International Mathematical Olympiad. Preliminary Selection Contest 2012 Hong Kong. Outline of Solutions

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1 Internatonal Mathematcal Olympad Prelmnary Selecton ontest Hong Kong Outlne of Solutons nswers: Solutons: Snce n s a two-dgt number, we have s less than 94 6 Snce the sum of dgts of n and so the sum of dgts of n n s equal to the square of the sum of dgts of n, the sum of dgts of n s less than 6 6 It remans to search through all the two-dgt numbers whose sum of dgts not greater than 5 There are 5 such numbers, namely,,,,, 4,,,,,,,, 4, 4, 5 y checkng these numbers one by one, we know that the answer s 7 Note that n n n ( )( ) We thus want one of n and n to be a prme and the other to be a product of two prmes ll of these prmes have to be odd snce n and n have the same party and there ests only one even prme all a postve nteger a semprme f t s the product of two dstnct odd prmes The frst fve semprmes are thus 5,,,, 9 We need to look for small prmes whch dffer from a semprme by We thus, get (, 5), (5, 7), (9, ), (, ) and (, ) as the fve smallest sets of ( n, n ) Therefore, the fve smallest good ntegers are 4, 6,, and Summng them up, we get the answer 4

2 The probablty that the montor can choose hs favourte seat depends on hs poston n the queue If he s frst n the queue, he must be able to get hs favourte seat; f he s the second n the queue, there s a probablty of that hs favourte seat has not been chosen by the frst person n the queue Smlarly, f he s k-th n the queue, the probablty that he gets hs favourte seat s 4 k Snce the montor s equally lkely to occupy each poston n the queue, the answer s 7 4 Snce a postve nteger leaves the same remander as ts sum of dgts when dvded by 9, ( ) s smply the remander when s dvded by 9 In other words, the answer s smply the remander when F (where F n s the n-th Fbonacc number) s dvded by 9 We compute the modulo 9 Fbonacc sequence (e the remanders when terms of the Fbonacc sequence are dvded by 9): we get,,,, 5,, 4,, 7,,, 9,,, 7, 6, 4,, 5, 6,,,, 9, after whch the net two terms are, and hence the sequence wll repeat every 4 terms Snce (mod 4), we have F F 6 (mod 9) 5 Let b and g (where 5) be the number of boys and grls n the -th class respectvely Snce there are 6 5 students n each class and at least boys and grls, each b and g s between and 7 nclusve onsder the matr: b b b b4 b5 g g g g4 g5 The sum of each row s, whle the sum of each column s If we crcle the smaller number n each column, the sum of the crcled numbers s the number of teams whch can be formed y the pgeonhole prncple, at least three numbers are crcled n the same row Wthout loss of generalty assume b, b and b are crcled Note that b b b 7 6 On the other hand, each crcled number n the fourth and ffth column s at least The sum of the crcled numbers s thus at least 6 9 Equalty s possble as the matr provdes one such eample The answer s thus 9

3 6 Note that 6 It follows that, or 4 Thus 5 ( ), or 4 ( 4 )( 4 4 7) all a postve nteger good f t can be epressed n the form a c b d where a, b, c, d are non-negatve ntegers learly, f n s good, then so s n because we can smply ncrease a and b by to double the value of a b c d 4 Note that,, 5, 7, 9 are good snce,, 5, Hence, 4, 6,, are also good by the remark n the prevous paragraph 4 a b k m ( ) Fnally, assume where m a b, n c d and k b d, wth m, n c d n n k m postve It follows that ( ) ( ) Snce the left hand sde s odd, we have k m n learly, nether m nor n can be equal to Thus (mod 4) s (mod 4), we get a contradcton as the left hand sde s congruent to but the rght hand sde s congruent to modulo 4 Thus s not good and so the answer s and y completng square we have f ( ) ( 6) 6 and so Smlarly, f f f ( ( ( ))) ( 6) Hence the answer s 4 f ( f ( )) ((( 6) 6) 6) 6 ( 6) 6 The equaton thus becomes 6 6 ( 6) 6, wth solutons 9 Ignorng the rule that no two adjacent letters be the same, the answer would be 9! as there!! are 9 letters ncludng two I s and two L s From ths we must count the number of permutatons wth two adjacent I s or two adjacent L s If the two I s are adjacent, we can treat them as one sngle letter and hence the number of permutatons would be! The same s true for permutatons wth two adjacent L s There are! overlappngs between these two types of permutatons though, as there are 7! permutatons n whch the two I s and the two L s are both adjacent Hence the answer s 9! 7! 5544!!!!

4 From the gven equatons we have 4sn 6sn y and 4cos cos y It follows that and so sn 4 4sn 4cos 6sn y cos y 6sn y ( sn y) y In the same way we get 7 sn, cos and 7 sn cos sn cos cos sn 49 () sn y cos y sn y cos y cos y sn y 5 cos y Hence Let d be the last dgt of, and wrte c d fter movng the last dgt to the front, the number becomes n n d c ccordng to the queston, we have d c ( c d), or n ( ) d 9 c omputng k modulo 9 for k =,,,, we get, 5,, 6,,, 5, 7,, 9, 4, 7,, 6,, 4, Hence the smallest possble value of n s 7 It remans to show that there s a -dgt number wth the gven property Snce c s to be a 7-dgt number, from c, we see that we should set d when n Ths corresponds to n ( ) d 9 c (hence ) It follows that the answer s Snce, the crcle has the same centre as the nscrbed crcle of If we let be the mdpont of, then s the pont where the nscrbed crcle of touches, and the same as true for and Snce, and, t s easy to fnd that, and Thus we have, and The area of s thus Smlarly, has area s 4 4 sn, the area of s 5 5 Fnally, snce the area of s 6, we have and hence 7 Obvously the negatve square root should be taken as 4

5 Let OP meet at Q Note that we have O O OP and hence OP P 9 Snce OP QP and P PQ (as P P P P 6, whch mples s concyclc), we have QP PQ 9 P Q Ths mples OQ s perpendcular to Yet t s also gven that O s perpendcular to Hence O must le on the straght lne P s O s the crcumcentre of P, t follows that P s a dameter of ts crcumcrcle Snce P 6, the crcumradus s O 4 Let be the common dfference of the arthmetc sequence a, b, c, d We have d a The queston s thus equvalent to countng the number of postve nteger solutons to the equaton ( a ) t When, the equaton becomes at and there are 9 solutons (correspondng to a =,,, 9) When, the equaton becomes at 7 and there are 6 solutons Lkewse, when =, 4, 5,, 669, 67, there are,, 997,, 5, solutons respectvely It follows that the answer s (9 )(67) Let the areas of the four trangles be n, n, n and n, where n s a postve nteger The area of the quadrlateral s thus 4n 6 Note that the area of s four tmes that of EF the area of s at most 6, whch s at least 4n Hence E F Equalty can be attaned when s an sosceles trapezum wth parallel sdes 6 and 4, and heght (We can check n ths case that the areas of EF, E, F and EF are,, and 4 respectvely, and has area 6) The answer s thus 6 5

6 6 Etend to E so that E Then E and are congruent so that we have E 6 and E It follows that,,, E are concyclc and so E E 9 Thus E 6 5 E M and hence the md-pont theorem asserts that M E 7 Let O be the md-pont of, whch s also the crcumcentre of Etend P to meet the crcumcrcle at Note that P s between O and (f P s between O and then P s less than 45 whle P s obtuse, contradctng P = P) Let P Then P P P and O It follows that O P, both beng (radus of the crcle) Usng the power chord theorem P P P P, we have P(7 P) 7 Solvng, we get P The postve square root s taken as P s between O and It follows that 7 P and so the answer s O P Note that, n order for the sum of any three adjacent ntegers after the rearrangement to be dvsble by, any three adjacent ntegers must be parwse dfferent modulo For the orgnal postons of,,, there are! 6 possbltes to arrange three numbers taken modulo (e eactly one of these postons s to be occuped by a number dvsble by, one by a number congruent to modulo, etc) The remanng numbers, taken modulo, are then fed (For eample, f the three numbers occupyng the orgnal postons of,, are, 9,, then the twelve numbers modulo n clockwse order startng from the orgnal poston of must be,,,,,,,,,,, ) For each such modulo arrangement, there are 4! 4 ways to arrange each of the 4 numbers congruent to, and modulo Hence the answer s

7 9 s 4 s of the form Indeed, we have , we naturally try to factorse ( ) [( ) 4 ( ) 4 ] 4 ( ) 4 4 ( ) 4 ( ) ( ) ( ) ( )( ) 4 4 y puttng, we get 4 Hence the answer s Let a, b, c (where ab c) be the lengths of the sdes of such a trangle y Heron s formula, we have whch smplfes to a b c a b c c a b b c a ( a b c), ( a b c)( c a b)( b c a) 64( a b c) Observe that a b c, c a b, b c a, ab c have the same party and hence must be even Set a b c r, c a b s, b c a t, where r, s, t are postve ntegers wth r s t The above equaton s then reduced to rst 6( r s t) 6( r s) t rs 6 rs dvdes 6( r s) We can then s t r s t t, we have 6 rs 4 lso, the above equaton mples Hence we need to fnd r and s so that 6 rs 4 and 6 lst out all such pars of (r, s) and compute the correspondng t, gettng dfferent solutons, namely, (r, s, t) = (, 6, 7), (, 7, ), (,, ), (,, ), (4, 5, 6), (4, 6, ), (4,, ) and (6, 7, ) Therefore there are such trangles 7

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