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Chile Classification NMO
2014 Chile Classification NMO Seniors
2014 Chile Classification NMO Seniors
Part of
Chile Classification NMO
Subcontests
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2014 Chile Classification / Qualifying NMO Seniors XXVI
p1. For a positive integer, let's call
<
a
>
<a>
<
a
>
the number obtained by multiplying each figure of a by
2
2
2
and writing the numbers thus obtained. For example
<
126
>
=
2412
<126> = 2412
<
126
>=
2412
and
<
809
>
=
16018
<809> = 16018
<
809
>=
16018
. Prove that it is not possible find two different positive integers
a
a
a
and
b
b
b
such that
<
a
>
=
<
b
>
<a> = <b>
<
a
>=<
b
>
. p2. In how many ways is it possible to cut a graph paper of
6
×
6
6\times 6
6
×
6
starting at the bottom of the paper and working to the top if it can only be cut on the lines of the grid, the two pieces in which it is divided must be equal and it can't be cut down (see figure)? https://cdn.artofproblemsolving.com/attachments/2/1/c9d6a5f0cf7230aff23cde16fd5308e08eac7e.pngNote: Two pieces are considered the same if you can place one over the other and they fit perfectly. p3. In an equilateral triangle
A
B
C
ABC
A
BC
with side
2
2
2
, side
A
B
AB
A
B
is extended to a point
D
D
D
so that
B
B
B
is the midpoint of
A
D
AD
A
D
. Let
E
E
E
be the point on
A
C
AC
A
C
such that
∠
A
D
E
=
1
5
o
\angle ADE = 15^o
∠
A
D
E
=
1
5
o
and take a point
F
F
F
on
A
B
AB
A
B
so that
∣
E
F
∣
=
∣
E
C
∣
|EF| = |EC|
∣
EF
∣
=
∣
EC
∣
. Determine the area of the triangle
A
F
E
AFE
A
FE
. p4. For each positive integer
n
n
n
we consider
S
(
n
)
S (n)
S
(
n
)
as the sum of its digits. For example
S
(
1234
)
=
1
+
2
+
3
+
4
=
10
S (1234) = 1 + 2 + 3 + 4 = 10
S
(
1234
)
=
1
+
2
+
3
+
4
=
10
. Calculate
S
(
1
)
−
S
(
2
)
+
S
(
3
)
−
S
(
4
)
+
.
.
.
−
S
(
2012
)
+
S
(
2013
)
−
S
(
2014
)
S (1)- S (2) + S (3)- S (4) +...- S (2012) + S (2013)- S (2014)
S
(
1
)
−
S
(
2
)
+
S
(
3
)
−
S
(
4
)
+
...
−
S
(
2012
)
+
S
(
2013
)
−
S
(
2014
)
p5. Given
102
102
102
points on a circle, next to one of them is written a
1
1
1
and next to each other a
0
0
0
. The allowed operation consists of choosing a point that has a
1
1
1
and change the number of that point, and also the number of its two neighbors, the one on the left and on the right (where there is a
1
1
1
, write
0
0
0
and where there is a
0
0
0
is written
1
1
1
). Show that it is impossible, with permitted operations, to achieve that all the points have a
0
0
0
. p6. Consider a convex figure
P
P
P
in the plane. For a point
Z
Z
Z
of the plane outside the figure we denote by
∣
Z
,
P
∣
|Z, P|
∣
Z
,
P
∣
the smallest length of the segments that join
Z
Z
Z
with some point of
P
P
P
. Consider a line
L
L
L
that does not intersect the figure
P
P
P
, two points
X
,
Y
X, Y
X
,
Y
over
L
L
L
, and
M
M
M
the midpoint between
X
X
X
and
Y
Y
Y
. Prove that
∣
X
,
P
∣
+
∣
Y
,
P
∣
2
≥
∣
M
,
P
∣
\frac{|X, P| + |Y, P|}{2} \ge |M, P|
2
∣
X
,
P
∣
+
∣
Y
,
P
∣
≥
∣
M
,
P
∣
.PS. Seniors P3, P4 were also proposed as [url=https://artofproblemsolving.com/community/c4h2690804p23355868]Juniors P3, easier P4.