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 n[r]
(1)Hanoi Mathematical Olympiad 2012
Senior Section
1. Letx =
√
6+2√5+√6−2√5 √
20 Find the value of(1 + x
− x7
)2012311
2. Arrange the numbersp = 2√2, q = 3, t = 21+√21 in increasing order.
3. LetABCD be a trapezoid with AD parallel to BC and BC = cm, DA = 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 = 2(
3
p
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= Find the value off (2012)
6. For everyn = 2, 3, , let
An=
1 − 1 + 21
×
1 −1 + + 31
× · · · ×
1 −
1 + + · · · + n
Determine all positive integersn such that
An is an integer
7. Prove thata = 1
| {z }
2012
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 = cm, BP = cm, find the value of AM/BN
10. Suppose that the equationx3
+ px2
+ qx + = 0, with p, q being some rational numbers, has three real roootsx1, x2, x3, wherex3 = +
√
5 Find the values of p, q
11. Suppose that the equationx3
+ px2
+ qx + r = has three real roots x1, x2, x3wherep, q, r
are integers :etSn= xn1 + xn2 + xn3, forn = 1, 2, , Prove that S2012is an integer
(2)12. 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 × × = 27 cubes with
sides of length cm If a denotes the number of small cubes of side-length 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 + = (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
(3)Solutions
1. Letx =
√
6+2√5+√6−2√5 √
20 Find the value of(1 + x
− x7
)2012311
Solution Notice that6 + 2√5 = (√5 + 1)2
and6 − 2√5 = (√5 − 1)2
,√20 = 2√5 then x = That is
(1 + x5− x7)2012311 =
2. Arrange the numbers p = 2√2
, q = 3, t = 21+
1
√2 in increasing order We have 21+ √2 ≥
21+12 =
2 = 2√2
Since√2 ≤ 32, then2√2
≤ 2√2 Notice that
t2
= 22+√2
≤ 22+3
2 ≤ 8√2
Thusq4
− t4
= 81 − 64 × < It follows that
p < t < q
3. LetABCD be a trapezoid with AD parallel to BC and BC = cm, DA = 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,=
1 3+
1
= Let M be the intersection of AC and BD
b b b b A D C B
By the Thales theorem we get OEBC + OF AD = OD BD + OC AC = OD BD + OB
BD = From this, OE =
1 BC +
1
AD Likewise,
1 OF =
1 BC +
1
AD Hence,OE = OF That is,
2 EF = OE = BC + AD = 3+ =
2 We getEF = cm
4. What is the largest integer less than or equal to4x3
−3x, where x = 2(
3
p
2 +√3+p3 2 −√3)
Solution By using the identitya3
+ b3
+ 3ab(a + b) = (a + b)3
, we get
(2x)3
=
3
q
2 +√3 + q
2 −√3 3
= + 6x
(4)5. Letf (x) be a function such that f (x) + 2fx+2010 x−1
= 4020 − x for all x 6= Find the value off (2012)
Solution Letu = x+2010x−1 thenx = u+2010u−1 Thus we have
f u + 2010 u −
+ 2f (u) = 4020 − u + 2010 u −
Interchangingu with x gives
f x + 2010 x −
+ 2f (x) = 4020 − x + 2010 x −
Leta = f (x), b = fx+2010 x−1
Solving the system
a + 2b = 4020 − x, b + 2a = 4020 − x + 2010 x − fora in terms of x gives
a = f (x) =
8040 − 4020 + 2x −2x + 4020 x −
=
4020 + 2x −4020 + 2x x −
Hence,
f (2012) =
8044 − 8044 2011
= 2680
6. For everyn = 2, 3, , let
An=
1 − 1 + 21
×
1 −1 + + 31
× · · · ×
1 −
1 + + · · · + n
Determine all positive integersn such that A1n is an integer
Solution Thek-th summand of the product has the form
ak= −
1
(k + 1)(k + 2) =
k(k + 3)
(k + 1)(k + 2), k = 1, 2, · · · , n −
from which we get
An=
n + 3n
and hence
An = −
6
(5)7. Prove thata = 1
| {z }
2012
5 | {z }
2011
6 is a perfect square
Solution Letp =
| {z }
2012
Then102012
= 9p + Hence,
a = p(9p + 1) + 5p + = (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
Solution We need to find the maximum value off (x, y)
f (x, y) = |x − y| + |x3− y3|
whenx, y vary satisfying the restriction x2
+ y2
=
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)
≤ − 2xy + + xy + + xy3 3
=
3
Hence,
f (x, y) ≤
r Equality occurs when
xy = −1 3, x
2
+ y2
=
This simultaneous equations are equivalent to
(6)Solving forx
x2−√x 3−
1 =
∆ = 3+
4 =
5 3, that is
x =
1 √
3− r
5
!
, x =
1 √
3 + r
5
!
Therefore, the value ofm3
is 53 q
5
3 Hence,mmax=
q
5
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 = cm, BP = cm, find the value of AM/BN
Solution Notice that ∠M P 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.N B = P M2
= P N2
Hence M AN B = M A2
M A.N B = M A2
P N2 =
P A2
P B2 =
32
42 =
9 16
10. Suppose that the equationx3
+ px2
+ qx + = 0, with p, q being some rational numbers, has three real roootsx1, x2, x3, wherex3 = +
√
5 Find the values of p, q
Solution Sincex = +√5 is one root of the equation, we get x − = √5 from which we
get a quadratic polynomialx2
− 4x − = by squaring
(x + α)(x2− 4x − 1) = x3+ px2+ qx + =
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 = has three real roots x1, x2, x3wherep, q, r
are integers LetSn= xn1 + x n + x
n
3, forn = 1, 2, , Prove that S2012is an integer
Solution By the Vieta theorem we getx1+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 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
(7)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 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 × × = 27 cubes with
sides of length cm If a denotes the number of small cubes of side-length 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.
Solution Just count from the diagram of the problem, we geta = 1, b = 4, c = 12, d =
Hence,a − b − c + d = −7.
14. Sovle the equation in the set of integers16x + = (x2
− y2)2.
Solution Since the right hand side is non-negative we have deduce that 16x + ≥ That
is,x take positive integers only Therefore, (x2
− y2
)2
≥ 1, or |x − y|2
|x + y|2
≥ That is, x2
≥
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 + ≥ Since x ∈ Z, we get x ≥ 0, which implies that 16x + ≥ Hence, (x2
− y2)2
≥ Thus, (x − y)2
≥ Now that
16x + = (x2
− y2
)2
= (x − y)2
(x + y)2
≥ x2
From this we obtain the inequality, x2
− 16x − < Solving this inequality gives x ∈ {0, 1, · · · , 16} In addition, 16x + is a perfect square, we get x ∈ {0, 3, 5, 14} Only x = 0; 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.