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Problems
Contests
National and Regional Contests
Vietnam Contests
VMEO = Vietnam Mathematical E-Olympiad
VMEO I 2004
VMEO I 2004
Part of
VMEO = Vietnam Mathematical E-Olympiad
Subcontests
(7)
4
1
Hide problems
\lambda_A KA +\lambda_B KB + \lambda_C KC+ \lambda_D KD=0, in vectors
In a quadrilateral
A
B
C
D
ABCD
A
BC
D
let
E
E
E
be the intersection of the two diagonals, I the center of the parallelogram whose vertices are the midpoints of the four sides of the quadrilateral, and K the center of the parallelogram whose sides pass through the points. divide the four sides of the quadrilateral into three equal parts (see illustration ). https://cdn.artofproblemsolving.com/attachments/1/c/8f2617103edd8361b8deebbee13c6180fa848b.png a) Prove that
E
K
→
=
4
3
E
I
→
\overrightarrow{EK} =\frac43 \overrightarrow{EI}
E
K
=
3
4
E
I
. b) Prove that
λ
A
K
A
→
+
λ
B
K
B
→
+
λ
C
K
C
→
+
λ
D
K
D
→
=
0
→
\lambda_A \overrightarrow{KA} +\lambda_B \overrightarrow{KB} + \lambda_C \overrightarrow{KC} + \lambda_D \overrightarrow{KD} = \overrightarrow{0}
λ
A
K
A
+
λ
B
K
B
+
λ
C
K
C
+
λ
D
KD
=
0
, where
λ
A
=
1
+
S
(
A
D
B
)
S
(
A
B
C
D
)
,
λ
B
=
1
+
S
(
B
C
A
)
S
(
A
B
C
D
)
,
λ
C
=
1
+
S
(
C
D
B
)
S
(
A
B
C
D
)
,
λ
D
=
1
+
S
(
D
A
C
)
S
(
A
B
C
D
)
\lambda_A=1+\frac{S(ADB)}{S(ABCD)},\lambda_B=1+\frac{S(BCA)}{S(ABCD)},\lambda_C=1+\frac{S(CDB)}{S(ABCD)},\lambda_D=1+\frac{S(DAC)}{S(ABCD)}
λ
A
=
1
+
S
(
A
BC
D
)
S
(
A
D
B
)
,
λ
B
=
1
+
S
(
A
BC
D
)
S
(
BC
A
)
,
λ
C
=
1
+
S
(
A
BC
D
)
S
(
C
D
B
)
,
λ
D
=
1
+
S
(
A
BC
D
)
S
(
D
A
C
)
, where
S
S
S
is the area symbol.
3
1
Hide problems
line construction such that intersection with angle has given area
In the plane, given an angle
A
x
y
Axy
A
x
y
. a) Given a triangle
M
N
P
MNP
MNP
of area
T
T
T
, describe how to construct a triangle of given area
T
T
T
and altitude
h
h
h
. Using this, describe how to construct parallelogram A
B
C
D
BCD
BC
D
with two sides lying on
A
x
Ax
A
x
and
A
y
Ay
A
y
, the area
T
T
T
and the distance between the two opposite sides equal to d given. b) From an arbitrary point
I
I
I
on the line
C
D
CD
C
D
, construct a line that intersects the lines
A
A
A
B,
B
C
BC
BC
,
A
D
AD
A
D
at
E
E
E
,
G
G
G
and
F
F
F
respectively so that the area of triangle
A
E
F
AEF
A
EF
is equal to the area of parallelogram
A
B
C
D
ABCD
A
BC
D
. c) Apply the above two sentences: Given any point
O
O
O
in the plane. From
O
O
O
, construct a line that intersects two rays
A
x
Ax
A
x
and
A
y
Ay
A
y
at
E
E
E
and
F
F
F
respectively so that the area of triangle
A
E
F
AEF
A
EF
is equal to the area of any given triangle.
2
1
Hide problems
Fibonacci numbers F_n is divisible by p^m
The Fibonacci numbers
(
F
n
)
n
=
1
∞
(F_n)_{n=1}^{\infty}
(
F
n
)
n
=
1
∞
are defined as follows:
F
1
=
F
2
=
1
,
F
n
=
F
n
−
2
+
F
n
−
1
,
n
=
3
,
4
,
.
.
.
F_1 = F_2 = 1, F_n = F_{n-2} + F_{n-1}, n = 3, 4, ...
F
1
=
F
2
=
1
,
F
n
=
F
n
−
2
+
F
n
−
1
,
n
=
3
,
4
,
...
Assume
p
p
p
is a prime greater than
3
3
3
. With
m
m
m
being a natural number greater than
3
3
3
, find all
n
n
n
numbers such that
F
n
F_n
F
n
is divisible by
p
m
p^m
p
m
.
7
1
Hide problems
product (4 sin^2 (2^{k}-1)\pi/2^{2005}} -3)
Calculate the following
P
=
(
4
sin
2
0
−
3
)
(
4
sin
2
π
2
2005
−
3
)
(
4
sin
2
2
π
2
2005
−
3
)
(
4
sin
2
3
π
2
2005
−
3
)
.
.
.
P=(4\sin^2{0} -3)(4\sin^2\frac{\pi}{2^{2005}} -3)(4\sin^2\frac{2\pi}{2^{2005}} -3)(4\sin^2\frac{3\pi}{2^{2005}} -3)...
P
=
(
4
sin
2
0
−
3
)
(
4
sin
2
2
2005
π
−
3
)
(
4
sin
2
2
2005
2
π
−
3
)
(
4
sin
2
2
2005
3
π
−
3
)
...
.
.
.
(
4
sin
2
(
2
2004
−
1
)
π
2
2005
−
3
)
(
4
sin
2
π
2
−
3
)
...\,\,\,\,(4\sin^2\frac{(2^{2004}-1)\pi}{2^{2005}} -3)(4\sin^2\frac{\pi}{2} -3)
...
(
4
sin
2
2
2005
(
2
2004
−
1
)
π
−
3
)
(
4
sin
2
2
π
−
3
)
6
1
Hide problems
no of binary sequences of length n that are not equivalent.
Consider all binary sequences of length
n
n
n
. In a sequence that allows the interchange of positions of an arbitrary set of
k
k
k
adjacent numbers, (
k
<
n
k < n
k
<
n
), two sequences are said to be equivalent if they can be transformed from one sequence to another by a finite number of transitions as above. Find the number of sequences that are not equivalent.
5
1
Hide problems
(x + y)(f (x)-f (y)) = f (x^2) - f (y^2)
Find all the functions
f
:
R
→
R
f:R \to R
f
:
R
→
R
satisfying
(
x
+
y
)
(
f
(
x
)
−
f
(
y
)
)
=
f
(
x
2
)
−
f
(
y
2
)
,
∀
x
,
y
∈
R
(x + y)(f (x)-f (y)) = f (x^2) - f (y^2),\, \forall x, y \in R
(
x
+
y
)
(
f
(
x
)
−
f
(
y
))
=
f
(
x
2
)
−
f
(
y
2
)
,
∀
x
,
y
∈
R
1
1
Hide problems
sum \sqrt{x+ (y-z)^2 /12 } <= \sqrt3 if x, y, z>=0, x + y + z = 1,
Let
x
,
y
,
z
x, y, z
x
,
y
,
z
be non-negative numbers, so that
x
+
y
+
z
=
1
x + y + z = 1
x
+
y
+
z
=
1
. Prove that
x
+
(
y
−
z
)
2
12
+
y
+
(
x
−
z
)
2
12
+
z
+
(
x
−
y
)
2
12
≤
3
\sqrt{x+\frac{(y-z)^2}{12}}+\sqrt{y+\frac{(x-z)^2}{12}}+\sqrt{z+\frac{(x-y)^2}{12}}\le \sqrt{3}
x
+
12
(
y
−
z
)
2
+
y
+
12
(
x
−
z
)
2
+
z
+
12
(
x
−
y
)
2
≤
3