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Problems
Contests
National and Regional Contests
Spain Contests
Spain Mathematical Olympiad
1971 Spain Mathematical Olympiad
1971 Spain Mathematical Olympiad
Part of
Spain Mathematical Olympiad
Subcontests
(8)
5
1
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1/(1-z) = lim }(1 + z)(1 + z^2)(1 + z^{2^2} ...
Prove that whatever the complex number
z
z
z
is, it is true that
(
1
+
z
2
n
)
(
1
−
z
2
n
)
=
1
−
z
2
n
+
1
.
(1 + z^{2^n})(1-z^{2^n})= 1- z^{2^{n+1}}.
(
1
+
z
2
n
)
(
1
−
z
2
n
)
=
1
−
z
2
n
+
1
.
Writing the equalities that result from giving
n
n
n
the values
0
,
1
,
2
,
.
.
.
0, 1, 2, . . .
0
,
1
,
2
,
...
and multiplying them, show that for
∣
z
∣
<
1
|z| < 1
∣
z
∣
<
1
holds
1
1
−
z
=
lim
k
→
∞
(
1
+
z
)
(
1
+
z
2
)
(
1
+
z
2
2
)
.
.
.
(
1
+
z
2
k
)
.
\frac{1}{1-z}= \lim_{k\to \infty}(1 + z)(1 + z^2)(1 + z^{2^2})...(1 + z^{2^k}).
1
−
z
1
=
k
→
∞
lim
(
1
+
z
)
(
1
+
z
2
)
(
1
+
z
2
2
)
...
(
1
+
z
2
k
)
.
4
1
Hide problems
(a A+bB+cC)/(a+b+c) >= \pi /3
Prove that in every triangle with sides
a
,
b
,
c
a, b, c
a
,
b
,
c
and opposite angles
A
,
B
,
C
A, B, C
A
,
B
,
C
, is fulfilled (measuring the angles in radians)
a
A
+
b
B
+
c
C
a
+
b
+
c
≥
π
3
\frac{a A+bB+cC}{a+b+c} \ge \frac{\pi}{3}
a
+
b
+
c
a
A
+
b
B
+
c
C
≥
3
π
Hint: Use
a
≥
b
≥
c
⇒
A
≥
B
≥
C
a \ge b \ge c \Rightarrow A \ge B \ge C
a
≥
b
≥
c
⇒
A
≥
B
≥
C
.
8
1
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n + 1 different numbers among 1-2n, exist >=2, one divides another
Among the
2
n
2n
2
n
numbers
1
,
2
,
3
,
.
.
.
,
2
n
1, 2, 3, . . . , 2n
1
,
2
,
3
,
...
,
2
n
are chosen in any way
n
+
1
n + 1
n
+
1
different numbers. Prove that among the chosen numbers there are at least two, such that one divides the other.
6
1
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the fastest path a submarine to move
The velocities of a submerged and surfaced submarine are, respectively,
v
v
v
and
k
v
kv
k
v
. It is situated at a point
P
P
P
at
30
30
30
miles from the center
O
O
O
of a circle of
60
60
60
mile radius. The surveillance of an enemy squadron forces him to navigate submerged while inside the circle. Discuss, according to the values of
k
k
k
, the fastest path to move to the opposite end of the diameter that passes through
P
P
P
. (Consider the case particular
k
=
5
k =\sqrt5
k
=
5
.)
1
1
Hide problems
sum_{k=5}^{k=49} 11_(k / 2\sqrt[3]{1331_(k} }
Calculate
∑
k
=
5
k
=
49
1
1
(
k
2
133
1
(
k
3
\sum_{k=5}^{k=49}\frac{11_(k}{2\sqrt[3]{1331_(k}}
k
=
5
∑
k
=
49
2
3
133
1
(
k
1
1
(
k
knowing that the numbers
11
11
11
and
1331
1331
1331
are written in base
k
≥
4
k \ge 4
k
≥
4
.
3
1
Hide problems
(px + qy)^2 <= px^2 + qy^2 if 0 < p, 0 < q, p +q < 1
If
0
<
p
0 < p
0
<
p
,
0
<
q
0 < q
0
<
q
and
p
+
q
<
1
p +q < 1
p
+
q
<
1
prove
(
p
x
+
q
y
)
2
≤
p
x
2
+
q
y
2
(px + qy)^2 \le px^2 + qy^2
(
p
x
+
q
y
)
2
≤
p
x
2
+
q
y
2
2
1
Hide problems
new axiomatic geometry, collinearity related
In a certain geometry we operate with two types of elements, points and lines, related to each other by the following axioms: I. Given two points
A
A
A
and
B
B
B
, there is a unique line
(
A
B
)
(AB)
(
A
B
)
that passes through both. II. There are at least two points on a line. There are three points not situated on a straight line. III. When a point
B
B
B
is located between
A
A
A
and
C
C
C
, then
B
B
B
is also between
C
C
C
and
A
A
A
. (
A
,
B
,
C
A, B, C
A
,
B
,
C
are three different points on a line.) IV. Given two points
A
A
A
and
C
C
C
, there exists at least one point
B
B
B
on the line
(
A
C
)
(AC)
(
A
C
)
of the form that C is between
A
A
A
and
B
B
B
. V. Among three points located on the same straight line, one at most is between the other two. VI. If
A
,
B
,
C
A, B, C
A
,
B
,
C
are three points not lying on the same line and a is a line that does not contain any of the three, when the line passes through a point on segment [AB] , then it goes through one of the
[
B
C
]
[BC]
[
BC
]
, or it goes through one of the [AC] . (We designate by [AB] the set of points that lie between
A
A
A
and
B
B
B
.)From the previous axioms, prove the following propositions: Theorem 1. Between points A and C there is at least one point
B
B
B
. Theorem 2. Among three points located on a line, one is always between the two others.
7
1
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invert 2 concentric circles into 2 equal circles
Transform by inversion two concentric and coplanar circles into two equal.