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Problems
Contests
National and Regional Contests
Serbia Contests
Federal Math Competition of Serbia and Montenegro
2002 Federal Math Competition of S&M
2002 Federal Math Competition of S&M
Part of
Federal Math Competition of Serbia and Montenegro
Subcontests
(4)
Problem 2
3
Hide problems
chords in a circle, show area inequality
Points
A
0
,
A
1
,
…
,
A
2
k
A_0,A_1,\ldots,A_{2k}
A
0
,
A
1
,
…
,
A
2
k
, in this order, divide a circumference into
2
k
+
1
2k+1
2
k
+
1
equal arcs. Point
A
0
A_0
A
0
is connected by chords to all the other points. These
2
k
2k
2
k
chords divide the interior of the circle into
2
k
+
1
2k+1
2
k
+
1
parts. These parts are alternately painted red and blue so that there are
k
+
1
k+1
k
+
1
red and
k
k
k
blue parts. Show that the blue area is larger than the red area.
cevian inequality
Let
O
O
O
be a point inside a triangle
A
B
C
ABC
A
BC
and let the lines
A
O
,
B
O
AO,BO
A
O
,
BO
, and
C
O
CO
CO
meet sides
B
C
,
C
A
BC,CA
BC
,
C
A
, and
A
B
AB
A
B
at points
A
1
,
B
1
A_1,B_1
A
1
,
B
1
, and
C
1
C_1
C
1
, respectively. If
A
A
1
AA_1
A
A
1
is the longest among the segments
A
A
1
,
B
B
1
,
C
C
1
AA_1,BB_1,CC_1
A
A
1
,
B
B
1
,
C
C
1
, prove that
O
A
1
+
O
B
1
+
O
C
1
≤
A
A
1
.
OA_1+OB_1+OC_1\le AA_1.
O
A
1
+
O
B
1
+
O
C
1
≤
A
A
1
.
triangle, sides √(fibonacci #s)
The (Fibonacci) sequence
f
n
f_n
f
n
is defined by
f
1
=
f
2
=
1
f_1=f_2=1
f
1
=
f
2
=
1
and
f
n
+
2
=
f
n
+
1
+
f
n
f_{n+2}=f_{n+1}+f_n
f
n
+
2
=
f
n
+
1
+
f
n
for
n
≥
1
n\ge1
n
≥
1
. Prove that the area of the triangle with the sides
f
2
n
+
1
,
f
2
n
+
2
,
\sqrt{f_{2n+1}},\sqrt{f_{2n+2}},
f
2
n
+
1
,
f
2
n
+
2
,
and
f
2
n
+
3
\sqrt{f_{2n+3}}
f
2
n
+
3
is equal to
1
2
\frac12
2
1
.
Problem 1
3
Hide problems
x that satisfy floor inequality
Determine all real numbers
x
x
x
such that
2002
⌊
x
⌋
⌊
−
x
⌋
+
x
>
⌊
2
x
⌋
x
−
⌊
1
+
x
⌋
.
\frac{2002\lfloor x\rfloor}{\lfloor-x\rfloor+x}>\frac{\lfloor2x\rfloor}{x-\lfloor1+x\rfloor}.
⌊
−
x
⌋
+
x
2002
⌊
x
⌋
>
x
−
⌊
1
+
x
⌋
⌊
2
x
⌋
.
inequality with natural parameters
For any positive numbers
a
,
b
,
c
a,b,c
a
,
b
,
c
and natural numbers
n
,
k
n,k
n
,
k
prove the inequality
a
n
+
k
b
n
+
b
n
+
k
c
n
+
c
n
+
k
a
n
≥
a
k
+
b
k
+
c
k
.
\frac{a^{n+k}}{b^n}+\frac{b^{n+k}}{c^n}+\frac{c^{n+k}}{a^n}\ge a^k+b^k+c^k.
b
n
a
n
+
k
+
c
n
b
n
+
k
+
a
n
c
n
+
k
≥
a
k
+
b
k
+
c
k
.
max{z} if x^2≥y+z & cyclic
Real numbers
x
,
y
,
z
x,y,z
x
,
y
,
z
satisfy the inequalities
x
2
≤
y
+
z
,
y
2
≤
z
+
x
z
2
≤
x
+
y
.
x^2\le y+z,\qquad y^2\le z+x\qquad z^2\le x+y.
x
2
≤
y
+
z
,
y
2
≤
z
+
x
z
2
≤
x
+
y
.
Find the minimum and maximum possible values of
z
z
z
.
Problem 4
3
Hide problems
L-tiling 2001x2003 rectangle
Is it possible to cut a rectangle
2001
×
2003
2001\times2003
2001
×
2003
into pieces of the form https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvNS82L2RjZTZjNzc0M2YxMzM1ZDIzZTY2Zjc2NGJlMWJlMWUwMmU2ZWRlLnBuZw==&rn=U2NyZWVuIFNob3QgMjAyMS0wNS0xMyBhdCAzLjQ2LjQ2IFBNLnBuZw== each consisting of three unit squares?
ranking players from 1-50, ranges
Each of the
15
15
15
coaches ranked the
50
50
50
selected football players on the places from
1
1
1
to
50
50
50
. For each football player, the highest and lowest obtained ranks differ by at most
5
5
5
. For each of the players, the sum of the ranks he obtained is computed, and the sums are denoted by
S
1
≤
S
2
≤
…
≤
S
50
S_1\le S_2\le\ldots\le S_{50}
S
1
≤
S
2
≤
…
≤
S
50
. Find the largest possible value of
S
1
S_1
S
1
.
2002!\cdot k without certain digits
Is there a positive integer
k
k
k
such that none of the digits
3
,
4
,
5
,
6
3,4,5,6
3
,
4
,
5
,
6
appear in the decimal representation of the number
2002
!
⋅
k
2002!\cdot k
2002
!
⋅
k
?
Problem 3
3
Hide problems
n choose k=2002
Find all pairs
(
n
,
k
)
(n,k)
(
n
,
k
)
of positive integers such that
(
n
k
)
=
2002
\binom nk=2002
(
k
n
)
=
2002
.
(2^m-1)^2|2n-1 iff m(2^m-1)|n
Let
m
m
m
and
n
n
n
be positive integers. Prove that the number
2
n
−
1
2n-1
2
n
−
1
is divisible by
(
2
m
−
1
)
2
(2^m-1)^2
(
2
m
−
1
)
2
if and only if
n
n
n
is divisible by
m
(
2
m
−
1
)
m(2^m-1)
m
(
2
m
−
1
)
.
Nice geometric inequality
Let
A
B
C
D
ABCD
A
BC
D
be a rhombus with \angle BAD \equal{} 60^{\circ}. Points
S
S
S
and
R
R
R
are chosen inside the triangles
A
B
D
ABD
A
B
D
and
D
B
C
DBC
D
BC
, respectively, such that \angle SBR \equal{} \angle RDS \equal{} 60^{\circ}. Prove that
S
R
2
≥
A
S
⋅
C
R
SR^2\geq AS\cdot CR
S
R
2
≥
A
S
⋅
CR
.