MathDB
Problems
Contests
National and Regional Contests
Bulgaria Contests
Bulgarian Spring Mathematical Competition
2023 Bulgarian Spring Mathematical Competition
2023 Bulgarian Spring Mathematical Competition
Part of
Bulgarian Spring Mathematical Competition
Subcontests
(12)
9.2
1
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Weird cyclic quadrilateral to prove
Given is triangle
A
B
C
ABC
A
BC
with angle bisector
C
L
CL
C
L
and the
C
−
C-
C
−
median meets the circumcircle
Γ
\Gamma
Γ
at
D
D
D
. If
K
K
K
is the midpoint of arc
A
C
B
ACB
A
CB
and
P
P
P
is the symmetric point of
L
L
L
with respect to the tangent at
K
K
K
to
Γ
\Gamma
Γ
, then prove that
D
L
C
P
DLCP
D
L
CP
is cyclic.
9.4
1
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Cycles in a directed graph
Given is a directed graph with
28
28
28
vertices, such that there do not exist vertices
u
,
v
u, v
u
,
v
, such that
u
→
v
u \rightarrow v
u
→
v
and
v
→
u
v \rightarrow u
v
→
u
. Every
16
16
16
vertices form a directed cycle. Prove that among any
17
17
17
vertices, we can choose
15
15
15
which form a directed cycle.
9.3
1
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Cute NT with binomials
Given a prime
p
p
p
, find
gcd
(
(
2
p
p
1
)
,
(
2
p
p
3
)
,
…
,
(
2
p
p
2
p
p
−
1
)
)
\gcd(\binom{2^pp}{1},\binom{2^pp}{3},\ldots, \binom{2^pp}{2^pp-1})
g
cd
(
(
1
2
p
p
)
,
(
3
2
p
p
)
,
…
,
(
2
p
p
−
1
2
p
p
)
)
.
10.2
1
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Easy isosceles triangle geo
An isosceles
△
A
B
C
\triangle ABC
△
A
BC
has
∠
B
A
C
=
∠
A
B
C
=
7
2
o
\angle BAC =\angle ABC =72^{o}
∠
B
A
C
=
∠
A
BC
=
7
2
o
. The angle bisector
A
L
AL
A
L
meets the line through
C
C
C
parallel to
A
B
AB
A
B
at
D
D
D
.
a
)
a)
a
)
Prove that the circumcenter of
△
A
D
C
\triangle ADC
△
A
D
C
lies on
B
D
BD
B
D
.
b
)
b)
b
)
Prove that
B
E
B
L
\frac {BE} {BL}
B
L
BE
is irrational.
10.3
1
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Operations on triangulated polygon
Given is a convex octagon
A
1
A
2
…
A
8
A_1A_2 \ldots A_8
A
1
A
2
…
A
8
. Given a triangulation
T
T
T
, one can take two triangles
△
A
i
A
j
A
k
\triangle A_iA_jA_k
△
A
i
A
j
A
k
and
△
A
i
A
k
A
l
\triangle A_iA_kA_l
△
A
i
A
k
A
l
and replace them with
△
A
i
A
j
A
l
\triangle A_iA_jA_l
△
A
i
A
j
A
l
and
△
A
j
A
l
A
k
\triangle A_jA_lA_k
△
A
j
A
l
A
k
. Find the minimal number of operations
k
k
k
we have to do so that for any pair of triangulations
T
1
,
T
2
T_1, T_2
T
1
,
T
2
, we can reach
T
2
T_2
T
2
from
T
1
T_1
T
1
using at most
k
k
k
operations.
10.4
1
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Good NT with orders
Find all positive integers
n
n
n
, such that there exists a positive integer
m
m
m
and primes
1
<
p
<
q
1<p<q
1
<
p
<
q
such that
q
−
p
∣
m
q-p \mid m
q
−
p
∣
m
and
p
,
q
∣
n
m
+
1
p, q \mid n^m+1
p
,
q
∣
n
m
+
1
.
11.1
1
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Equation with powers of 3 and trig
Find all real
a
a
a
such that the equation
3
cos
(
2
x
)
+
1
−
(
a
−
5
)
3
cos
2
(
2
x
)
=
7
3^{\cos (2x)+1}-(a-5)3^{\cos^2(2x)}=7
3
c
o
s
(
2
x
)
+
1
−
(
a
−
5
)
3
c
o
s
2
(
2
x
)
=
7
has a real root. This was the statement given at the contest, but there was actually a typo and the intended equation was
3
cos
(
2
x
)
+
1
−
(
a
−
5
)
3
cos
2
(
x
)
=
7
3^{\cos (2x)+1}-(a-5)3^{\cos^2(x)}=7
3
c
o
s
(
2
x
)
+
1
−
(
a
−
5
)
3
c
o
s
2
(
x
)
=
7
, which is much easier.
11.2
1
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weird ratio in a cyclic quad
Given is a cyclic quadrilateral
A
B
C
D
ABCD
A
BC
D
and a point
E
E
E
lies on the segment
D
A
DA
D
A
such that
2
∠
E
B
D
=
∠
A
B
C
2\angle EBD = \angle ABC
2∠
EB
D
=
∠
A
BC
. Prove that
D
E
=
A
C
.
B
D
A
B
+
B
C
DE= \frac {AC.BD}{AB+BC}
D
E
=
A
B
+
BC
A
C
.
B
D
.
11.3
1
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Positive integers representable as sum and difference of powers of 2
A positive integer
b
b
b
is called good if there exist positive integers
1
=
a
1
,
a
2
,
…
,
a
2023
=
b
1=a_1, a_2, \ldots, a_{2023}=b
1
=
a
1
,
a
2
,
…
,
a
2023
=
b
such that
∣
a
i
+
1
−
a
i
∣
=
2
i
|a_{i+1}-a_i|=2^i
∣
a
i
+
1
−
a
i
∣
=
2
i
. Find the number of the good integers.
11.4
1
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Non-existence of a special vertex implies bound for the longest path of a tree
Given is a tree
G
G
G
with
2023
2023
2023
vertices. The longest path in the graph has length
2
n
2n
2
n
. A vertex is called good if it has degree at most
6
6
6
. Find the smallest possible value of
n
n
n
if there doesn't exist a vertex having
6
6
6
good neighbors.
12.3
1
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Interesting variation of Brocard's equation
Given is a polynomial
f
f
f
of degree
m
m
m
with integer coefficients and positive leading coefficient. A positive integer
n
n
n
is
good
for
f(x)
\textit {good for f(x)}
good for f(x)
if there exists a positive integer
k
n
k_n
k
n
, such that
n
!
+
1
=
f
(
n
)
k
n
n!+1=f(n)^{k_n}
n
!
+
1
=
f
(
n
)
k
n
. Prove that there exist only finitely many integers good for
f
f
f
.
12.4
1
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Find large subset of A not containing entirely any A_i
Given is a set
A
A
A
of
n
n
n
elements and positive integers
k
,
m
k, m
k
,
m
such that
4
≤
k
<
n
4 \leq k <n
4
≤
k
<
n
and
m
≤
min
{
k
−
3
,
n
2
}
m \leq \min \{k-3, \frac {n} {2}\}
m
≤
min
{
k
−
3
,
2
n
}
. Let
A
1
,
A
2
,
…
,
A
l
A_1, A_2, \ldots, A_l
A
1
,
A
2
,
…
,
A
l
be subsets of
A
A
A
, all with size
k
k
k
, such that
∣
A
i
∩
A
j
∣
≤
m
|A_i \cap A_j| \leq m
∣
A
i
∩
A
j
∣
≤
m
for all
i
≠
j
i \neq j
i
=
j
. Prove that there exists a subset
B
B
B
of
A
A
A
with at least
n
m
+
1
+
m
\sqrt[m+1]{n}+m
m
+
1
n
+
m
elements which doesn't contain entirely any of the subsets
A
1
,
A
2
,
…
,
A
l
A_1, A_2, \ldots, A_l
A
1
,
A
2
,
…
,
A
l
.