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National and Regional Contests
Greece Contests
Greece Junior Math Olympiad
1997 Greece Junior Math Olympiad
1997 Greece Junior Math Olympiad
Part of
Greece Junior Math Olympiad
Subcontests
(4)
4
1
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sum of 100 signed numbers in 10 concentric circles equals zero
Consider ten concentric circles and ten rays as in the following figure. At the points where the inner circle is intersected by the rays write successively, in direction clockwise, the numbers
1
,
2
,
3
,
4
,
5
,
6
,
7
,
8
,
9
,
10
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
1
,
2
,
3
,
4
,
5
,
6
,
7
,
8
,
9
,
10
. In the next circle we write the numbers
11
,
12
,
13
,
14
,
15
,
16
,
17
,
18
,
19
,
20
11, 12, 13, 14, 15, 16, 17, 18, 19,20
11
,
12
,
13
,
14
,
15
,
16
,
17
,
18
,
19
,
20
successively, and so on successively until the last round were we write the numbers
91
,
92
,
93
,
94
,
95
,
96
,
97
,
98
,
99
,
100
91, 92, 93, 94, 95, 96, 97, 98, 99, 100
91
,
92
,
93
,
94
,
95
,
96
,
97
,
98
,
99
,
100
successively. In this orde, the numbers
1
,
11
,
21
,
31
,
41
,
51
,
61
,
71
,
81
,
91
1, 11, 21, 31, 41, 51, 61, 71, 81, 91
1
,
11
,
21
,
31
,
41
,
51
,
61
,
71
,
81
,
91
are in the same ray, and similarly for the other rays. In front of
50
50
50
of those
100
100
100
numbers, we use the sign ''
−
-
−
'' such as: a) in each of the ten rays, exist exactly
5
5
5
signs ''
−
-
−
'' , and also b) in each of the ten concentric circles, to be exactly
5
5
5
signs ''
−
-
−
''. Prove that the sum of the
100
100
100
signed numbers that occur, equals zero. https://cdn.artofproblemsolving.com/attachments/9/d/ffee6518fcd1b996c31cf06d0ce484a821b4ae.gif
3
1
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Combinatorics
Establish if we can rewrite the numbers
1
,
2
,
3
,
4
,
5
,
6
,
7
,
8
,
9
,
10
1,2,3,4,5,6,7,8,9,10
1
,
2
,
3
,
4
,
5
,
6
,
7
,
8
,
9
,
10
in a row in such a way that: (a) The sum of any three consecutive numbers (in the new order) does not exceed
16
16
16
. (b) The sum of any three consecutive numbers (in the new order) does not exceed
15
15
15
.
2
1
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Number Theory
Determine all natural numbers n for which the number
A
=
n
4
+
4
n
3
+
5
n
2
+
6
n
A = n^4 + 4n^3 +5n^2 + 6n
A
=
n
4
+
4
n
3
+
5
n
2
+
6
n
is a perfect square of a natural number.
1
1
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BE=EZ=ZC, ratio of areas, equilateral (Greece Juniors 1997 p1)
Let
A
B
C
ABC
A
BC
be an equilateral triangle whose angle bisectors of
B
B
B
and
C
C
C
intersect at
D
D
D
. Perpendicular bisectors of
B
D
BD
B
D
and
C
D
CD
C
D
intersect
B
C
BC
BC
at points
E
E
E
and
Z
Z
Z
respectively. a) Prove that
B
E
=
E
Z
=
Z
C
BE=EZ=ZC
BE
=
EZ
=
ZC
. b) Find the ratio of the areas of the triangles
B
D
E
BDE
B
D
E
to
A
B
C
ABC
A
BC