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