MathDB
Problems
Contests
National and Regional Contests
Bulgaria Contests
Bulgaria EGMO TST
2016 Bulgaria EGMO TST
2016 Bulgaria EGMO TST
Part of
Bulgaria EGMO TST
Subcontests
(3)
3
2
Hide problems
Magician's epicness
The eyes of a magician are blindfolded while a person
A
A
A
from the audience arranges
n
n
n
identical coins in a row, some are heads and the others are tails. The assistant of the magician asks
A
A
A
to write an integer between
1
1
1
and
n
n
n
inclusive and to show it to the audience. Having seen the number, the assistant chooses a coin and turns it to the other side (so if it was heads it becomes tails and vice versa) and does not touch anything else. Afterwards, the bandages are removed from the magician, he sees the sequence and guesses the written number by
A
A
A
. For which
n
n
n
is this possible?[hide=Spoiler hint] The original formulation asks: a) Show that if
n
n
n
is possible, so is
2
n
2n
2
n
; b) Show that only powers of
2
2
2
are possible; I have omitted this from the above formulation, for the reader's interest.
Hard functional inequality
Prove that there is no function
f
:
R
+
→
R
+
f:\mathbb{R}^{+} \to \mathbb{R}^{+}
f
:
R
+
→
R
+
such that
f
(
x
)
2
≥
f
(
x
+
y
)
(
f
(
x
)
+
y
)
f(x)^2 \geq f(x+y)(f(x)+y)
f
(
x
)
2
≥
f
(
x
+
y
)
(
f
(
x
)
+
y
)
for all
x
,
y
∈
R
+
x,y \in \mathbb{R}^{+}
x
,
y
∈
R
+
.
2
1
Hide problems
A disrespectfully easy geo
Let
A
B
C
ABC
A
BC
be a right triangle with
∠
A
C
B
=
9
0
∘
\angle ACB = 90^{\circ}
∠
A
CB
=
9
0
∘
and centroid
G
G
G
. The circumcircle
k
1
k_1
k
1
of triangle
A
G
C
AGC
A
GC
and the circumcircle
k
2
k_2
k
2
of triangle
B
G
C
BGC
BGC
intersect
A
B
AB
A
B
at
P
P
P
and
Q
Q
Q
, respectively. The perpendiculars from
P
P
P
and
Q
Q
Q
respectively to
A
C
AC
A
C
and
B
C
BC
BC
intersect
k
1
k_1
k
1
and
k
2
k_2
k
2
at
X
X
X
and
Y
Y
Y
. Determine the value of
C
X
⋅
C
Y
A
B
2
\frac{CX \cdot CY}{AB^2}
A
B
2
CX
⋅
C
Y
.
1
2
Hide problems
Old fashioned number theory equation
Find all positive integers
x
x
x
such that
3
x
+
x
2
+
135
3^x + x^2 + 135
3
x
+
x
2
+
135
is a perfect square.
Weird partition of the integers
Is it possible to partition the set of integers into three disjoint sets so that for every positive integer
n
n
n
the numbers
n
n
n
,
n
−
50
n-50
n
−
50
and
n
+
1987
n+1987
n
+
1987
belong to different sets?