
Z
∥
C
A
XZ \| CA
XZ
∥
C
A
. Show that
X
,
Y
X,Y
X
,
Y
, and
Z
Z
Z
are the midpoints of the respective sides of
A
B
C
ABC
A
BC
.