Document Text Contents
Page 1
1
Iranian Team Members (from left to right):
• Saeed Tayyebi
• Javad Abedi
• Morteza Ashrafi-jou
• Sina Rezaei
• Hessamoddin Rajab-Zadeh
• Masoud Shafaei
This booklet is prepared by Meysam Aghighi and Morteza Saghafian.
Copyright © Young Scholars Club 2009-2010. All rights reserved.
Page 2
2
27th
Iranian Mathematical Olympiad
2009-2010
Contents
Problems
FIRST ROUND ................................................................................................................ 4
SECOND ROUND .......................................................................................................... 6
THIRD ROUND .............................................................................................................. 9
Solutions
FIRST ROUND .............................................................................................................. 12
SECOND ROUND ........................................................................................................ 15
THIRD ROUND ............................................................................................................ 20
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13
Now, noting that 50 50 2500, and using the above
inequalities, we have 2 2 3 2 32 2 4 2 2 2 2 2500 2 2 2 52 1000
So the total number of empty pieces are not greater than 1000.
3. Since is the bisector, we have
2
So the circumcircle of is tangent to in . Hence if the measure of the
arc in this circle be , we have
2 2
Similarly we have . Thus 180° 180°
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4. Suppose that it can be done for . We claim that it can be done for 2, and
since it can't be done for 1, must be even. Let up, down, left and right be the
four directions. Note that if columns have moved to 2 columns, then the
first row must've moved down, the last row must've moved up, the leftmost
column must've moved left, and the rightmost column must've moved right.
Now, consider the remaining soldiers, they are 2 columns with soldiers
in each column, that have moved in columns with 2 soldiers in each.
Thus, if it can be done for , then must be even.
5. For all distinct , that , we have | , and
On the other hand all of the above terms are divisors of , so if the divisors of
in decreasing order be , then . Since we
know that , we have
1 1 1
And the problem is solved.
6. First divide the table into 11 equal arcs. Now if the cards , be on the points , on the table, we define the distance between these two cards, the number
of arcs between , on the table. (the smaller one) For example, the distance
between , is 5 in the following figure:
Now, after each step we sum the distances between every two cards with
consecutive numbers. Easily it can be proved that after each step this sum
either remains invariant or varies by 2 . Because if the place of card be
between 1 and 1, this sum doesn’t change, otherwise changes by 2. So
the parity of this sum remains invariant. At first it is odd ( 11), and if they are
all in one place this sum would be even, so it is impossible.
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11. Suppose that be an arithmetic progression with common
difference . Since is increasing, must be positive. (even if for one , we
have , the problem is easy) so we know that
. Now by using Chinese Remainder Theorem, we
conclude that exists such that and .
This means that can be displayed as both forms and .
Now it is seen that is in both progressions, so:
So . And hence for every , and also is constant for every .
Name this constant value ( ). From the other hand we can find
such that , it means appears in all of the progressions.
Also we can find such that , it means . (
are found using the Chinese Remainder Theorem) so for every , and
are in both progressions, hence:
And this means that if , then form
an arithmetic progression.
12. Since is on both circumcircles of and , so there is a spiral
similarity about carrying one of these circles to another such that it carries
to , and to . Suppose this similarity carries to a point . It is enough
that we prove that the points are collinear, because then, considering the
similarity of and we have:
Now we prove that , are collinear:
The spiral similarity about , carries to respectively. So the
triangles and are similar. From the other hand: (let denote
the measure of the arc )
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Thus the triangles and are similar, and hence and are
similar. So , also . From which we
deduce that and are similar. So , and:
1
Iranian Team Members (from left to right):
• Saeed Tayyebi
• Javad Abedi
• Morteza Ashrafi-jou
• Sina Rezaei
• Hessamoddin Rajab-Zadeh
• Masoud Shafaei
This booklet is prepared by Meysam Aghighi and Morteza Saghafian.
Copyright © Young Scholars Club 2009-2010. All rights reserved.
Page 2
2
27th
Iranian Mathematical Olympiad
2009-2010
Contents
Problems
FIRST ROUND ................................................................................................................ 4
SECOND ROUND .......................................................................................................... 6
THIRD ROUND .............................................................................................................. 9
Solutions
FIRST ROUND .............................................................................................................. 12
SECOND ROUND ........................................................................................................ 15
THIRD ROUND ............................................................................................................ 20
Page 13
13
Now, noting that 50 50 2500, and using the above
inequalities, we have 2 2 3 2 32 2 4 2 2 2 2 2500 2 2 2 52 1000
So the total number of empty pieces are not greater than 1000.
3. Since is the bisector, we have
2
So the circumcircle of is tangent to in . Hence if the measure of the
arc in this circle be , we have
2 2
Similarly we have . Thus 180° 180°
Page 14
14
4. Suppose that it can be done for . We claim that it can be done for 2, and
since it can't be done for 1, must be even. Let up, down, left and right be the
four directions. Note that if columns have moved to 2 columns, then the
first row must've moved down, the last row must've moved up, the leftmost
column must've moved left, and the rightmost column must've moved right.
Now, consider the remaining soldiers, they are 2 columns with soldiers
in each column, that have moved in columns with 2 soldiers in each.
Thus, if it can be done for , then must be even.
5. For all distinct , that , we have | , and
On the other hand all of the above terms are divisors of , so if the divisors of
in decreasing order be , then . Since we
know that , we have
1 1 1
And the problem is solved.
6. First divide the table into 11 equal arcs. Now if the cards , be on the points , on the table, we define the distance between these two cards, the number
of arcs between , on the table. (the smaller one) For example, the distance
between , is 5 in the following figure:
Now, after each step we sum the distances between every two cards with
consecutive numbers. Easily it can be proved that after each step this sum
either remains invariant or varies by 2 . Because if the place of card be
between 1 and 1, this sum doesn’t change, otherwise changes by 2. So
the parity of this sum remains invariant. At first it is odd ( 11), and if they are
all in one place this sum would be even, so it is impossible.
Page 25
25
11. Suppose that be an arithmetic progression with common
difference . Since is increasing, must be positive. (even if for one , we
have , the problem is easy) so we know that
. Now by using Chinese Remainder Theorem, we
conclude that exists such that and .
This means that can be displayed as both forms and .
Now it is seen that is in both progressions, so:
So . And hence for every , and also is constant for every .
Name this constant value ( ). From the other hand we can find
such that , it means appears in all of the progressions.
Also we can find such that , it means . (
are found using the Chinese Remainder Theorem) so for every , and
are in both progressions, hence:
And this means that if , then form
an arithmetic progression.
12. Since is on both circumcircles of and , so there is a spiral
similarity about carrying one of these circles to another such that it carries
to , and to . Suppose this similarity carries to a point . It is enough
that we prove that the points are collinear, because then, considering the
similarity of and we have:
Now we prove that , are collinear:
The spiral similarity about , carries to respectively. So the
triangles and are similar. From the other hand: (let denote
the measure of the arc )
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Thus the triangles and are similar, and hence and are
similar. So , also . From which we
deduce that and are similar. So , and:
