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]
(1)Hanoi 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 The sum of all even positive intergers less than 100 those are not divisible by is
(A): 938; (B): 940; (C): 1634; (D): 1638; (E): None of the above Answer: C
Question 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 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 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
(2)Question Let a, b, c∈[−1,1] such that + 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 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 findx= as a solution If [x] =−1,then from (2) we find x=−1 as a solution
Now we suppose thatx2 >2.It follows from (2), [x]>0 and thenx >√2.Hence
x2(x2−2) = [x]
x ≤1 and x
2−2≤
x <1.It follows x < √
3, i.e √2< x <√3 It means that [x] = and then x=p1 +√2 is a solution of the equation Question 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
(3)Factoring 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 = is a polynomial of degree 6, then it has at most 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 Let a, b, c be positive numbers withabc= 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
c = max
n1
a, b,
1 c
o
or
b = max
n1
a, b,
1 c
o
,then
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 cm2.
(4)Solution We haveOJ =OD =OG= radius of the circle Let the sides of ∆ABC be a, b, c.Then
OJ2 =OM2+M J2 =b+a
2
+b
2
=b2+ab+1 4(a
2+b2).
OD2 =ON2+N D2 =a+ b
2
+a
2
=a2+ab+1 4(a
2+b2).
OG2 =OL2+LG2 =c2 +c
2
= 4c
2 =
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
2ab= ⇔a=b = √
2, c= units
Question 11 Given a convex quadrilateralABCD LetO 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 followsBI ×AD≤AO×BD (AE ≤AO) and BI.AD≤AC×BD× AO
AC Moreover, we haveOK kCH then AO
AC = OK
CH andBI×AD≤AC×BD× OK CH It followsBI ×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 perpendicular 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 LetK be intersection point of DB and AC Since BE⊥CD; CK⊥CD then BE kCK
In ∆DAC we see M E kAC so M E
AC = DM
AD (8)
In ∆DAK we see M B kAK so M B
AK = DM
(5)From (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
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 Letm be given odd number, and leta, bdenote the roots of equation x2 +mx−1 = and c = a2014 +b2014, d = a2015 +b2015 Prove that c and d are
relatively prime numbers
Solution Since a2+ma−1 = then a6= 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 allxn are integers Hence, c, dare integers
Now we prove (xn;xn+1) = for every n ∈N
For n= 0, x0 = and m is odd then (x0;x1) = (2;−m) =
Suppose that (xk;xk+1) = for k ≥ and (xk+1;xk+2) > Let p be a prime
factor ofxk+1 andxk+2, then fromxk=xk+2+mxk+1, it followspis a prime factor of
xk It means that p|(xk;xk+1) = 1, absurd Hence (xk+1;xk+2) = and (c, d) =
Question 14 Determine all pairs of integers (x;y) such that 2xy2+x+y+ =x2+ 2y2+xy
Solution We have
2xy2+x+y+ =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−xare divisors of -1.
Case
(
x−1 =
2y2−y−x=−1 ⇔
(
x= y=
x= y=−1
(6)- Case
(
x−1 =−1
2y2−y−x= 1 ⇔
(
x= y=
x= y =−1
2(absurd) Hence all integral pairs (x;y) are (2; 1); (0 ; 1)
Question 15 Let the numbersa, 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,