A right-angled triangle has property that, when a square is drawn externally on each side of the triangle, the six vertices of the squares that are not vertices of the triangle are concy[r]
Trang 1Hanoi Open Mathematical Competition 2015
Senior Section Important:
Answer to all 15 questions
Write your answers on the answer sheets provided
For the multiple choice questions, stick only the letters (A, B, C, D or E) of your choice
No calculator is allowed
Question 1 The sum of all even positive intergers less than 100 those are not divisible by 3 is
(A): 938; (B): 940; (C): 1634; (D): 1638; (E): None of the above
Answer: C
Question 2 A regular hexagon and an equilateral triangle have equal perimeter
If the area of the triangle is 4√
3 square units, the area of the hexagon is (A): 5√
3; (B): 6√
3; (C): 7√
3; (D): 8√
3; (E): None of the above
Answer: B
Question 3 Suppose that a > b > c > 1 One of solutions of the equation
(x − a)(x − b) (c − a)(c − b) +
(x − b)(x − c) (a − b)(a − c) +
(x − c)(x − a) (b − c)(b − a) = x
is
(A): -1; (B): -2; (C): 0; (D): 1; (E): None of the above
Answer: D
Question 4 Let a, b, c and m (0 ≤ m ≤ 26) be integers such that
a + b + c = (a − b)(b − c)(c − a) = m (mod 27)
then m is
(A): 0; (B): 1; (C): 25; (D): 26; (E): None of the above
Answer: A
Question 5 The last digit of number 20172017− 20132015 is
(A): 2; (B): 4; (C): 6; (D): 8; (E): None of the above
Answer: E
Trang 2Question 6 Let a, b, c ∈ [−1, 1] such that 1 + 2abc ≥ a2+ b2+ c2 Prove that
1 + 2a2b2c2 ≥ a4+ b4+ c4
Solution The constraint can be written as
(a − bc)2 ≤ (1 − b2)(1 − c2) (1) Using the Cauchy inequality, we have
(a + bc)2 ≤ (|a| + |bc|)2 ≤ (1 + |b||c|)2 ≤ (1 + b2)(1 + c2)
Multiplying by (1), we get
(a − bc)2(a + bc)2 ≤ (1 − b2)(1 + b2)(1 − c2)(1 + c2)
⇔ (a2 − b2c2)2 ≤ (1 − b2)(1 + b2)(1 − c2)(1 + c2)
⇔ (a2− b2c2)2 ≤ (1 − b4)((1 − c4)
⇔ 1 + 2a2b2c2 ≥ a4+ b4+ c4
Question 7 Solve equation
x4 = 2x2+ [x], (2) where [x] is an integral part of x
Solution
We have
(2) ⇔ [x] = x2 x2− 2 Consider the case x2 ≤ 2, then −√2 ≤ x ≤ √
2 and [x] ≤ 0 It follows [x] ∈ {−1; 0}
If [x] = 0, then from (2) we find x = 0 as a solution
If [x] = −1, then from (2) we find x = −1 as a solution
Now we suppose that x2 > 2 It follows from (2), [x] > 0 and then x >√
2 Hence
x2(x2− 2) = [x]
x ≤ 1 and x2− 2 ≤ 1
x < 1 It follows x <
√
3, i.e √
2 < x <√
3
It means that [x] = 1 and then x =p1 +√
2 is a solution of the equation Question 8 Solve the equation
(x + 1)3(x − 2)3+ (x − 1)3(x + 2)3 = 8(x2− 2)3 (3)
Solution Rewrite equation (1) in the form
(x2 − x − 2)3+ (x2+ x − 2)3 = (2x2− 4)3 (4)
Trang 3Factoring the sum of cubes on the right side of the equation (1), we find that one factor is (2x2− 4), thus, two solutions of the equation is x = ±√2
Now we rewrite the equation (2) as
(x2− x − 2)3 = (2x2− 4)3− (x2+ x − 2)3 (5) Factoring the difference of cubes on the left side of the equation (), we find that one factor is (x2 − x − 2), thus, two solution of the equation is x = −1, x = 2
Finally, we rewrite (4) as
(x2+ x − 2)3 = (2x2− 4)3 − (x2− x − 2)3 Factoring again, we see that one factor is (x2 + x − 2) Thus, two solutions of the equation are x = 1, x = −2
Since the left hand side of the equation 8(x2− 2)3− (x + 1)3(x − 2)3− (x − 1)3(x + 2)3 = 0 is a polynomial of degree 6, then it has at most 6 roots, and we have them Hence,
8(x2− 2)3− (x + 1)3(x − 2)3 − (x − 1)3(x + 2)3 = 6(x2− 2)(x2− 1)(x2− 4)
Question 9 Let a, b, c be positive numbers with abc = 1 Prove that
a3+ b3 + c3+ 2[(ab)3 + (bc)3+ (ca)3] ≥ 3(a2b + b2c + c2a)
Solution Asume that a = max{a, b, c} then a ≥ b ≥ c > 0 or a ≥ c ≥ b > 0 and
a3+ b3+ c3− (a2b + b2c + c2a) = (a − b)(a2− c2) + (b − c)(b2− c2) ≥ 0 Hence
a3+ b3+ c3 ≥ a2b + b2c + c2a (6) Since 1
c = max
n1
a,
1
b,
1 c
o
or 1
b = max
n1
a,
1
b,
1 c
o , then
1
c3 + 1
b3 + 1
a3 ≥ 1
c2b +
1
b2a +
1
a2c. Since abc = 1, this can be written as
(ab)3+ (bc)3 + (ca)3 ≥ a2b + b2c + c2a (7) (6) and (7) together imply the proposed inequality
Question 10 A right-angled triangle has property that, when a square is drawn externally on each side of the triangle, the six vertices of the squares that are not vertices of the triangle are concyclic Assume that the area of the triangle is 9 cm2 Determine the length of sides of the triangle
Trang 4Solution We have OJ = OD = OG = radius of the circle Let the sides of ∆ABC
be a, b, c Then
OJ2 = OM2+ M J2 =b + a
2
2
+b 2
2
= b2+ ab +1
4(a
2+ b2)
OD2 = ON2+ N D2 =a + b
2
2
+a 2
2
= a2+ ab + 1
4(a
2+ b2)
OG2 = OL2+ LG2 = c2 +c
2
2
= 5
4c
2 = 5
4(a
2+ b2) = a2+ b2+1
4(a
2+ b2) Comparing these right-hand sides, we get
b2+ ab = a2+ ab = a2 + b2 ⇔ a = b
It means that the given triangle with the desired property is the isosceles right triangle and then 1
2ab = 9 ⇔ a = b = 3
√
2, c = 6 units
Question 11 Given a convex quadrilateral ABCD Let O be the intersection point
of diagonals AC and BD and let I, K, H be feet of perpendiculars from B, O, C to
AD, respectively Prove that
AD × BI × CH ≤ AC × BD × OK
Solution Draw AE⊥BD (E ∈ BD)
We have SABD = BI × AD
2 =
AE × BD
2 Then BI × AD = AE × BD It follows BI × AD ≤ AO × BD (AE ≤ AO) and BI.AD ≤ AC × BD ×AO
AC. Moreover, we have OK k CH then AO
AC =
OK
CH and BI ×AD ≤ AC ×BD ×
OK
CH.
It follows BI × AD × CH ≤ AC × BD × OK
The equality holds if quadrilateral ABCD has two perpendicular diagonals
Question 12 Give an isosceles triangle ABC at A Draw ray Cx being perpen-dicular to CA, BE perpenperpen-dicular to Cx (E ∈ Cx) Let M be the midpoint of BE, and D be the intersection point of AM and Cx Prove that BD ⊥ BC
Solution Let K be intersection point of DB and AC
Since BE⊥CD; CK⊥CD then BE k CK
In ∆DAC we see M E k AC so
M E
AC =
DM
In ∆DAK we see M B k AK so
M B
AK =
DM
Trang 5From (8) and (9), we get M E
AC =
M B
AK This and equality M B = M E together imply
AK = AC = AB and then BA = CK
2 . Note that BA is a median line ∆BKC and BA = CK
2 then ∆BKC is a right triangle at B Hence BD⊥BC
Question 13 Let m be given odd number, and let a, b denote the roots of equation
x2 + mx − 1 = 0 and c = a2014 + b2014, d = a2015 + b2015 Prove that c and d are relatively prime numbers
Solution Since a2+ ma − 1 = 0 then a 6= 0 and
an+2 = −man+1+ an ∀n ∈ N
Similarly, bn+2 = −mbn+1+ bn; ∀n ∈ N
Hence, the sequence xn, n ∈ N are defined as
(
x0 = 2, x1 = −m
xn+2= −mxn+1+ xn ∀n ∈ N
It is easy to see all xn are integers Hence, c, d are integers
Now we prove (xn; xn+1) = 1 for every n ∈ N
For n = 0, x0 = 2 and m is odd then (x0; x1) = (2; −m) = 1
Suppose that (xk; xk+1) = 1 for k ≥ 0 and (xk+1; xk+2) > 1 Let p be a prime factor of xk+1 and xk+2, then from xk= xk+2+mxk+1, it follows p is a prime factor of
xk It means that p | (xk; xk+1) = 1, absurd Hence (xk+1; xk+2) = 1 and (c, d) = 1 Question 14 Determine all pairs of integers (x; y) such that
2xy2+ x + y + 1 = x2+ 2y2+ xy
Solution We have
2xy2+ x + y + 1 = x2+ 2y2+ xy
⇔ 2y2(x − 1) − y(x − 1) − x(x − 1) = −1
⇔ (x − 1)(2y2− y − x) = −1
Since x; y are integers then x − 1 and 2y2− y − x are divisors of -1
Case 1
(
x − 1 = 1 2y2− y − x = −1 ⇔
(
x = 2
y = 1
x = 2
y = −1
2(absurd)
Trang 6- Case 2.
(
x − 1 = −1 2y2− y − x = 1 ⇔
x = 0
y = 1
x = 0
y = −1
2(absurd) Hence all integral pairs (x; y) are (2; 1); (0 ; 1)
Question 15 Let the numbers a, b, c, d satisfy the relation a2+ b2+ c2+ d2 ≤ 12 Determine the maximum value of
M = 4(a3+ b3+ c3+ d3) − (a4+ b4+ c4+ d4)
Solution Note that x2(x − 2)2 ≥ 0 for each real x This inequality can be rewritten as 4x3− x4 ≤ 4x2 It follows that
(4a3− a4) + (4b3− b4) + (4c3− c4) + (4d3− d4) ≤ 4(a2+ b2+ c2+ d2) = 48, The equality holds for (a, b, c, d) = (2, 2, 2, 0).(2, 2, 0, 2), (2, 0, 2, 2), (0, 2, 2, 2) Hence, max M = 48