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Problems
Contests
National and Regional Contests
Paraguay Contests
Paraguay Mathematical Olympiad
2013 Paraguay Mathematical Olympiad
2013 Paraguay Mathematical Olympiad
Part of
Paraguay Mathematical Olympiad
Subcontests
(5)
5
1
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Problem 5
Let
A
B
C
ABC
A
BC
be an obtuse triangle, with
A
B
AB
A
B
being the largest side. Draw the angle bisector of
∡
B
A
C
\measuredangle BAC
∡
B
A
C
. Then, draw the perpendiculars to this angle bisector from vertices
B
B
B
and
C
C
C
, and call their feet
P
P
P
and
Q
Q
Q
, respectively.
D
D
D
is the point in the line
B
C
BC
BC
such that
A
D
⊥
A
P
AD \perp AP
A
D
⊥
A
P
. Prove that the lines
A
D
AD
A
D
,
B
Q
BQ
BQ
and
P
C
PC
PC
are concurrent.
4
1
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Problem 4
Pedro and Juan are playing the following game:
−
-
−
There are
2
2
2
piles of rocks, with
X
X
X
rocks in one pile and
Y
Y
Y
rocks in the other pile (
X
<
12
,
Y
<
11
X < 12, Y < 11
X
<
12
,
Y
<
11
).
−
-
−
Each player can draw: --
1
1
1
rock from one of the piles, or --
2
2
2
rocks from one of the piles, or --
1
1
1
rock from each pile, or --
2
2
2
rock from one pile and
1
1
1
from the other pile. Each player must perform one of these four operations in their turns. The looser is the one who takes the last rock. Pedro plays first and has a winning strategy. What are the three maximum possible values of (
X
+
Y
X+Y
X
+
Y
)?
3
1
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Problem 3
We divide a natural number
N
N
N
, that has
k
k
k
digits, by
19
19
19
and get a residue of
1
0
k
−
2
−
Q
10^{k-2} -Q
1
0
k
−
2
−
Q
, where
Q
Q
Q
is the quotient and
Q
<
101
Q < 101
Q
<
101
. Also,
1
0
k
−
2
−
Q
10^{k-2}-Q
1
0
k
−
2
−
Q
is larger than
0
0
0
. How many possible values of
N
N
N
are there?
2
1
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Problem 2
Let
A
B
C
ABC
A
BC
be a triangle with area
9
9
9
, and let
M
M
M
and
N
N
N
be the midpoints of sides
A
B
AB
A
B
and
A
C
AC
A
C
, respectively. Let
P
P
P
be the point in side
B
C
BC
BC
such that
P
C
=
1
3
B
C
PC = \frac{1}{3}BC
PC
=
3
1
BC
. Let
O
O
O
be the intersection point between
P
N
PN
PN
and
C
M
CM
CM
. Find the area of the quadrilateral
B
P
O
M
BPOM
BPOM
.
1
1
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Problem 1
Evaluate the following expression:
201
3
2
+
201
1
2
+
…
+
5
2
+
3
2
−
201
2
2
−
201
0
2
−
…
−
4
2
−
2
2
2013^2 + 2011^2 + … + 5^2 + 3^2 -2012^2 -2010^2-…-4^2-2^2
201
3
2
+
201
1
2
+
…
+
5
2
+
3
2
−
201
2
2
−
201
0
2
−
…
−
4
2
−
2
2