(a) On a long pavement, a sequence of $999$ integers is written in chalk. The numbers need not be in increasing order and need not be distinct. Merlijn encircles $500$ of the numbers with red chalk. From left to right, the numbers circled in red are precisely the numbers $1, 2, 3, ...,499, 500$. Next, Jeroen encircles $500$ of the numbers with blue chalk. From left to right, the numbers circled in blue are precisely the numbers $500, 499, 498, ...,2,1$. Prove that the middle number in the sequence of $999$ numbers is circled both in red and in blue. (b) Merlijn and Jeroen cross the street and find another sequence of $999$ integers on the pavement. Again Merlijn circles $500$ of the numbers with red chalk. Again the numbers circled in red are precisely the numbers $1, 2, 3, ...,499, 500$ from left to right. Now Jeroen circles $500$ of the numbers, not necessarily the same as Merlijn, with green chalk. The numbers circled in green are also precisely the numbers $1, 2, 3, ...,499, 500$ from left to right. Prove: there is a number that is circled both in red and in green that is not the middle number of the sequence of $999$ numbers.

For an integer $n \ge 1$ we consider sequences of $2n$ numbers, each equal to $0, -1$ or $1$. The sum product value of such a sequence is calculated by first multiplying each pair of numbers from the sequence, and then adding all the results together. For example, if we take $n = 2$ and the sequence $0,1, 1, -1$, then we find the products $0\cdot 1, 0\cdot 1, 0\cdot -1, 1\cdot 1, 1\cdot -1, 1\cdot -1$. Adding these six results gives the sum product value of this sequence: $0+0+0+1+(-1)+(-1) = -1$. The sum product value of this sequence is therefore smaller than the sum product value of the sequence $0, 0, 0, 0$, which equals $0$. Determine for each integer $n \ge 1$ the smallest sum product value that such a sequence of $2n$ numbers could have. Attention: you are required to prove that a smaller sum product value is impossible.

Find all possible triples $(a, b, c)$ of positive integers with the following properties: • $gcd(a, b) = gcd(a, c) = gcd(b, c) = 1$, • $a$ is a divisor of $a + b + c$, • $b$ is a divisor of $a + b + c$, • $c$ is a divisor of $a + b + c$. (Here $gcd(x,y)$ is the greatest common divisor of $x$ and $y$.)

In a quadrilateral $ABCD$ the intersection of the diagonals is called $P$. Point $X$ is the orthocentre of triangle $PAB$. (The orthocentre of a triangle is the point where the three altitudes of the triangle intersect.) Point $Y$ is the orthocentre of triangle $PCD$. Suppose that $X$ lies inside triangle $PAB$ and $Y$ lies inside triangle $PCD$. Moreover, suppose that $P$ is the midpoint of line segment $XY$ . Prove that $ABCD$ is a parallelogram. [asy][asy] import geometry; unitsize (1.5 cm); pair A, B, C, D, P, X, Y; A = (0,0); B = (2,-0.5); C = (3.5,2.2); D = A + C - B; P = (A + C)/2; X = orthocentercenter(A,B,P); Y = orthocentercenter(C,D,P); draw(A--B--C--D--cycle); draw(A--C); draw(B--D); draw(A--extension(A,X,B,P), dotted); draw(B--extension(B,X,A,P), dotted); draw(P--extension(P,X,A,B), dotted); draw(C--extension(C,Y,D,P), dotted); draw(D--extension(D,Y,C,P), dotted); draw(P--extension(P,Y,C,D), dotted); dot("$A$", A, W); dot("$B$", B, S); dot("$C$", C, E); dot("$D$", D, N); dot("$P$", P, E); dot("$X$", X, NW); dot("$Y$", Y, SE); [/asy][/asy]

In the acute triangle $ABC$, the midpoint of side $BC$ is called $M$. Point $X$ lies on the angle bisector of $\angle AMB$ such that $\angle BXM = 90^o$. Point $Y$ lies on the angle bisector of $\angle AMC$ such that $\angle CYM = 90^o$. Line segments $AM$ and $XY$ intersect in point $Z$. Prove that $Z$ is the midpoint of $XY$ . [asy][asy] import geometry; unitsize (1.2 cm); pair A, B, C, M, X, Y, Z; A = (0,0); B = (4,1.5); C = (0.5,3); M = (B + C)/2; X = extension(M, incenter(A,B,M), B, B + rotate(90)*(incenter(A,B,M) - M)); Y = extension(M, incenter(A,C,M), C, C + rotate(90)*(incenter(A,C,M) - M)); Z = extension(A,M,X,Y); draw(A--B--C--cycle); draw(A--M); draw(M--interp(M,X,2)); draw(M--interp(M,Y,2)); draw(B--X, dotted); draw(C--Y, dotted); draw(X--Y); dot("$A$", A, SW); dot("$B$", B, E); dot("$C$", C, N); dot("$M$", M, NE); dot("$X$", X, NW); dot("$Y$", Y, NE); dot("$Z$", Z, S); [/asy][/asy]

Bas has coloured each of the positive integers. He had several colours at his disposal. His colouring satises the following requirements: • each odd integer is coloured blue, • each integer $n$ has the same colour as $4n$, • each integer $n$ has the same colour as at least one of the integers $n+2$ and $n + 4$. Prove that Bas has coloured all integers blue.