MathDB
Problems
Contests
National and Regional Contests
Greece Contests
Greece Team Selection Test
2013 Greece Team Selection Test
2013 Greece Team Selection Test
Part of
Greece Team Selection Test
Subcontests
(4)
4
2
Hide problems
Counting Regions
Given are
n
n
n
different concentric circles on the plane.Inside the disk with the smallest radius (strictly inside it),we consider two distinct points
A
,
B
A,B
A
,
B
.We consider
k
k
k
distinct lines passing through
A
A
A
and
m
m
m
distinct lines passing through
B
B
B
.There is no line passing through both
A
A
A
and
B
B
B
and all the lines passing through
k
k
k
intersect with all the lines passing through
B
B
B
.The intersections do not lie on some of the circles.Determine the maximum and the minimum number of regions formed by the lines and the circles and are inside the circles.
Tiling with trapezoids in an equilateral grid
Let
n
n
n
be a positive integer. An equilateral triangle with side
n
n
n
will be denoted by
T
n
T_n
T
n
and is divided in
n
2
n^2
n
2
unit equilateral triangles with sides parallel to the initial, forming a grid. We will call "trapezoid" the trapezoid which is formed by three equilateral triangles (one base is equal to one and the other is equal to two). Let also
m
m
m
be a positive integer with
m
<
n
m<n
m
<
n
and suppose that
T
n
T_n
T
n
and
T
m
T_m
T
m
can be tiled with "trapezoids". Prove that, if from
T
n
T_n
T
n
we remove a
T
m
T_m
T
m
with the same orientation, then the rest can be tiled with "trapezoids".
3
2
Hide problems
Maximum Value
Find the largest possible value of
M
M
M
for which
x
1
+
y
z
x
+
y
1
+
z
x
y
+
z
1
+
x
y
z
≥
M
\frac{x}{1+\frac{yz}{x}}+\frac{y}{1+\frac{zx}{y}}+\frac{z}{1+\frac{xy}{z}}\geq M
1
+
x
yz
x
+
1
+
y
z
x
y
+
1
+
z
x
y
z
≥
M
for all
x
,
y
,
z
>
0
x,y,z>0
x
,
y
,
z
>
0
with
x
y
+
y
z
+
z
x
=
1
xy+yz+zx=1
x
y
+
yz
+
z
x
=
1
Concurrent Diagonals Of Hexagon
Given is a triangle
A
B
C
ABC
A
BC
.On the extensions of the side
A
B
AB
A
B
we consider points
A
1
,
B
1
A_1,B_1
A
1
,
B
1
such that
A
B
1
=
B
A
1
AB_1=BA_1
A
B
1
=
B
A
1
(with
A
1
A_1
A
1
lying closer to
B
B
B
).On the extensions of the side
B
C
BC
BC
we consider points
B
4
,
C
4
B_4,C_4
B
4
,
C
4
such that
C
B
4
=
B
C
4
CB_4=BC_4
C
B
4
=
B
C
4
(with
B
4
B_4
B
4
lying closer to
C
C
C
).On the extensions of the side
A
C
AC
A
C
we consider points
C
1
,
A
4
C_1,A_4
C
1
,
A
4
such that
A
C
1
=
C
A
4
AC_1=CA_4
A
C
1
=
C
A
4
(with
C
1
C_1
C
1
lying closer to
A
A
A
).On the segment
A
1
A
4
A_1A_4
A
1
A
4
we consider points
A
2
,
A
3
A_2,A_3
A
2
,
A
3
such that
A
1
A
2
=
A
3
A
4
=
m
A
1
A
4
A_1A_2=A_3A_4=mA_1A_4
A
1
A
2
=
A
3
A
4
=
m
A
1
A
4
where
0
<
m
<
1
2
0<m<\frac{1}{2}
0
<
m
<
2
1
.Points
B
2
,
B
3
B_2,B_3
B
2
,
B
3
and
C
2
,
C
3
C_2,C_3
C
2
,
C
3
are defined similarly,on the segments
B
1
B
4
,
C
1
C
4
B_1B_4,C_1C_4
B
1
B
4
,
C
1
C
4
respectively.If
D
≡
B
B
2
∩
C
C
2
,
E
≡
A
A
3
∩
C
C
2
,
F
≡
A
A
3
∩
B
B
3
D\equiv BB_2\cap CC_2 \ , \ E\equiv AA_3\cap CC_2 \ , \ F\equiv AA_3\cap BB_3
D
≡
B
B
2
∩
C
C
2
,
E
≡
A
A
3
∩
C
C
2
,
F
≡
A
A
3
∩
B
B
3
,
G
≡
B
B
3
∩
C
C
3
,
H
≡
A
A
2
∩
C
C
3
\ G\equiv BB_3\cap CC_3 \ , \ H\equiv AA_2\cap CC_3
G
≡
B
B
3
∩
C
C
3
,
H
≡
A
A
2
∩
C
C
3
and
I
≡
A
A
2
∩
B
B
2
I\equiv AA_2\cap BB_2
I
≡
A
A
2
∩
B
B
2
,prove that the diagonals
D
G
,
E
H
,
F
I
DG,EH,FI
D
G
,
E
H
,
F
I
of the hexagon
D
E
F
G
H
I
DEFGHI
D
EFG
H
I
are concurrent. [asy]import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -7.984603447540051, xmax = 21.28710511372557, ymin = -6.555010307713199, ymax = 10.006614273002825; /* image dimensions */ pen aqaqaq = rgb(0.6274509803921569,0.6274509803921569,0.6274509803921569); pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); draw((1.1583842866003107,4.638449718549554)--(0.,0.)--(7.,0.)--cycle, aqaqaq); /* draw figures */ draw((1.1583842866003107,4.638449718549554)--(0.,0.), uququq); draw((0.,0.)--(7.,0.), uququq); draw((7.,0.)--(1.1583842866003107,4.638449718549554), uququq); draw((1.1583842866003107,4.638449718549554)--(1.623345080409327,6.500264738079558)); draw((0.,0.)--(-0.46496079380901606,-1.8618150195300045)); draw((-3.0803965232149757,0.)--(0.,0.)); draw((7.,0.)--(10.080396523214976,0.)); draw((1.1583842866003107,4.638449718549554)--(0.007284204967787214,5.552463941947242)); draw((7.,0.)--(8.151100081632526,-0.9140142233976905)); draw((-0.46496079380901606,-1.8618150195300045)--(8.151100081632526,-0.9140142233976905)); draw((-3.0803965232149757,0.)--(0.007284204967787214,5.552463941947242)); draw((10.080396523214976,0.)--(1.623345080409327,6.500264738079558)); draw((0.,0.)--(3.7376079411107392,4.8751985535596685)); draw((-0.7646359770779035,4.164347956460432)--(7.,0.)); draw((1.1583842866003107,4.638449718549554)--(5.997084862772141,-1.150964422430769)); draw((0.,0.)--(7.966133662513563,1.6250661845198895)); draw((-2.308476341169285,1.3881159854868106)--(7.,0.)); draw((1.1583842866003107,4.638449718549554)--(1.6890544250513695,-1.624864820496926)); draw((2.0395968109217,2.660375186246903)--(2.9561195753832448,0.6030390855677443), linetype("2 2")); draw((3.4388364046369224,1.909931693481981)--(1.4816619768719694,0.8229159040072803), linetype("2 2")); draw((1.3969966570225139,1.8221911417546572)--(4.301698851378541,0.8775330211014288), linetype("2 2")); /* dots and labels */ dot((1.1583842866003107,4.638449718549554),linewidth(3.pt) + dotstyle); label("
A
A
A
", (0.6263408942608304,4.2), NE * labelscalefactor); dot((0.,0.),linewidth(3.pt) + dotstyle); label("
B
B
B
", (-0.44658827292841696,0.04763072114368767), NE * labelscalefactor); dot((7.,0.),linewidth(3.pt) + dotstyle); label("
C
C
C
", (7.008893888822507,0.18518574257820614), NE * labelscalefactor); dot((1.623345080409327,6.500264738079558),linewidth(3.pt) + dotstyle); label("
B
1
B_1
B
1
", (1.7267810657369815,6.6777827542874775), NE * labelscalefactor); dot((-0.46496079380901606,-1.8618150195300045),linewidth(3.pt) + dotstyle); label("
A
1
A_1
A
1
", (-1.1068523758141076,-1.6305405403574376), NE * labelscalefactor); dot((10.080396523214976,0.),linewidth(3.pt) + dotstyle); label("
B
4
B_4
B
4
", (10.062615364668826,-0.612633381742001), NE * labelscalefactor); dot((-3.0803965232149757,0.),linewidth(3.pt) + dotstyle); label("
C
4
C_4
C
4
", (-3.3077327187664096,-0.612633381742001), NE * labelscalefactor); dot((0.007284204967787214,5.552463941947242),linewidth(3.pt) + dotstyle); label("
C
1
C_1
C
1
", (0.1036318128096586,5.714897604245849), NE * labelscalefactor); dot((8.151100081632526,-0.9140142233976905),linewidth(3.pt) + dotstyle); label("
A
4
A_4
A
4
", (8.521999124602214,-1.1903644717669786), NE * labelscalefactor); dot((-2.308476341169285,1.3881159854868106),linewidth(3.pt) + dotstyle); label("
C
3
C_3
C
3
", (-2.9776006673235647,1.7808239912186203), NE * labelscalefactor); dot((-0.7646359770779035,4.164347956460432),linewidth(3.pt) + dotstyle); label("
C
2
C_2
C
2
", (-1.1618743843879151,4.504413415622086), NE * labelscalefactor); dot((1.6890544250513695,-1.624864820496926),linewidth(3.pt) + dotstyle); label("
A
2
A_2
A
2
", (1.6167370485893664,-2.125738617521704), NE * labelscalefactor); dot((5.997084862772141,-1.150964422430769),linewidth(3.pt) + dotstyle); label("
A
3
A_3
A
3
", (6.211074764502297,-1.603029536070534), NE * labelscalefactor); dot((7.966133662513563,1.6250661845198895),linewidth(3.pt) + dotstyle); label("
B
3
B_3
B
3
", (8.081823056011753,1.7808239912186203), NE * labelscalefactor); dot((3.7376079411107392,4.8751985535596685),linewidth(3.pt) + dotstyle); label("
B
2
B_2
B
2
", (3.8451283958285725,5.027122497073257), NE * labelscalefactor); dot((2.0395968109217,2.660375186246903),linewidth(3.pt) + dotstyle); label("
D
D
D
", (1.7542920700238853,2.991308179842383), NE * labelscalefactor); dot((3.4388364046369224,1.909931693481981),linewidth(3.pt) + dotstyle); label("
E
E
E
", (3.542507348672631,2.083445038374561), NE * labelscalefactor); dot((4.301698851378541,0.8775330211014288),linewidth(3.pt) + dotstyle); label("
F
F
F
", (4.22,0.93), NE * labelscalefactor); dot((2.9561195753832448,0.6030390855677443),linewidth(3.pt) + dotstyle); label("
G
G
G
", (2.909754250073844,0.10265272971749505), NE * labelscalefactor); dot((1.4816619768719694,0.8229159040072803),linewidth(3.pt) + dotstyle); label("
H
H
H
", (0.9839839499905795,0.43278478116033936), NE * labelscalefactor); dot((1.3969966570225139,1.8221911417546572),linewidth(3.pt) + dotstyle); label("
I
I
I
", (0.9839839499905795,1.8908680083662353), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]
2
2
Hide problems
Concyclic points
Let
A
B
C
ABC
A
BC
be a non-isosceles,aqute triangle with
A
B
<
A
C
AB<AC
A
B
<
A
C
inscribed in circle
c
(
O
,
R
)
c(O,R)
c
(
O
,
R
)
.The circle
c
1
(
B
,
A
B
)
c_{1}(B,AB)
c
1
(
B
,
A
B
)
crosses
A
C
AC
A
C
at
K
K
K
and
c
c
c
at
E
E
E
.
K
E
KE
K
E
crosses
c
c
c
at
F
F
F
and
B
O
BO
BO
crosses
K
E
KE
K
E
at
L
L
L
and
A
C
AC
A
C
at
M
M
M
while
A
E
AE
A
E
crosses
B
F
BF
BF
at
D
D
D
.Prove that: i)
D
,
L
,
M
,
F
D,L,M,F
D
,
L
,
M
,
F
are concyclic. ii)
B
,
D
,
K
,
M
,
E
B,D,K,M,E
B
,
D
,
K
,
M
,
E
are concyclic.
Find the pairs of integers...
For the several values of the parameter
m
∈
N
∗
m\in \mathbb{N^{*}}
m
∈
N
∗
,find the pairs of integers
(
a
,
b
)
(a,b)
(
a
,
b
)
that satisfy the relation
[
a
,
m
]
+
[
b
,
m
]
(
a
+
b
)
m
=
10
11
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{[a,m]+[b,m]}{(a+b)m}=\frac{10}{11}
(
a
+
b
)
m
[
a
,
m
]
+
[
b
,
m
]
=
11
10
,and,moreover,on the Cartesian plane
O
x
y
Oxy
O
x
y
the lie in the square
D
=
{
(
x
,
y
)
:
1
≤
x
≤
36
,
1
≤
y
≤
36
}
D=\{(x,y):1\leq x\leq 36,1\leq y\leq 36\}
D
=
{(
x
,
y
)
:
1
≤
x
≤
36
,
1
≤
y
≤
36
}
.Note:
[
k
,
l
]
[k,l]
[
k
,
l
]
denotes the least common multiple of the positive integers
k
,
l
k,l
k
,
l
.
1
2
Hide problems
Equation in natural numbers.
Find all pairs of non-negative integers
(
m
,
n
)
(m,n)
(
m
,
n
)
satisfying
n
(
n
+
2
)
4
=
m
4
+
m
2
−
m
+
1
\frac{n(n+2)}{4}=m^4+m^2-m+1
4
n
(
n
+
2
)
=
m
4
+
m
2
−
m
+
1
Determine whether it is irreducible
Determine whether the polynomial
P
(
x
)
=
(
x
2
−
2
x
+
5
)
(
x
2
−
4
x
+
20
)
+
1
P(x)=(x^2-2x+5)(x^2-4x+20)+1
P
(
x
)
=
(
x
2
−
2
x
+
5
)
(
x
2
−
4
x
+
20
)
+
1
is irreducible over
Z
[
X
]
\mathbb{Z}[X]
Z
[
X
]
.