Tài liệu Đề thi toán bằng tiếng anh 2012 pdf

Hanoi Open Mathematical Olympiad • • • www.hexagon.edu.vn
H
E
XAGON®
inspiring minds always
Hanoi Mathematical Olympiad 2012
Senior Section
1. Let x =
√
6+2
√
5+
√
6−2
√
5
√
20
. Find the value of (1 + x
5
− x
7
)
2012
311
2. Arrange the numbers p = 2
√
2
, q = 3, t = 2
1+
1

√
2
in increasing order.
3. Let ABCD be a trapezoid with AD parallel to BC and BC = 3 cm, DA = 6 cm. Find the
length of the line segment EF parallel to the two bases and passing through the intersection of
the two diagonals AC, BD, E is on CD, F on AB.
4. What is the largest integer less than or equal to 4x
3
−3x, where x =
1
2
(
3

2 +
√
3+
3

2 −
√
3).
5. Let f(x) be a function such that f(x) + 2f

x+2010
x−1

= 4020 −x for all x = 1. Find the value
of f(2012).
6. For every n = 2, 3, . . . , let

A
n
=

1 −
1
1 + 2

×

1 −
1
1 + 2 + 3

× ··· ×

1 −
1
1 + 2 + ··· + n

.
Determine all positive integers n such that
1
A
n
is an integer.
7. Prove that a =
1 . . . 1



2012
5 . . . 5


2011
6 is a perfect square.
8. Determine the greatest number m such that the system
x
2
+ y
2
= 1, |x
3
−y
3
| + |x − y| = m
3
has a solution.
9. Let P be the intersection of the three internal angle bisectors of a triangle ABC. The line
passing through P and perpendicular to CP intersects AC and BC at M, N respectively. If
AP = 3 cm, BP = 4 cm, find the value of AM/BN .
10. Suppose that the equation x
3
+ px
2
+ qx + 1 = 0, with p, q being some rational numbers, has
three real rooots x
1
, x
2

, x
3
, where x
3
= 2 +
√
5. Find the values of p, q.
11. Suppose that the equation x
3
+ px
2
+ qx + r = 0 has three real roots x
1
, x
2
, x
3
where p, q, r
are integers. :et S
n
= x
n
1
+ x
n
2
+ x
n
3
, for n = 1, 2, . . . ,. Prove that S

2012
is an integer.
Copyright
c
 2011 H
E
XAGON
1
Hanoi Open Mathematical Olympiad • • • www.hexagon.edu.vn
12. Let M be a point on the side BC of an isosceles triangle ABC with BC = BA. Let O be the
circumcenter of the triangle and S be its incenter. Suppose that SM is parallel to AC. Prove
that OM ⊥ BS.
13. A cube with sides of length 3 cm is painted red and then cut into 3 × 3 × 3 = 27 cubes with
sides of length 1 cm. If a denotes the number of small cubes of side-length 1 cm that are not
painted at all, b the number of cubes painted on one side, c the number of cubes painted on two
sides, and d the number of cubes painted on three sides, determine the value of a − b − c + d.
14. Sovle the equation in the set of integers 16x + 1 = (x
2
− y
2
)
2
.
15. Determine the smallest value of the expression s = xy−yz−zx, where x, y, z are real numbers
satisfying the condition x
2
+ 2y
2
+ 5z
2

= 22.
2
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Solutions
1. Let x =
√
6+2
√
5+
√
6−2
√
5
√
20
. Find the value of (1 + x
5
− x
7
)
2012
311
Solution. Notice that 6 + 2
√
5 = (
√
5 + 1)
2
and 6 − 2
√

5 = (
√
5 − 1)
2
,
√
20 = 2
√
5 then
x = 1. That is
(1 + x
5
− x
7
)
2012
311
= 1.
2. Arrange the numbers p = 2
√
2
, q = 3, t = 2
1+
1
√
2
in increasing order. We have 2
1+
1
√

2
≥
2
1+
1
2
= 2
3
2
= 2
√
2.
Since
√
2 ≤
3
2
, then 2
√
2
≤ 2
√
2. Notice that
t
2
= 2
2+
√
2
≤ 2

2+
3
2
≤ 8
√
2.
Thus q
4
−t
4
= 81 −64 × 2 < 0. It follows that
p < t < q.
3. Let ABCD be a trapezoid with AD parallel to BC and BC = 3 cm, DA = 6 cm. Find the
length of the line segment EF parallel to the two bases and passing through the intersection of
the two diagonals AC, BD, E is on CD, F on AB.
Hint. Making use of the similarity of triangles. The line segment is the harmonic means of the
two bases, =
2
1
3
+
1
6
= 4. Let M be the intersection of AC and BD.
A D
CB
By the Thales theorem we get
OE
BC
+

OF
AD
=
OD
BD
+
OC
AC
=
OD
BD
+
OB
BD
= 1. From this,
1
OE
=
1
BC
+
1
AD
. Likewise,
1
OF
=
1
BC
+

1
AD
. Hence, OE = OF . That is,
2
EF
=
1
OE
=
1
BC
+
1
AD
=
1
3
+
1
6
=
1
2
. We get EF = 4 cm.
4. What is the largest integer less than or equal to 4x
3
−3x, where x =
1
2
(

3

2 +
√
3+
3

2 −
√
3).
Solution. By using the identity a
3
+ b
3
+ 3ab(a + b) = (a + b)
3
, we get
(2x)
3
=

3

2 +
√
3 +
3

2 −
√

3

3
= 4 + 6x.
Thus 4x
3
− 3x = 2. That is, the largest integer desired is 2.
3
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5. Let f(x) be a function such that f(x) + 2f

x+2010
x−1

= 4020 −x for all x = 1. Find the value
of f(2012).
Solution. Let u =
x+2010
x−1
then x =
u+2010
u−1
. Thus we have
f

u + 2010
u − 1

+ 2f (u) = 4020 −
u + 2010

u − 1
.
Interchanging u with x gives
f

x + 2010
x − 1

+ 2f (x) = 4020 −
x + 2010
x − 1
.
Let a = f(x), b = f

x+2010
x−1

. Solving the system
a + 2b = 4020 −x, b + 2a = 4020 −
x + 2010
x − 1
for a in terms of x gives
a = f(x) =
1
3

8040 − 4020 + 2x −
2x + 4020
x − 1


=
1
3

4020 + 2x −
4020 + 2x
x − 1

.
Hence,
f(2012) =
1
3

8044 −
8044
2011

= 2680.
6. For every n = 2, 3, . . . , let
A
n
=

1 −
1
1 + 2

×


1 −
1
1 + 2 + 3

× ··· ×

1 −
1
1 + 2 + ··· + n

.
Determine all positive integers n such that
1
A
n
is an integer.
Solution. The k-th summand of the product has the form
a
k
= 1 −
1
(k + 1)(k + 2)
=
k(k + 3)
(k + 1)(k + 2)
, k = 1, 2, ··· , n −1
from which we get
A
n
=

n + 2
3n
and hence
1
A
n
= 3 −
6
n + 2
. It follows that 1/A
n
is an integer if and only if n + 2 is positive
factor of 6. Notice that n ≥ 2, we get n = 4.
4
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7. Prove that a = 1 . . . 1


2012
5 . . . 5


2011
6 is a perfect square.
Solution. Let p = 1 . . . 1


2012
. Then 10
2012

= 9p + 1. Hence,
a = p(9p + 1) + 5p + 1 = (3p + 1)
2
,
which is a perfect square.
8. Determine the greatest number m such that the system
x
2
+ y
2
= 1, |x
3
−y
3
| + |x − y| = m
3
has a solution.
Solution. We need to find the maximum value of f(x, y)
f(x, y) = |x − y| + |x
3
− y
3
|
when x, y vary satisfying the restriction x
2
+ y
2
= 1.
Rewriting this as
f(x, y) = |x − y|(1 + x

2
+ xy + y
2
) = |x − y|(2 + xy).
from which we square to arrive at
f
2
(x, y) = (x −y)
2
(2 + xy)
2
= (1 −2xy)(2 + xy)
2
.
By the AM-GM inequality we get
f
2
(x, y) = (1 − 2xy)(2 + xy)
2
= (1 −2xy)(2 + xy)(2 + xy)
≤

1 − 2xy + 2 + xy + 2 + xy
3

3
=

5
3


3
.
Hence,
f(x, y) ≤
5
3
.

5
3
.
Equality occurs when
xy = −
1
3
, x
2
+ y
2
= 1.
This simultaneous equations are equivalent to
xy = −
1
3
, x + y =
1
√
3
.

5
Hanoi Open Mathematical Olympiad • • • www.hexagon.edu.vn
Solving for x
x
2
−
x
√
3
−
1
3
= 0.
∆ =
1
3
+
4
3
=
5
3
, that is
x =
1
2

1
√
3

−

5
3

, x =
1
2

1
√
3
+

5
3

.
Therefore, the value of m
3
is
5
3

5
3
. Hence, m
max
=


5
3
.
9. Let P be the intersection of the three internal angle bisectors of a triangle ABC. The line
passing through P and perpendicular to CP intersects AC and BC at M, N respectively. If
AP = 3 cm, BP = 4 cm, find the value of AM/BN .
Solution. Notice that ∠MP A = ∠AP C − ∠MP C =

90
◦
+
∠ABC
2

− 90
◦
=
∠ABC
2
=
∠P BN . Similarly, ∠N P B = ∠P AM. The triangle AP M is similar to triangle P BN. Since
P M = P N, we get M A.NB = P M
2
= PN
2
. Hence
M A
N B
=
M A

2
M A.N B
=
M A
2
P N
2
=
P A
2
P B
2
=
3
2
4
2
=
9
16
.
10. Suppose that the equation x
3
+ px
2
+ qx + 1 = 0, with p, q being some rational numbers, has
three real rooots x
1
, x
2

, x
3
, where x
3
= 2 +
√
5. Find the values of p, q.
Solution. Since x = 2 +
√
5 is one root of the equation, we get x − 2 =
√
5 from which we
get a quadratic polynomial x
2
− 4x −1 = 0 by squaring.
(x + α)(x
2
− 4x −1) = x
3
+ px
2
+ qx + 1 = 0.
Expanding the left hand side and comparing the coefficients give α = −1 and hence
p = −3, q = −5.
11. Suppose that the equation x
3
+ px
2
+ qx + r = 0 has three real roots x
1

, x
2
, x
3
where p, q, r
are integers. Let S
n
= x
n
1
+ x
n
2
+ x
n
3
, for n = 1, 2, . . . ,. Prove that S
2012
is an integer.
Solution. By the Vieta theorem we get x
1
+x
2
+x
3
= −p, x
1
x
2
+x

2
x
3
+x
3
x
1
= q, x
1
x
2
x
3
=
−r for p, q, r ∈ Z. We can prove the following recursive relation
S
n
= −p.S
n−1
− qS
n−2
− rS
n−3
.
From this and mathematical induction, by virtue of S
1
= −p ∈ Z, we get the desired result.
12. Let M be a point on the side BC of an isosceles triangle ABC with AC = BC. Let O be the
circumcenter of the triangle and S be its incenter. Suppose that SM is parallel to AC. Prove
that OM ⊥ BS.

6
Hanoi Open Mathematical Olympiad • • • www.hexagon.edu.vn
Solution. Let OM meet SB at H. N is the midpoint of AB. Since O is the circumcenter
of triangle OBC which is isosceles with CA = CB and SM  AC we have ∠SOB =
2∠OCB = ∠ACB = ∠SMB. It follows that quadrilateral OMBS is concyclic. Hence,
∠HOS = ∠SBM = ∠SBN which implies the concyclicity of HOBN. Hence, ∠OHB =
∠ONB = 90
◦
, as desired.
13. A cube with sides of length 3 cm is painted red and then cut into 3 × 3 × 3 = 27 cubes with
sides of length 1 cm. If a denotes the number of small cubes of side-length 1 cm that are not
painted at all, b the number of cubes painted on one side, c the number of cubes painted on two
sides, and d the number of cubes painted on three sides, determine the value of a − b − c + d.
Solution. Just count from the diagram of the problem, we get a = 1, b = 4, c = 12, d = 8.
Hence, a − b −c + d = −7.
14. Sovle the equation in the set of integers 16x + 1 = (x
2
− y
2
)
2
.
Solution. Since the right hand side is non-negative we have deduce that 16x + 1 ≥ 0. That
is, x take positive integers only. Therefore, (x
2
− y
2
)
2
≥ 1, or |x − y|

2
|x + y|
2
≥ 1. That is,
x
2
≥ 1.
It is evident that if (x, y) is a solution of the equation, then (x, −y) is also its solution. Hence,
it is sufficient to consider y ≥ 0.
From the right hand side of the equation, we deduce that 16x + 1 ≥ 0. Since x ∈ Z, we get
x ≥ 0, which implies that 16x + 1 ≥ 1. Hence, (x
2
−y
2
)
2
≥ 1. Thus, (x −y)
2
≥ 1. Now that
16x + 1 = (x
2
− y
2
)
2
= (x −y)
2
(x + y)
2
≥ x

2
.
From this we obtain the inequality, x
2
− 16x − 1 < 0. Solving this inequality gives x ∈
{0, 1, ··· ,16}. In addition, 16x + 1 is a perfect square, we get x ∈ {0, 3, 5, 14}. Only x = 0; 5
give integer value of y.
The equation has solutions (0; 1), (0; −1), (5; 4), (5; −4).
15. Determine the smallest value of the expression s = xy−yz−zx, where x, y, z are real numbers
satisfying the condition x
2
+ 2y
2
+ 5z
2
= 22.
Solution.
7