Hanoi Opening Mathematical Olympiad

File này là đề thi Olympiad Toán Hà Nội Mở rộng năm 2014 dành cho các em học sinh lớp 8 và lớp 10. Nếu cảm thấy tự tin về vốn kiến thức toán học cũng như anh ngữ của mình, kì thi Hanoi Opening Mathematical Olympiad, hay còn dc viết tắt là HOMC, sẽ là 1 cơ hội thuận lợi cho các bạn cọ xát, thử thách bản thân và đồng thời cũng có thể thấy dc vốn kiến thức của mình đang ở ngưỡng nào mà phấn đấu thêm. Không những thế, kì thi HOMC còn giúp cho các bạn làm quen với cách học và giải toán bằng tiếng anh, từ đó các bạn sẽ cảm thấy tự tin hơn khi đối mặt với các kì thi toán quốc tế, hoặc đơn giản hơn là khi đọc các tài liệu nước ngoài Chúc các bạn sẽ vững tin và bay cao trong sự nghiệp của mình

Hanoi Open Mathematical Competitions 2014

Junior Section Answers and solutions

Sunday, 23 March 2014 08h30-11h30

Q1 Let the numbers x and y satisfy the conditions

(

x2 + y2− xy = 2

x4 + y4+ x2y2 = 8

The value of P = x8+ y8+ x2014y2014 is

(A): 46; (B): 48; (C): 50; (D): 52; (E) None of the above

Answer (B)

We have

x2+ y2 = 2 + xy

It follows

x4+ y4+ 2x2y2 = x2y2+ 4xy + 4

Equivalently,

x4+ y4+ x2y2 = 4(xy + 1),

8 = 4(xy + 1)) ⇔ xy + 1 = 2 ⇔ xy = 1 and then x4+ y4 = 7, xy = 1

Hence x8 + y8 = (x4+ y4)2− 2x2y2 = 49 − 2 = 47 and we find P = 47 + 1 = 48

Q2 How many diagonals does 11-sided convex polygon have?

(A): 43; (B): 44; (C): 45; (D): 46; (E) None of the above

Answer (B)

The number of diagonals of 11-sided convex polynom is difined by

11(11 − 1)

2 − 11 = 44

Q3 How many zeros are there in the last digits of the following number

P = 11 × 12 × · · · × 88 × 89

(A): 16; (B): 17; (C): 18; (D): 19; (E) None of the above

Answer (C)

We have

P = 518· A, where A is the integer containing sufficiently many even factors Thus, the answer is 18

Q4 If p is a prime number such that there exist positive integers a and b such that

1

p =

1

a2 + 1

b2,

then p is

(A): 3; (B): 5; (C): 11; (D): 7; (E) None of the above

Answer (E) Write m := (a, b) It implies

(

a = mr

b = ms

We then have

1

p =

1

m2r2 + 1

m2s2

It follows

p(r2+ s2) = mr2s2

Since r2+ s2 is not divisable by neither r2 nor s2, we have p .r2 and p .s2 As p is a prime number, r2 = s2 = 1 We find m = 2 and p = 2 So, the answer is (E)

Q5 The first two terms of a sequence are 2 and 3 Each next term thereafter is the sum

of the nearestly previous two terms if their sum is not greather than 10, 0 otherwise The 2014th term is

(A): 0; (B): 8; (C): 6; (D): 4; (E) None of the above

Answer (B)

We have the sequence as follows

2, 3, 5, 8, 0, 8, 8, 0, 8, 8,

Starting from the fifth term, the sequence is periodic with the period 3 We then have

an=

(

0, if n − 4 = 1 (mod 3)

8, otherwise

Note that 2014 − 4 = 2010 = 0 (mod 3) Thus, the answer is 8

Q6 Let a, b, c be the length sides of a given triangle and x, y, z be the sides length of bisectrices, respectively Prove the following inequality

1

x +

1

y +

1

z >

1

a +

1

b +

1

c. Solution

Let AD be bisector of angle BAC and D belongs to BC Draw DE k AB, E belongs

to AC Then triangle EAD is iscoceles and DE = EA = d We have ED

AB =

EC

CA and then AE

AC +

ED

AB =

EC

CA +

AE

CA = 1 It follows

d

b +

d

c = 1 ⇔

1

b +

1

c =

1

d Hence 1

2

1

b +

1

c



= 1 2d <

1

x (1) since 2d > AD.

Similarly,

1

2

1

c +

1

a



< 1

y, (2) 1

2

1

c +

1

a



< 1

y. (3) From (1), (2) and (3) we find

1

x +

1

y +

1

z >

1

a +

1

b +

1

c, which was to be proved

Q7 Determine the integral part of A, where

A = 1

672 +

1

673 + · · · +

1

2014. Solution Consider the sum

S =

3n+1

X

k=n+1

1

k.

Note that there are 2n + 1 terms in the sum and the middle term is 1

2n + 1 So we can write the sum in the form

S = 1 2n + 1+

n

X

k=1

 1 2n + 1 + k +

1 2n + 1 − k



= 1 2n + 1+

2 2n + 1

n

X

k=1

1

1 − k 2n + 1

2

On the other hand, using the inequalities

1 + a < 1

1 − a < 1 + 2a for 0 < a <

1

2,

we get

1

1 − k 2n + 1

2 > 1 + k

2n + 1

2

and

1

1 − k 2n + 1

2 < 1 + 2 k

2n + 1

2

Hence

1

2n + 1+

2 2n + 1

n

X

k=1

h

1 + k 2n + 1

2i

< S < 1

2n + 1 +

2 2n + 1

n

X

k=1

h

1 + 2 k

2n + 1

2i

⇔ 1 + 2

(2n + 1)3

n

X

k=1

k2 < S < 1 + 4

(2n + 1)3

n

X

k=1

k2

⇔ 1 + n(n + 1)

3(2n + 1)2 < S < 1 +2

3

n(n + 1) (2n + 1)2

It is easy to check that

2

9 ≤ n(n + 1) (2n + 1)2 < 1

4, ∀n ≥ 1.

This leads to 29

27 < S <

7

6 and then [S] = 1 and [A] = 1.

Q8 Let ABC be a triangle Let D, E be the points out side of the triangle so that

AD = AB, AC = AE and ∠DAB = ∠EAC = 90◦ Let F be at the same side of the line

BC as A so that F B = F C and ∠BF C = 90◦ Prove that triangle DEF is a right-isoceles triangle

Solution

Note that CD ⊥ BE and ∆F EB = ∆F DC Hence, F E = F D (1)

On the other hand, ∠DF C = ∠EF B then ∠DF E = 900 (1)

(1) and (2) together imply |deltaDEF | is right-iscoceles, q.e.d

Q9 Determine all real numbers a, b, c such that the polynomial f (x) = ax2 + bx + c satisfies simultaneously the folloving conditions

(

|f (x)| ≤ 1 for |x| ≤ 1

f (x) ≥ 7 for x ≥ 2

Figure 1:

Solution Let g(x) = 2x2 − 1 − f (x), then deg g ≤ 2 Note that

g(−1) = 1 − f (−1) ≥ 0, g(0) = −1 − f (0) ≤ 0, g(1) = 1 − f (1) ≥ 0

and g(2) = 7 − f (2) ≤ 7 − 7 = 0 Hence the equation g(x) = 0 has at least 3 roots It follows g(x) ≡ 0 and then f (x) = 2x2− 1

Thus (a, b, c) = (2, 0, −1)

Q10 Let S be area of the given parallelogram ABCD and the points E, F belong to BC and AD, respectively, such that BC = 3CE, 3AD = 2AF Let O be the intersection of

AE and BF The straightlines AE and BF meet CD at points M and N, respectively Determine area of triangle M ON

Solution

Let a = AB and h be the height of the parallelogram Since ∆EAB ∼ ∆EM C,

it follows CM

AB =

EC

EB =

1

2 and CM =

a

2 Similarly, we have N D = a, then M N =

a + a +a

2 =

5a

2 . Let h, h1, h2 be heights of the given parallelogram, ∆OM N and ∆OAB, respectively, then

h1

h2

= M N

AB =

5

2.

It follows

h1

h =

h1

h1+ h2 =

5

7.

Hence h1 = 5h

7 These follow that

SM ON = 1

2× M N × h1 = 1

2 × 5a

2 × 5h

7 =

25ah

28 =

25

28S.

Q11 Find all pairs of integers (x, y) satisfying the following equality

8x2y2+ x2+ y2 = 10xy

Solution We have 8x2y2+ x2 + y2 = 10xy ⇔ 8xy (xy − 1) + (x − y)2 = 0 (1)

Consequently, if x; y is an integral root of the equation then xy (xy − 1) ≤ 0 ⇒ 0 ≤

xy ≤ 1

Because x; y are integral, there are two possibilities:

- If xy = 0 then from (1) we have a root x = y = 0

- If xy = 1 then from (1) we have other roots x = y = 1; x = y = −1

Thus the equation has three solutions (x, y) = (0, 0), (x, y) = (1, 1), (x, y) = (−1, −1)

Q12 Find a polynomial Q(x) such that (2x2 − 6x + 5)Q(x) is a polynomial with all positive coefficients

Solution Note that

(2x2− 6x + 5)(2x2+ 6x + 5)(4x4+ 6x2+ 25) = 16x8+ 164x4+ 625

Hence

(2x2−6x+5)(2x2+6x+5)(4x4+6x2+25)(x3+x2+x+1) = (16x8+164x4+625)(x3+x2+x+1)

= 16x11+16x10+16x9+16x8+164x7+164x6+164x5+164x4+625x3+625x2+625x+625

So we can choose Q(x) = (2x2+ 6x + 5)(4x4+ 6x2+ 25)(x3+ x2+ x + 1)

Q13 Let a, b, c > 0 and abc = 1 Prove that

a − 1

c +

c − 1

b +

b − 1

a ≥ 0

Solution We prove the following inequality

a

c +

c

b + b

a ≥ ab + bc + ca (1)

Indeed, from inequality (c − 1)2(2c + 1) ≥ 0 for c > 0, it follows

2c3+ 1 ≥ 3c2 ⇔ 1

c + 2c

2 ≥ 3c ⇔ ab + 2c2 ≥ 3abc2 ⇔ a

c +

2c

b ≥ 3ac

Similarly,

c

b +

2b

a ≥ 3bc, b

a +

2a

c ≥ 3ab

These three inequalities imply

a

c +

c

b +

b

a ≥ ab + bc + ca = 1

c +

1

a +

1

b. Hence

a

c +

c

b +

b

a ≥ 1

c +

1

a +

1

b, which is

a − 1

c +

c − 1

b +

b − 1

a ≥ 0

The equality holds iff a = b = c = 1

Q14 Let be given a < b < c and f (x) = c(x − a)(x − b)

(c − a)(c − b) +

a(x − b)(x − c) (a − b)(a − c) +

b(x − c)(x − a) (b − c)(b − a) . Determine f (2014)?

Solution Let g(x) = f (x) − x Then deg g ≤ 2 and g(a) = g(b) = g(c) = 0 Hence g(x) ≡ 0 and f (x) = x for all x ∈ R It follows f (2014) = 2014

Q15 Let a1, a2, , a9 ≥ −1 and a3

1+ a32+ · · · + a39 = 0 Determine the maximal value of

M = a1+ a2+ · · · + a9

Solution

For a ≥ −1, we find (a + 1)a − 1

2

2

≥ 0 It follows

a3− 3

4a +

1

4 ≥ 0 ⇔ 3a ≤ 4a3+ 1, ∀a ≥ −1

Hence

3(a1+ a2+ · · · + a9) ≤ 4(a31+ a32+ · · · + a39) + 9 = 9

So M ≤ 3 The equality holds for a1, a2, , a9 ∈n− 1,1

2

o For example, we can choose

a1 = −1, a2 = a3 = · · · = a9 = 1

2, then a

3

1+ a3

2+ · · · + a3

9 = 0 and a1+ a2+ · · · + a9 = 3 Thus, max M = 3

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