Saturday, May 29, 2010

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 (xi - a)2 + (yi - 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.  


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 Pi be the point (xi,yi). 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. 


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