1. Let P be the intersection of lines l1 and l2. Let S1 and S2 be two circles externally tangent at P and both tangent to l1, and let T1
and T2be two circles externally tangent at P and both tangent to l2. Let A be the second intersection of S1 and T1, B that of S1and T2, C that of S2 and T1, and D that of S2and T2. Show that the points A, B, C, D are concyclic if and only if l1 and l2 are perpendicular.
Solution: Invert around P ; S1, l1, S2 become parallel lines, as do T1, l2, T2. Thus ABCD inverts to a parallelogram and we need that a parallelogram is cyclic iff it is a rectangle, which is obvious.
2. Let m, n, p, q be positive integers, and consider an m×n checkerboard with a checker on each of its mn squares. A piece can be moved from (x, y) to (x0, y0) if and only if|x−x0| = p and |y −y0| = q. How many ways can all of the pieces be moved simultaneously so that one piece ends up on each square?
Solution: Answer: 1 if 2p| m and 2q | n, else 0.
It will be convenient to first prove this in the case p = q = 1. Color the grid in a checkerboard pattern, with black in the upper left hand corner. If m is odd, there will be more black squares in the top row than the second, and the move will be impossible; similarly if n is odd. Finally, if both m and n are even, all markers in the top row must go to the second and all the markers in the second are needed to fill the top. This can happen in exactly one way (divide the top two rows into two by two blocks, each marker switches with the one diagonally opposite it). This now reduces to the m× (n − 2) case.
Now we tackle the general case, with p and q arbitrary. The key observation is that the squares can be divided into pq categories, based on their x coordinate mod p and y coordinate mod q. The movements on each grid are independent, and each one acts like a grid with p = q = 1. The horizontal side of these grids is either bm/pc or bm/pc + 1 and similarly for the vertical size. The only way all of these can have even size is if 2p| m and 2q | n.
3. [Corrected] On a blackboard are written the numbers 49/k with k = 1, 2, . . . , 97. At each step, two numbers a, b are erased and
2ab− a − b + 1 is written in their place. After 96 steps, a single number remains on the blackboard. Determine all possible such numbers.
Solution: The only such number is 1. Note that 2(2ab− a − b + 1) − 1 = (2a − 1)(2b − 1),
so the product of 2a− 1 over the numbers a on the board never changes. Thus the final number N must satisfy
2N−1 = (2·49/1−1) · · · (2·49/97−1) = (97/1)·(96/2) · · · (1/97) = 1 and so N = 1.
4. In a convex quadrilateral ABCD, the sides AB and CD are parallel, the diagonals AC and BD intersect at E, and the triangles EBC and EAD have respective orthocenters F and G. Prove that the midpoint of GF lies on the line through E perpendicular to AB.
Solution: Let H be the orthocenter of ABE and I the ortho-center of CDE; we must show the midpoint of F G is on HI. We will show the stronger result that F HGI is a parallelogram. Now F is on the perpendicular from B to CE and H is on the perpendicular from B to AE. But A, E, C are collinear so F H is perpendicular to AE and GI is similarly perpendicular. Hence they are parallel; a similar argument applies to HG and IF .
5. Let p1, p2, p3, p4 be distinct primes. Prove there does not exist a cubic polynomial Q(x) with integer coefficients such that
|Q(p1)| = |Q(p2)| = |Q(p3)| = |Q(p4)| = 3.
Solution: WLOG, we can consider two cases.
Case 1: Q(p1) = Q(p2) = Q(p3) =−3. Then Q(x) = a(x − p1)(x− p2)(x− p3)− 3 and we have (p4− p1)(p4− p2)(p4− p3)|6. If p4> 2, there are at least two even factors on the left and this is impossible;
if p4= 2 the left has absolute value at least (3−2)(5−2)(7−2) = 15, which is also impossible.
Case 2: Q(p1) = Q(p2) =−3, Q(p3) = Q(p4) = 3 and p1, p2 and p3
aren’t 2. Then as before, (p3− p1)(p3− p2)|6, but the left hand side has two even factors.
6. Prove there does not exist f : Z→ Z such that f(x+f(y)) = f(x)−y for all integers x, y.
Solution: First note that f (x + nf (y)) = f (x)− ny. (Proof:
this is obvious for n = 0, and the functional equation gives f (x + (n + 1)f (y)) = f (x + nf (y))− y from which we can induct in both directions.) Therefore, f (1+f (1)·f(1)) = 0. Put k = 1+f(1)·f(1)k and note k > 0. Then f (x) = f (x + f (k)) = f (x)− k, contradiction.
7. (a) Prove that for all p, q∈ R, p2+ q2+ 1 > p(q + 1).
(b) Determine the largest real number b such that p2+ q2+ 1 ≥ bp(q + 1) for all p, q∈ R.
(c) Determine the largest real number b such that p2+ q2+ 1 ≥ bp(q + 1) for all p, q∈ Z.
Solution:
(a) See below.
(b) Answer: b =√
2. WLOG, assume p, q≥ 0. Now just observe p2+ q2+ 1≥ p2+ 2((q + 1)/2)2≥ 2p
p2(q + 1)2/2 =√
2p(q + 1).
Equality is obtained for p =√
2, q = 1, proving the maximality.
(c) Answer: 3/2. Note this is achieved by p = q = 1, so 3/2 gives an upper bound for the maximum b. To see that 3/2 works, again assume p, q ≥ 0. If q ≥ 3, then p(3 − q) − 2 < 0 ≤ 2(p − q)2, so p2+ q2+ 1 > 32p(q + 1). If q = 0, we need p2+ 1≥ 32p for all real p. If q = 1, we need p2+ 2≥ 3p or (p − 1)(p − 2) ≥ 0, which holds for p ∈ Z. If q = 2, we need p2+ 5 ≥ 92p, or (p− 2)(p − 5/2) ≥ 0, which again holds for all p ∈ Z.
8. Let n be a natural number and M a set with n elements. Find the biggest integer k such that there exists a k-element family of three-element subsets of M , no two of which are disjoint.
Solution: For n ≤ 5, we take all three-element subsets and get k = n3. We will prove that k = n−12 for n ≥ 6; this bound is achieved by taking all sets containing some fixed element x of M . The base case n = 6 holds because we can pair each set with its complement, and only one of each pair can occur.
Let {1, 2, 3} be one of the sets in the family. If every other set contained two of 1, 2, 3, there would be at most 1+3(n−3) < n−12 +1 sets in the family, so assume there exists some set containing only one of 1, 2, 3; suppose it is{1, 4, 5}. We win by induction if 7 belongs to n− 2 or fewer sets (there are at most n−22 sets not containing 7), so assume it belongs to n− 1 sets. In particular, 7 belongs to a set not containing 1, say{2, 4, 7}. Likewise, we win unless 6 belongs to n− 1 ≥ 6 sets; in fact, 6 can only belong to the 6 sets
{1, 2, 6}, {1, 4, 6}, {2, 4, 6}, {1, 6, 7}, {3, 4, 6}, {2, 5, 6}.
We are done unless n = 7 and 6 belongs to all six of the above sets; in that case, note that 7 can now only belong to the sets {1, 6, 7}, {2, 4, 7}, {3, 5, 7}, so we win again.
9. Let P be a parallelepiped with volume V and surface area S, and let L be the sum of the lengths of the edges of P . For t≥ 0, let Pt
be the set of points which lie at distance at most t from some point of P . Prove that the volume of Ptis
V + St +π
4Lt2+4π 3 t3.
Solution: At each edge, raise a plane containing that edge and perpendicular to each face containing that edge. These divide Pt into four types of pieces: the original parallelepiped, six rectangu-lar prisms, twelve sections of cylinders and eight pieces of spheres.
The volume of the parallelepiped is V and the combined volume of the rectangular prisms is St. For any four parallel edges, the angles of the cylinders add up to 360 so the volume of those four cylin-ders adds to πt2 times the length of the edge; all cylinders together give πt2L/4. Finally, the spherical pieces fit together exactly, giving 4πt3/3.