2nd Vietnam Mathematical Olympiad 1963 Problems

2nd Vietnam Mathematical Olympiad 1963 Problems

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 60^o 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: 7½^o + k90^o or 37½^o + k90^o

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 = 30^o + k360^o or 150^o + k360^o. So x = 7½^o + k90^o or 37½^o + k90^o.

Source: Nguyễn Thị Lan Phương, http://www.kidsmathbooks.com