International Mathematical Olympiad. Preliminary Selection Contest 2017 Hong Kong. Outline of Solutions 5. 3*

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1 International Mathematical Olympiad Preliminary Selection Contest Hong Kong Outline of Solutions Answers: * *See the remar after the solution Solutions: We consider the remainders when a 0, a, a, are divided by 7 Note that when we compute the remainder when a n is divided by 7, it suffices to replace an and an by the respective remainders in the equation a a ( a ) (eg once we now a and a 7 6 n n n (mod 7), then we have a a a 6 6 (mod 7)) Thus it is easy to find that the remainders are respectively,,, 6, 6, 0, 6,, 0,,,,, 6,, which repeat every 0 terms The remainder when a is divided by 7 is therefore the same as that when a 7 is divided by 7, which is from the above list From x ( ) ( ), we have x x x y y x y y or x y 0 The former is the same as y or x y Each of these equations represents a straight line Therefore, we can draw the figure below In particular, the lines x y 0 and x y are parallel and hence they have no intersection One easily counts that there are regions in total

2 Note that for each integer n greater than, the expression for f( n ) consists of 06 terms We consider the contribution from each of these terms Let S be the contribution from the term n where Then It follows that S f () f () f () S S S We need to count the number of positive integer solutions to the equation ab c 600 such that ab c Then a 00, and we note that each even a (say, a ) leads to 0 solutions (eg when a 00, there are solutions with ( bc, ) (00, 00), (0, 99),, (0,0); each odd a (say, a ) leads to 0 solutions (eg when a 99, there are solutions with ( bc, ) (99, 0), (00, 0),, (0, ) Thus the answer is thus (0 ) (0 ) a c Clearly n If n, we may let x and x where a, b, c, d are positive integers b d Then x x implies ( ad) ( bc) ( bd), contradicting Fermat s Last Theorem (which says that when n is an integer greater than the equation n n n x y z has no positive integer

3 solution) Finally, as 6, we have possible It follows that the answer is and so n is Remar In the live paper, the condition less than was accidentally missing That would mae the problem trivial with answer Both and were accepted as correct during the contest 6 First note that b cannot be, so there are 99 possible values for b Now the equation can be log a log a 06 rewritten as, ie (log a) (log a)(log b) If log a 0, log b log b which means a, then any of the 99 values of b would wor If log a 0, we can simplify the equation as log a (log b) This equation has two solutions in a for each of the 99 possibilities for b Hence the total number of solutions is As 0 and a b, each of x, y, z is of the form, where each of 000 a, b is, or Furthermore, among the three a s chosen, one of them must be and one of them must be, leading to choices for the three a s (including 6 permutations of (,,), permutations of (,,) and permutations of (,,)) By the same argument there are choices for the three b s, leading to a total of choices However, because of the requirement x y z, many of these have to be discarded In most cases, out of 6 will wor because of the permutations of the values of x, y and z In some cases two of x, y, z are equal (note that x, y, z cannot be all equal), leading to only permutations There are sets of ( x, y, z ) for which two of x, y, z are equal namely, ( xz, ) (, ) and ( xz, ) (, ), with y being equal to either x or z Hence, among the choices mentioned in the previous paragraph, the number of choices satisfying x y z is The condition implies f ( x) ( x ) g( x) for some constant This gives x x bx 00 x c x ( a ) x ( a) x (0 ) x 0 By comparing the coefficient of x, we get 90 This implies a 89 Hence f () ( ) g() ( )( a ) 7007

4 9 Suppose the extension of AB and CD meet at P From ABD BCD, we find that PBC ~ PDB Then PB BC As PB PA AB PD 8, we obtain PD DB PC PD 0 Using the similar triangles again, we have PB This implies PC Hence CD 0 0 Rotate P about B by 90 clocwise to obtain point Q Then PBQ is right-angled and isosceles From BA BC, BP BQ and ABP 90PBC CBQ, we have ABP CBQ This implies CQ AP Also, we have PQ BP BQ 8 As CQ PQ 9 PC, PQC is right-angled at Q Hence we have APB CQB 90, and so it follows that AB ()() cos The area of ABCD is thus ( AD BC) AB 6 AB If Ann is to win, then one the following cases must happen If the first votes all go to Ann, she wins The probability for this to happen is 6 Suppose exactly out of the first votes are given to Ann (with probability C ) Ann wins if the next votes are both go to her, with probability 6 If she gets exactly vote among the next, she has to get both of the remaining votes to win, with probability C Suppose exactly out of the first votes are given to Ann (with probability C ) 8 Then the remaining votes should all go to Ann in order that she can win The probability for this to happen is 8 8 The answer is thus

5 Without loss of generality, assume d d dn Note that n is even and dd j n j! for n j For convenience, we write m! We find that d j dn j m d m d m ( d m)( d m) j n j j n j d d m m d d m m m j n j ( j n j ) n Hence the summands in the expression can form pairs so that the sum of each pair is m As 8! 7, we have n (8 )( )( )( )( ) 0 and so the answer is 0 77! 66 If, we have x 6 ; if, we have x, which should be rejected When, the equation is quadratic and can be factorised as [( ) x ][( ) x 6] 0 6 The solutions are and Let m where m is a positive integer Then 6 m For to be a positive integer, the denominator must be positive and so m 6m m is at most Furthermore, we need 6 m to divide m, and we chec that only m =,, wor, and these corresponds to, and 7 respectively The answer is thus 7 7 Note that two of the solutions come from Qx ( ) while the other two come from Qx ( ) Let c be a positive integer solution to Qx ( ) Then the other root is a c from the sum of roots Also, we have c( a c) b from the product of roots Similarly, let d be a root to Qx ( ) Then the other root is a d and we have d( a d) b Using these to eliminate b, we obtain d( a d) c( a c), which implies ( c d)( a c d) Note that c, d and a c are positive integers Hence, we have ( c d, a c d) (,), (,), (, ) or (, ) This shows that a is odd Also, from ac 0, we now that a is positive ( c d) ( a c d) a Let a (m ) where m Since c, we chec from the above four cases that c is either m (in the first two cases) or m (in the last two cases) In either case we have b c a c m m ( ), and it can be checed that the solutions to

6 Qx ( ) are always m, m, m and m Since these are positive integers, we must have m Also, from 6, we get m 9 It is easy to see that each of, b m m,, 9 is a possible value of m, and each corresponds to one set of ( ab, ) The answer is thus 7 Applying cosine formula on ADE and ABC, we get n ( n ) n cos A m n( n) ()() This simplifies to give n ( m ) 7 From n AC and n AD DE AE n, we get n 9 or When n 9, m 7 has no integer solution Thus we try n, which gives m 0 as the only possible solution 6 Suppose there are n families in total and suppose there are m children There are m choices for the best child, then n choices for the best mother (since the best mother cannot be from the same family as the best child) and similarly n choices for the best father Hence we have m( n )( n ) 7770 As m n, we have 7770 ( )( ) n n n n This implies n Since ( n)( n ) is a factor of , it is easy to chec that the only possibility is n 6, which corresponds to m 7 7 Consider the homothety with centre A and ratio Then points B and C are mapped to the midpoints M and N of AB and AC respectively Let the tangents at M and N to the circumcircle of AMN intersect at a point F Then F is the image of D under the homothety, and so F lies on BC by the given condition that A and D are equidistant from BC Note that BMF ANM ACB from tangent properties and the fact that MN and BC are parallel This shows that A, M, F, C are concyclic Hence we have BF BC BM BA 8 Similarly, we have CF CB CN CA 8 Adding these, we obtain BC BF BC CF CB 6 Thus BC 6 6

7 8 For each X i, there are C 6 blue lines not passing through it Hence, there are 6 0 red 0 lines in total They form C intersections (including intersections of parallel lines and counting multiplicities when three or more lines meet at a point) However, the following should be discounted 6 For each X i, there are 6 red lines passing through it The C intersections formed from these lines coincide Hence, ( ) 70 intersections need to be discounted For each of the C 0 blue lines, there are red lines which are perpendicular to it These red lines are parallel and hence do not intersect Thus 0C 0 intersections need to be discounted C triangles formed from The three altitudes of a triangle are concurrent There are 0 the points Hence another 0 0 intersections should be discounted The other intersections do not coincide in general Hence, the maximum number of points of intersection is Let a, b, c be the lengths of BC, CA, AB respectively Let r and s be the inradius and semi-perimeter of ABC By considering the area of ABC, we have rs s( s a)( s b)( s c) Hence we have rs BD DC ( s b)( s c) s a UV Applying the extended sine law, we have AX sin A This gives sin A As the triangle is acute, this A sin A implies r AE tan AE cos A Let H be the foot of altitude from A to BC and let Y be point of tangency of the two circles Consider the homothety about Y that sends ' to As the tangent at A to ' is parallel to the tangent at D to, the points A, Y, D are collinear From XYD XHD 90, points X, H, 9 D, Y are concyclic It follows that AX AH AY AD AE This gives AH, a s 8 and from rs AH we get a It follows that 8 8 BD rs DC s a 0 Suppose n is a good number with respect to a set X with elements Then both n and n belong to X, so we must have and n The remaining elements can be 0 chosen from the remaining 0 elements Thus there are C such sets X, and this gives 7

8 C C C C n sets for which n is good Note that n is at most 0 Summing over all n, the total number of good positive integers is Using the relation C m r S 0C 0C 0C C Adding these, we obtain m Cm r , we get S 0C 0C 0C C S 0( C C C C ) 0 Hence, the required expected value is S

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