Find the length of the line segmentEF parallel to the two bases and passing through the intersection of the two diagonalsAC, BD, E is on CD, F on AB.. Let P be the intersection of the th
Trang 1Hanoi Mathematical Olympiad 2012
Senior Section
1. Letx =
√
6+2 √ 5+√
6−2√5
√
20 Find the value of(1 + x5
− x7
)2012 311
2. Arrange the numbersp = 2√2, q = 3, t = 21+
1
√2 in increasing order
3. LetABCD be a trapezoid with AD parallel to BC and BC = 3 cm, DA = 6 cm Find the length of the line segmentEF parallel to the two bases and passing through the intersection of the two diagonalsAC, BD, E is on CD, F on AB
4. What is the largest integer less than or equal to4x3
−3x, where x = 1
2(p3 2 +√
3+p3 2 −√3)
5. Letf (x) be a function such that f (x) + 2fx+2010
x−1
= 4020 − x for all x 6= 1 Find the value
off (2012)
6. For everyn = 2, 3, , let
An=
1 − 1 + 21
×
1 −1 + 2 + 31
× · · · ×
1 + 2 + · · · + n
Determine all positive integersn such that 1
A n is an integer
7. Prove thata = 1 1
| {z }
2012
5 5
| {z }
2011
6 is a perfect square
8. Determine the greatest numberm 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 equationx3
+ px2
+ qx + 1 = 0, with p, q being some rational numbers, has three real roootsx1, x2, x3, wherex3 = 2 +√
5 Find the values of p, q
11. Suppose that the equationx3
+ px2
+ qx + r = 0 has three real roots x1, x2, x3wherep, q, r are integers :etSn= xn
1 + xn
2 + xn
3, forn = 1, 2, , Prove that S2012is an integer
Copyright c
Trang 212. LetM 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 thatOM ⊥ BS
13. A cube with sides of length3 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, andd the number of cubes painted on three sides, determine the value of a − b − c + d
14. Sovle the equation in the set of integers16x + 1 = (x2
− y2)2
15. Determine the smallest value of the expressions = xy−yz−zx, where x, y, z are real numbers satisfying the conditionx2
+ 2y2
+ 5z2
= 22
Trang 31. Letx =
√
6+2 √
5+√
6−2√5
√
20 Find the value of(1 + x5
− x7
)2012 311
5 = (√
5 + 1)2
and6 − 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 = 21+
1
√2 in increasing order We have 21+
1
√2 ≥
21+1 = 23 = 2√
2
Since√
2 ≤ 3
2, then2√2
≤ 2√2 Notice that
t2
= 22+ √
2
≤ 22+3
2 ≤ 8√2
Thusq4
− t4
= 81 − 64 × 2 < 0 It follows that
p < t < q
3. LetABCD be a trapezoid with AD parallel to BC and BC = 3 cm, DA = 6 cm Find the length of the line segmentEF parallel to the two bases and passing through the intersection of the two diagonalsAC, 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
C B
By the Thales theorem we get OEBC + 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 getEF = 4 cm.
4. What is the largest integer less than or equal to4x3
−3x, where x = 1
2(p3 2 +√
3+p3 2 −√3)
+ b3
+ 3ab(a + b) = (a + b)3
, we get (2x)3
=
3
q
2 +√
3 + 3
q
2 −√3
3
= 4 + 6x
Trang 45. Letf (x) be a function such that f (x) + 2fx+2010x−1 = 4020 − x for all x 6= 1 Find the value
off (2012)
f u + 2010
u − 1
+ 2f (u) = 4020 − u + 2010
u − 1 . Interchangingu with x gives
f x + 2010
x − 1
+ 2f (x) = 4020 − x + 2010
x − 1 . Leta = f (x), b = fx+2010
x−1
Solving the system
a + 2b = 4020 − x, b + 2a = 4020 − x + 2010
x − 1 fora 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 everyn = 2, 3, , let
An=
1 − 1 + 21
×
1 −1 + 2 + 31
× · · · ×
1 + 2 + · · · + n
Determine all positive integersn such that A1
n is an integer
ak= 1 − (k + 1)(k + 2)1 = 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/Anis an integer if and only ifn + 2 is positive factor of6 Notice that n ≥ 2, we get n = 4
Trang 57. Prove thata = 1 1
| {z }
2012
5 5
| {z }
2011
6 is a perfect square
| {z }
2012
Then102012
= 9p + 1 Hence,
a = p(9p + 1) + 5p + 1 = (3p + 1)2, which is a perfect square
8. Determine the greatest numberm such that the system
x2+ y2= 1, |x3− y3| + |x − y| = m3 has a solution
f (x, y) = |x − y| + |x3− y3| whenx, 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.
r 5
3. Equality occurs when
xy = −1
3, x
2
+ y2
= 1
This simultaneous equations are equivalent to
xy = −13, x + y = √1
3.
Trang 6Solving forx
x2−√x
3−13 = 0
∆ = 1
3+ 4
3 = 5
3, that is
x = 1 2
1
√
3−
r 5 3
! , x = 1 2
1
√
3 +
r 5 3
!
Therefore, the value ofm3
is 53
q
5
3 Hence,mmax=q5
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
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.N B = P M2
= P N2
Hence M AN B = M A.N BM A2 = M AP N22 = P BP A22 =
3 2
4 2 = 9
16
10. Suppose that the equationx3
+ px2
+ qx + 1 = 0, with p, q being some rational numbers, has three real roootsx1, x2, x3, wherex3 = 2 +√
5 Find the values of p, q
5 is one root of the equation, we get x − 2 = √5 from which we get a quadratic polynomialx2
− 4x − 1 = 0 by squaring
(x + α)(x2− 4x − 1) = x3+ px2+ qx + 1 = 0
Expanding the left hand side and comparing the coefficients giveα = −1 and hence
p = −3, q = −5
11. Suppose that the equationx3
+ px2
+ qx + r = 0 has three real roots x1, x2, x3wherep, q, r are integers LetSn= xn
1 + xn
2 + xn
3, forn = 1, 2, , Prove that S2012is an integer
−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 ofS1 = −p ∈ Z, we get the desired result
12. LetM 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 thatOM ⊥ BS
Trang 7Solution 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 k 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 =
∠ON B = 90◦, as desired
13. A cube with sides of length3 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, andd the number of cubes painted on three sides, determine the value of a − b − c + d
Hence,a − b − c + d = −7
14. Sovle the equation in the set of integers16x + 1 = (x2
− y2)2
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 considery ≥ 0
From the right hand side of the equation, we deduce that16x + 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 ofy
The equation has solutions(0; 1), (0; −1), (5; 4), (5; −4)
15. Determine the smallest value of the expressions = xy−yz−zx, where x, y, z are real numbers satisfying the conditionx2+ 2y2+ 5z2 = 22
Solution.