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National and Regional Contests
Hungary Contests
Durer Math Competition
2022 Durer Math Competition (First Round)
2022 Durer Math Competition (First Round)
Part of
Durer Math Competition
Subcontests
(5)
5
2
Hide problems
game master divides a group of 12 players into 2 teams of 6
a) A game master divides a group of
12
12
12
players into two teams of six. The players do not know what the teams are, however the master gives each player a card containing the names of two other players: one of them is a teammate and the other is not, but the master does not tell the player which is which. Can the master write the names on the cards in such a way that the players can determine the teams? (All of the players can work together to do so.)b) On the next occasion, the game master writes the names of
3
3
3
teammates and
1
1
1
opposing player on each card (possibly in a mixed up order). Now he wants to write the names in such away that the players together cannot determine the two teams. Is it possible for him to achieve this?c) Can he write the names in such a way that the players together cannot determine the two teams, if now each card contains the names of
4
4
4
teammates and
1
1
1
opposing player (possibly in a mixed up order)?
a_i = -a_{n+1-i} if sum a_i^{2k+1} = 0 when a_1 <= a_2 <= ... <= a_n
Let
a
1
≤
a
2
≤
.
.
.
≤
a
n
a_1 \le a_2 \le ... \le a_n
a
1
≤
a
2
≤
...
≤
a
n
be real numbers for which
∑
i
=
1
n
a
i
2
k
+
1
=
0
\sum_{i=1}^{n} a_i^{2k+1} = 0
i
=
1
∑
n
a
i
2
k
+
1
=
0
holds for all integers
0
≤
k
<
n
0 \le k < n
0
≤
k
<
n
. Show that in this case,
a
i
=
−
a
n
+
1
−
i
a_i = -a_{n+1-i}
a
i
=
−
a
n
+
1
−
i
holds for all
1
≤
i
≤
n
1 \le i \le n
1
≤
i
≤
n
.
4
2
Hide problems
min no of partitions of 1-100 , each group every 2 are coprime, or any 2 are not
We want to partition the integers
1
,
2
,
3
,
.
.
.
,
100
1, 2, 3, . . . , 100
1
,
2
,
3
,
...
,
100
into several groups such that within each group either any two numbers are coprime or any two are not coprime. At least how many groups are needed for such a partition?We call two integers coprime if they have no common divisor greater than
1
1
1
.
intersection of circles (M,ME) and (N,NF) lies on AB
Let
A
B
C
ABC
A
BC
be an acute triangle, and let
F
A
F_A
F
A
and
F
B
F_B
F
B
be the midpoints of sides
B
C
BC
BC
and
C
A
CA
C
A
, respectively. Let
E
E
E
and
F
F
F
be the intersection points of the circle centered at
F
A
F_A
F
A
and passing through
A
A
A
and the circle centered at
F
B
F_B
F
B
and passing through
B
B
B
. Prove that if segments
C
E
CE
CE
and
C
F
CF
CF
have midpoints
N
N
N
and
M
M
M
, respectively, then the intersection points of the circle centered at
M
M
M
and passing through
E
E
E
and the circle centered at
N
N
N
and passing through
F
F
F
lie on the line
A
B
AB
A
B
.
3
2
Hide problems
Paraflea jumpinng from (x, y) to (x + p, y + p^2)
Paraflea makes jumps on the plane, starting from the origin
(
0
,
0
)
(0, 0)
(
0
,
0
)
. From point
(
x
,
y
)
(x, y)
(
x
,
y
)
it may jump to another point of the form
(
x
+
p
,
y
+
p
2
)
(x + p, y + p^2)
(
x
+
p
,
y
+
p
2
)
, where
p
p
p
is any positive real number. (The value of
p
p
p
may differ for each jump.)a) Is there any point in quadrant
I
I
I
which cannot be reached by the flea? (Quadrant
I
I
I
contains points
(
x
,
y
)
(x, y)
(
x
,
y
)
for which
x
x
x
and
y
y
y
are positive real numbers.)b) What is the minimum number of jumps that the flea must make from the origin so that it gets to the point
(
100
,
1
)
(100, 1)
(
100
,
1
)
?
game master divides a group of 40 players into 4 teams of 10
a) A game master divides a group of
40
40
40
players into four teams of ten. The players do not know what the teams are, however the master gives each player a card containing the names of two other players: one of them is a teammate and the other is not, but the master does not tell the player which is which. Can the master write the names on the cards in such a way that the players can determine the teams? (All of the players can work together to do so.)b) On the next occasion, the game master writes the names of
7
7
7
teammates and
2
2
2
opposing players on each card (possibly in a mixed up order). Now he wants to write the names in such a way that the players together cannot determine the four teams. Is it possible for him to achieve this?c) Can he write the names in such a way that the players together cannot determine the four teams, if now each card contains the names of
6
6
6
teammates and
2
2
2
opposing players (possibly in a mixed up order)?
2
2
Hide problems
R^2 = OP x OQ
In the acute triangle
A
B
C
ABC
A
BC
the circle through
B
B
B
touching the line
A
C
AC
A
C
at
A
A
A
has centre
P
P
P
, the circle through
A
A
A
touching the line
B
C
BC
BC
at
B
B
B
has centre
Q
Q
Q
. Let
R
R
R
and
O
O
O
be the circumradius and circumcentre of triangle
A
B
C
ABC
A
BC
, respectively. Show that
R
2
=
O
P
⋅
O
Q
R^2 = OP \cdot OQ
R
2
=
OP
⋅
OQ
.
all triangles that can be split into 2 congruent pieces by 1 cut
Determine all triangles that can be split into two congruent pieces by one cut. A cut consists of segments
P
1
P
2
P_1P_2
P
1
P
2
,
P
2
P
3
P_2P_3
P
2
P
3
, . . . ,
P
n
−
1
P
n
P_{n-1}P_n
P
n
−
1
P
n
where points
P
1
,
P
2
,
.
.
.
,
P
n
P_1, P_2, . . . , P_n
P
1
,
P
2
,
...
,
P
n
are distinct, points
P
1
P_1
P
1
and
P
n
P_n
P
n
lie on the perimeter of the triangle and the rest of the points lie in the interior of the triangle such that the segments are disjoint except for the endpoints.
1
1
Hide problems
cylindershaped cake for 15 guests
Dorothy organized a party for the birthday of Duck Mom and she also prepared a cylindershaped cake. Since she was originally expecting to have
15
15
15
guests, she divided the top of the cake into this many equal circular sectors, marking where the cuts need to be made. Just for fun Dorothy’s brother Donald split the top of the cake into
10
10
10
equal circular sectors in such a way that some of the radii that he marked coincided with Dorothy’s original markings. Just before the arrival of the guests Douglas cut the cake according to all markings, and then he placed the cake into the fridge. This way they forgot about the cake and only got to eating it when only
6
6
6
of them remained. Is it possible for them to divide the cake into
6
6
6
equal parts without making any further cuts?