MathDB
Problems
Contests
National and Regional Contests
North Macedonia Contests
North Macedonia National Olympiads
1996 North Macedonia National Olympiad
1996 North Macedonia National Olympiad
Part of
North Macedonia National Olympiads
Subcontests
(5)
2
1
Hide problems
4 conditions' mapping from set of polygons
Let
P
P
P
be the set of all polygons in the plane and let
M
:
P
→
R
M : P \to R
M
:
P
→
R
be a mapping that satisfies: (i)
M
(
P
)
≥
0
M(P) \ge 0
M
(
P
)
≥
0
for each polygon
P
P
P
, (ii)
M
(
P
)
=
x
2
M(P) = x^2
M
(
P
)
=
x
2
if
P
P
P
is an equilateral triangle of side
x
x
x
, (iii) If a polygon
P
P
P
is partitioned into polygons
S
S
S
and
T
T
T
, then
M
(
P
)
=
M
(
S
)
+
M
(
T
)
M(P) = M(S)+ M(T)
M
(
P
)
=
M
(
S
)
+
M
(
T
)
, (iv) If polygons
P
P
P
and
T
T
T
are congruent, then
M
(
P
)
=
M
(
T
)
M(P) = M(T )
M
(
P
)
=
M
(
T
)
. Determine
M
(
P
)
M(P)
M
(
P
)
if
P
P
P
is a rectangle with edges
x
x
x
and
y
y
y
.
4
1
Hide problems
good polygons (3 conditions given)
A polygon is called good if it satisfies the following conditions: (i) All its angles are in
(
0
,
π
)
(0,\pi)
(
0
,
π
)
or in
(
π
,
2
π
)
(\pi ,2\pi)
(
π
,
2
π
)
, (ii) It is not self-intersecing, (iii) For any three sides, two are parallel and equal. Find all
n
n
n
for which there exists a good
n
n
n
-gon.
3
1
Hide problems
1/sina+ 1/sinb \ge 8/( 3+2 cos c)
Prove that if
α
,
β
,
γ
\alpha, \beta, \gamma
α
,
β
,
γ
are angles of a triangle, then
1
sin
α
+
1
sin
β
≥
8
3
+
2
c
o
s
γ
\frac{1}{\sin \alpha}+ \frac{1}{\sin \beta} \ge \frac{8}{ 3+2 \ cos\gamma}
s
i
n
α
1
+
s
i
n
β
1
≥
3
+
2
cos
γ
8
.
5
1
Hide problems
n concurrent lines in space, any 2of them form the same angle
Find the greatest
n
n
n
for which there exist
n
n
n
lines in space, passing through a single point, such that any two of them form the same angle.
1
1
Hide problems
equal angels wanted, parallelogram and two perpendicularities given
Let
A
B
C
D
ABCD
A
BC
D
be a parallelogram which is not a rectangle and
E
E
E
be the point in its plane such that
A
E
⊥
A
B
AE \perp AB
A
E
⊥
A
B
and
C
E
⊥
C
B
CE \perp CB
CE
⊥
CB
. Prove that
∠
D
E
A
=
∠
C
E
B
\angle DEA = \angle CEB
∠
D
E
A
=
∠
CEB
.