MathDB
Problems
Contests
National and Regional Contests
Bulgaria Contests
Bulgaria EGMO TST
2019 Bulgaria EGMO TST
2019 Bulgaria EGMO TST
Part of
Bulgaria EGMO TST
Subcontests
(3)
3
1
Hide problems
Products and sums of digits
In terms of the fixed non-negative integers
α
\alpha
α
and
β
\beta
β
determine the least upper bound of the ratio (or show that it is unbounded)
S
(
n
)
S
(
2
α
5
β
n
)
\frac{S(n)}{S(2^{\alpha}5^{\beta}n)}
S
(
2
α
5
β
n
)
S
(
n
)
as
n
n
n
varies through the positive integers, where
S
(
⋅
)
S(\cdot)
S
(
⋅
)
denotes sum of digits in decimal representation.
1
2
Hide problems
Triangle from bits
Let
x
1
,
…
,
x
n
x_1,\ldots,x_n
x
1
,
…
,
x
n
be a sequence with each term equal to
0
0
0
or
1
1
1
. Form a triangle as follows: its first row is
x
1
,
…
,
x
n
x_1,\ldots,x_n
x
1
,
…
,
x
n
and if a row if
a
1
,
a
2
,
…
,
a
m
a_1, a_2, \ldots, a_m
a
1
,
a
2
,
…
,
a
m
, then the next row is
a
1
+
a
2
,
a
2
+
a
3
,
…
,
a
m
−
1
+
a
m
a_1 + a_2, a_2 + a_3, \ldots, a_{m-1} + a_m
a
1
+
a
2
,
a
2
+
a
3
,
…
,
a
m
−
1
+
a
m
where the addition is performed modulo
2
2
2
(so
1
+
1
=
0
1+1=0
1
+
1
=
0
). For example, starting from
1
1
1
,
0
0
0
,
1
1
1
,
0
0
0
, the second row is
1
1
1
,
1
1
1
,
1
1
1
, the third one is
0
0
0
,
0
0
0
and the fourth one is
0
0
0
.A sequence is called good it is the same as the sequence formed by taking the last element of each row, starting from the last row (so in the above example, the sequence is
1010
1010
1010
and the corresponding sequence from last terms is
0010
0010
0010
and they are not equal in this case). How many possibilities are there for the sequence formed by taking the first element of each row, starting from the last row, which arise from a good sequence?
Lovely starting geo
Determine the length of
B
C
BC
BC
in an acute triangle
A
B
C
ABC
A
BC
with
∠
A
B
C
=
4
5
∘
\angle ABC = 45^{\circ}
∠
A
BC
=
4
5
∘
,
O
G
=
1
OG = 1
OG
=
1
and
O
G
∥
B
C
OG \parallel BC
OG
∥
BC
. (As usual
O
O
O
is the circumcenter and
G
G
G
is the centroid.)
2
2
Hide problems
Hardcore algebra sequence
The sequence of real numbers
(
a
n
)
n
≥
0
(a_n)_{n\geq 0}
(
a
n
)
n
≥
0
is such that
a
0
=
1
a_0 = 1
a
0
=
1
,
a
1
=
a
>
2
a_1 = a > 2
a
1
=
a
>
2
and
a
n
+
1
=
(
(
a
n
a
n
−
1
)
2
−
2
)
a
n
\displaystyle a_{n+1} = \left(\left(\frac{a_n}{a_{n-1}}\right)^2 -2\right)a_n
a
n
+
1
=
(
(
a
n
−
1
a
n
)
2
−
2
)
a
n
for every positive integer
n
n
n
. Prove that
∑
i
=
0
k
1
a
i
<
2
+
a
−
a
2
−
4
2
\displaystyle \sum_{i=0}^k \frac{1}{a_i} < \frac{2+a-\sqrt{a^2-4}}{2}
i
=
0
∑
k
a
i
1
<
2
2
+
a
−
a
2
−
4
for every positive integer
k
k
k
.
Geometry full of stuff
Let
A
B
C
D
ABCD
A
BC
D
be a cyclic quadrilateral with circumcircle
ω
\omega
ω
centered at
O
O
O
, whose diagonals intersect at
H
H
H
. Let
O
1
O_1
O
1
and
O
2
O_2
O
2
be the circumcenters of triangles
A
H
D
AHD
A
HD
and
B
H
C
BHC
B
H
C
. A line through
H
H
H
intersects
ω
\omega
ω
at
M
1
M_1
M
1
and
M
2
M_2
M
2
and intersects the circumcircles of triangles
O
1
H
O
O_1HO
O
1
H
O
and
O
2
H
O
O_2HO
O
2
H
O
at
N
1
N_1
N
1
and
N
2
N_2
N
2
, respectively, so that
N
1
N_1
N
1
and
N
2
N_2
N
2
lie inside
ω
\omega
ω
. Prove that
M
1
N
1
=
M
2
N
2
M_1N_1 = M_2N_2
M
1
N
1
=
M
2
N
2
.