MathDB
Problems
Contests
National and Regional Contests
Canada Contests
Canada National Olympiad
1981 Canada National Olympiad
1981 Canada National Olympiad
Part of
Canada National Olympiad
Subcontests
(5)
5
1
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11 theatrical groups at a festival watch each other
11
11
11
theatrical groups participated in a festival. Each day, some of the groups were scheduled to perform while the remaining groups joined the general audience. At the conclusion of the festival, each group had seen, during its days off, at least
1
1
1
performance of every other group. At least how many days did the festival last?
4
1
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Two polynomials, prove P(P(x))=Q(Q(x)) has no real solution
P
(
x
)
,
Q
(
x
)
P(x),Q(x)
P
(
x
)
,
Q
(
x
)
are two polynomials such that
P
(
x
)
=
Q
(
x
)
P(x)=Q(x)
P
(
x
)
=
Q
(
x
)
has no real solution, and
P
(
Q
(
x
)
)
≡
Q
(
P
(
x
)
)
∀
x
∈
R
P(Q(x))\equiv Q(P(x))\forall x\in\mathbb{R}
P
(
Q
(
x
))
≡
Q
(
P
(
x
))
∀
x
∈
R
. Prove that
P
(
P
(
x
)
)
=
Q
(
Q
(
x
)
)
P(P(x))=Q(Q(x))
P
(
P
(
x
))
=
Q
(
Q
(
x
))
has no real solution.
3
1
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Lines and Circles on a Plane
Given a finite collection of lines in a plane
P
P
P
, show that it is possible to draw an arbitrarily large circle in
P
P
P
which does not meet any of them. On the other hand, show that it is possible to arrange a countable infinite sequence of lines (first line, second line, third line, etc.) in
P
P
P
so that every circle in
P
P
P
meets at least one of the lines. (A point is not considered to be a circle.)
2
1
Hide problems
Maximum Area of Triangle PQR
Given a circle of radius
r
r
r
and a tangent line
ℓ
\ell
ℓ
to the circle through a given point
P
P
P
on the circle. From a variable point
R
R
R
on the circle, a perpendicular
R
Q
RQ
RQ
is drawn to
ℓ
\ell
ℓ
with
Q
Q
Q
on
ℓ
\ell
ℓ
. Determine the maximum of the area of triangle
P
Q
R
PQR
PQR
.
1
1
Hide problems
Solve Equation with Floor Function Sum
For any real number
t
t
t
, denote by
[
t
]
[t]
[
t
]
the greatest integer which is less than or equal to
t
t
t
. For example:
[
8
]
=
8
[8] = 8
[
8
]
=
8
,
[
π
]
=
3
[\pi] = 3
[
π
]
=
3
, and
[
−
5
/
2
]
=
−
3
[-5/2] = -3
[
−
5/2
]
=
−
3
. Show that the equation
[
x
]
+
[
2
x
]
+
[
4
x
]
+
[
8
x
]
+
[
16
x
]
+
[
32
x
]
=
12345
[x] + [2x] + [4x] + [8x] + [16x] + [32x] = 12345
[
x
]
+
[
2
x
]
+
[
4
x
]
+
[
8
x
]
+
[
16
x
]
+
[
32
x
]
=
12345
has no real solution.