Title Singapore Mathematical Olympiads (2011) 2.8 MB 64
Junior Section
First Round
Solutions
Second Round
Solutions
Senior Section
First Round
Solutions
Second Round
Solutions
Open Section
First Round
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Second Round
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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,

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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

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