# International Mathematical Olymp

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```					International Mathematical Olympiad 2000 Hong Kong Team Selection Test 1 (Solution)

1.

Find the remainder of 19992000 when it is divided by 31.

Solution. By Fermat Little Theorem, a 30  1 (mod 31) if gcd(31, a)  1 .
19992000  15 2000  (1530 ) 66  15 20  15 20  (15 2 )10  810  (8 2 ) 5  25  1 (mod 31)

2.

Let a, b, c be positive and abc  1 . Prove that
1  ab 2 1  bc 2 1  ca 2 18    3 . 3 3 3 c a b a  b3  c3

Determine when equality is attained.

Solution. Apply Cauchy-Schwarz Inequality, we have
1  ab 2 1  bc 2 1  ca 2 3 (   )( c  a 3  b 3 )  ( 1  ab 2  1  bc 2  1  ca 2 ) 2 3 3 3 c a b

It remains to prove

1  ab 2  1  bc 2  1  ca 2  18 . The proof goes as follows.
(Triangular Inequality)

1  ab 2  1  bc 2  1  ca 2  (1  1  1) 2  ( ab 2  bc 2  ca 2 ) 2  9  (3 abc ) 2  18

( A.M .  G.M . Inequality)

This proved the desired inequality. Equality holds if and only if a  b  c  1.

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3.

On a table there are 1999 tea cups with their mouths facing upward initially. In each move, 100 of them are turned upside down. After a number of moves, can they be turned so that all their mouths face downward? Why? Answer the above two questions for the case where the number of cups is 1998?

Solution. Case 1 (number of cups is 1999): It is impossible to make all the cups downward. For each cup, we assign the number 1 to it if it is facing upward, otherwise we assign the number 0 to it. Denote S be the sum of these numbers. At first, every cup is facing upward, so S  1999 which is an odd number. In each move, 100 cups are turned upside down. The parity of S remains unchanged. Hence, it is impossible to make all the cups downward because the corresponding value of S is zero, which is an even number. Case 2 (number of cups is 1998): It is possible to make all the cups downward. We are going to prove there is a procedure to turn 2 cups upside down without affecting the others. Label the cups by 1, 2, …, 1998. Assume we want to change cup 1 and cup 2, we may first change the cups 1, 3, 4, 5, …, 101, and then the cups 2, 3, 4, 5, …, 101. After this procedure, only cup 1 and cup 2 have been changed. Similarly we can change any two cups without affecting the others. The result follows.

4.

ABC is a triangle with BC  CA  AB . D is a point on side BC, and E is a point on BA produced beyond A so that BD  BE  CA . Let P be a point on side AC such that E, B, D, P are concyclic, and let Q be the second intersection point of BP with the circumcircle of ABC . Prove that AQ  CQ  BP .
E

A

Q P B D C

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Solution.
E

A

Q P B D C

We claim that AQC ~ EPD . This is because CAQ  CBQ  DEP and AQC  180  ABD  EPD. On the other hand, by Ptolemy’s Theorem, we have

BP  DE  BE  DP  BD EP .
So BP  BE 
DP EP CQ AQ  BD   CA   CA   AQ  CQ . DE DE CA CA

5.

Let G be a connected simple graph which has 2 p vertices (where p is a positive integer), but contains no triangles. Prove that the number # E of its edges satisfies: # E  p2 .

Solution. We will prove the statement by induction on p. The case p  1 is clear. We assume the statement is true for some positive integer p, and then consider a graph G with 2 p  2 vertices containing no triangle. Take any two connected vertices, says u and v. Note that for any other vertex w, it may adjacent with u or v, but not both! Otherwise a triangle exist in the graph. This tell us the number of edges incident to u or v not exceeding 2 p  1 .
w v u

We remove all these edges(and also vertices u and v) from the graph. This gives a new graph G  with 2 p vertices containing no triangle, by assumption it has at most p 2 edges. Together with the edges we removed, the number of edges in graph G is at most p 2  2 p  1  ( p  1) 2 . The result follows. Page 3 of 4

6.

Determine all primes of the form n n  1 , which are less than 1019 (n is a positive integer).

Solution. The solutions are 1, 5, 257 for which n  1, 2, 4 . The proof goes follows. Clearly we have solution n  1, we may assume n  1 in the following argument. Note that if n has odd factor greater than 1, then n n  1 can be factorized. For example, if n  nk for odd k  1 , then n n  1  (n n ) k  1k which is divisible by n n  1 . So n must be power of 2. Next we claim that n  16 , this is because
16 16  1  2 64  2 4  (210 ) 6  16  1000 6  1.6  10 19  10 19 .

By direct checking, n  1, 2, 4 satisfy the condition. For n  8 , 88  1  (2 8 ) 3  1 which is divisible by 28  1  257 . To conclude, the solutions are 1, 5, 257 for which n  1, 2, 4 .

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