MathDB
Problems
Contests
National and Regional Contests
Mexico Contests
Mexico National Olympiad
2012 Mexico National Olympiad
2012 Mexico National Olympiad
Part of
Mexico National Olympiad
Subcontests
(6)
6
1
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Two incenters on a parallel line
Consider an acute triangle
A
B
C
ABC
A
BC
with circumcircle
C
\mathcal{C}
C
. Let
H
H
H
be the orthocenter of
A
B
C
ABC
A
BC
and
M
M
M
the midpoint of
B
C
BC
BC
. Lines
A
H
AH
A
H
,
B
H
BH
B
H
and
C
H
CH
C
H
cut
C
\mathcal{C}
C
again at points
D
D
D
,
E
E
E
, and
F
F
F
respectively; line
M
H
MH
M
H
cuts
C
\mathcal{C}
C
at
J
J
J
such that
H
H
H
lies between
J
J
J
and
M
M
M
. Let
K
K
K
and
L
L
L
be the incenters of triangles
D
E
J
DEJ
D
E
J
and
D
F
J
DFJ
D
F
J
respectively. Prove
K
L
KL
K
L
is parallel to
B
C
BC
BC
.
5
1
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Jumping colored frog puzzle
Some frogs, some red and some others green, are going to move in an
11
×
11
11 \times 11
11
×
11
grid, according to the following rules. If a frog is located, say, on the square marked with # in the following diagram, then[*]If it is red, it can jump to any square marked with an x. [*]if it is green, it can jump to any square marked with an o. \begin{tabular}{| p{0.08cm} | p{0.08cm} | p{0.08cm} | p{0.08cm} | p{0.08cm} | p{0.08cm} | p{0.08cm} | p{0.08cm} | p{0.08cm} | l} \hline &&&&&&\\ \hline &&x&&o&&\\ \hline &o&&&&x&\\ \hline &&&\small{\#}&&&\\ \hline &x&&&&o&\\ \hline &&o&&x&&\\ \hline &&&&&&\\ \hline \end{tabular} We say 2 frogs (of any color) can meet at a square if both can get to the same square in one or more jumps, not neccesarily with the same amount of jumps.[*]Prove if 6 frogs are placed, then there exist at least 2 that can meet at a square. [*]For which values of
k
k
k
is it possible to place one green and one red frog such that they can meet at exactly
k
k
k
squares?
4
1
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Operations on positive integers
The following process is applied to each positive integer: the sum of its digits is subtracted from the number, and the result is divided by
9
9
9
. For example, the result of the process applied to
938
938
938
is
102
102
102
, since
938
−
(
9
+
3
+
8
)
9
=
102.
\frac{938-(9 + 3 + 8)}{9} = 102.
9
938
−
(
9
+
3
+
8
)
=
102.
Applying the process twice to
938
938
938
the result is
11
11
11
, applied three times the result is
1
1
1
, and applying it four times the result is
0
0
0
. When the process is applied one or more times to an integer
n
n
n
, the result is eventually
0
0
0
. The number obtained before obtaining
0
0
0
is called the house of
n
n
n
.How many integers less than
26000
26000
26000
share the same house as
2012
2012
2012
?
3
1
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Relative primes
Prove among any
14
14
14
consecutive positive integers there exist
6
6
6
which are pairwise relatively prime.
2
1
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Numbers in an n x n grid
Let
n
≥
4
n \geq 4
n
≥
4
be an even integer. Consider an
n
×
n
n \times n
n
×
n
grid. Two cells (
1
×
1
1 \times 1
1
×
1
squares) are neighbors if they share a side, are in opposite ends of a row, or are in opposite ends of a column. In this way, each cell in the grid has exactly four neighbors. An integer from 1 to 4 is written inside each square according to the following rules:[*]If a cell has a 2 written on it, then at least two of its neighbors contain a 1. [*]If a cell has a 3 written on it, then at least three of its neighbors contain a 1. [*]If a cell has a 4 written on it, then all of its neighbors contain a 1.Among all arrangements satisfying these conditions, what is the maximum number that can be obtained by adding all of the numbers on the grid?
1
1
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Two circles
Let
C
1
\mathcal{C}_1
C
1
be a circumference with center
O
O
O
,
P
P
P
a point on it and
ℓ
\ell
ℓ
the line tangent to
C
1
\mathcal{C}_1
C
1
at
P
P
P
. Consider a point
Q
Q
Q
on
ℓ
\ell
ℓ
different from
P
P
P
, and let
C
2
\mathcal{C}_2
C
2
be the circumference passing through
O
O
O
,
P
P
P
and
Q
Q
Q
. Segment
O
Q
OQ
OQ
cuts
C
1
\mathcal{C}_1
C
1
at
S
S
S
and line
P
S
PS
PS
cuts
C
2
\mathcal{C}_2
C
2
at a point
R
R
R
diffferent from
P
P
P
. If
r
1
r_1
r
1
and
r
2
r_2
r
2
are the radii of
C
1
\mathcal{C}_1
C
1
and
C
2
\mathcal{C}_2
C
2
respectively, Prove
P
S
S
R
=
r
1
r
2
.
\frac{PS}{SR} = \frac{r_1}{r_2}.
SR
PS
=
r
2
r
1
.