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Problems
Contests
National and Regional Contests
Romania Contests
District Olympiad
2000 District Olympiad (Hunedoara)
2000 District Olympiad (Hunedoara)
Part of
District Olympiad
Subcontests
(4)
3
3
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find a particular set of points on a plane
Let
α
\alpha
α
be a plane and let
A
B
C
ABC
A
BC
be an equilateral triangle situated on a parallel plane whose distance from
α
\alpha
α
is
h
.
h.
h
.
Find the locus of the points
M
∈
α
M\in\alpha
M
∈
α
for which
∣
M
A
∣
2
+
h
2
=
∣
M
B
∣
2
+
∣
M
C
∣
2
.
\left|MA\right| ^2 +h^2 = \left|MB\right|^2 +\left|MC\right|^2.
∣
M
A
∣
2
+
h
2
=
∣
MB
∣
2
+
∣
MC
∣
2
.
limit of an interesting sequence
Let be two distinct natural numbers
k
1
k_1
k
1
and
k
2
k_2
k
2
and a sequence
(
x
n
)
n
≥
0
\left( x_n \right)_{n\ge 0}
(
x
n
)
n
≥
0
which satisfies x_nx_m +k_1k_2\le k_1x_n +k_2x_m, \forall m,n\in\{ 0\}\cup\mathbb{N}. Calculate
lim
n
→
∞
n
!
⋅
(
−
1
)
1
+
n
⋅
x
n
2
n
n
.
\lim_{n\to\infty}\frac{n!\cdot (-1)^{1+n}\cdot x_n^2}{n^n} .
lim
n
→
∞
n
n
n
!
⋅
(
−
1
)
1
+
n
⋅
x
n
2
.
primitivableness on contest
Let be a function
f
:
R
⟶
R
f:\mathbb{R}\longrightarrow\mathbb{R}
f
:
R
⟶
R
such that: \text{(i)} f(0)=0 \text{(ii)} f'(x)\neq 0, \forall x\in\mathbb{R} \text{(iii)} \left. f''\right|_{\mathbb{R}}\text{ exists and it's continuous} Demonstrate that the function
g
:
R
⟶
R
g:\mathbb{R}\longrightarrow\mathbb{R}
g
:
R
⟶
R
defined as
g
(
x
)
=
{
cos
1
f
(
x
)
,
e
m
s
p
;
x
≠
0
0
,
e
m
s
p
;
x
=
0
g(x)=\left\{\begin{matrix}\cos\frac{1}{f(x)},  x\neq 0\\ 0,  x=0\end{matrix}\right.
g
(
x
)
=
{
cos
f
(
x
)
1
,
0
,
e
m
s
p
;
x
=
0
e
m
s
p
;
x
=
0
is primitivable.
4
3
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pyramid problem
Consider the pyramid
V
A
B
C
D
,
VABCD,
V
A
BC
D
,
where
V
V
V
is the top and
A
B
C
D
ABCD
A
BC
D
is a rectangular base. If
∠
B
V
D
=
∠
A
V
C
,
\angle BVD = \angle AVC,
∠
B
V
D
=
∠
A
V
C
,
then prove that the triangles
V
A
C
VAC
V
A
C
and
V
B
D
VBD
V
B
D
share the same perimeter and area.
vectorial geometry
Let be a circle centeted at
O
,
O,
O
,
and
A
,
B
,
C
,
A,B,C,
A
,
B
,
C
,
points situated on this circle. Show that if
∣
O
A
→
+
O
B
→
∣
=
∣
O
B
→
+
O
C
→
∣
=
∣
O
C
→
+
O
A
→
∣
,
\left|\overrightarrow{OA} +\overrightarrow{OB}\right| = \left|\overrightarrow{OB} +\overrightarrow{OC}\right| = \left|\overrightarrow{OC} +\overrightarrow{OA}\right| ,
O
A
+
OB
=
OB
+
OC
=
OC
+
O
A
,
then
A
=
B
=
C
,
A=B=C,
A
=
B
=
C
,
or
A
B
C
ABC
A
BC
is an equilateral triangle.
calculate limit using integral
Let
f
:
[
0
,
1
]
⟶
R
+
∗
f:[0,1]\longrightarrow\mathbb{R}_+^*
f
:
[
0
,
1
]
⟶
R
+
∗
be a Riemann-integrable function. Calculate
lim
n
→
∞
(
−
n
+
∑
i
=
1
n
e
1
n
⋅
f
(
i
n
)
)
.
\lim_{n\to\infty}\left(-n+\sum_{i=1}^ne^{\frac{1}{n}\cdot f\left(\frac{i}{n}\right)}\right) .
lim
n
→
∞
(
−
n
+
∑
i
=
1
n
e
n
1
⋅
f
(
n
i
)
)
.
2
3
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Romanian District Olympiad 2000, Grade IX, Problem 2
a) Let
a
,
b
a,b
a
,
b
two non-negative integers such that
a
2
>
b
.
a^2>b.
a
2
>
b
.
Show that the equation
⌊
x
2
+
2
a
x
+
b
⌋
=
x
+
a
−
1
\left\lfloor\sqrt{x^2+2ax+b}\right\rfloor =x+a-1
⌊
x
2
+
2
a
x
+
b
⌋
=
x
+
a
−
1
has an infinite number of solutions in the non-negative integers. Here,
⌊
α
⌋
\lfloor\alpha\rfloor
⌊
α
⌋
denotes the floor of
α
.
\alpha.
α
.
b) Find the floor of
m
=
2
+
2
+
⋯
⏟
n times
+
2
,
m=\sqrt{2+\sqrt{2+\underbrace{\cdots}_{\text{n times}}+\sqrt{2}}} ,
m
=
2
+
2
+
n times
⋯
+
2
,
where
n
n
n
is a natural number. Justify.
matrix determinant problem
Calculate the determinant of the
n
×
n
n\times n
n
×
n
complex matrix
(
a
j
i
)
1
≤
j
≤
n
1
≤
i
≤
n
\left(a_j^i\right)_{1\le j\le n}^{1\le i\le n}
(
a
j
i
)
1
≤
j
≤
n
1
≤
i
≤
n
defined by
a
j
i
=
{
1
+
x
2
,
e
m
s
p
;
i
=
j
x
,
e
m
s
p
;
∣
i
−
j
∣
=
1
0
,
e
m
s
p
;
∣
i
−
j
∣
≥
2
,
a_j^i=\left\{\begin{matrix} 1+x^2,  i=j\\x,  |i-j|=1\\0,  |i-j|\ge 2\end{matrix}\right. ,
a
j
i
=
⎩
⎨
⎧
1
+
x
2
,
x
,
0
,
e
m
s
p
;
i
=
j
e
m
s
p
;
∣
i
−
j
∣
=
1
e
m
s
p
;
∣
i
−
j
∣
≥
2
,
where
n
n
n
is a natural number greater than
2.
2.
2.
basic complex algebra
Let
z
1
,
z
2
,
z
3
∈
C
z_1,z_2,z_3\in\mathbb{C}
z
1
,
z
2
,
z
3
∈
C
such that \text{(i)} \left|z_1\right| = \left|z_2\right| = \left|z_3\right| = 1 \text{(ii)} z_1+z_2+z_3\neq 0 \text{(iii)} z_1^2 +z_2^2+z_3^2 =0. Show that
∣
z
1
3
+
z
2
3
+
z
3
3
∣
=
1.
\left| z_1^3+z_2^3+z_3^3\right| = 1.
z
1
3
+
z
2
3
+
z
3
3
=
1.
1
4
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