International Mathematical Olympiad. Preliminary Selection Contest 2009 Hong Kong. Outline of Solutions

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Edward Caldwell
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1 International Matheatical Olypiad Preliinary Selection ontest 009 Hong Kong Outline of Solutions nswers: Solutions:. We have and so setting a 009 gives the answer 3 a ( a ) a a a a a a ( a ) a a a 3 ( a a 3a a ) ( a a) ( a a) a a a a Reark. It would be easier to start by plugging in sall values of a and then try to look for a pattern.

2 . Let P be the intersection of D and E. Then P = PED = PD = PD = = 3, and so P = P = 7. Hence P = = and P = 9 E 3 = 0. It follows that P is equilateral and P = PE. Thus PE = 08 o 0 = and EP =, and so E = ED = + 3 = 0. P D 3. Suppose Donald, Henry and John started with n, n and n candies respectively. fter Donald gave John 0 candies and Henry shared candies with John, they had n 0, 3n 0 and 3n 0 candies respectively. Hence we have n 0 3(3n 0), or n 0. In the end, Donald had n 0 x 0 x candies while Henry had 3n 0 x 0 x candies. Hence we have 0 x (0 x), giving x 0.. Note that 9997 n (0000 3) n 0000( n ) ( n). Thus n ust be odd to ensure that the last digit of 9997n is odd, and hence n is even. If n 333, then n 0000 and hence the ten thousands digit of 9997n is the sae as the unit digit of 0000( n ), which is even. Therefore 9997n contains an even digit. If n 333, we find that It follows that the sallest possible value of n is For n, we have Since xn 0, we have x n n n k k k x x ( x x ) n n n x k ( k ) k 3 ( n) n n n (note that this also holds for n ). Hence xn 009 n nn n

3 Reark. One could easily see a pattern and ake a guess for the answer by coputing x, x x, x x x3 and so on.. Note that HK is perpendicular to, PH is perpendicular to and PK is perpendicular to. Hence it is easy to see that PHK ~. Let M be the idpoint of (which is also the idpoint of HK HK). Then PM M 3. 0, and so P = PM + M =. + = P H. M K 7. We have D ()() cos 0 3. The area of D is ()()sin 0 3, and siilarly D has area 3 3. Recall that the radius of the inscribed circle of a triangle is equal to twice the area divided by the perieter. (This can be seen by connecting the in-centre to the three vertices and then considering area.) It follows that rs D 8. Let R S n and DQ k. Since RX ~ RD, we have n n or n. Q Since QDY ~ Q, we have k k or k. S Since RS ~ RDQ, we have n n or k n. k n D Y R Fro the above, we get. Solving this equation subject to the condition 0, we get 3. X P 3

4 9. Note that on the line segent joining (a, b) and (c, d), where a, b, c, d are integers, the nuber of lattice points (excluding the endpoints) is exactly less than the H..F. of c a and d b. Hence there is an even nuber of lattice points on the line segent (excluding the endpoints) joining (0, 00) and (a, b) if and only if the H..F. of a and 00 b is odd, which is true if and only if at least one of a, b is odd. Now there are 00 choices for each of a and b (naely,,,, 00). Hence the probability that both are even is 3, and so the required probability is. 0. Randoly arrange the players in a row (this can be done in 8! ways) so that the two leftost players copete against each other, the next two copete against each other and so on. Now suppose we want no two atheaticians to play against each other. Then there are 8 possible positions for the first atheatician, then position for the second atheatician, and siilarly for the third and for the last. fter the positions of the atheaticians are fixed, there are! ways to arrange the positions of the non-atheaticians. Hence the answer is 8! 8. 8! 3. Note that (k) (k) tan tan tan (k) tan (k). Hence k (k )(k ) tan tan tan tan tan tan(tan ( 009 ) tan ) tan tan ( 009 ) Reark. One could easily see a pattern and ake a guess for the answer by coputing tan tan, tan tan tan, tan tan tan tan and so on. 3. olour the cells by black and white alternately. It is clear that the nubers of coins in cells of sae colour are of the sae parity and the nubers of coins in cells of different colours are of different parity. Hence n ust be odd, for otherwise the total nuber of coins ust be even since there are an even nuber of white cells and an even nuber of black cells. Now we ay assue there are n black cells and n white cells. Then we ust have

5 n n 009 and 009. Hence n 3. It is possible for n 3, by putting one coin 3 in each of the 98 black cells, we then put two coins on each of the first white cells of the first row. Together we have coins on the chessboard. It follows that the answer is We denote by [XYZ] the area of XYZ. y scaling we ay assue E D and E x. Since DG G ~ G, we have G x. Note that ED is a parallelogra, so F Fro this we get F and hence FG [ ( x)] x. ( x) x FG [ D ] [ DFG] x [ D] [ E] x[ E] ( x)( x) x x 8 x x This is axiu when x x, or x, in which case x 3 D. D E x F G. Let be the nuber of people taking the quiz. Then we have 009 n ( n ) ( n ) ( n ) n ( n ) ( n ) 009 Hence n, where is a factor of 009. We check that gives the sallest value of n, which is 89.. Note that For 08, we have Hence each suand is equal to 009 plus a decial part. Suing up all the decial parts, we have

6 and so the answer is Note that if (a, b, c) is a positive integer solution to the equation x 3y z 009, then ( a, b, c ) is a non-negative integer solution to the equation x 3y z 000. onversely, every positive integer solution to x 3y z 000 corresponds to a positive integer solution to x 3y z 009. Hence positive integer solutions to the two equations are in one-to-one correspondence except for solutions to x 3y z 009 with at least one variable equal to, in which case the corresponding solutions to x 3y z 000 have at least one variable being 0. The nuber of such solutions is exactly equal to the value of f(009) f(000). To count this nuber, we note that if x, the equation becoes 3yz 00 and so y ay be equal to, 3,,, 7 so that z is a positive integer, i.e. there are 7 33 solutions in this case; if y, the equation becoes xz 00, or xz 003, which has 0 positive integer solutions (corresponding to x =,,, 0); and if z, the equation becoes x3y 007 and so y ay be equal to,, 9,, so that z is a positive integer, i.e. there are 7 solutions in this case. Two of these solutions have been double-counted (corresponding to x y and yz ). It follows that the answer is Note that at least three colours have to be used. So we have four possibilities. If exactly three colours are used, there are 3 0 pair of opposite faces ust be painted in the one colour, and there is only one way of ways to choose colours. Then each colouring (up to rotation). Hence there are 0 colourings in this case. If exactly four colours are used, there are exactly one pair of opposite faces with different colours, and this pair can be coloured in ways to choose colours. There ust be ways. fter that the colours of the other four faces are fixed (up to rotation) as each pair of opposite faces ust receive one colour. Hence there are 90 colourings in this case. If exactly five colours are used, there are ways to choose colours. There ust be exactly one pair of opposite faces with the sae colour, and there are choices of colours for this pair (say, top and botto faces). fter that, there are four faces in a ring shape to

7 be assigned the reaining four colours. We only have to split the four colours into two pairs so that each pair of colours belongs to a pair of adjacent faces. There are 3 ways to do so (just fix one colour and consider which colour to pair up). Hence there are 3 90 colourings in this case. If all six colours are used, fix any colour and assue (rotate suitably if necessary) that it is for the botto face. Then there are choices for the colour of the top face. There are now ways to assign the reaining four colours to the four lateral faces. (Not 3 ways as in the previous case, because the top and botto faces now have different colours, so switching the colours of a pair of opposite lateral faces results in a different colouring.) Hence there are 30 colourings in this case. obining the four cases, the answer is Let P be the iage of P by an anticlockwise rotation about through 90 so that goes to. s P + P = P + P = 80, PP is a cyclic quadrilateral. So the circucentre R is the iage of the circucentre Q under the rotation. Thus QR is also a rightangled isosceles triangle, and so QPR is a square and P = QR =. Let = x. pplying the cosine law in P, we have P R P x ( ) x( )cos. This gives x x 0 and hence x 3. It follows that P and P Q 9. Expressing w, z and y in ters of t and x, we have xt w, x z xt t x xt and y 3 xt t xt xt t x. Substituting the last expression into x t, we have y xt t x x 3 xt t xt 3 3 xt x t t xt x t t 0 t t xt x ( )( ) 0 Note that the last factor on the left hand side of the last row is non-zero, for otherwise we get 7 t

8 x t x x y which contradicts the fact that x y. Since t 0, we ust have t. Reark. Owing to soe typo,, and were all accepted as correct answers in the live paper. 0. Note that each of a, b, c, d is of the for 37 n where is one of 0,,, 3 and n is one of 0,,, 3,,. Furtherore, at least two of the four s are equal to 3 and at least two of the four n s are equal to. We first count the nuber of ways of choosing the four s. There are three possibilities: If all four of the are 3, there is choice. If exactly three of the are 3, there are ways to choose the outlier and 3 ways to choose the value of the outlier (naely, 0, or ), giving 3 choices. If exactly two of the are 3, there are ways to locate the 3 s, then choose the value of the other two s, giving 9 choices. 3 9 ways to Hence there are 7 ways to choose the values of the s. Likewise, there are 7 ways to choose the values of the n s. It follows that the answer is

Taiwan International Mathematics Competition 2012 (TAIMC 2012)

Section. (5 points each) orrect nswers: Taiwan International Mathematics ompetition 01 (TIM 01) World onference on the Mathematically Gifted Students ---- the Role of Educators and Parents Taipei, Taiwan,