Đề thi Hanoi Mathematical Olympiad 2012

ion
1. Let x =
√
6+2
√
5+
√
6−2
√
5√
20
. Find the value of (1 + x5 − x7)2012311
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 4x3−3x, 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 6= 1. Find the value
of f(2012).
6. For every n = 2, 3, . . . , let
An =
(
1− 1
1 + 2
)
×
(
1− 1
1 + 2 + 3
)
× · · · ×
(
1− 1
1 + 2 + · · ·+ n
)
.
Determine all positive integers n such that 1
An
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
x2 + y2 = 1, |x3 − y3|+ |x− y| = m3
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 x3 + px2 + qx+ 1 = 0, with p, q being some rational numbers, has
three real rooots x1, x2, x3, where x3 = 2 +
√
5. Find the values of p, q.
11. Suppose that the equation x3 + px2 + qx+ r = 0 has three real roots x1, x2, x3 where p, q, r
are integers. :et Sn = x
n
1 + x
n
2 + x
n
3 , for n = 1, 2, . . . ,. Prove that S2012 is an integer.
Copyright c© 2011 HEXAGON
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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 = (x2 − y2)2.
15. Determine the smallest value of the expression s = xy−yz−zx, where x, y, z are real numbers
satisfying the condition x2 + 2y2 + 5z2 = 22.
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Solutions
1. Let x =
√
6+2
√
5+
√
6−2
√
5√
20
. Find the value of (1 + x5 − x7)2012311
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 + x5 − x7)2012311 = 1.
2. Arrange the numbers p = 2
√
2, q = 3, t = 2
1+
1√
2 in increasing order. We have 2
1+
1√
2 ≥
21+
1
2 = 2
3
2 = 2
√
2.
Since
√
2 ≤ 3
2
, then 2
√
2 ≤ 2√2. Notice that
t2 = 22+
√
2 ≤ 22+ 32 ≤ 8
√
2.
Thus q4 − t4 = 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, = 21
3
+
1
6
= 4. Let M be the intersection of AC and BD.
b b
bb
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 4x3−3x, where x = 1
2
(
3
√
2 +
√
3+
3
√
2−√3).
Solution. By using the identity a3 + b3 + 3ab(a+ b) = (a+ b)3, we get
(2x)3 =
(
3
√
2 +
√
3 +
3
√
2−
√
3
)3
= 4 + 6x.
Thus 4x3 − 3x = 2. That is, the largest integer desired is 2.
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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 6= 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
An =
(
1− 1
1 + 2
)
×
(
1− 1
1 + 2 + 3
)
× · · · ×
(
1− 1
1 + 2 + · · ·+ n
)
.
Determine all positive integers n such that 1
An
is an integer.
Solution. The k-th summand of the product has the form
ak = 1− 1
(k + 1)(k + 2)
=
k(k + 3)
(k + 1)(k + 2)
, k = 1, 2, · · · , n− 1
from which we get
An =
n+ 2
3n
and hence
1
An
= 3 − 6
n+ 2
. It follows that 1/An is an integer if and only if n + 2 is positive
factor of 6. Notice that n ≥ 2, we get n = 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 102012 = 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
x2 + y2 = 1, |x3 − y3|+ |x− y| = m3
has a solution.
Solution. We need to find the maximum value of f(x, y)
f(x, y) = |x− y|+ |x3 − y3|
when x, y vary satisfying the restriction x2 + y2 = 1.
Rewriting this as
f(x, y) = |x− y|(1 + x2 + xy + y2) = |x− y|(2 + xy).
from which we square to arrive at
f2(x, y) = (x− y)2(2 + xy)2 = (1− 2xy)(2 + xy)2.
By the AM-GM inequality we get
f2(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
, x2 + y2 = 1.
This simultaneous equations are equivalent to
xy = −1
3
, x+ y =
1√
3
.
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Solving for x
x2 − 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 m3 is 5
3
√
5
3
. Hence, mmax =
√
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 ∠MPA = ∠APC − ∠MPC = (90◦ + ∠ABC
2
) − 90◦ = ∠ABC
2
=
∠PBN . Similarly, ∠NPB = ∠PAM . The triangle APM is similar to triangle PBN . Since
PM = PN , we get MA.NB = PM2 = PN2. Hence MA
NB
= MA
2
MA.NB
= MA
2
PN2
= PA
2
PB2
=
32
42
= 9
16
.
10. Suppose that the equation x3 + px2 + qx+ 1 = 0, with p, q being some rational numbers, has
three real rooots x1, x2, x3, where x3 = 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 x2 − 4x− 1 = 0 by squaring.
(x+ α)(x2 − 4x− 1) = x3 + px2 + 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 x3 + px2 + qx+ r = 0 has three real roots x1, x2, x3 where p, q, r
are integers. Let Sn = x
n
1 + x
n
2 + x
n
3 , for n = 1, 2, . . . ,. Prove that S2012 is an integer.
Solution. By the Vieta theorem we get x1+x2+x3 = −p, x1x2+x2x3+x3x1 = q, x1x2x3 =
−r for p, q, r ∈ Z. We can prove the following recursive relation
Sn = −p.Sn−1 − qSn−2 − rSn−3.
From this and mathematical induction, by virtue of S1 = −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.
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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 = (x2 − y2)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, (x2 − y2)2 ≥ 1, or |x − y|2|x + y|2 ≥ 1. That is,
x2 ≥ 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, (x2 − y2)2 ≥ 1. Thus, (x− y)2 ≥ 1. Now that
16x+ 1 = (x2 − y2)2 = (x− y)2(x+ y)2 ≥ x2.
From this we obtain the inequality, x2 − 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 x2 + 2