HOMCS2014 sol CS

Hanoi Open Mathematical Competition 2014
Senior Section
Answers and solutions

Q1. Let a and b satisfy the conditions
a3 − 6a2 + 15a = 9
b3 − 3b2 + 6b = −1
The value of (a − b)2014 is
(A): 1; (B): 2; (C): 3; (D): 4; (E) None of the above.
Answer. (A)
Write the system in the form
(a − 2)3 = 1 − 3a
(b − 1)3 = −2 − 3b

(1)
(2)

Subtracting (2) from (1) we find
(a − 2)3 − (b − 1)3 = 3 − 3(a − b)
and then
(a − b − 1)[(a − 2)2 + (a − 2)(b − 1) + (b − 1)2 + 3] = 0.
It follows a − b = 1 and A = 1.

Q2. How many integers are there in {0, 1, 2, . . . , 2014} such that
x
999
C2014
≥ C2014
.

(A): 15; (B): 16; (C): 17; (D): 18; (E) None of the above.

Answer. (C)
Note that
x
2014−x
C2014
= C2014
x+1
x
and C2014
< C2014
for x = 0, 1, . . . , 1006.
x
999
These imply C2014 ≥ C2014
that is equivalent to 999 ≤ x ≤ 1015.

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Q3. How many 0’s are there in the sequence x1 , x2 , . . . , x2014 , where
n
n+1
− √
, n = 1, 2, . . . , 2014.
xn = √
2015
2015
(A): 1128; (B): 1129; (C): 1130; (D): 1131; (E) None of the above.
Answer. (E)
It is easy to check that 0 ≤ xn ≤ 1. Hence xn ∈ {0, 1} and the number of all 1 in the

√
2015
1
given sequence is x1 + x2 + · · · + x2014 = √
− √
=
2015 . It easy
2015
2015
√
to check that 144 < 2015 < 145. Hence, the number of all 1 in the given sequence
is 144.
Thus, the number of all 0 in the given sequence 2014 − 144 > 1130.

Q4. Find the smallest positive integer n such that the number 2n + 28 + 211 is a
perfect square.
(A): 8; (B): 9; (C): 11; (D): 12; (E) None of the above.
Answer. (D)
2
2
For n > 8 we find 28 + 211 + 2n = (24 ) (1 + 8 + 2n−8 ) = (24 ) (9 + 2n−8 ) Hence,
we find n such that 2n−8 + 9 is a perfect square. Putting 2n−8 + 9 = k 2 , we get
(k − 3)(k + 3) = 2n−8 . It follows k − 3 and k + 3 are the powers of 2 and their
k+3=8
then k = 5 and n = 12.
difference is 6. So
k−3=2
Indded, 28 + 211 + 212 = 802 is a perfect square.
For n ≤ 8,
n ∈ {1; 2; 3; 4; 5; 6; 7; 8} .

The corresponding values 28 + 211 + 2n are not perfect squares.

Q5. The first two terms of a sequence are 2 and 3. Each next term thereafter is the
sum of the nearest previous two terms if their sum is not greater than 10 and is 0
otherwise. The 2014th term is
(A): 0; (B): 8; (C): 6; (D): 4; (E) None of the above.
Answer. (B)
The corresponding sequence is
2, 3, 5, 8, 0, 8, 8, 0, 8, 8, . . .
2


0 if n − 4 = 1(mod 3)
8 otherwise
Hence 2014th term is 8 for 2014 − 4 = 2010 = 0(mod 3).

So an =

Q6. Let S be area of the given parallelogram ABCD and the points E, F belong
to BC and AD, respectively, such that BC = 3BE, 3AD = 4AF. Let O be the
intersection of AE and BF. Each straightline of AE and BF meets that of CD at
points M and N, respectively. Determine area of triangle M ON.
Solution.

Let a = AB. Note that

EAB ∼

EM C. It follows

CM
CE
=
= 2.
AB
BE
This implies CM = 2a. By the same argument as above we obtain
DN = a/3.
Hence,
M N = a/3 + a + 2a = 10a/3.

(1)

Let h denote the height of the paralelogram, and let h1 and h2 the heights of
triangles OM N and OAB, respectively. We have
h1
MN
10
=
= .
h2
AB
3
We deduce

h1
h1
10
=
= .

h2
h1 + h2
3
3


We then have
h1 =

10
h
3

(2)

Combining (1) and (2) we receive
1 10a 10h
50
50
1
.
= ah = S.
SM ON = M N.h1 = .
2
2 3 13
39
39
The answer is

50

S.
39

Q7. Let two circles C1 , C2 with different radius be externally tangent at a point T .
Let A be on C1 and B be on C2 , with A, B = T such that ∠AT B = 900 .
(a) Prove that all such lines AB are concurrent.
(b) Find the locus of the midpoints of all such segments AB.
Solution.

Let O1 , O2 denote the centres of C1 , C2 whose radii are r1 , r2 , respectively. Without loss of generality we can assume that r1 < r2 . We have
∠AT B = 900 .
It implies
∠O1 T A + ∠O2 T B = 900 .
We deduce
∠O1 AB + O2 BA = ∠O1 AT + ∠T AB + ∠ABT + ∠O2 BT = 1800 .
Hence, O1 A

O2 B. Let X denote the intersection of AB and O1 O2 . We then have
XO1
r1
=
XO2
r2
4


which proves the conclusion.
b) Let M, N denote the midpoints of AB, O1 O2 , respectively. We have
1
r1 + r2

O1 A + O2 B =
.
2
2

MN =

We deduce that M is on the circle whose centre is N and the radius equals to
except for O1 , O2 .

r1 + r2
,
2

Q8. Determine the integral part of A, where
A=

1
1
1
+
+ ··· +
.
672 673
2014

Solution. Consider the sum

3n+1

S=

1
.
k
k=n+1

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

1
1
1
+
+
S=
2n + 1 k=1 2n + 1 + k 2n + 1 − k
n

1
2
=
+
2n + 1 2n + 1

k=1

1
k

1−
2n + 1

2

.

On the other hand, using the inequalities
1+a<

1
1
< 1 + 2a for 0 < a < ,
1−a
2

we get
1
1−
and

k
2n + 1

2

1
k
1−
2n + 1

2

k
2n + 1

2

>1+

k
2n + 1

2

<1+2

5

.

1
. So we
2n + 1


Hence
1
2
+

2n + 1 2n + 1

n

k
2n + 1

1+
k=1

2
⇔1+
(2n + 1)3
⇔1+

2


1
2
+
2n + 1 2n + 1

n

4
k (2n + 1)3
k=1

n

1+2
k=1

k
2n + 1

2

n

2

k2
k=1

n(n + 1)
2 n(n + 1)
.
2
3(2n + 1)
3 (2n + 1)2

It is easy to check that
n(n + 1)
2
1

≤
< , ∀n ≥ 1.
2
9
(2n + 1)
4
29
7
< S < and then [S] = 1 and [A] = 1.
27
6
Remark. Note that
This leads to

A<

1
1343
× (2014 − 672 + 1) =
< 2.
672
672

On the other hand,
A= 1+

1
1
1
1

+ ··· +
− 1 + + ··· +
2
2014
2
671

1 1 1 1
1 1 1 1
1
1
1
1
− − +
+
− − +
+· · ·+
−
−
+
> 1.
2 3 3 4
5 6 6 7
2012 2013 2013 2014
Hence 1 < A < 2.
= 1+

Q9. Solve the system
16x3 + 4x = 16y + 5
16y 3 + 4y = 16x + 5

Solution. Subtracting the second equation for the first, we have
16(x − y)(x2 + xy + y 2 ) + (x − y) = 16(y − x).
Therefore
4(x − 4)(4x2 + 4xy + 4y 2 + 5) = 0.

6


Hence, y = x. We then have
5
1
1
⇔ 4x3 − 3x =
2+
.
4
2
2
1 √
1
3
The last equation has a unique solution x =
.
2+ √
3
2
2
4x3 − 3x =

Q10. Find all pairs of integers (x, y) satisfying the condition

12x2 + 6xy + 3y 2 = 28(x + y).
Solution.
12x2 + 6xy + 3y 2 = 28 (x + y)
⇔ 3 4x2 + 2xy + y 2 = 28 (x + y) .

(1)

.
Since 3 and 28 are prime relative integers then x + y ..3 and then x + y = 3k
with k ∈ Z. Form (1), we get 3x2 + (x + y)2 = 28k and then 3x2 + 9k 2 = 28k. It
.
follows k ..3 and k = 3n, n ∈ Z. Hence, x2 + 3k 2 = 28n and x2 + 27n2 = 28n. Thus
28
28
− n ≥ 0 and 0 ≤ n ≤
. It follows n = 0 and
x2 = n (28 − 27n) ≥ 0 ⇔ n
27
27
n = 1.
x=0
For n = 0 we have
x+y =0
It follows x = y = 0.
x2 = 1
x = 1; y = 8
For n = 1 then k = 3 and we have
⇒
x
= −1; y = 10

x+y =9
Hence the equation has 3 integral roots x = y = 0; x = 1, y = 8 and x = −1, y =
10.

Q11. Determine all real numbers a, b, c, d such that the polynomial f (x) = ax3 +
bx2 + cx + d satisfies simultaneously the folloving conditions
|f (x)| ≤ 1 for |x| ≤ 1
f (2) = 26
Consider the polynomial g(x) = 4x3 − 3x − f (x) of degree ≤ 3 and
1
1
g(−1) ≤ 0, g −
≥ 0, g
≤ 0, g(1) ≥ 0, then g(x) = 0 has at least 3 real roots
2
2
Solution.

7


in [−1, 1]. On the other hand, g(2) = 26 = 26 + f (2). It follows g(x) has at least 4
roots and then g(x) ≡ 0 and f (x) = 4x3 − 3x.
Hence (a, b, c, d) = (4, 0, −3, 0).

Q12. Given a rectangle paper of size 15 cm × 20 cm, fold it along a diagonal.
Determine the area of the common part of two halfs of the paper?
Solution.

We have AC = 25cm. EC 2 = CH.CA then 225 = CH.25 and CH = 9cm.

It is easy to see that DE AC, AK = CH = 9cm. Hence DE = AC2 × CH =
2518 = 7 cm.
Note that
SACF
FC
=
= df rac257
SAF D
FD
SACF
then
= df rac2532. It follows
SACD
SACF =

25 1
25 1
1875 2
25
× SACD =
× SABCD =
× × 15 × 20 =
cm .
32
32 2
32 2
16

Q13. Let a, b, c satisfies the conditions



5 ≥ a ≥ b ≥ c ≥ 0
a+b≤8


a + b + c = 10
Prove that a2 + b2 + c2 ≤ 38.
8


Solution. Using the inequality x2 ≥ y 2 + 2y(x − y) for all x, y ∈ R we find

2
2

5 ≥ a + 2a(5 − 2)
32 ≥ b2 + 2b(3 − b)

 2
2 ≥ c2 + 2c(2 − c)
Hence
38 = 52 + 32 + 22 ≥ a2 + b2 + c2 + 2[a(5 − a) + b(3 − b) + c(2 − c)]
= a2 + b2 + c2 + 2[(a − b)(5 − a) + (b − c)(3 − b) + (a + b + c)(2 − c)] ≥ a2 + b2 + c2 ,
which was to be proved.
Remark. We can use the equality a2 +b2 +c2 = a(a−b)+(b−c)(a+b)+c(a+b+c) to
establish the inequality a2 +b2 +c2 ≤ 5(a−b)+8(b−c)+10c = 2(a+b+c)+(a+b)+2a ≤
2 × 10 + 8 + 2 × 5 = 38.

Q14. Let ω be a circle with centre O, and let be a line that does not intersect ω.
Let P be an arbitrary point on . Let A, B denote the tangent points of the tangent

lines from P . Prove that AB passes through a point being independent of choosing
P.
Solution.

We have Draw the perpendicular from O to the line . Let M be

intersection of AB and OQ. Note that A, P, Q, B and O are concyclic. Hence
∠OQB = ∠BAO = ∠ABO.
Consider triangles OBM and OBQ, we have ∠BOQ is common, ∠OBM =
∠OQB, then triangle OBM ∼ ∆OQB then OM × OQ = OB 2 constant then the
point M is independent of the choice of P , q.e.d

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Q15. Let a1 , a2 , . . . , a9 ≥ −1 and a31 + a32 + · · · + a39 = 0. Determine the greatest
value of M = a1 + a2 + · · · + a9 .
Solution. For a ≥ −1, we find (a + 1) a −

1
2

2

≥ 0. It follows

1
3
a3 − a + ≥ 0 ⇔ 3a ≤ 4a3 + 1, ∀a ≥ −1.
4

4
Hence
3(a1 + a2 + · · · + a9 ) ≤ 4(a31 + a32 + · · · + a39 ) + 9 = 9.
1
So M ≤ 3. The equality holds for a1 , a2 , . . . , a9 ∈ − 1,
. For example, we can
2
1
choose a1 = −1, a2 = a3 = · · · = a9 = , then a31 + a32 + · · · + a39 = 0 and
2
a1 + a2 + · · · + a9 = 3.
Thus, max M = 3.
—————————————–

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