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.
1
Q3. How many 0’s are there in the sequence x1 , x2 , . . . , x2014 , where
h n+1 i h n i
− √
, 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
h 2015 i h 1 i h√
i
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)
So an =
8 otherwise
Hence 2014th term is 8 for 2014 − 4 = 2010 = 0(mod 3).
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 4EAB ∼ 4EM 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 6= 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 k 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
r +r
1
1
2
M N k=
O1 A + O2 B =
.
2
2
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
S=
3n+1
X
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
X
1
1
1
+
+
S=
2n + 1 k=1 2n + 1 + k 2n + 1 − k
n
1
2 X
=
+
2n + 1 2n + 1 k=1
1
k 2
1−
2n + 1
.
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
1
1−
k
2n + 1
2 > 1 +
2 < 1 + 2
5
k 2
2n + 1
k 2
.
2n + 1
1
. So we
2n + 1
Hence
n
n
k 2 i
k 2 i
1
2 Xh
1
2 Xh
+
1+
<S<
+
1+2
2n + 1 2n + 1 k=1
2n + 1
2n + 1 2n + 1 k=1
2n + 1
n
n
X
X
2
4
2
⇔1+
k <S <1+
k2
(2n + 1)3 k=1
(2n + 1)3 k=1
⇔1+
n(n + 1)
2 n(n + 1)
<S <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,
1 1 1
1
− 1 + + ··· +
A = 1 + + ··· +
2
2014
2
671
1 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.
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
.
2
2
2
2
follows k ..3 and k = 3n, n ∈ Z. Hence,
x + 3k = 28n and x + 27n = 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 k 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
9
Q15. Let a1 , a2 , . . . , a9 ≥ −1 and a31 + a32 + · · · + a39 = 0. Determine the greatest
value of M = a1 + a2 + · · · + a9 .
1 2
Solution. For a ≥ −1, we find (a + 1) a −
≥ 0. It follows
2
1
3
a3 − a + ≥ 0 ⇔ 3a ≤ 4a3 + 1, ∀a ≥ −1.
4
4
Hence
3(a1 + a2 + · · · + a9 ) ≤ 4(a31 + a32 + · · · + a39 ) + 9 = 9.
n
1o
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.
—————————————–
10
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 + y 2 − xy = 2
x4 + y 4 + x2 y 2 = 8
The value of P = x8 + y 8 + x2014 y 2014 is
(A): 46; (B): 48; (C): 50; (D): 52; (E) None of the above.
Answer. (B)
We have
x2 + y 2 = 2 + xy.
It follows
x4 + y 4 + 2x2 y 2 = x2 y 2 + 4xy + 4.
Equivalently,
x4 + y 4 + x2 y 2 = 4(xy + 1),
8 = 4(xy + 1)) ⇔ xy + 1 = 2 ⇔ xy = 1
and then x4 + y 4 = 7, xy = 1.
Hence x8 + y 8 = (x4 + y 4 )2 − 2x2 y 2 = 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)
− 11 = 44.
2
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.
1
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
1
1
= 2 + 2,
p
a
b
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
1
1
= 2 2 + 2 2.
p
mr
ms
It follows
p(r2 + s2 ) = mr2 s2 .
.
.
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
(
0, if n − 4 = 1 (mod 3)
an =
8, otherwise.
Note that 2014 − 4 = 2010 = 0 (mod 3). Thus, the answer is 8.
2
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 1 1
1 1 1
+ + > + + .
x y z
a b c
Solution.
Let AD be bisector of angle BAC and D belongs to BC. Draw DE k AB, E belongs
EC
ED
=
and
to AC. Then triangle EAD is iscoceles and DE = EA = d. We have
AB
CA
AE ED
EC AE
d d
1 1
1
then
+
=
+
= 1. It follows + = 1 ⇔ + = . Hence
AC
AB
CA CA
b c
b c
d
11 1
1
1
+
=
<
(1) since 2d > AD.
2 b c
2d
x
Similarly,
11 1 1
+
< ,
(2)
2 c a y
1
1 1 1
+
< .
(3)
2 c a
y
From (1), (2) and (3) we find
1 1 1
1 1 1
+ + > + + ,
x y z
a b c
which was to be proved.
Q7. Determine the integral part of A, where
1
1
1
A=
+
+ ··· +
.
672 673
2014
Solution. Consider the sum
3n+1
X 1
S=
.
k
k=n+1
Note that there are 2n + 1 terms in the sum and the middle term is
write the sum in the form
n
X
1
1
1
S=
+
+
2n + 1 k=1 2n + 1 + k 2n + 1 − k
3
1
. So we can
2n + 1
n
=
1
2 X
+
2n + 1 2n + 1 k=1
1
k 2
1−
2n + 1
.
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
1
1−
k
2n + 1
2 > 1 +
2 < 1 + 2
k 2
2n + 1
k 2
.
2n + 1
Hence
n
n
k 2 i
k 2 i
2 Xh
1
2 Xh
1
+
1+
<S<
+
1+2
2n + 1 2n + 1 k=1
2n + 1
2n + 1 2n + 1 k=1
2n + 1
n
n
X
X
2
4
2
⇔1+
k <S <1+
k2
3
3
(2n + 1) k=1
(2n + 1) k=1
⇔1+
2 n(n + 1)
n(n + 1)
<S <1+
.
2
3(2n + 1)
3 (2n + 1)2
It is easy to check that
2
n(n + 1)
1
≤
< , ∀n ≥ 1.
2
9
(2n + 1)
4
This leads to
7
29
< S < and then [S] = 1 and [A] = 1.
27
6
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.
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.
(1).
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
4
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
EC
1
CM
it follows
=
=
and CM
AB
EB
2
a
5a
a+a+ = .
2
2
Let h, h1 , h2 be heights of the given
then
h1
h2
of the parallelogram. Since ∆EAB ∼ ∆EM C,
a
= . Similarly, we have N D = a, then M N =
2
parallelogram, ∆OM N and ∆OAB, respectively,
=
MN
5
= .
AB
2
5
It follows
h1
h1
5
=
= .
h
h1 + h2
7
Hence h1 =
5h
. These follow that
7
SM ON =
1
1 5a 5h
25ah
25
× M N × h1 = ×
×
=
= S.
2
2
2
7
28
28
Q11. Find all pairs of integers (x, y) satisfying the following equality
8x2 y 2 + x2 + y 2 = 10xy.
Solution. We have 8x2 y 2 + x2 + y 2 = 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−1 b−1
+
+
≥ 0.
c
b
a
Solution. We prove the following inequality
a c b
+ + ≥ ab + bc + ca.
c b a
6
(1)
Indeed, from inequality (c − 1)2 (2c + 1) ≥ 0 for c > 0, it follows
2c3 + 1 ≥ 3c2 ⇔
1
a 2c
+ 2c2 ≥ 3c ⇔ ab + 2c2 ≥ 3abc2 ⇔ +
≥ 3ac.
c
c
b
Similarly,
c 2b
b 2a
+
≥ 3bc,
+
≥ 3ab.
b
a
a
c
These three inequalities imply
a c b
1 1 1
+ + ≥ ab + bc + ca = + + .
c b a
c a b
Hence
a c b
1 1 1
+ + ≥ + + ,
c b a
c a b
which is
a−1 c−1 b−1
+
+
≥ 0.
c
b
a
The equality holds iff a = b = c = 1.
Q14. Let be given a < b < c and f (x) =
c(x − a)(x − b) a(x − b)(x − c) b(x − c)(x − a)
+
+
.
(c − a)(c − b)
(a − b)(a − c)
(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 a31 + a32 + · · · + a39 = 0. Determine the maximal value of
M = a1 + a2 + · · · + a9 .
Solution.
1 2
For a ≥ −1, we find (a + 1) a −
≥ 0. It follows
2
3
1
a3 − a + ≥ 0 ⇔ 3a ≤ 4a3 + 1, ∀a ≥ −1.
4
4
Hence
3(a1 + a2 + · · · + a9 ) ≤ 4(a31 + a32 + · · · + a39 ) + 9 = 9.
n
1o
So M ≤ 3. The equality holds for a1 , a2 , . . . , a9 ∈ − 1,
. For example, we can choose
2
1
a1 = −1, a2 = a3 = · · · = a9 = , then a31 + a32 + · · · + a39 = 0 and a1 + a2 + · · · + a9 = 3.
2
Thus, max M = 3.
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