Download Singapore Mathematical Olympiads (2011) PDF

TitleSingapore Mathematical Olympiads (2011)
File Size2.8 MB
Total Pages64
Table of Contents
                            Junior Section
	First Round
		Solutions
	Second Round
		Solutions
Senior Section
	First Round
		Solutions
	Second Round
		Solutions
Open Section
	First Round
		Solutions
	Second Round
		Solutions
                        
Document Text Contents
Page 33

Moreover,

2x

SO X > 3. 5.

8. Ans wer: (B)

2(log 3 +log 5� +log 7i)
log 2

log(9 x 5) +log(49i) log(45 x 27 i)
-------- > ----,----

log2 log 2
log( 45 x 3)

>
log(128)

= 7
�g2 �g2

'

Note that 74-1 = 2400, s o that 74n-1 is divis ible by 100 for any n E z+. Now,

where

756 7(756-1 -1 + 1)
7(756-1-1) + 7
7(74n-1) + 7 ,

n =
56-1 '77+

4
E !LJ .

Since 7(74n -1) is divis ible by 100, its las t two digits are 00. It follows that the las t
two digits of 756 are 07.

9. Ans wer: (A)

logx 2011 +logy 2011
logxy 2011

(
log 2011

+
log 2011

) . (
log xy

)
log x logy log 2011
1 1

(- + -) . (logx +logy)log x logy
logx logy

2+--+--
logy logx

> 4 (us ing AM 2:: G M),

and the equality is attained when log x = logy, or equivalently, x = y.

10. Ans wer: (C)
The roots of the equation x2- (c-1)x + c2- 7c + 14 = 0 are a and b, which are
real. Thus the dis criminant of the equation is non-negative. In other words ,

(c-1)2- 4(c2- 7c + 14) = -3c2 + 26c- 55= (-3c + 11)(c- 5) 2:: 0.

So we have
11 < c < 5. Together with the equalities
3 - -

(a+ b?- 2ab
(c- 1)2-2(c2- 7c + 14)
-c2 + 12c- 27 = 9 - ( c- 6)2,

32

Page 63

which leads to a contradiction. On the other hand, s uppos e that r1, r3, r5, r7, r9 2:: 6.
Then the s um of any 2 cons ecutive ri's is :::; 9. Again we get a contradiction as

(r1 + r2) + · · · + (r7 + rs) + rg :::; 4 x 9 + 9 = 45.

3. Let r = 1/x, s = 1/y, t = 1/ z. There exis ts a < 1 s uch that r + s + t = a2rst
or a(r + s + t) = a3rst. Let a= ar, b =as, c =at. Write a= tanA, b = tanB,
c = tan C, then A+ B + C = 1r. It is clear that

1 1 1 1
- X LHS = + + -----;==::;;;: 2 y1 + r2 v1 + s2 v1 + t2

1 1 1
< + + -----;==::;;;:

v1 + a2 v1 + b2 v1 + c2
= cos A+ cos B + cos B

< 3 cos (A + B + C) = � = � x RHS.- 3 2 2
2nd soln: Note that

Hence

1 1 1 1
-+ - + - < - :::::? xy + yz + xz < 1.
x y z xyz

2x 2x 2x
----r===;;;= < -r-::::::=======
y1 + x2 Jx2 + xy + xz + yz J(x + y)(x + z)

By AM-GM we have

Similarly,

2x X X
----r-.;======:=;===;= < --+ --.
J(x+y)(x+z) - x+y x+z

2z z z 2y < _Y_+_Y_
J(y + z)(y + x) - y + z y + x'

----r-.;======:=;===;= < -- + --.V ( Z + X) (z + y) - Z + X Z + y
The des ired inequality then follows by adding up the three inequalities .

4. Let P(x) = anxn + · · · + a1x + a0. Define Q(x) = P(x + 1) - P(x). Then Q(x) is
of degree n- 1. We'll prove by contradiction that IQ(x)l :::; 3 for all x. This will imply
that n :::; 1. As s ume that IQ(a)i > 3 for s ome a E R Then IP(a + 1) - P(a)l > 3.
Thus there are 3 integers between P(a) and P(a + 1). Hence there exis ts three values

62

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