International Mathematical Olympiad. Preliminary Selection Contest 2011 Hong Kong. Outline of Solutions

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1 International Mathematical Olympiad Preliminary Selection Contest Hong Kong Outline of Solutions Answers: Solutions:. Such fractions include. They can be grouped into pairs each with sum (i.e. on). Hence the answer is. and so. The sum of consecutive positive integers has unit digit equal to that of ( ) which is. Hence f ( n) f( n) for all n. y direct computation we have f() f() f() Similarly we have f() f() f() 7 f() f() f(6) 7 and so on until f() f() f() 7. Altogether we have groups of 7. Finally since

2 f() f() f() f() f() f() the answer is We must have k and so k 7. On the other hand the product of digits of k must not exceed k (to see this suppose k is an n-digit number with leftmost digit b; then k b n but the product of digits of k is at most k b ) so we have k k and thus k. Since k being the product of digits of k must be an integer k is divisible by. Hence the product of digits of k is also even i.e. is odd. Therefore k (mod ) so the possible k values of k include 76 and. It is easy to check that only works.. Note that 6 and. Hence 6 6 ( ) ( )( ) and so is a possible answer. Remark. It can be checked that the answer is unique.. Note that there are different letters with I and M occurring twice. If the three letters are distinct there are C combinations. Otherwise there are either two I s or two M s plus one of the remaining letters. There are 6 such combinations. Hence the answer is We have y ax. Note that a b c is equal to the value of y when x i.e. a abca. Hence we need to look for the smallest positive a for 6

3 which a is divisible by 6. This corresponds to a 6 or a. 7. The four points form a rectangle if and only they are two pairs of diametrically opposite points. Hence among the C possible outcomes there are C 6 favourable outcomes (by choosing out of the 6 pairs of diametrically opposite points). Therefore the answer is.. Let D be a point on C such that DA D. Then 7 cosdac cos( A ). Let D DA x. Then DC x and applying cosine law in DAC gives x A A ( x) x ( x)() 7 or x i.e. DA and DC cosine law in DAC again we have. Applying x D x C cosc. ()() 6. Let P = ( a) and Q = ( b) where each of a and b is randomly chosen from to. Since each circle has radius the two circles intersect if and only if ba. ( ) ( ) PQ b a or On the Cartesian plane (with axes labelled a and b) the set of possible outcomes is the square bounded by the lines a and b. The set of favourable outcomes is the region of the square between the lines ba and ba as shown. b a b Consider the non-shaded triangular part of the square in the lower right hand corner. It is easy to find that the coordinates of the vertices of the triangle are ( ) ( ) and ( ). Thus it is a right- a angled isosceles triangle with leg and whose area is 7. The same is true for the non-shaded triangular part of the square in the b a

4 upper left hand corner. Hence the required probability is 7. D A. y the angle bisector theorem DC AC 7 and so D. y the power chord 7 theorem D M A and so M. Finally MN is parallel to C since A DM MAD DAN DMN. Hence we have AM A so that MN 6. MN or C MN M D N C. Since the lengths of the altitudes are inversely proportional to the side lengths the sides of the triangle are in ratio : : ::. Let the side lengths be k k k. Then the triangle has 6 perimeter k and Heron s formula asserts that it has area k.(. )(. )(. ) k. Since the in-radius of a triangle is equal to twice the area divided by the perimeter we have k k 6 which gives k. Let be the angle opposite the shortest side and R be the radius of the circumcircle. Then the sine law and cosine law give and hence 6 R. k k k R sin cos

5 . Since PQC is a cyclic quadrilateral with C = we have PQ =. Also P PQ since PC = PCQ =. y Ptolemy s Theorem we have A PQC PQC PC Q. Since Q P PQ and C this P simplifies to QC PC. Let CQ y. Then y PC and considering the area of CPQ gives y 6 y sin or (y)(y) so that y. D Q C. It suffices to consider the last three digits. There are 76 6 possibilities and we need to count how many of these are divisible by. To be divisible by the unit digit must be 6 or. We consider these cases one by one. If the unit digit is the tens digit must be or 7 in order for the number to be divisible by. If the tens digit is or there are choices of hundreds digit (e.g. if the tens digit is then the hundreds digit must be or 7 as and 7 are divisible by ); if the tens digit is or 7 there are also choices (namely 6 ). Hence there are possibilities in this case. If the unit digit is the tens digit must be 6 or leading to possibilities namely The case where the unit digit is 6 is essentially the same as the first case; possibilities. The case where the unit digit is leads to possibilities (it is slightly different from the second case) namely and 6. It follows that the answer is. 6. Consider any consecutive cards. There can be at most one since there are at least three cards between any two s. On the other hand there is at least one for otherwise they have to be s and subject to the required conditions which can easily be seen to be impossible. In other words there is exactly one among any consecutive cards. Since.7 the number of s must either be or. oth bounds can be achieved (for the former case consider ; for the latter case consider ). Hence mm.

6 . Divide the students into groups according to their surname. If there are k students in a group then every student in the group will write k for the question on surnames. The same is true if we divide the students into groups according to birthdays. Since each of has appeared as an answer there is at least one group of each of the sizes. As each student belongs to two groups (one surname group and one birthday group ) the number of students must be at least ( ). Finally it is possible that there are students when for instance the answers to the surname question are all (i.e. there are four surname groups with students) and the answers to the birthday question are all 6 7 (i.e. there are seven birthday groups with 6 7 students). Hence the answer is. 6. Let m be the common root of x ax and x bxc. Clearly m. Since c c we must have a b as m amm bmc. This also gives m. Similarly if a b n is the common root of a b x xa and x cxb then n. It follows that c mn. As m is a root of the equation root of this equation. Now n is a common root of the equations x ax and x ax which has product of roots n is also a x x a so we have n ann na which simplifies to ( a)( n). Clearly a for otherwise the equation x ax will have no real root. Thus we must have n and n na gives a. Since mn we have m and m bmc gives bc. Thus abc. 7. y scaling we may assume AC and A. Then the cosine law asserts that C cos6. Since C A C AC CP and CQ it can be C C shown that P is between and Q. Since A P A C( C CP) C P we have. C A Together with the common we have AC ~ PA and so AP = CA. It follows that APQ AP AP AP CA 6. On the other hand we have AQP since P A Q C A Q ( AC) ( C CQ) AC C C CQ CQ AC CQ. Therefore PAQ 6. 6

7 . We consider two cases. Case : No green ball and blue ball are adjacent to each other In this case the red balls must either occupy the st rd th th positions or the nd th 6th th positions. In each case there are C ways to arrange the green and blue balls. Hence there are altogether ways in this case. Case : A green ball and a blue ball are adjacent to each other We place the green-blue pair first. Since half of the total number of balls are red and that no two red balls may be adjacent the number of empty spaces before and after this pair must both be odd. Hence there are choices of positions of this pair (nd and rd th and th. th and th). Of course there are ways to arrange the two balls among the pair. After this pair is placed the positions of the red balls are fixed (e.g. if the pair is placed in the 6th and 7th positions then the red balls must be placed at the st rd th th th th and th positions) and there are C 7 ways to arrange the remaining green balls and blue balls. Hence there are altogether 7 6 ways in this case. Combining the two cases the answer is There are possible 6-digit passwords and we shall count how many of these have at least three identical consecutive digits. Let A (resp. C D) denote the set of passwords in which digits to (resp. to to to 6) are identical. Then we have A C D A C CD AC D AD AC CD AD ACD ACD It follows that ACD ( ) ( ) 6 and so the answer is Dividing the first equation by the second we get and denominator on the left hand side by x y xy. Dividing both the numerator x y t or y and writing t for x y we have t t t t (*) t 7

8 x Clearly every solution ( x y ) to the original system gives rise to a solution t to (*). y Moreover two different solutions ( x y ) and ( x' y ') to the original system give rise to x x ' two different solutions t and t ' to (*). (This is because once t is fixed then so y y ' are x and y ; for instance the first equation together with x ty imply y Writing t x i i we need to find the value of yi t t t t t t t t t t t ( t )( t )( t ) where each t i is a different solution to (*) and hence ( )( )( ). Consequently we () () () have ( t)( t)( t) 7..)