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Contests
National and Regional Contests
Canada Contests
Canadian Mathematical Olympiad Qualification Repechage
2021 Canadian Mathematical Olympiad Qualification
2021 Canadian Mathematical Olympiad Qualification
Part of
Canadian Mathematical Olympiad Qualification Repechage
Subcontests
(8)
8
1
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Circular permutations without succession
King Radford of Peiza is hosting a banquet in his palace. The King has an enormous circular table with
2021
2021
2021
chairs around it. At The King's birthday celebration, he is sitting in his throne (one of the
2021
2021
2021
chairs) and the other
2020
2020
2020
chairs are filled with guests, with the shortest guest sitting to the King's left and the remaining guests seated in increasing order of height from there around the table. The King announces that everybody else must get up from their chairs, run around the table, and sit back down in some chair. After doing this, The King notices that the person seated to his left is different from the person who was previously seated to his left. Each other person at the table also notices that the person sitting to their left is different.Find a closed form expression for the number of ways the people could be sitting around the table at the end. You may use the notation
D
n
,
D_{n},
D
n
,
the number of derangements of a set of size
n
n
n
, as part of your expression.
7
1
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Symmetrical three variable cosine equation
If
A
,
B
A, B
A
,
B
and
C
C
C
are real angles such that
cos
(
B
−
C
)
+
cos
(
C
−
A
)
+
cos
(
A
−
B
)
=
−
3
/
2
,
\cos (B-C)+\cos (C-A)+\cos (A-B)=-3/2,
cos
(
B
−
C
)
+
cos
(
C
−
A
)
+
cos
(
A
−
B
)
=
−
3/2
,
find
cos
(
A
)
+
cos
(
B
)
+
cos
(
C
)
\cos (A)+\cos (B)+\cos (C)
cos
(
A
)
+
cos
(
B
)
+
cos
(
C
)
6
1
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Four variable Diophantine equation with many squares
Show that
(
w
,
x
,
y
,
z
)
=
(
0
,
0
,
0
,
0
)
(w, x, y, z)=(0,0,0,0)
(
w
,
x
,
y
,
z
)
=
(
0
,
0
,
0
,
0
)
is the only integer solution to the equation
w
2
+
11
x
2
−
8
y
2
−
12
y
z
−
10
z
2
=
0
w^{2}+11 x^{2}-8 y^{2}-12 y z-10 z^{2}=0
w
2
+
11
x
2
−
8
y
2
−
12
yz
−
10
z
2
=
0
5
1
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Combinatorial game: cutting rectangles with integer side lengths
Alphonse and Beryl are playing a game. The game starts with two rectangles with integer side lengths. The players alternate turns, with Alphonse going first. On their turn, a player chooses one rectangle, and makes a cut parallel to a side, cutting the rectangle into two pieces, each of which has integer side lengths. The player then discards one of the three rectangles (either the one they did not cut, or one of the two pieces they cut) leaving two rectangles for the other player. A player loses if they cannot cut a rectangle.Determine who wins each of the following games:(a) The starting rectangles are
1
×
2020
1 \times 2020
1
×
2020
and
2
×
4040
2 \times 4040
2
×
4040
. (b) The starting rectangles are
100
×
100
100 \times 100
100
×
100
and
100
×
500
100 \times 500
100
×
500
.
4
1
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Relating the circumdiameter to a triangle in the incircle
Let
O
O
O
be the centre of the circumcircle of triangle
A
B
C
ABC
A
BC
and let
I
I
I
be the centre of the incircle of triangle
A
B
C
ABC
A
BC
. A line passing through the point
I
I
I
is perpendicular to the line
I
O
IO
I
O
and passes through the incircle at points
P
P
P
and
Q
Q
Q
. Prove that the diameter of the circumcircle is equal to the perimeter of triangle
O
P
Q
OPQ
OPQ
.
3
1
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Circles centred at vertices of regular pentagon
A
B
C
D
E
ABCDE
A
BC
D
E
is a regular pentagon. Two circles
C
1
C_1
C
1
and
C
2
C_2
C
2
are drawn through
B
B
B
with centers
A
A
A
and
C
C
C
respectively. Let the other intersection of
C
1
C_1
C
1
and
C
2
C_2
C
2
be
P
P
P
. The circle with center
P
P
P
which passes through
E
E
E
and
D
D
D
intersects
C
2
C_2
C
2
at
X
X
X
and
A
E
AE
A
E
at
Y
Y
Y
. Prove that
A
X
=
A
Y
AX = AY
A
X
=
A
Y
.
2
1
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Symmetrical system of equations
Determine all integer solutions to the system of equations: \begin{align*} xy + yz + zx &= -4 \\ x^2 + y^2 + z^2 &= 24 \\ x^{3} + y^3 + z^3 + 3xyz &= 16 \end{align*}
1
1
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Functional equation with real polynomials
Determine all real polynomials
p
p
p
such that
p
(
x
+
p
(
x
)
)
=
x
2
p
(
x
)
p(x+p(x))=x^2p(x)
p
(
x
+
p
(
x
))
=
x
2
p
(
x
)
for all
x
x
x
.