ingapore Mathematical Society
Singapore Mathematical Olympiad (SMO) 2009
(Junior Section Solutions)
Answer: (C)
As seen from the diagram below, there are 6 circles that are tangent to both C)
and Cy
Answer: (A)
Rotate AOAB 90° anticlockwise about the point O to overlap with AODC.
@ @
A
Shaded area =} (64m — 162) = 121.
10 Answer: (D)
(fx-2 + V7-x) = 54+2V@-207 = = 542 625-(-45).
Hence maximum value of k=1f 5 +2(2.5) = 10.
Answer: (B)
Let P be the midpoint of XY and Q be the point where the line 17Z meets the third
circle.
PB_ 20
Then by similar As, G5 = 99 =P
Answer: (D)
‘The maximum value of y is ~5.
Answer: (C)
Let by = dy ~ ay 0 (50)(51)(101) = 42925,
On the other hand, ¥ by = aso ~ ay => asy = 42925 + ay = 44934.
uw 10
u
Answer: (B)
&
Consider the equilateral triangle formed by joining the centres of 3 adjacent coins.
It is easy to see that the required percentage is given by the percentage of the
triangle covered by these three coins.
2
We have F73» 100% =
Answer: (E)
P+ =6=9 (e+ pV = Dy +6. So2x+ (x+y)! -6} +2v=22+3V2)
=> (etyP + 2+ yt 1H 114692 > +y+1P=-B+V2).
Answer: (E)
y= QP = 16) (7 =1
So minimum value of.
452x? + 50176 = (x? — 226)' ~ 900.
900.
Answer: (A)
(a, b, c,d) = (2,3,7,42), (2,3,8.24), (2,3,9,18), (2,3,10,15), (2.4,5,20), (2,4,6,12).
Answer: (337)
16 16x
12
20 7 20 7 7
Standard Widescreen
Area of Standard __(16)(12) _ 337
2 (On? =202 ae
(lex) + Oxy" = 20° 3374" = 400. ~ Area of Widescreen ~ (16x)(9x) ~ 300
2 12 Answer: (47)
Let the area of the pentagon and the rectangle be P and R respectively.
Tp mined
3 p_2 m
We have 76 P= 9 8. So =
13. Answer: (513)
‘Minimum number of answer scripts is 2” + 1 = 513.
14 Answer: (2520)
First we arrange the 6 girls in 6! ways. Next, there are 7 spaces between the 6
girls to insert the 5 boys. Hence k= "Cs x 5! = 2520.
15 Answer: (6)
We rotate ABAC 90? anticlockwise about the point B to get ABC'G.
C'G=BC=\ 26-8 =V18.
G
D
EA c
The area of ADBG 5 x DBxC'G
16 Answer: (173)
1 ay 1 1
Note that 7+ 1)x (+2) Graces (et x@+2)"
1 1 1 1 I
Hence 75354 t 3x45 *4x5x6* "13x ldx 5 * Mx 15x16
“GS wo) ie
“202%37 15x16) 160
13 17
18
19
20
a
2
1 (ey #0), then x—145=
= 0S sy= 1 ab-a+b=2.
5
Answer: (2012)
Suppose y > 0. From |y|—y+x=7 we have x=7 and from |x| +x + 5y=2we
2
have y=-"F <0 (€). Soy <0.
2
$>0(94).Sox>0.
B 4
Suppose x < 0. From |x| +2 + Sy =2 we have
Hence the two equations become -2y + x= 7 and 2x + 5y=2=> x=
Answer: (7)
Since 3p + 3q = 6n— 27, p +g = 2n—9 which is odd. So p = 2, q = 3 and n= 7.
Answer: (6)
Note that xy + yx +f 2009xy — 2009 — 2009p ~ 2009
= (fx + Vy — 4.2009 af xy --V 2009 ), so xy = 2009.
(x, ») = (1, 2009), (7, 287), (41, 49), (49, 41), (287, 7), (2009, 1).
Answer: (286)
aj, tt tt
Let L=3993 * 3004 * 2005 * 3006 * 2007 * 2008 * 2009
1 7 1a
Clearly 3999 <2
(BCEF) + (HIK) = (ACD) + (DEW) + (FGH)
= 500 + (HIK) = 22 + 482 +22 = (HIK) =26.
4 23
24
25
26
27
Answer: (7)
[77-2098 .y=V 77+ 20913 and A=X+¥.
Since X°+ Y°= 154, X¥ = 77— 2013 = 9, AP= (X4 YP =P YH AV(XE DY)
=> A= 1544274 => AP -27A — 154 =O (4-74? + 74 +22) = 0 > ART.
Let.
Answer: (133)
We consider 3 sets of integers:
(a) 0001 ~ 0999, We count the number of nonnegative solutions of a+ 6 +¢= 11.
Itis "Cs, = 8C2 = 78, Note that 11 should not be split as 0+0+11 nor 0+1+10.
So the number of solutions is 78 ~ 3 — 6 = 69.
(b) 1001 ~ 1999. We count the number of nonnegative solutions of a + b +.¢ = 10.
It is !"'C,_, = Cy = 66. Note that 10 should not be split as 0+0+10. So the
number of solutions is 66 ~ 3 = 63.
(6) 2001 ~ 2009. We see that only 2009 satisfies the property.
. The total number of integers satisfying the property is 69 + 63 + 1 = 133.
Answer: (5)
Let x= I, we have ay +a; +az+..taq=142424 0.42752
Also, ay=142+3+..40= 222,
506 MED, MOEN, | amt 3 oy as
Answer: (26)
Note that area of OPQ =4 « 4m x 13 em=26 em’,
Answer: (7)
From given equation, xj + x2 +3 + 44 = 0, =x) x2(x3 +4) +23 x4 (81 +2) = 90
and x1 x23.x4 = ~2009. Now x1 x2 = 49 => 234 = “41. So ay +42 = Las ta4=—1.
Hence k= xy x2 + x1 x3 +1 X4-+ 0.85 + xoay +A5.xg = 49 + (1) x (1) 41 =7,
15 28
29
30
Answer: (1052)
Note that (Area of AQABY' = (Area of AOAC)*+(Area of AOBC.
So (Area of AOAB)’ + (Area of AOAC)’ + (Area of AOBC)” + (Area of AABC)*
= 2x ((Area of AAG) + (Area of SOB) + (Area of MBO}
= 2x { (2x 7x 6) + (1/2 x 7 x 2) + (1/2 x 2 x 6)? } = 1052.
Answer: (111)
malt 101
Consider 49 = 9+ 5-19
is a non-zero reducible fraction,
on 1
then =~ 59 is also a non-zero reducible fraction — Least positive integer n= 111
10
Answer: (90)
x? + Am + 5)x + (100m + 9) = 0 => x = —(m + 5) +m — 45) — 2009. This
yields integer solutions if and only if (m — 45)’ — 2009 is a perfect square, say n°
Hence (m — 45) — 1? = 2009= 7° x 41 =>
|m-45|+n=2009 and |m—45|~n=1, or
m—45|+n=287 and |m—45|~n=7, or
m—45\+n=49 and |m—45|-n=41
Solving, n = 45 + 1005, 45 + 147, 45 + 45 => The smallest positive 1 is 90.
Answer: (5)
Area of the triangle is } x AD x BC =} AC x BE= 4% AB x CF
Since AD = 4 and BE = 12, BC: AC=3: 1. Let AC=x, then BC = 3x.
Using Triangle Inequality, 4B < AC + BC and BC < AB + AC => 2x < AB < 4x.
From CF = “25, 3 < CF <6 =» The largest integer value for CF is 5.
ae
16
(Area of AABCY. 32
33
34
Answer: (97)
We let the 2 distinct digits be A and B with 1 <4<9,0
‘There are 81 — 12 = 69.
Case 2: AABB = 11 x (1004 + B). There are 11 possibilities that are factors of 7.
(A,B) = (1,5) , 2,3), 3,1), (3,8), 4,6), (5,4), (6,2), (6,9), (7,0), (8,5), (9,3).
Case 3: ABBA = 11 x (914 + 10B). B must be 0 or 7. There are 17 possibilities.
Total such possible number = 69 + 11 + 17=97,
Answer: (64)
Ifn=33,m
Ifm = 33,
12,3,» 32, so there are 32 pairs.
33, 34, .. . 40, so there are 8 pairs.
If-n, m #33, we have 2 cases:
Case 1: m= 3a,n=11b (a4 11 and b #3), thus | <3a< 11b $40,
So if b=1,a= 1,2, 3 and if b=2,a=1, 2,3, ..., 7. There are 10 pairs.
Case 2: m= 1a, n=3b (a#3 and b # 11), thus 1 $ Has 36 <40.
So if a=, b=4, 5, 6, 7, 8, 9, 10, 12, 13 and if a= 2, b=8, 9, 10, 12, 13.
There are 9+ 5 = 14 pairs.
Hence we have altogether 32 + 8 + 10 + 14 = 64 pairs.
Answer: (3402)
We call a sequence that satisfy the condition a “good” sequence.
Let A, denote the number of “good” sequence that end in either 0 or 4,
B, denote the number of “good” sequence that end in either 1 or 3,
C, denote the number of “good” sequence that end in 2.
We have
(1) Anyi = By because each sequence in 4,1 can be converted to a sequence
in B, by deleting its last digit.
17 35
(2) Bry =An + 2Cy because each sequence in 4, can be converted into a
sequence in B,,, by adding a 1 (if it ends with a 0) or a 3 (if it ends with a
4) to its end, and each sequence in C,, can be converted into a sequence in
By by adding a 1 or a3 to it,
(3) Covi = By because each sequence in C, .1 can be converted to a sequence
in B, by deleting the 2 at its end.
Hence we can show that By .1= 3B, —1 for m > 2.
Check that By = 2 and By =4, 50 Boe, 1= 2 3" and By, =4x 3"!
So Ais + Bis + Cis = 2Bi2 + Bip = 2x4 x 3° +2 x 3° = 3402.
Answer: (79497)
Clearly, m and n are both 5-digit numbers.
Next, it would be helpful that we know mn =2 x 3° x 5x 7x 1? x 19 x37.
Now since the last digit of mn is 0, we may assume 5|m and 2| n, But the first
digit of mn is 1 => Last digit of m is 5 (not 0) and last digit of n is 2 (not 4, 6 or 8),
Also, 3° | mn, so 9 divides at least one of m and n, On the other hand, 9| m => 9|
Similarly 11 | m => 11].
Set = 198k. Then the last digit of k is 4 or 9.
Since the remaining factors 3, 7, 19, 37 are odd, the last digit of k must be 9.
We have only the following combinations: k = 7 x 37 or 3 x 7 x 19 or 3 x 19 x 37.
Recall that the first digit of n is 5, so 50000 < 198k < 60000 => k= 7 x 37.
Hence n= 198k = 51282 and m = 28215 => m +n = 79497. Singapore Mathematical Society
Singapore Mathematical Olympiad (SMO) 2009
(Junior Section, Round 2)
Singapore Mathematical Society
Singapore Mathematical Olympiad (SMO) 2009
(Junior Section, Round 2 solutions)
1. Let AD be the angle bisector of ZA. Then AABC
AC/DC = BC/AC. Let BD = x and DC = y. Then
cb = ax, 6? = ay. Thus b* + cb = az + ay and hence b(5 + ¢
ADAC. Thus AB/DA =
b/y'= a/b. ‘Thus
‘We solve the general case of 2n-digit integers where n > 2. There are 10%" — 10?”
In-digit integers. There are 10" — 10"! n-digit integers. Consider all the products of
pairs of n-digit integers. The total number P of such products satisfies
(10" — 10"4)(10" — 10"! — 1)
vi a
P<10"—10" ' +
102" — 102-1 — (102-1 — 10}
yan yq2n—1
2
<
These products include all the numbers in M. Thus |M| < |N|-
3. 'Two-digit numbers which contain three distinct prime factors are:
30 = 2-3-5,42 = 2-3-7,60 = 4-3-5,66 = 2-3-11,70 = 2-5-7, 78 = 2-3-13, 84 = 4-3-7
20 From here, we conclude that a; = 6 for i= 1,2,..., 2007 and ago9s is either 6 or 0,
4. If z is the smallest integer in S such that « =i (mod 3), then x + 3k € S and
2 —3(k +1) ¢ S$ for all k > 0. We have 3 is the smallest multiple of 3 that is in $; 50
is smallest number in § that is (mod 3) and 100 is the smallest number in $ that
is = 1 (mod 3). Thus the positive numbers not in § are 1,4,...,97 and 2,5,...,47.
‘Their sum is
3397 +1) , 16(2 +47)
2 + 2
= 2009.
5. The left hand side is
ary x9...25 +a°b(wyrongxy + ie2@grts +++ + Toxgr405)
+ 08b"(xy rors + eyS204 +++ + Taras)
+ oP (ayaa +103 +--+ + ats) + abA(ay + a2 +--+ +25) + 0°
> a +5a%b + 1056? + 10076? + Sab! + 6° = (a+b)? = 1.
‘The last is true since by AM-GM inequality,
ayryayts + eimarges +--+ maracas > 5(ei22e5rar5)"/*
zyxors + 2itoe4 $+ + agz4r5 > 10(e 2203045)" = 10
2129 + aytg +--+ tars > 10(a 122032475)! = 10
ay +p +--+ +25 > B(aiteeszaas) =5
(Note. For a proof of the general case, see Senior Q4)
21