9th Vietnam Mathematical Olympiad 1970

A1.  ABC is a triangle. Show that sin A/2 sin B/2 sin C/2 < 1/4.  

A2.  Find all positive integers which divide 1890·1930·1970 and are not divisible by 45. 

A3.  The function f(x, y) is defined for all real numbers x, y. It satisfies f(x,0) = ax (where a is a non-zero constant) and if (c, d) and (h, k) are distinct points such that f(c, d) = f(h, k), then f(x, y) is constant on the line through (c, d) and (h, k). Show that for any real b, the set of points such that f(x, y) = b is a straight line and that all such lines are parallel. Show that f(x, y) = ax + by, for some constant b.  

B1.  AB and CD are perpendicular diameters of a circle. L is the tangent to the circle at A. M is a variable point on the minor arc AC. The ray BM, DM meet the line L at P and Q respectively. Show that AP·AQ = AB·PQ. Show how to construct the point M which gives BQ parallel to DP. If the lines OP and BQ meet at N find the locus of N. The lines BP and BQ meet the tangent at D at P' and Q' respectively. Find the relation between P' and Q'. The lines DP and DQ meet the line BC at P" and Q" respectively. Find the relation between P" and Q". 

B2.  A plane p passes through a vertex of a cube so that the three edges at the vertex make equal angles with p. Find the cosine of this angle. Find the positions of the feet of the perpendiculars from the vertices of the cube onto p. There are 28 lines through two vertices of the cube and 20 planes through three vertices of the cube. Find some relationship between these lines and planes and the plane p. 

Solutions

A1.

Put x = A/2, y = B/2. We have sin C/2 = sin(90^o-x-y) = cos(x+y). So we need to show that sin x sin y cos(x+y) < 1/4, or (cos(x-y) - cos(x+y) )cos(x+y) < 1/2, or 2 cos(x-y) cos(x+y) < 1 + 2 cos²(x+y). But 2 cos(x-y) cos(x+y) ≤ cos²(x+y) + cos²(x-y) ≤ 1 + cos²(x+y) < 1 + 2 cos²(x+y) (since 0 < x,y < 90^o. 

A2. Answer: k·2^a7^b193^c197^d, where k = 1, 3, 3², 3³, 5, 3·5, a = 0, 1, 2, or 3, b = 0 or 1, c = 0 or 1, d = 0 or 1 (192 solutions in all) 

Solution

1890 = 2·3³5·7, 1930 = 2·5·193, 1970 = 2·5·197 (and 193 and 197 are prime). So 1890·1930·1970 = 2³3³5³7·193·197.

8th Vietnam Mathematical Olympiad 1969

1.  A graph G has n + k points. A is a subset of n points and B is the subset of the other k points. Each point of A is joined to at least k - m points of B where nm < k. Show that there is a point in B which is joined every point in A. 

2.  Find all real x such that 0 < x < π and 8/(3 sin x - sin 3x) + 3 sin²x ≤ 5.  

3.  The real numbers x₁, x₄, y₁, y₂ are positive and the real numbers x₂, x₃, y₃, y₄ are negative. We have (x_i - a)² + (y_i - b)² ≤ c² for i = 1, 2, 3, 4. Show that a² + b² ≤ c². State the result in geometric language. 

4.  Two circles centers O and O', radii R and R', meet at two points. A variable line L meets the circles at A, C, B, D in that order and AC/AD = CB/BD. The perpendiculars from O and O' to L have feet H and H'. Find the locus of H and H'. If OO'² < R² + R'², find a point P on L such that PO + PO' has the smallest possible value. Show that this value does not depend on the position of L. Comment on the case OO'² > R² + R'².  

Solutions

2. Answer: π/2

We have 3 sin x - sin 3x = 4 sin³x. Put s = sin x. Then we want 2/s³ + 3s² ≤ 5. Note that since 0 < x < π we have s positive. But by AM/GM we have 1/s³ + 1/s³ + s² + s² + s² > 5 with equality iff s = 1, so we must have sin x = 1 and hence x = π/2. 

3. Stated geometrically, the result is: if a disk includes a point in each quadrant, then it must also include the origin. We use the fact that a disk is convex. Let P_i be the point (x_i,y_i). The segment P₁P₂ must intersect the positive x-axis. By convexity, the point of intersection, call it X, must lie in the disk. Similarly, P₃P₄ must intersect the negative x-axis at some point Y, which must be in the disk. Then all points of the segment XY are in the disk and hence, in particular, the origin. 

Source: http://321math.blogspot.com

4th Vietnam Mathematical Olympiad 1965

1.  At time t = 0, a lion L is standing at point O and a horse H is at point A running with speed v perpendicular to OA. The speed and direction of the horse does not change. The lion's strategy is to run with constant speed u at an angle 0 < φ < π/2 to the line LH. What is the condition on u and v for this strategy to result in the lion catching the horse? If the lion does not catch the horse, how close does he get? What is the choice of φ required to minimise this distance?

2.  AB and CD are two fixed parallel chords of the circle S. M is a variable point on the circle. Q is the intersection of the lines MD and AB. X is the circumcenter of the triangle MCQ. Find the locus of X. What happens to X as M tends to (1) D, (2) C? Find a point E outside the plane of S such that the circumcenter of the tetrahedron MCQE has the same locus as X.

3.  m an n are fixed positive integers and k is a fixed positive real. Show that the minimum value of x1m + x2m + x3m + ... + xnm for real xi satisfying x1 + x2 + ... + xn = k occurs at x1 = x2 = ... = xn.

Source: Nguyễn Thị Lan Phương, http://321math.blogspot.com

3rd Vietnam Mathematical Olympiad 1964

1.  Find cos x + cos(x + 2π/3) + cos(x + 4π/3) and sin x + sin(x + 2π/3) + sin(x + 4π/3).
2.  Draw the graph of the functions y = | x2 - 1 | and y = x + | x2 - 1 |. Find the number of roots of the equation x + | x2 - 1 | = k, where k is a real constant.
3.  Let O be a point not in the plane p and A a point in p. For each line in p through A, let H be the foot of the perpendicular from O to the line. Find the locus of H.
4.  Define the sequence of positive integers fn by f0 = 1, f1 = 1, fn+2 = fn+1 + fn. Show that fn = (an+1 - bn+1)/√5, where a, b are real numbers such that a + b = 1, ab = -1 and a > b.  Solutions 1. Using cos(A+B) = cos A cos B - sin A sin B, we have cos(x + 2π/3) = -(1/2) cos x + (√3)/2 sin x, cos(x + 4π/3) = -(1/2) cos x - (√3)/2 sin x. Hence cos x + cos(x + 2π/3) + cos(x + 4π/3) = 0. Similarly, sin(x + 2π/3) = -1/2 sin x + (√3)/2 cos x, sin(x + 4π/3) = -1/2 sin x - (√3)/2 cos x, so sin x + sin(x + 2π/3) + sin(x + 4π/3) = 0. 2. Answer 0 for k < -1 1 for k = -1 2 for -1 < k < 1 3 for k = 1 4 for 1 < k < 5/4 3 for k = 5/4 2 for k > 5/4 It is clear from the graph that there are no roots for k < -1, and one root for k = -1 (namely x = -1). Then for k > -1 there are two roots except for a small interval [1, 1+h]. At k = 1, there are 3 roots (x = -2, 0, 1). The upper bound is at the local maximum between 0 and 1. For such x, y = x + 1 - x2 = 5/4 - (x - 1/2)2, so the local maximum is at 5/4. Thus there are 3 roots at k = 5/4 and 4 roots for k ∈ (1, 5/4). 3. Answer: circle diameter AB, where OB is the normal to p Let B be the foot of the perpendicular from O to p. We claim that the locus is the circle diameter AB. Any line in p through A meets this circle at one other point K (except for the tangent to the circle at A, but in that case A is obviously the foot of the perpendicular from O to the line). Now BK is perpendicular to AK, so OK is also perpendicular to AK, and hence K must be the foot of the perpendicular from O to the line. 4. Put a = (1+√5)/2, b = (1-√5)/2. Then a, b are the roots of x2 - x - 1 = 0 and satisfy a + b = 1, ab = -1. We show by induction that fn = (an+1 - bn+1)/√5. We have f0 = (a-b)/√5 = 1, f1 = (a2-b2)/√5 = (a+1 - b-1)/√5 = 1, so the result is true for n = 0, 1. Finally, suppose fn = (an+1 - bn+1)/√5 and fn+1 = (an+2 - bn+2)/√5. Then fn+2 = fn+1 + fn = (1/√5)(an+1(a+1) - bn+1(b+1) ) = (an+1a2 - bn+1b2)/√5, so the result is true for n+1.

Source: http://321math.blogspot.com

2nd Vietnam Mathematical Olympiad 1963

1.  A conference has 47 people attending. One woman knows 16 of the men who are attending, another knows 17, and so on up to the last woman who knows all the men who are attending. Find the number of men and women attending the conference.

2.  For what values of m does the equation x2 + (2m + 6)x + 4m + 12 = 0 has two real roots, both of them greater than -2.

3.  Solve the equation sin3x cos 3x + cos3x sin 3x = 3/8.

4.  The tetrahedron SABC has the faces SBC and ABC perpendicular. The three angles at S are all 60o and SB = SC = 1. Find its volume.

5.  The triangle ABC has perimeter p. Find the side length AB and the area S in terms of ∠A, ∠B and p. In particular, find S if p = 23.6, A = 52.7 deg, B = 46 4/15 deg. Solutions 1. Suppose there are m women. Then the last woman knows 15+m men, so 15+2m = 47, so m = 16. Hence there are 31 men and 16 women. 2. Answer: m ≤ -3 For real roots we must have (m+3)2 ≥ 4m+12 or (m-1)(m+3) ≥ 0, so m ≥ 1 or m ≤ -3. If m ≥ 1, then -(2m+6) ≤ -8, so at least one of the roots is < -2. So we must have m ≤ -3. The roots are -(m+3) ±√(m2+2m-3). Now -(m+3) ≥ 0, so -(m+3) + √(m2+2m-3) ≥ 0 > -2. So we need -(m+3) - √(m2+2m-3) > -2, or √(m2+2m-3) < -m-1 = √(m2+2m+1), which is always true. 3. Answer: 7½o + k90o or 37½o + k90o We have sin 3x = 3 sin x - 4 sin3x, cos 3x = 4 cos3x - 3 cos x. So we need 4 sin3x cos3x - 3 sin3x cos x + 3 sin x cos3x - 4 sin3x cos3x = 3/8 or 8 sin x cos x(cos2x - sin2x) = 1, or 4 sin 2x cos 2x = 1 or sin 4x = 1/2. Hence 4x = 30o + k360o or 150o + k360o. So x = 7½o + k90o or 37½o + k90o.

Source: Nguyễn Thị Lan Phương, http://321math.blogspot.com

1st Vietnam Mathematical Olympiad 1962

1.  Prove that 1/(1/a + 1/b) + 1/(1/c + 1/d) ≤ 1/(1/(a+c) + 1/(b+d) ) for positive reals a, b, c, d.
2.  f(x) = (1 + x)(2 + x2)1/2(3 + x3)1/3. Find f '(-1).
3.  ABCD is a tetrahedron. A' is the foot of the perpendicular from A to the opposite face, and B' is the foot of the perpendicular from B to the opposite face. Show that AA' and BB' intersect iff AB is perpendicular to CD. Do they intersect if AC = AD = BC = BD?
4.  The tetrahedron ABCD has BCD equilateral and AB = AC = AD. The height is h and the angle between ABC and BCD is α. The point X is taken on AB such that the plane XCD is perpendicular to AB. Find the volume of the tetrahedron XBCD.
5.  Solve the equation sin6x + cos6x = 1/4. Solutions 1. A straightforward, if inelegant, approach is to multiply out and expand everything. All terms cancel except four and we are left with 2abcd ≤ a2d2 + b2c2, which is obviously true since (ad - bc)2 ≥ 0.  2. Differentiating gives f '(x) = (2 + x2)1/2(3 + x3)1/3 + terms with factor (1 + x). Hence f '(-1) =31/221/3.  3. Let the ray AB' meet CD at X and the ray BA' meet CD at Y. If AB' and A'B intersect, then X = Y. Let L be the line through A' parallel to CD. Then L is perpendicular to AA'. Hence CD is perpendicular to AA'. Similarly, let L' be the line through B' parallel to CD. Then L' is perpendicular to BB', and hence CD is perpendicular to BB'. So CD is perpendicular to two non-parallel lines in the plane ABX. Hence it is perpendicular to all lines in the plane ABX and, in particular, to AB. Suppose conversely that AB is perpendicular to CD. Consider the plane ABY. CD is perpendicular to AB and to AA', so CD is perpendicular to the plane. Similarly CD is perpendicular to the plane ABX. But it can only be perpendicular to a single plane through AB. Hence X = Y and so AA' and BB' belong to the same plane and therefore meet. 4. Put a = sin2x, b = cos2x. Then a and b are non-negative with sum 1, so we may put a = 1/2 + h, b = 1/2 - h. Then a3 + b3 = 1/4 + 3h2 ≥ 1/4 with equality iff h = 0. Hence x is a solution of the equation given iff sin2x = cos2x = 1/2 or x is an odd multiple of π/4.  Source: Nguyễn Thị Lan Phương, http://321math.blogspot.com