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Problems
Contests
National and Regional Contests
Vietnam Contests
VMEO = Vietnam Mathematical E-Olympiad
VMEO III 2006
VMEO III 2006
Part of
VMEO = Vietnam Mathematical E-Olympiad
Subcontests
(12)
11.2
1
Hide problems
locus of midpoints wanted, symmetric rays wrt line
Let
A
B
C
D
ABCD
A
BC
D
be an isosceles trapezoid, with a large base
C
D
CD
C
D
and a small base
A
B
AB
A
B
. Let
M
M
M
be any point on side
A
B
AB
A
B
and
(
d
)
(d)
(
d
)
be the line through
M
M
M
and perpendicular to
A
B
AB
A
B
. Two rays
M
x
Mx
M
x
and
M
y
My
M
y
are said to satisfy the condition
(
T
)
(T)
(
T
)
if they are symmetric about each other through
(
d
)
(d)
(
d
)
and intersect the two rays
A
D
AD
A
D
and
B
C
BC
BC
at
E
E
E
and
F
F
F
respectively. Find the locus of the midpoint of the segment
E
F
EF
EF
when the two rays
M
x
Mx
M
x
and
M
y
My
M
y
change and satisfy condition
(
T
)
(T)
(
T
)
.
10.2
1
Hide problems
sum of digits of one number divisible by 12 among 39 consecutive
Prove that among
39
39
39
consecutive natural numbers, there is always a number that has sum of its digits divisible by
12
12
12
. Is it true if we replace
39
39
39
with
38
38
38
?
12.1
1
Hide problems
concurrent wanted, 3 pairs of similar triangles exterior to sides of ABC
Given a triangle
A
B
C
ABC
A
BC
and a point
K
K
K
. The lines
A
K
AK
A
K
,
B
K
BK
B
K
,
C
K
CK
C
K
hit the opposite side of the triangle at
D
,
E
,
F
D,E,F
D
,
E
,
F
respectively. On the exterior of
A
B
C
ABC
A
BC
, we construct three pairs of similar triangles:
B
D
M
BDM
B
D
M
,
D
C
N
DCN
D
CN
on
B
D
BD
B
D
,
D
C
DC
D
C
,
C
E
P
CEP
CEP
,
E
A
Q
EAQ
E
A
Q
on
C
E
CE
CE
,
E
A
EA
E
A
, and
A
F
R
AFR
A
FR
,
F
B
S
FBS
FBS
on
A
F
AF
A
F
,
F
B
FB
FB
. The lines
M
N
MN
MN
,
P
Q
PQ
PQ
,
R
S
RS
RS
intersect each other form a triangle
X
Y
Z
XYZ
X
Y
Z
. Prove that
A
X
AX
A
X
,
B
Y
BY
B
Y
,
C
Z
CZ
CZ
are concurrent.
11.4
2
Hide problems
marbles on lattice points on infinite grid
On an infi nite grid, a square with four vertices lie at
(
m
,
n
)
(m, n)
(
m
,
n
)
,
(
m
−
1
,
n
)
(m-1, n)
(
m
−
1
,
n
)
,
(
m
,
n
−
1
)
(m,n-1)
(
m
,
n
−
1
)
,
(
m
−
1
,
n
−
1
)
(m-1, n-1)
(
m
−
1
,
n
−
1
)
is denoted as cell
(
m
,
n
)
(m,n)
(
m
,
n
)
(
m
,
n
∈
Z
)
(m, n \in Z)
(
m
,
n
∈
Z
)
. Some marbles are dropped on some cell. Each cell may have more than one marble or have no marble at all. Consider a "move" can be conducted in one of two following ways: i) Remove one marble from cell
(
m
,
n
)
(m,n)
(
m
,
n
)
(if there is marble at that cell), then add one marble to each of cell
(
m
−
1
,
n
−
2
)
(m - 1, n- 2)
(
m
−
1
,
n
−
2
)
and cell
(
m
−
2
,
n
−
1
)
(m -2, n - 1)
(
m
−
2
,
n
−
1
)
. ii) Remove two marbles from cell
(
m
,
n
)
(m,n)
(
m
,
n
)
(if there is marble at that cell), then add one marble to each of cell
(
m
+
1
,
n
−
2
)
(m +1, n - 2)
(
m
+
1
,
n
−
2
)
and cell
(
m
−
2
,
n
+
1
)
(m - 2, n +1)
(
m
−
2
,
n
+
1
)
. Assume that initially, there are
n
n
n
marbles at the cell
(
1
,
n
)
,
(
2
,
n
−
1
)
,
.
.
.
,
(
n
,
1
)
(1,n), (2,n - 1),..., (n, 1)
(
1
,
n
)
,
(
2
,
n
−
1
)
,
...
,
(
n
,
1
)
(each cell contains one marble). Can we conduct an finite amount of moves such that both cells
(
n
+
1
,
n
)
(n + 1, n)
(
n
+
1
,
n
)
and
(
n
,
n
+
1
)
(n, n + 1)
(
n
,
n
+
1
)
have marbles?
sequence with coprime terms wanted
Given an integer
a
>
1
a>1
a
>
1
. Let
p
1
<
p
2
<
.
.
.
<
p
k
p_1 < p_2 <...< p_k
p
1
<
p
2
<
...
<
p
k
be all prime divisors of
a
a
a
. For each positive integer
n
n
n
we define:
C
0
(
n
)
=
a
2
n
,
C
1
(
n
)
=
a
2
n
p
1
2
,
.
.
.
.
,
C
k
(
n
)
=
a
2
n
p
k
2
C_0(n) = a^{2n}, C_1(n) =\frac{a^{2n}}{p^2_1}, .... , C_k(n) =\frac{a^{2_n}}{p^2_k}
C
0
(
n
)
=
a
2
n
,
C
1
(
n
)
=
p
1
2
a
2
n
,
....
,
C
k
(
n
)
=
p
k
2
a
2
n
A
=
a
2
+
1
A = a^2 + 1
A
=
a
2
+
1
T
(
n
)
=
A
C
0
(
n
)
−
1
T(n) = A^{C_0(n)} - 1
T
(
n
)
=
A
C
0
(
n
)
−
1
M
(
n
)
=
L
C
M
(
a
2
n
+
2
,
A
C
1
(
n
)
−
1
,
.
.
.
,
A
C
k
(
n
)
−
1
)
M(n) = LCM(a^{2n+2}, A^{C_1(n)} - 1, ..., A^{C_k(n)} - 1)
M
(
n
)
=
L
CM
(
a
2
n
+
2
,
A
C
1
(
n
)
−
1
,
...
,
A
C
k
(
n
)
−
1
)
A
n
=
T
(
n
)
M
(
n
)
A_n =\frac{T(n)}{M(n)}
A
n
=
M
(
n
)
T
(
n
)
Prove that the sequence
A
1
,
A
2
,
.
.
.
A_1, A_2, ...
A
1
,
A
2
,
...
satisfies the properties: (i) Every number in the sequence is an integer greater than
1
1
1
and has only prime divisors of the form
a
m
+
1
am + 1
am
+
1
. (ii) Any two different numbers in the sequence are coprime.
10.1
2
Hide problems
<PBC + <PCB = <BAC if PB/PC = PC_1/PB_1 = AB/AC
Given a triangle
A
B
C
ABC
A
BC
(
A
B
≠
A
C
AB \ne AC
A
B
=
A
C
). Let
P
P
P
be a point in the plane containing triangle
A
B
C
ABC
A
BC
satisfying the following property: If the projections of
P
P
P
onto
A
B
AB
A
B
,
A
C
AC
A
C
are
C
1
C_1
C
1
,
B
1
B_1
B
1
respectively, then
P
B
P
C
=
P
C
1
P
B
1
=
A
B
A
C
\frac{PB}{PC}=\frac{PC_1}{PB_1}=\frac{AB}{AC}
PC
PB
=
P
B
1
P
C
1
=
A
C
A
B
or
P
B
P
C
=
P
B
1
P
C
1
=
A
B
A
C
\frac{PB}{PC}=\frac{PB_1}{PC_1}=\frac{AB}{AC}
PC
PB
=
P
C
1
P
B
1
=
A
C
A
B
.Prove that
∠
P
B
C
+
∠
P
C
B
=
∠
B
A
C
\angle PBC + \angle PCB = \angle BAC
∠
PBC
+
∠
PCB
=
∠
B
A
C
.
concurrent wanted; arc midpoints who are also segment midpoints
Let
A
B
C
ABC
A
BC
be a triangle inscribed in a circle with center
O
O
O
. Let
A
1
A_1
A
1
be a point on arc
B
C
BC
BC
that does not contain
A
A
A
such that the line perpendicular to
O
A
OA
O
A
at
A
1
A_1
A
1
intersects the lines
A
B
AB
A
B
and
A
C
AC
A
C
at two points and the line segment joining those two points has as midpoint
A
1
A_1
A
1
. Points
B
1
B_1
B
1
,
C
1
C_1
C
1
are determined similarly. Prove that the lines
A
A
1
AA_1
A
A
1
,
B
B
1
BB_1
B
B
1
,
C
C
1
CC_1
C
C
1
are concurrent.
12.2
2
Hide problems
m^2 =\sqrt{n} +\sqrt{2n + 1}
Find all positive integers
(
m
,
n
)
(m, n)
(
m
,
n
)
that satisfy
m
2
=
n
+
2
n
+
1
.
m^2 =\sqrt{n} +\sqrt{2n + 1}.
m
2
=
n
+
2
n
+
1
.
complete graph of n vertices, 3 colors
A complete graph of
n
n
n
vertices is a set of
n
n
n
vertices and those vertices are connected in pairs by edges. Suppose the graph has
n
n
n
vertices
A
1
,
A
2
,
.
.
.
,
A
n
A_1, A_2, ..., A_n
A
1
,
A
2
,
...
,
A
n
, the cycle is a set of edges of the form
A
i
1
A
i
2
,
A
i
2
A
i
3
,
.
.
.
,
A
i
m
A
i
1
A_{i_1}A_{i_2}, A_{i_2}A_{i_3},..., A_{i_m}A_{i_1}
A
i
1
A
i
2
,
A
i
2
A
i
3
,
...
,
A
i
m
A
i
1
with
i
1
,
i
2
,
.
.
.
,
i
m
∈
1
,
2
,
.
.
.
,
n
i_1, i_2, ..., i_m \in {1, 2, ..., n}
i
1
,
i
2
,
...
,
i
m
∈
1
,
2
,
...
,
n
double one different. We call
m
m
m
the length of this cycle. Find the smallest positive integer
n
n
n
such that for every way of coloring all edges of a complete graph of
n
n
n
vertices, each edge filled with one of three different colors, there is always a cycle of even length with the same color. PS. The same problem with another wording [url=https://artofproblemsolving.com/community/c6h151391p852296]here .
11.1
2
Hide problems
Calculate $P(\alpha^2+\alpha+1)$
Given a polynomial
P
(
x
)
=
x
4
+
x
3
+
3
x
2
−
6
x
+
1
P(x)=x^4+x^3+3x^2-6x+1
P
(
x
)
=
x
4
+
x
3
+
3
x
2
−
6
x
+
1
. Calculate
P
(
α
2
+
α
+
1
)
P(\alpha^2+\alpha+1)
P
(
α
2
+
α
+
1
)
where
α
=
1
+
5
2
3
+
1
−
5
2
3
\alpha=\sqrt[3]{\frac{1+\sqrt{5}}{2}}+\sqrt[3]{\frac{1-\sqrt{5}}{2}}
α
=
3
2
1
+
5
+
3
2
1
−
5
11 contestants solve 9 math problems
In a contest, there are
11
11
11
contestants to solve
9
9
9
math problems. After the end of the contest, it was found that any two contestants solved no more than
1
1
1
problem together. Find the largest positive integer
k
k
k
such that each problem can be solved by at least
k
k
k
candidates.
12.4
1
Hide problems
A binary serie problem
Given a binary serie
A
=
a
1
a
2
.
.
.
a
k
A=a_1a_2...a_k
A
=
a
1
a
2
...
a
k
is called "symmetry" if
a
i
=
a
k
+
1
−
i
a_i=a_{k+1-i}
a
i
=
a
k
+
1
−
i
for all
i
=
1
,
2
,
3
,
.
.
.
,
k
i=1,2,3,...,k
i
=
1
,
2
,
3
,
...
,
k
, and
k
k
k
is the length of that binary serie. If
A
=
11...1
A=11...1
A
=
11...1
or
A
=
00...0
A=00...0
A
=
00...0
then it is called "special". Find all positive integers
m
m
m
and
n
n
n
such that there exist non "special" binary series
A
A
A
(length
m
m
m
) and
B
B
B
(length
n
n
n
) satisfying when we place them next to each other, we receive a "symmetry" binary serie
A
B
AB
A
B
11.3
2
Hide problems
VMEO III Problem 3
Given a prime
p
p
p
in the form
4
m
+
1
4m+1
4
m
+
1
(
m
∈
Z
m\in\mathbb{Z}
m
∈
Z
). Prove that the number
216
p
3
216p^3
216
p
3
can't be represented in the form
x
2
+
y
2
+
z
9
x^2+y^2+z^9
x
2
+
y
2
+
z
9
,
x
,
y
,
z
∈
Z
x,y,z\in\mathbb{Z}
x
,
y
,
z
∈
Z
\sqrt[3]{x - y + z^3} + \sqrt[3]{y - z + x^3} + \sqrt[3]{z - x + y^3} <= 1
Let
x
,
y
,
z
x, y, z
x
,
y
,
z
be non-negative real numbers whose sum is
1
1
1
. Prove that:
x
−
y
+
z
3
3
+
y
−
z
+
x
3
3
+
z
−
x
+
y
3
3
≤
1
\sqrt[3]{x - y + z^3} + \sqrt[3]{y - z + x^3} + \sqrt[3]{z - x + y^3} \le 1
3
x
−
y
+
z
3
+
3
y
−
z
+
x
3
+
3
z
−
x
+
y
3
≤
1
10.3
2
Hide problems
A nica inequality
Prove that for all non negative real numbers
a
,
b
,
c
a,b,c
a
,
b
,
c
we have
a
2
+
b
2
+
c
2
≤
(
b
2
−
b
c
+
c
2
)
(
c
2
−
c
a
+
a
2
)
+
(
c
2
−
c
a
+
a
2
)
(
a
2
−
a
b
+
b
2
)
+
(
a
2
−
a
b
+
b
2
)
(
b
2
−
b
x
+
c
2
)
a^2+b^2+c^2\leq\sqrt{(b^2-bc+c^2)(c^2-ca+a^2)}+\sqrt{(c^2-ca+a^2)(a^2-ab+b^2)}+\sqrt{(a^2-ab+b^2)(b^2-bx+c^2)}
a
2
+
b
2
+
c
2
≤
(
b
2
−
b
c
+
c
2
)
(
c
2
−
c
a
+
a
2
)
+
(
c
2
−
c
a
+
a
2
)
(
a
2
−
ab
+
b
2
)
+
(
a
2
−
ab
+
b
2
)
(
b
2
−
b
x
+
c
2
)
f(x^2 + f(y) - y) = (f(x))^2
Find all functions
f
:
R
→
R
f : R \to R
f
:
R
→
R
that satisfy
f
(
x
2
+
f
(
y
)
−
y
)
=
(
f
(
x
)
)
2
f(x^2 + f(y) - y) = (f(x))^2
f
(
x
2
+
f
(
y
)
−
y
)
=
(
f
(
x
)
)
2
for all
x
,
y
∈
R
x,y \in R
x
,
y
∈
R
.
10.4
2
Hide problems
ax^2 + bx + c = 0 has real root in [\beta, \beta + \alpha]
Find the least real number
α
\alpha
α
such that there is a real number
β
\beta
β
so that for all triples of real numbers
(
a
,
b
,
c
)
(a, b,c)
(
a
,
b
,
c
)
satisfying
2006
a
+
10
b
+
c
=
0
2006a + 10b + c = 0
2006
a
+
10
b
+
c
=
0
, the equation
a
x
2
+
b
x
+
c
=
0
ax^2 + bx + c = 0
a
x
2
+
b
x
+
c
=
0
always has real root in the interval
[
β
,
β
+
α
]
[\beta, \beta + \alpha]
[
β
,
β
+
α
]
.
VMEO III problem 4, October 2006
Given a convex polygon
G
G
G
, show that there are three vertices of
G
G
G
which form a triangle so that it's perimeter is not less than 70% of the polygon's perimeter.
12.3
1
Hide problems
VMEO Month 12 - 2006
Let
a
,
b
,
c
,
d
a,b,c,d
a
,
b
,
c
,
d
be positive real numbers such that
(
a
+
b
+
c
+
d
)
(
1
a
+
1
b
+
1
c
+
1
d
)
=
20.
(a+b+c+d)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)=20.
(
a
+
b
+
c
+
d
)
(
a
1
+
b
1
+
c
1
+
d
1
)
=
20.
Prove that
(
a
2
+
b
2
+
c
2
+
d
2
)
(
1
a
2
+
1
b
2
+
1
c
2
+
1
d
2
)
≥
36.
\left(a^{2}+b^{2}+c^{2}+d^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}\right)\ge 36.
(
a
2
+
b
2
+
c
2
+
d
2
)
(
a
2
1
+
b
2
1
+
c
2
1
+
d
2
1
)
≥
36.
There are two solutions, one by Phan Thanh Nam, one by me, which are very nice.