International Mathematical Olympiad 2003 HK Preliminary Selection Contest (May 26, 2002)
Solutions
1. We only need to consider the case 10n = 2n ×5n . (In other cases, there will be a zero as unit digit for one of the two factors.) For n ≤ 7 , 2 and n 5 do not contain the digit 0 . For n n = 8,
390625
58 = . 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.
3. The sum of the numbers from 700 to 799 is 699 700 74950 . 2
1 800 799 2 1
= ×
× − ×
× The
sum of the numbers from 70 to 79 is 69 70 745 . 2
1 80 79 2
1 × × − × × =
Hence all numbers
that end from 70 to 79 (excluding those starting from 7, as we have already counted those from 700
to 799) is 7 4 5× +9 1 0 1 0 0
(
+ 2 0 0+L + 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(
)
(
)
9 7+1 7+L + 6 7+ 8 7 + 9 7 + 9 1 0 0 + 2 0 0+L + 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
positive integers containing no “7”s in their digits. Clearly 999 1000 499500. 2
P= × = On the
other hand, we can easily see Q is the sum of 93= 729 numbers. Considering the unit digits of
these numbers, there should be 81( 729) 9
4. If a student answers 100 questions correctly, he gets 400 marks. If he answers 99 questions
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.
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
Alternate Solution
The problem is a matter of ratios of areas, and we may take any triangle satisfying the prescribed
conditions. Suppose we take A(0, 1), B(-2,0), C(1,0), then we have Y(0,0), M
of the areas of the quadrilateralSRNM over the triangle is
42
Remark: The problem may also be solved using the Menelaus’ theorem.
Alternate Solution
From x12 +5x22 =10, we may have let x1 = 10cosα, x2 = 2sinα. Similarly if we let
β
cos
1 r
y = and sinβ
5
2
r
y = , then y12 +5y22 =r2. From x2y1−x1y2 =5 and
105 5 2 2
1
1y + x y =
x , we have
2 sin cos 10 cos sin 2 sin( ) 5
5
. 10 cos cos 5 2 sin sin 10 cos( ) 105
5 r
r r
r
r r
α β α β α β
α β α β α β
⋅ − ⋅ = + =
⋅ + ⋅ = − =
This implies 23
10 105
2
5 2 2
2 =
+ = r
Alternate Solution 2
Apparently any special case should work. Try x1= 10,x2 =0. From x y2 1−x y1 2 =5, get
2
5 . 10
y = − From x y1 1+5x y2 2 = 105, get 1
105 . 10
y = Hence 12 22
105 5 25
5 23.
10 10 y + y = + ⋅ =
10. Since 15=1×15=3×15 are the only factorizations of 15, so a positive integer has exactly 15
positive integer factors if and only if it is of the form p14 for some prime p, or p2q4 for two
distinct primes p and q. Now 214 >500, so there are no numbers of the first form between 1 and 500. The only one of the second form are 32×24 =144, 40052×24 = , and 22×34 =324, hence 3 is the answer.
11. Note that
(
n+1)
2 −n2 = 2n+1, hence every odd positive integer is the difference of squares ofintegers. Also
(
+2)
2 − 2 =4 +4=4(
+1)
n n
n
n , hence any integer divisible by 4 is the difference of squares of integers. Now for a number of the form 4n+2, suppose 4n+2=x2 −y2, then x and y must be of the same parity, hence x2−y2 is divisible by 4, while 4n+2 is not. So the positive integers that cannot be expressed as the difference of two square integers are exactly those of the
12. Let AC and BD intersect at O . From ∆ABC , get
OC OB OB OA =
, from ∆BCD , get OD OC OC OB =
. If OA=a,
ar
OB= , then OC=ar2, OD=ar3 . By Pythagorean theorem ,AB=a 1+r2 , AD=a 1+r6 .
Then 1.
1 1 11
1001
91 2 4 2
6 2
2
+ − = + + = =
= r r
r r AB
AD
Hence r4−r2 −90=
(
r2−10)(
r2 +9)
=0, or r2=10, r=10. Since 11= AB=a 1+r2 , get a=1, and110
2
2 + =
= OB OC
BC
Alternate Solution
Assign the coordinates (0, 11), (0, 0), (1, 0) andA B C D c d( , ).
As AC⊥BD, we have 11=−1
−
c c
d
, or c2 = 11d.
Hence, 2 = 2+( − 11)2
d c AD
2
2 2
( 11) 1001.
11
c c
= + − =
Simplify to get c4 −11c2 −10890=0.
Let c2 =s, and observe that 10890 = 99×110, have(s−110)(s+99)=0, or s=110,
implying c= 110..
A
B C
O D
ar3
a
ar ar2
E
B (0, 0) C (c,0) D (c,d)
A (0, 11)
1001
13. Apply cosine rule to ∆BCP , we have
simplifications, get
Alternate Solution From the first equality, we have x
x
From the second equality, we have
15. As shown in the figure, AB=10, BC=2x and CA=2y. 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
property of centroid = = GE AG GD CG
2 = GF BG
, hence CG=6, GD=3. Let GE=a, then AG=2a.
By Pythagorean theorem, we have 2 36 2
x
a + = ,
( )
2a 2 +36=( )
2y 2, 9+( )
2a 2 = 25. If follows that a=2 , x=2 10 and y= 13 . Letθ
=∠BCA. Apply cosine rule to ∆BCA , get( ) ( )
2 2 2( )( )
2 2 cosθ100= y 2+ x 2 − x y , so xycosθ =14 . Apply cosine rule to ∆BCF , get
( )
2 2 2( )
2 cos 13 160 4 14 1172
2 = + − θ = + − × =
x y x y
BF . Hence BF = 117 =3 13 .
Alternate Solution
Choose the coordinates of A, B and C as (0,0), (10,0) and (b, ) respectively. Then the coordinates c
of D, E and F are (5, 0), (10 , )
2 2
b c +
and ( , ) 2 2 b c
respectively.
As CD= 9, hence
(
b−5)
2 +c2 =81, or b2+c2=10b + 56.……(1) Also CD⊥ AE, get10 5
c c
b b⋅
+ − = -1, or
2 2
c
b + = -5b + 50.……(2)
(1) and (2) imply
5 2 − = b .
Finally 2 2
2 2
2
) 20 ( 4 1 4 1 10 2
2 = + −
− +
= c b c b
BF
100 10
) (
4
1 2 + 2 − +
= b c b = (52) 4 100 117 4
1 + + =
.
Hence BF = 117= 3 13
E
G
A B
C
F
D
A (0,0) B (10,0) C (b,c)
D (5,0) F (
2 , 2
c
b ) E (
16. Let x be the number of blue vertices and y be the number of blue faces in a coloring. If x=0, 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
answer is 5+8+10+8+5=36.
17. Rewrite the recurrence relation as
1 1
3 1 1
3 1
− −
− + =
n n n
a a
a . Define θn =tan−1an , then the above is
equivalent to
+
= −
+
= −
− −
6 tan
tan 6 tan 1
6 tan tan
tan 1
1
1 π
θ θ
π π θ
θ n
n n
n . Thus
6
1
π θ
θn = n− + (“modulo” π if
necessary) and thus an+6 =tanθn+6 =tan
(
θn +π)
=tanθn =an . If follows that(
)
( )
( )
( )
0
2002 4 4 0
0
2
tan tan 2 3
2 3 5 3 8
tan tan
2
3 1 2 2 11
1 tan tan 3
a a
π θ
π
θ θ
π θ
+ + − −
= = = + = = =
− − −
.
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 si . Hence the answer is at least
(
)
7 286 2877 1 2 286 287 1001 287287
2
× ×
× + + +L = = × = . Indeed this is possible if we assign
the number 1 7 i +
19. Let the five given points be A, B, C, D and E. The number of straight lines constructed in step 1 is 10
5 2 =
C . From each of the five pints A, B, C, D and E, a total of C24 =6 perpendicular lines can be drawn. It seems that the number of intersection points should be C230=435. However for any straight line, say AB, the three perpendicular lines drawn from C, D and E are parallel, thus will not meet. Hence the number of intersection points should be cut down by 10×C23 =30. Next for any of the C35 =10 triangle, formed, say ∆ABC, there are three perpendicular lines which are concurrent. Hence the number of intersection points should be further cut down by
(
1)
2010× C23− = . 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
(
1)
705× C26 − = . The required number of points of intersection is therefore 315
70 20 30
435− − − = .
20. Let ABCD be the piece of paper, with
BC b a
AB= ≤ = . Let O be the center of the rectangle, B be folded to coincide with D, and POQ
be the crease, with P on BC and Q on DA. Then PQ is the perpendicular bisector of BD , and
DBC BPO ∆
∆ ~ . Let x be the length of PQ, then
a b
x b a
= +
2 1 2
1 2 2
, or a2 b2 b
a
x= + . Now
= a2+b2 =65 b
a
x , thus
2 2
2
65 a a b
−
= . For
b to be an integer, the denominator 2 2
65 −a must be an integer. Hence we try 652 as a sum of two squares. Using 52 =32+42 and 2 2 2
12 5
13 = + we have
(
) (
2) (
2) (
2)
2 25 12 5 5 4 13 3 13
65 = × + × = × + × =
(
3×12+4×5) (
2 + 4×12−3×5)
2 =(
3×12−4×5)
2(
)
25 3 12 4× + ×
+ . 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
possibility. Thus the perimeter is 2×
(
60+144)
=408.~ End of Paper ~
D A
P
B C
Q
O
a
