International Mathematical Olympiad 2003 HK Preliminary Selection Contest (May 26, 2002) Solutions

  HK Preliminary Selection Contest (May 26, 2002) _{n n n} Solutions

  1. We only need to consider the case 10 = 2 × _{n n} 5 . (In other cases, there will be a zero as unit digit for one of the two factors.) For n ≤ 7 , 2 and 5 do not contain the digit 0 . For n = 8 ,

  8 5 = 390625 . So answer is 8.

  2. Between 1:00 am and 1:00 pm, the hands overlap 11 times. Between two overlappings there are 11 times when the distance is an integer (2 times for distances equal 2, 3, 4, 5, 6, and 1 time for distance equals 7). After the last overlapping (12:00 noon), and before 1:00 pm, again 11 times when distances are integers. Thus altogether

  11 + 11 × +

  10 11 = 132 times that distances are integers.

  1

  1

  3. The sum of the numbers from 700 to 799 is × 799 × 800 − × 699 × 700 = 74950 . The

  2

  2

  1

  1 × 79 × 80 − × 69 × 70 = 745 . sum of the numbers from 70 to 79 is Hence all numbers

  2

  2 that end from 70 to 79 (excluding those starting from 7, as we have already counted those from 700 L to 799) is 7 4 5 × +

  

( )

  9 1 0 1 0 0 2 0 0 6 0 0 8 0 0 9 0 0 = 4 4 7 0 5 . The sum of all + + + + +

  numbers ending in 7, (excluding the previous two cases), is L L 9 7 1 7 6 7 8 7 9 7 9 1 0 0 2 0 0 6 0 0 8 0 0 9 0 0 = 3 8 1 8 7 . So

  ( ) ( ) + + the total sum of numbers containing a “ 7 ” is 7 4 9 5 0 4 7 7 0 5 3 8 1 8 7 = 1 5 7 8 4 2.

  Alternate Solution

  Let P and Q be respectively the sum of all positive integers less than 1000 and the sum of all correctly, depending if he answers or does not answer the remaining question, he gets 396 or 395 marks. If a student answers 98 questions correctly, depending if he answers or does not answer the remaining 2 questions, he gets 390, 391, or 392 marks. Hence the total marks cannot be 389, 393, 394, 397, 398, 399. If a student answers 97 (or fewer) questions correctly, he get at most 388 marks, by adjusting the questions he answers wrongly, he gets everything possible “ in between ”. Thus

  = + the number of different “ total marks ” 100 401 − 6 = 495 .

  5. Each 4-digit palindrome is of the form abba , where a ranges from 1 to 9 and b from 0 to 9. Hence _{9 9} the sum of all 4-digit palindromes is equal to 1001 m 110 n _{n m 1}

• ₉

  ( ) ∑ ∑

  = =

  45 ×

  10 45 × + 9 × 110 = 11000 45 × = 495000.

  ∑ _{n =}

  45 110 n × + = ( 1001 × 9 ) = 1001 ×

  6. The 1000 lines through B divide the plane into 2000 regions. For the first line drawn through A, the number of new regions formed is 1001. For each new line drawn through A, the number of new regions formed is 1002. Therefore the total number of regions formed is 1002 1002 1 2000 1006003. × − + =

  7. Denote by [ BMN ] the area of ∆ BMN , then

  A E D

  1 BMN = . Join YN , then YN // BA and

  [ ] M F

  3 S

  N RN YN CY

  1 BR

  3 = = = , hence = . Draw

  R BR AB CB

  3 BN

  4 B C

  X Y AD and CD such that AD // BC , CD // AB ;

  extend BM and BN to meet AD and CD

  AE AM

  1 at E and F respectively. We have = = , so E is the midpoint of AD and

  BC MC

  2

  2

  4

  2 BS BY

  4

  4 BS

  6

  3

  7 BM = BE . Also = = = and so . Finally we have . Now BS = BE = = The problem is a matter of ratios of areas, and we may take any triangle satisfying the prescribed

  1

  2   conditions. Suppose we take A(0, 1), B(-2,0), C(1,0), then we have Y(0,0), M , and  

  3

  3  

  2

  1

  4

  1      

  

N , . AY and BM meet at S , ; also AY and BN meet at R , . The area of the

       

  3

  3

  7

  4      

  A(0,1) M(1/3,2/3) S N(2/3,1/3)

  R C(1,0)

B(-2,0) Y(0,0)

  4

  7

  1

  4

  1

  1

  4

  4

  1

  1

  1

  1

  15

  2 1      

• quadrilateral SRNM is = − − = ( )

  45 = . Hence, the ratio

       

  2

  2

  9

  21

  6

  9 2 126

  84

  3

  3

       

  1

  2

  3

  3

  4

  7

  15

  3

  5

  of the areas of the quadrilateral SRNM over the triangle is /

  =

  84

  2

  42 Remark : The problem may also be solved using the Menelaus’ theorem.

  8. Check that log 12 = 2 log 2 log 3 , so we have 1 . 0791 = 2 × . 3010 . 4771 < log 12 <

  1 . 0794 , which implies

  39 ₃₇ 3 log 2 < 39 . 9267 < 37 log 12 < 39 . 9378 < 39 + + 2 log 3 = 39 log 9 . Hence the leftmost digit of 12 is 8.

• 2 . 3011 . 4772 =

  39 log + × + 8 =

  Alternate Solution _{2 2} From x + x

  5 = 10 , we may have let x = 10 cos , x = 2 sin . Similarly if we let _{1 2 1} α α ₂

  r ^{2 2}

  5 . From and ₁ β β y y = r x y − y x = _{2 1}

₂

² _{2 1 1 2}

• y = r cos and y = sin , then

  5

  5 _{1 1} + x y 5 x y = 105 , we have _{2 2}

  r

   2 sin r cos 10 cos sin 2 sin( r )

  5 α ⋅ β − α ⋅ β = α β + =

  

  5 . 

  r

   10 cos α ⋅ r cos β 5 2 sin α + ⋅ sin β = 10 cos( r α β − ) = 105

  5  _{2 2}

     5  105 ² +

  This implies r =   =

  23  

  2

  10  

    Alternate Solution 2 Apparently any special case should work. Try x = 10, x =

  0. From x y x y 5, get _{1 2 2 1} − = _{1 2} 5 105 105 5 25 ⋅

• y = − . From x y
_{2 1 1} 5 x y = 105, get y = . Hence y _{2 2 1 1} ² 5 y = = ₂ ² + + 23.

  10

  10

  10

  10

  10. Since

  15 = 1 × 15 = 3 × 15 are the only factorizations of 15, so a positive integer has exactly 15 _{14 2 4}

  positive integer factors if and only if it is of the form p for some prime p , or p q for two ₁₄ distinct primes p and q . Now 2 > 500 , so there are no numbers of the first form between 1 and _{2 4 2 4 2 4}

  500. The only one of the second form are 3 × 2 = 144 , 400 5 × 2 = , and 2 × 3 = 324 , hence 3 is the answer. _{2 2}

  11. Note that n

  1 − n = 2 n + + 1 , hence every odd positive integer is the difference of squares of ( ) _{2 2}

  integers. Also ( n

• 2 ) + + − = n 4 n

  4 = 4 ( n 1 ) , hence any integer divisible by 4 is the difference of _{2 2}

  squares of integers. Now for a number of the form

  4 n 2 = x − y , then x and

  4 n + + 2 , suppose

  OB OC OA = , from ∆ BCD , get OB = . If OA = a , D OB OC OC OD _{2 3} OB = ar , then OC = ar , OD = ar . By Pythagorean _{2 6 3} theorem , AB = a 1 r , AD = a 1 r .

  A ar ^{2 6}

  1001 AD 1 r ^{4 2} + Then 91 = = = = r − r 1 .

• a

  O ^{2 2}

  1 r

• 11 AB

  Hence r − r − ^{4 2} 90 = r − ² 10 r ² 9 = , or r = ² 10 , r = 10 . ²

• ar

  ( )( ) ar ²

  11 = AB = a 1 r , get a =

• Since

  1 , and C B ^{2 2}

• BC = OB OC = 110

  Alternate Solution Assign the coordinates (0, 11), (0, 0), (1, 0) and A B C D c d ( , ).

   

  d

  11   ² As AC ⊥ BD , we have − = − 1 , or c = 11 d .

     

  c c

     

  c , d

  Hence, AD = c ( d − _{2 c 2} ^{2 2} ₂ 11 ) ²

• D ( )

  = + c ( − 11) = 1001.

  1001

  11 _{4 2} A (0, 11 )

  Simplify to get c − c ₂ 11 − 10890 = .

  Let , and observe that 10890 =

  99 110 , E c = s ×

  11

  99 ) , or 110 ,

  s − s = s = implying c 110. .

• have ( 110 )(

  =

  C ( ) A B

  BE PE − PB ^{2 2 2}

• cos ∠ PEB = ,

  2 BE ⋅ PE ^{2 2 2}

• PE EC − PC

  3 cos ∠ PEC = . Using the fact P

  2 PE ⋅ EC

  cos ∠ PEB = − cos ∠ PEC , and after _{2 2 EC}

  E simplifications, get PE = PB ⋅

  BC

  2

  2

• PC ⋅ − EC ⋅ BE , or PE = PB
^{2 BE 2 2} BC

  5 D C

  3 ²

• PC − 6 …(1) Apply cosine rule to

  5 _{2 2 2 2 2 2} BP PC − BC DP PC − CD + +

  

∆ BCD , we have cos ∠ CPB = , cos ∠ CPD = . Again by

  2 BP ⋅ PC

  2 DP ⋅ PC

  5 2 , after simplifications, get

  cos ∠ CPB = − cos ∠ CPD , BC = CD =

5 , BP =

  2

  2

2 PC = BC − BP ⋅ DP = BC − BP BD − BP =

  25 − BP

  5 2 − BP …(2). Put (2) in (1), get

  ( ) _{2 2} ( ) _{2 2} PE PB = − + PC = +

  3 ₂

  2 BP 9 … (3). Hence PE PB − ₂

  3

  2 BP

  5

  2 BP

  25  

  9 PB − + + +

  3

  9

  5

  2

  25  

  2 PB − + + = + . Take PB x , then PE PC corresponds to the + = PB −  

   

  2

  2

  2

  2  

     

  3

  3

  5 2 −

  5

  2   2 ,

  2 , and , . For PE PC to be smallest

• sum of distances from (x, 0) to

     

  2

  2

  2

  2  

     

  3

  5

  2

  5

  2 2 − −  

  2

  2  

  2 (x, 0) lies on the line formed by these two points. Hence = , solve to get

  3

  5

  2

  5

  2 2 x − −

  2

  2

  2

  15

  2

  x = . (Corresponds to PE + PC = 34 ).

  8

  Alternate Solution

  Clearly PC = PA, hence PE + PC is smallest when PE + PA is smallest, i.e. when A, P, E are collinear.

  A B

  Assign the coordinates A(0,5), B(5,5), C(5,0), D(0,0),

  3

• then E = (5, 2), and the equation of the line AE is y = − x 5.

  5

₂

₂ P E

  25

  25 ²

  25

  25     5  −

• This means P = , and PB =  −

  5      

  8

  8

  8

  8      

  D C

  450

  15 = or PB =

  2

  64

  8

  14. From C construct ∠ ACD = ∠ BAC , with D on AB . Then ∠BDC =

  ∠ BCD = 2 ∠ DAC . Hence BC = BD = 5 , AD = CD = 6 . Apply cosine rule to ² +

  5 5 − ²

  6 ²

  14

  ∆ BDC , get ∠DBC = = . Apply cosine rule to ∆ ABC , get _{2 2 2 154 576} 2 × 5 ×

  5

  50

  24

  5 AC = +

  11 5 −

  2

  5 11 cos ∠ B = 146 − = , Hence AC = 24 = .

  ( )( )

  5

  5

  5

  5 C

  A B D

  Alternate Solution

  11

  5 AC

  o

  Let ∠ CAB = x < 90 . By the sine rule, we have = = . sin 3 x sin x sin(180 − x ) 11 sin

  3 x ²

  1

  2 From the first equality, we have = = 3 − 4 sin x , hence sin x = , and cos x = . 5 sin x

  5

  5

  15. As shown in the figure, AB =

  10 , BC = 2 x and CA =

  2 y . D , E and F are midpoints of AB ,

  BC and CA respectively, G is the centroid, and further CD ⊥ AE , we have CD = 9 , from the

CG AG BG

  property of centroid = = = 2 , hence CG =

  6 , GD = 3 . Let GE = a , then AG = 2 a .

GD GE GF

  ₂

₂

_{2 2 2} By Pythagorean theorem, we have a

  ( ) ( ) ( )

  36 = x , 2 a + + 36 = 2 y , 9 + a 2 = 25 . If follows

  that a

  2 , = and = . Let = ∠ BCA . Apply cosine rule to BCA , get = x 2 10 y 13 θ ∆

  2 y 2 x

• 100

  14 . Apply cosine rule to ∆ BCF , get

  ( ) ( ) ( )( ) ^{2 2 2} + BF y = θ + 2 x − 2 y 2 x cos = 13 160 − 4 × 14 = 117 . Hence = = .

  2 2 x 2 y cos = − θ , so xy cos θ = ^{2 2}

  ( ) ( ) BF 117

  3

  13 C E F G B A D

  Alternate Solution Choose the coordinates of A , B and C as (0,0), (10,0) and ( b , ) respectively. Then the coordinates c

• 10 b c b c of D, E and F are (5, 0), ( , ) and ( , ) respectively.

  2 2 2 2 ^{2 2 2 2}

  5 ) c = 81 , or b + =10b + 56.……(1) c

• As CD = 9, hence ( b −

  c c _{2 2} Also CD ⊥ AE , get ⋅ = -1, or b + = -5b + 50.……(2) c

• 10 b b −

  5 C ( )

  b , c

  2 (1) and (2) imply b = − .

  5

  10 b c E ( ) _{2 2 , ,} F ( )

• b c

  2

  2 _{2 c b}

  2

  2

  1 ²

  1 ²

  16. Let x be the number of blue vertices and y be the number of blue faces in a coloring. If x = , we have 5 different ways to color the faces, namely y = 0, 1, 2, 3, 4. If x =

  1 , there are 2 ways to color the face opposite to the blue vertex and the remaining three faces may have 0, 1, 2 or 3 blue faces.

  Thus there are

  2 × 4 = 8 different ways in this case. If x = 2 , call the edge connecting the two blue

  vertices the “ main edge ” and the two faces common to this edge the “ main faces ”. There are 3 ways to color the main faces, namely, both red, both blue, and one red and one blue. In the former two cases there are 3 ways to color the remaining faces, while in the last case there are four because reversing the color of the two faces does matter. Hence there are 10 ways in this case. By symmetry, there are 8 ways corresponding to the case x =

  3 and 5 ways corresponding to x = 4 . Hence the

  5

  10

  8 5 = 36 .

  8 + + + + answer is

  1 _{n − 1} + a ₁ 3 −

  17. Rewrite the recurrence relation as a = . Define θ = tan a , then the above is _{n n n}

  1 1 − a _{n 1}

  −

  3 π tan θ + tan _{n 1}

  −

   π  π

  6 π if equivalent to tan θ = = tan θ . Thus θ + = θ + (“modulo” _{n  n 1  n n 1}

  − −

  π

  6

  6   1 − tan tan θ _{n 1}

  −

  6 necessary) and thus a = tan θ = tan ( θ π ) θ + = tan = a . If follows that _{n 6 n 6 n n n}

  2 π tan tan

  θ + 2 + − 3 2 π ( ) 5 3 8 −

   

  3 a = = + a tan = tan = = = . _{2002 4} θ θ ₄   2 π

  3  

  11   1 −

  2 −

  2

  ( ) ( )

  1 − tan tan θ

  ( )

   

  3  

  18. Since 2002 =

  2 × 7 × 11 × 13 , we can choose certain vertices among the 2002 vertices to form

  regular 7-, 11-, 13-, L polygons (where the number of sides runs through divisors of 2002 greater than 2 and less than 1001). Since 2002 =

  7 × 286 , so there are at least 286 different positive

  integers among the ' a s . Hence the answer is at least _i 7 286 287 × × 7 × + + + 1 2 L 286 = = 287 1001 × = 287287 . Indeed this is possible if we assign

  ( )

  2

  5 ⁴ C = ₂ 10 . From each of the five pints A, B, C, D and E, a total of C = ₂ 6 perpendicular lines can be ₃₀ drawn. It seems that the number of intersection points should be C = 435 . However for any ₂ straight line, say AB, the three perpendicular lines drawn from C, D and E are parallel, thus will ₃ not meet. Hence the number of intersection points should be cut down by ₅

  10 ×C = ₂ 30 . Next for

  10

  any of the C = triangle, formed, say ∆ ABC, there are three perpendicular lines which are ₃ concurrent. Hence the number of intersection points should be further cut down by ₃ 10 × C − ₂ 1 = 20 . Finally the points A, B, C, D and E, each represents the intersection of six lines

  ( )

  (lines drawn from each of them), hence the number of intersection points should be cut down by ₆ 5 × C − ₂ 1 = 70 . The required number of points of intersection is therefore

  ( )

  435 − 30 − 20 − 70 = 315 .

  20. Let ABCD be the piece of paper, with

  AB = a ≤ b = BC . Let O be the center of the A D

  rectangle, B be folded to coincide with D , and POQ be the crease, with P on BC and Q on DA . Then

  P PQ is the perpendicular bisector of BD , and ∆ BPO ~ ∆ DBC . Let x be the length of PQ , then b

  O

  1

• a b
^{2 2} b a

  2 ^{2 2}

  = , or x = a b . Now

• Q

  1

  a b x

  2 C B ₂ a

  a a

  2 x a b

  65 = = , thus b = . For _{2 2}

• 2

  b

  65 − a _{2 2 2}

  a

  65 b to be an integer, the denominator ₂ 65 − must be an integer. Hence we try as a sum of ₂

₂

_{2 2 2}

  two squares. Using _{2 2 2 2} 5 =

  3 ₂ + + 4 and ₂ 13 =

  5 12 we have _{2 2}

  65

  13

  3

  13

  4

  5

  5

  12

  5

  3

  12

  4

• 3 ×

  12 4 ×

  5 = ( × ) ( × ) ( = × ) ( × ) =

  5 + + 4 × 12 − 3 × 5 = ( × − × ) ₂ ( ) ( ) ( + × +

  4

  12 3 × 5 ) . Consider all these a which may be 16, 25, 33, 39, 52, 56, 60 or 63. Put these to

  find b , and bear in mind b is an integer with b ≥ a , we get a =

  60 , and b = 144 the only