MathDB
Problems
Contests
National and Regional Contests
Poland Contests
Poland - Second Round
1981 Poland - Second Round
1981 Poland - Second Round
Part of
Poland - Second Round
Subcontests
(6)
6
1
Hide problems
lateral surface area of pyramid, sphere inscribed
The surface areas of the bases of a given truncated triangular pyramid are equal to
B
1
B_1
B
1
and
B
2
B_2
B
2
. This pyramid can be cut with a plane parallel to the bases so that a sphere can be inscribed in each of the obtained parts. Prove that the lateral surface area of the given pyramid is
(
B
1
+
B
2
)
(
B
1
4
+
B
2
4
)
2
(\sqrt{B_1} + \sqrt{B_2})(\sqrt[4]{B_1} + \sqrt[4]{B_2})^2
(
B
1
+
B
2
)
(
4
B
1
+
4
B
2
)
2
.
5
1
Hide problems
2 sets of n points each
In the plane there are two disjoint sets
A
A
A
and
B
B
B
, each of which consists of
n
n
n
points, and no three points of the set
A
∪
B
A \cup B
A
∪
B
lie on one straight line. Prove that there is a set of
n
n
n
disjoint closed segments, each of which has one end in the set
A
A
A
and the other in the set
B
B
B
.
4
1
Hide problems
2 sequences with quotients and remainders
The given natural numbers are
k
,
n
k, n
k
,
n
. We inductively define two sequences of numbers
(
a
j
)
(a_j)
(
a
j
)
and
(
r
j
)
(r_j)
(
r
j
)
as follows: Step one: we divide
k
k
k
by
n
n
n
and get the quotient
a
1
a_1
a
1
and the remainder
r
i
r_i
r
i
, step j: we divide
k
+
r
j
−
1
k+r_{j-1}
k
+
r
j
−
1
by
n
n
n
and get the quotient
a
j
a_j
a
j
and the remainder
r
j
r_j
r
j
. Calculate the sum of
a
1
+
…
+
a
n
a_1 + \ldots + a_n
a
1
+
…
+
a
n
.
3
1
Hide problems
f(f(x)) = - x continuous
Prove that there is no continuous function
f
:
R
→
R
f: \mathbb{R} \to \mathbb{R}
f
:
R
→
R
satisfying the condition
f
(
f
(
x
)
)
=
−
x
f(f(x)) = - x
f
(
f
(
x
))
=
−
x
for every
x
x
x
.
2
1
Hide problems
angle bisector wanted, 2 internally tangent circles
Two circles touch internally at point
P
P
P
. A line tangent to one of the circles at point
A
A
A
intersects the other circle at points
B
B
B
and
C
C
C
. Prove that the line
P
A
PA
P
A
is the bisector of the angle
B
P
C
BPC
BPC
.
1
1
Hide problems
\sqrt{\sum (x_j^2+y_j^2)} <=1 / \sqrt{2} \sum \sqrt{x_j^2+y_j^2}.
Prove that for any real numbers
x
1
,
x
2
,
…
,
x
1981
x_1, x_2, \ldots, x_{1981}
x
1
,
x
2
,
…
,
x
1981
,
y
1
,
y
2
,
…
,
y
1981
y_1, y_2, \ldots, y_{1981}
y
1
,
y
2
,
…
,
y
1981
such that
∑
j
=
1
1981
x
j
=
0
\sum_{j=1}^{1981} x_j = 0
∑
j
=
1
1981
x
j
=
0
,
∑
j
=
1
1981
y
j
=
0
\sum_{j=1}^{1981} y_j = 0
∑
j
=
1
1981
y
j
=
0
the inequality occurs
∑
j
=
1
1981
(
x
j
2
+
y
j
2
)
≤
1
2
∑
j
=
1
1981
x
j
2
+
y
j
2
.
\sqrt{\sum_{j=1}^{1981} (x_j^2+y_j^2)} \leq \frac{1}{\sqrt{2}} \sum_{j=1}^{1981} \sqrt{x_j^2+y_j^2}.
j
=
1
∑
1981
(
x
j
2
+
y
j
2
)
≤
2
1
j
=
1
∑
1981
x
j
2
+
y
j
2
.