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Contests
National and Regional Contests
Canada Contests
Canadian Mathematical Olympiad Qualification Repechage
2011 Canadian Mathematical Olympiad Qualification Repechage
2011 Canadian Mathematical Olympiad Qualification Repechage
Part of
Canadian Mathematical Olympiad Qualification Repechage
Subcontests
(8)
8
1
Hide problems
Determine all pairs (m,n) - [Canadian Repêchage 2011]
Determine all pairs
(
n
,
m
)
(n,m)
(
n
,
m
)
of positive integers for which there exists an infinite sequence
{
x
k
}
\{x_k\}
{
x
k
}
of
0
0
0
's and
1
1
1
's with the properties that if
x
i
=
0
x_i=0
x
i
=
0
then
x
i
+
m
=
1
x_{i+m}=1
x
i
+
m
=
1
and if
x
i
=
1
x_i = 1
x
i
=
1
then
x
i
+
n
=
0.
x_{i+n} = 0.
x
i
+
n
=
0.
7
1
Hide problems
What is the minimum? - [Canadian Repêchage 2011]
One thousand students participate in the
2011
2011
2011
Canadian Closed Mathematics Challenge. Each student is assigned a unique three-digit identification number
a
b
c
,
abc,
ab
c
,
where each of
a
,
b
a, b
a
,
b
and
c
c
c
is a digit between
0
0
0
and
9
,
9,
9
,
inclusive. Later, when the contests are marked, a number of markers will be hired. Each of the markers will be given a unique two-digit identification number
x
y
,
xy,
x
y
,
with each of
x
x
x
and
y
y
y
a digit between
0
0
0
and
9
,
9,
9
,
inclusive. Marker
x
y
xy
x
y
will be able to mark any contest with an identification number of the form
x
y
A
xyA
x
y
A
or
x
A
y
xAy
x
A
y
or
A
x
y
,
Axy,
A
x
y
,
for any digit
A
.
A.
A
.
What is the minimum possible number of markers to be hired to ensure that all contests will be marked?
6
1
Hide problems
There exists exactly one alpha - [Canadian Repêchage 2011]
In the diagram,
A
B
D
F
ABDF
A
B
D
F
is a trapezoid with
A
F
AF
A
F
parallel to
B
D
BD
B
D
and
A
B
AB
A
B
perpendicular to
B
D
.
BD.
B
D
.
The circle with center
B
B
B
and radius
A
B
AB
A
B
meets
B
D
BD
B
D
at
C
C
C
and is tangent to
D
F
DF
D
F
at
E
.
E.
E
.
Suppose that
x
x
x
is equal to the area of the region inside quadrilateral
A
B
E
F
ABEF
A
BEF
but outside the circle, that y is equal to the area of the region inside
△
E
B
D
\triangle EBD
△
EB
D
but outside the circle, and that
α
=
∠
E
B
C
.
\alpha = \angle EBC.
α
=
∠
EBC
.
Prove that there is exactly one measure
α
,
\alpha,
α
,
with
0
∘
≤
α
≤
9
0
∘
,
0^\circ \leq \alpha \leq 90^\circ,
0
∘
≤
α
≤
9
0
∘
,
for which
x
=
y
x = y
x
=
y
and that this value of
1
2
<
sin
α
<
1
2
.
\frac 12 < \sin \alpha < \frac{1}{\sqrt 2}.
2
1
<
sin
α
<
2
1
.
[asy] import graph; size(150); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen qqttff = rgb(0,0.2,1); pen fftttt = rgb(1,0.2,0.2); draw(circle((6.04,2.8),1.78),qqttff); draw((6.02,4.58)--(6.04,2.8),fftttt); draw((6.02,4.58)--(6.98,4.56),fftttt); draw((6.04,2.8)--(8.13,2.88),fftttt); draw((6.98,4.56)--(8.13,2.88),fftttt); dot((6.04,2.8),ds); label("
B
B
B
", (5.74,2.46), NE*lsf); dot((6.02,4.58),ds); label("
A
A
A
", (5.88,4.7), NE*lsf); dot((6.98,4.56),ds); label("
F
F
F
", (7.06,4.6), NE*lsf); dot((7.39,3.96),ds); label("
E
E
E
", (7.6,3.88), NE*lsf); dot((8.13,2.88),ds); label("
D
D
D
", (8.34,2.56), NE*lsf); dot((7.82,2.86),ds); label("
C
C
C
", (7.5,2.46), NE*lsf); clip((-4.3,-10.94)--(-4.3,6.3)--(16.18,6.3)--(16.18,-10.94)--cycle); [/asy]
5
1
Hide problems
Gold and Black Vertices - [Canadian Repêchage 2011]
Each vertex of a regular
11
11
11
-gon is colored black or gold. All possible triangles are formed using these vertices. Prove that there are either two congruent triangles with three black vertices or two congruent triangles with three gold vertices.
4
1
Hide problems
Winning Strategy? - [Canadian Repêchage 2011]
Alphonse and Beryl play a game starting with a blank blackboard. Alphonse goes first and the two players alternate turns. On Alphonse's first turn, he writes the integer
1
0
2011
10^{2011}
1
0
2011
on the blackboard. On each subsequent turn, each player can do exactly one of the following two things:(i) replace any number
x
x
x
that is currently on the blackboard with two integers a and b greater than
1
1
1
such that
x
=
a
b
,
x = ab,
x
=
ab
,
or(ii) erase one or two copies of a number
y
y
y
that appears at least twice on the blackboard.Thus, there may be many numbers on the board at any time. The first player who cannot do either of these things loses. Determine which player has a winning strategy and explain the strategy.
3
1
Hide problems
Solve the system of equations - [Canadian Repêchage 2011]
Determine all solutions to the system of equations:
x
2
+
y
2
+
x
+
y
=
12
x^2 + y^2 + x + y = 12
x
2
+
y
2
+
x
+
y
=
12
x
y
+
x
+
y
=
3
xy + x + y = 3
x
y
+
x
+
y
=
3
[This is the exact form of problem that appeared on the paper, but I think it means to solve in
R
.
\mathbb R.
R
.
]
2
1
Hide problems
What are the five numbers? - [Canadian Repêchage 2011]
Brennan chooses a set
A
=
{
a
,
b
,
c
,
d
,
e
}
A = \{a, b,c, d, e \}
A
=
{
a
,
b
,
c
,
d
,
e
}
of five real numbers with
a
≤
b
≤
c
≤
d
≤
e
.
a \leq b \leq c \leq d \leq e.
a
≤
b
≤
c
≤
d
≤
e
.
Delaney determines the subsets of
A
A
A
containing three numbers and adds up the numbers in these subsets. She obtains the sums
0
,
3
;
4
,
8
;
9
,
10
,
11
,
12
,
14
,
19.
0, 3; 4, 8; 9, 10, 11, 12, 14, 19.
0
,
3
;
4
,
8
;
9
,
10
,
11
,
12
,
14
,
19.
What are the five numbers in Brennan's set?
1
1
Hide problems
Cetermine the length of AB - [Canadian Repêchage 2011]
In the diagram, the circle has radius
7
\sqrt 7
7
and and centre
O
.
O.
O
.
Points
A
,
B
A, B
A
,
B
and
C
C
C
are on the circle. If
∠
B
O
C
=
12
0
∘
\angle BOC=120^\circ
∠
BOC
=
12
0
∘
and
A
C
=
A
B
+
1
,
AC = AB + 1,
A
C
=
A
B
+
1
,
determine the length of
A
B
.
AB.
A
B
.
[asy] import graph; size(120); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen qqttff = rgb(0,0.2,1); pen xdxdff = rgb(0.49,0.49,1); pen fftttt = rgb(1,0.2,0.2); draw(circle((2.34,2.4),2.01),qqttff); draw((2.34,2.4)--(1.09,0.82),fftttt); draw((2.34,2.4)--(4.1,1.41),fftttt); draw((1.09,0.82)--(1.4,4.18),fftttt); draw((4.1,1.41)--(1.4,4.18),fftttt); dot((2.34,2.4),ds); label("
O
O
O
", (2.1,2.66),NE*lsf); dot((1.09,0.82),ds); label("
B
B
B
", (0.86,0.46),NE*lsf); dot((4.1,1.41),ds); label("
C
C
C
", (4.2,1.08),NE*lsf); dot((1.4,4.18),ds); label("
A
A
A
", (1.22,4.48),NE*lsf); clip((-4.34,-10.94)--(-4.34,6.3)--(16.14,6.3)--(16.14,-10.94)--cycle); [/asy]