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.