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Problems
Contests
International Contests
Czech-Polish-Slovak Match
2008 Czech-Polish-Slovak Match
2008 Czech-Polish-Slovak Match
Part of
Czech-Polish-Slovak Match
Subcontests
(3)
3
2
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Wolstenholme's Theorem
Find all primes
p
p
p
such that the expression
(
p
1
)
2
+
(
p
2
)
2
+
⋯
+
(
p
p
−
1
)
2
\binom{p}1^2+\binom{p}2^2+\cdots+\binom{p}{p-1}^2
(
1
p
)
2
+
(
2
p
)
2
+
⋯
+
(
p
−
1
p
)
2
is divisible by
p
3
p^3
p
3
.
m^2-n^2=pk
Find all triplets
(
k
,
m
,
n
)
(k, m, n)
(
k
,
m
,
n
)
of positive integers having the following property: Square with side length
m
m
m
can be divided into several rectangles of size
1
×
k
1\times k
1
×
k
and a square with side length
n
n
n
.
2
2
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Similar to Brianchon's Theorem
A
B
C
D
E
F
ABCDEF
A
BC
D
EF
is a convex hexagon, such that
∣
∠
F
A
B
∣
=
∣
∠
B
C
D
∣
=
∣
∠
D
E
F
∣
|\angle FAB| = |\angle BCD| =|\angle DEF|
∣∠
F
A
B
∣
=
∣∠
BC
D
∣
=
∣∠
D
EF
∣
and
∣
A
B
∣
=
∣
B
C
∣
,
|AB| =|BC|,
∣
A
B
∣
=
∣
BC
∣
,
∣
C
D
∣
=
∣
D
E
∣
|CD| = |DE|
∣
C
D
∣
=
∣
D
E
∣
,
∣
E
F
∣
=
∣
F
A
∣
|EF| = |FA|
∣
EF
∣
=
∣
F
A
∣
. Prove that the lines
A
D
AD
A
D
,
B
E
BE
BE
and
C
F
CF
CF
are concurrent.
ABCDE- regualar pentagon. Find smallest value of expression
A
B
C
D
E
ABCDE
A
BC
D
E
is a regular pentagon. Determine the smallest value of the expression
∣
P
A
∣
+
∣
P
B
∣
∣
P
C
∣
+
∣
P
D
∣
+
∣
P
E
∣
,
\frac{|PA|+|PB|}{|PC|+|PD|+|PE|},
∣
PC
∣
+
∣
P
D
∣
+
∣
PE
∣
∣
P
A
∣
+
∣
PB
∣
,
where
P
P
P
is an arbitrary point lying in the plane of the pentagon
A
B
C
D
E
ABCDE
A
BC
D
E
.
1
2
Hide problems
All solutions for (x,y,z)
Determine all triples
(
x
,
y
,
z
)
(x, y, z)
(
x
,
y
,
z
)
of positive real numbers which satisfies the following system of equations
2
x
3
=
2
y
(
x
2
+
1
)
−
(
z
2
+
1
)
,
2x^3=2y(x^2+1)-(z^2+1),
2
x
3
=
2
y
(
x
2
+
1
)
−
(
z
2
+
1
)
,
2
y
4
=
3
z
(
y
2
+
1
)
−
2
(
x
2
+
1
)
,
2y^4=3z(y^2+1)-2(x^2+1),
2
y
4
=
3
z
(
y
2
+
1
)
−
2
(
x
2
+
1
)
,
2
z
5
=
4
x
(
z
2
+
1
)
−
3
(
y
2
+
1
)
.
2z^5=4x(z^2+1)-3(y^2+1).
2
z
5
=
4
x
(
z
2
+
1
)
−
3
(
y
2
+
1
)
.
Prime divisors of k^2+k+n are greater than 2008
Prove that there exists a positive integer
n
n
n
, such that the number
k
2
+
k
+
n
k^2+k+n
k
2
+
k
+
n
does not have a prime divisor less than
2008
2008
2008
for any integer
k
k
k
.