**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.

**3.**The real numbers x

_{1}, x

_{4}, y

_{1}, y

_{2}are positive and the real numbers x

_{2}, x

_{3}, y

_{3}, y

_{4}are negative. We have (x

_{i}- a)

^{2}+ (y

_{i}- b)

^{2}≤ c

^{2}for i = 1, 2, 3, 4. Show that a

^{2}+ b

^{2}≤ c

^{2}. 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}< R

^{2}+ 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}> R

^{2}+ R'

^{2}.

**Solutions****2. Answer**: π/2

We have 3 sin x - sin 3x = 4 sin

^{3}x. Put s = sin x. Then we want 2/s^{3}+ 3s^{2}≤ 5. Note that since 0 < x < π we have s positive. But by AM/GM we have 1/s^{3}+ 1/s^{3}+ s^{2}+ s^{2}+ s^{2}> 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 P

_{1}P

_{2}must intersect the positive x-axis. By convexity, the point of intersection, call it X, must lie in the disk. Similarly, P

_{3}P

_{4}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**
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