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Problems
Contests
National and Regional Contests
Canada Contests
Canada National Olympiad
1998 Canada National Olympiad
1998 Canada National Olympiad
Part of
Canada National Olympiad
Subcontests
(5)
5
1
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Equation and sequence
Let
m
m
m
be a positive integer. Define the sequence
a
0
,
a
1
,
a
2
,
⋯
a_0, a_1, a_2, \cdots
a
0
,
a
1
,
a
2
,
⋯
by
a
0
=
0
,
a
1
=
m
,
a_0 = 0,\; a_1 = m,
a
0
=
0
,
a
1
=
m
,
and
a
n
+
1
=
m
2
a
n
−
a
n
−
1
a_{n+1} = m^2a_n - a_{n-1}
a
n
+
1
=
m
2
a
n
−
a
n
−
1
for
n
=
1
,
2
,
3
,
⋯
n = 1,2,3,\cdots
n
=
1
,
2
,
3
,
⋯
. Prove that an ordered pair
(
a
,
b
)
(a,b)
(
a
,
b
)
of non-negative integers, with
a
≤
b
a \leq b
a
≤
b
, gives a solution to the equation
a
2
+
b
2
a
b
+
1
=
m
2
{\displaystyle \frac{a^2 + b^2}{ab + 1} = m^2}
ab
+
1
a
2
+
b
2
=
m
2
if and only if
(
a
,
b
)
(a,b)
(
a
,
b
)
is of the form
(
a
n
,
a
n
+
1
)
(a_n,a_{n+1})
(
a
n
,
a
n
+
1
)
for some
n
≥
0
n \geq 0
n
≥
0
.
4
1
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Angle chasing in triangle
Let
A
B
C
ABC
A
BC
be a triangle with
∠
B
A
C
=
4
0
∘
\angle{BAC} = 40^{\circ}
∠
B
A
C
=
4
0
∘
and
∠
A
B
C
=
6
0
∘
\angle{ABC}=60^{\circ}
∠
A
BC
=
6
0
∘
. Let
D
D
D
and
E
E
E
be the points lying on the sides
A
C
AC
A
C
and
A
B
AB
A
B
, respectively, such that
∠
C
B
D
=
4
0
∘
\angle{CBD} = 40^{\circ}
∠
CB
D
=
4
0
∘
and
∠
B
C
E
=
7
0
∘
\angle{BCE} = 70^{\circ}
∠
BCE
=
7
0
∘
. Let
F
F
F
be the point of intersection of the lines
B
D
BD
B
D
and
C
E
CE
CE
. Show that the line
A
F
AF
A
F
is perpendicular to the line
B
C
BC
BC
.
3
1
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Very classical inequatily with natural numbers
Let
n
n
n
be a natural number such that
n
≥
2
n \geq 2
n
≥
2
. Show that \frac {1}{n \plus{} 1} \left( 1 \plus{} \frac {1}{3} \plus{} \cdot \cdot \cdot \plus{} \frac {1}{2n \minus{} 1} \right) > \frac {1}{n} \left( \frac {1}{2} \plus{} \frac {1}{4} \plus{} \cdot \cdot \cdot \plus{} \frac {1}{2n} \right).
2
1
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Equation of exponents
Find all real numbers
x
x
x
such that:
x
=
x
−
1
x
+
1
−
1
x
x = \sqrt{ x - \frac{1}{x} } + \sqrt{ 1 - \frac{1}{x} }
x
=
x
−
x
1
+
1
−
x
1
1
1
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[x] equation
Determine the number of real solutions
a
a
a
to the equation:
[
1
2
a
]
+
[
1
3
a
]
+
[
1
5
a
]
=
a
.
\left[\,\frac{1}{2}\;a\,\right]+\left[\,\frac{1}{3}\;a\,\right]+\left[\,\frac{1}{5}\;a\,\right] = a.
[
2
1
a
]
+
[
3
1
a
]
+
[
5
1
a
]
=
a
.
Here, if
x
x
x
is a real number, then
[
x
]
[\,x\,]
[
x
]
denotes the greatest integer that is less than or equal to
x
x
x
.