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Problems
Contests
National and Regional Contests
Greece Contests
Greece Team Selection Test
2008 Greece Team Selection Test
2008 Greece Team Selection Test
Part of
Greece Team Selection Test
Subcontests
(3)
2
2
Hide problems
Tourism In Five Villages
In a village
X
0
X_0
X
0
there are
80
80
80
tourists who are about to visit
5
5
5
nearby villages
X
1
,
X
2
,
X
3
,
X
4
,
X
5
X_1,X_2,X_3,X_4,X_5
X
1
,
X
2
,
X
3
,
X
4
,
X
5
.Each of them has chosen to visit only one of them.However,there are cases when the visit in a village forces the visitor to visit other villages among
X
1
,
X
2
,
X
3
,
X
4
,
X
5
X_1,X_2,X_3,X_4,X_5
X
1
,
X
2
,
X
3
,
X
4
,
X
5
.Each tourist visits only the village he has chosen and the villages he is forced to.If
X
1
,
X
2
,
X
3
,
X
4
,
X
5
X_1,X_2,X_3,X_4,X_5
X
1
,
X
2
,
X
3
,
X
4
,
X
5
are totally visited by
40
,
60
,
65
,
70
,
75
40,60,65,70,75
40
,
60
,
65
,
70
,
75
tourists respectively,then find how many tourists had chosen each one of them and determine all the ordered pairs
(
X
i
,
X
j
)
:
i
,
j
∈
{
1
,
2
,
3
,
4
,
5
}
(X_i,X_j):i,j\in \{1,2,3,4,5\}
(
X
i
,
X
j
)
:
i
,
j
∈
{
1
,
2
,
3
,
4
,
5
}
which are such that,the visit in
X
i
X_i
X
i
forces the visitor to visit
X
j
X_j
X
j
as well.
Six Lines Are Concurrent
The bisectors of the angles
∠
A
,
∠
B
,
∠
C
\angle{A},\angle{B},\angle{C}
∠
A
,
∠
B
,
∠
C
of a triangle
△
A
B
C
\triangle{ABC}
△
A
BC
intersect with the circumcircle
c
1
(
O
,
R
)
c_1(O,R)
c
1
(
O
,
R
)
of
△
A
B
C
\triangle{ABC}
△
A
BC
at
A
2
,
B
2
,
C
2
A_2,B_2,C_2
A
2
,
B
2
,
C
2
respectively.The tangents of
c
1
c_1
c
1
at
A
2
,
B
2
,
C
2
A_2,B_2,C_2
A
2
,
B
2
,
C
2
intersect each other at
A
3
,
B
3
,
C
3
A_3,B_3,C_3
A
3
,
B
3
,
C
3
(the points
A
3
,
A
A_3,A
A
3
,
A
lie on the same side of
B
C
BC
BC
,the points
B
3
,
B
B_3,B
B
3
,
B
on the same side of
C
A
CA
C
A
,and
C
3
,
C
C_3,C
C
3
,
C
on the same side of
A
B
AB
A
B
).The incircle
c
2
(
I
,
r
)
c_2(I,r)
c
2
(
I
,
r
)
of
△
A
B
C
\triangle{ABC}
△
A
BC
is tangent to
B
C
,
C
A
,
A
B
BC,CA,AB
BC
,
C
A
,
A
B
at
A
1
,
B
1
,
C
1
A_1,B_1,C_1
A
1
,
B
1
,
C
1
respectively.Prove that
A
1
A
2
,
B
1
B
2
,
C
1
C
2
,
A
A
3
,
B
B
3
,
C
C
3
A_1A_2,B_1B_2,C_1C_2,AA_3,BB_3,CC_3
A
1
A
2
,
B
1
B
2
,
C
1
C
2
,
A
A
3
,
B
B
3
,
C
C
3
are concurrent.[asy]import graph; size(11cm); 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 = -9.26871978147865, xmax = 19.467150423463277, ymin = -6.150626456647122, ymax = 10.10782642246474; /* image dimensions */ pen aqaqaq = rgb(0.6274509803921569,0.6274509803921569,0.6274509803921569); pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); draw((1.0409487561836381,4.30054785243355)--(0.,0.)--(6.,0.)--cycle, aqaqaq); /* draw figures */ draw((1.0409487561836381,4.30054785243355)--(0.,0.), uququq); draw((0.,0.)--(6.,0.), uququq); draw((6.,0.)--(1.0409487561836381,4.30054785243355), uququq); draw(circle((3.,1.550104087253063), 3.376806580383107)); draw(circle((1.9303371951242874,1.5188413314630436), 1.5188413314630436)); draw((1.0226422135625703,7.734611112525813)--(1.0559139088339535,1.4932847901569466), linetype("2 2")); draw((-1.2916762981259242,-1.8267024931300444)--(1.0559139088339535,1.4932847901569466), linetype("2 2")); draw((-0.2820306621765219,2.344520485530311)--(1.0559139088339535,1.4932847901569466), linetype("2 2")); draw((1.0559139088339535,1.4932847901569466)--(5.212367857300808,4.101231513568902), linetype("2 2")); draw((1.0559139088339535,1.4932847901569466)--(3.,-1.8267024931300442), linetype("2 2")); draw((12.047991949367804,-1.8267024931300444)--(1.0559139088339535,1.4932847901569466), linetype("2 2")); draw((1.0226422135625703,7.734611112525813)--(-1.2916762981259242,-1.8267024931300444)); draw((-1.2916762981259242,-1.8267024931300444)--(12.047991949367804,-1.8267024931300444)); draw((12.047991949367804,-1.8267024931300444)--(1.0226422135625703,7.734611112525813)); /* dots and labels */ dot((1.0409487561836381,4.30054785243355),linewidth(3.pt) + dotstyle); label("
A
A
A
", (0.5889800538632699,4.463280489351154), NE * labelscalefactor); dot((0.,0.),linewidth(3.pt) + dotstyle); label("
B
B
B
", (-0.5723380089304358,-0.10096957139619551), NE * labelscalefactor); dot((6.,0.),linewidth(3.pt) + dotstyle); label("
C
C
C
", (6.233525986976863,0.06107480945873997), NE * labelscalefactor); label("
c
1
c_1
c
1
", (1.9663572911302232,5.111458012770896), NE * labelscalefactor); dot((3.,-1.8267024931300442),linewidth(3.pt) + dotstyle); label("
A
2
A_2
A
2
", (2.9386235762598374,-2.3155761097469805), NE * labelscalefactor); dot((5.212367857300808,4.101231513568902),linewidth(3.pt) + dotstyle); label("
B
2
B_2
B
2
", (5.315274495465561,4.274228711687063), NE * labelscalefactor); dot((-0.2820306621765219,2.344520485530311),linewidth(3.pt) + dotstyle); label("
C
2
C_2
C
2
", (-0.9234341674494632,2.6807922999468636), NE * labelscalefactor); dot((1.0226422135625703,7.734611112525813),linewidth(3.pt) + dotstyle); label("
A
3
A_3
A
3
", (1.1291279900463889,7.893219884113956), NE * labelscalefactor); dot((-1.2916762981259242,-1.8267024931300444),linewidth(3.pt) + dotstyle); label("
B
3
B_3
B
3
", (-1.8146782621516093,-1.4783468086631473), NE * labelscalefactor); dot((12.047991949367804,-1.8267024931300444),linewidth(3.pt) + dotstyle); label("
C
3
C_3
C
3
", (12.148145888182015,-1.6673985863272387), NE * labelscalefactor); dot((1.9303371951242874,1.5188413314630436),linewidth(3.pt) + dotstyle); label("
I
I
I
", (2.047379481557691,1.681518618008095), NE * labelscalefactor); dot((1.9303371951242878,0.),linewidth(3.pt) + dotstyle); label("
A
1
A_1
A
1
", (1.4532167517562602,-0.5600953171518461), NE * labelscalefactor); label("
c
2
c_2
c
2
", (1.5072315453745722,3.247947632939138), NE * labelscalefactor); dot((2.9254299438737803,2.666303492733126),linewidth(3.pt) + dotstyle); label("
B
1
B_1
B
1
", (2.8576013858323694,3.1129106488933584), NE * labelscalefactor); dot((0.45412477306806903,1.8761589424582812),linewidth(3.pt) + dotstyle); label("
C
1
C_1
C
1
", (0,2.3296961414278368), NE * labelscalefactor); dot((1.0559139088339535,1.4932847901569466),linewidth(3.pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]
4
1
Hide problems
It almost never has solutions
Given is the equation
x
2
+
y
2
−
a
x
y
+
2
=
0
x^2+y^2-axy+2=0
x
2
+
y
2
−
a
x
y
+
2
=
0
where
a
a
a
is a positive integral parameter.
i
.
i.
i
.
Show that,for
a
≠
4
a\neq 4
a
=
4
there exist no pairs
(
x
,
y
)
(x,y)
(
x
,
y
)
of positive integers satisfying the equation.
i
i
.
ii.
ii
.
Show that,for
a
=
4
a=4
a
=
4
there exist infinite pairs
(
x
,
y
)
(x,y)
(
x
,
y
)
of positive integers satisfying the equation,and determine those pairs.
1
1
Hide problems
Polynomial Divisibility
Find all possible values of
a
∈
R
a\in \mathbb{R}
a
∈
R
and
n
∈
N
∗
n\in \mathbb{N^*}
n
∈
N
∗
such that
f
(
x
)
=
(
x
−
1
)
n
+
(
x
−
2
)
2
n
+
1
+
(
1
−
x
2
)
2
n
+
1
+
a
f(x)=(x-1)^n+(x-2)^{2n+1}+(1-x^2)^{2n+1}+a
f
(
x
)
=
(
x
−
1
)
n
+
(
x
−
2
)
2
n
+
1
+
(
1
−
x
2
)
2
n
+
1
+
a
is divisible by
ϕ
(
x
)
=
x
2
−
x
+
1
\phi (x)=x^2-x+1
ϕ
(
x
)
=
x
2
−
x
+
1