1. Prove that for all natural numbers n ≥ 3 there exist odd natural numbers xn, yn such that 7x2n+ y2n= 2n.
Solution: For n = 3 we have x3 = y3 = 1. Now suppose that for a given natural number n we have odd natural numbers xn, yn such that 7x2n+ y2n = 2n; we shall exhibit a pair (X, Y ) such that 7X2+ Y2= 2n+1. In fact,
7 xn± yn
2
2
+ 7xn∓ yn
2
2
= 2(7x2n+ yn2) = 2n+1. One of (xn+ yn)/2 and|xn− yn|/2 is odd (as their sum is the larger of xn and yn, which is odd), giving the desired pair.
2. The circles k1 and k2 with respective centers O1 and O2 are exter-nally tangent at the point C, while the circle k with center O is externally tangent to k1and k2. Let ` be the common tangent of k1
and k2at the point C and let AB be the diameter of k perpendicular to `. Assume that O and A lie on the same side of `. Show that the lines AO2, BO1, ` have a common point.
Solution: Let r, r1, r2 be the respective radii of k, k1, k2. Also let M and N be the intersections of AC and BC with k. Since AM B is a right triangle, the triangle AM O is isosceles and
∠AM O = ∠OAM = ∠O1CM = ∠CM O1.
Therefore O, M, O1are collinear and AM/M C = OM/M O1= r/r1. Similarly O, N, O2are collinear and BN/N C = ON/N O2= r/r2. Let P be the intersection of ` with AB; the lines AN, BM, CP con-cur at the orthocenter of ABC, so by Ceva’s theorem, AP/P B = (AM/M C)(CN/N B) = r2/r1. Now let D1 and D2 be the intersec-tions of ` with BO1 and AO2. Then CD1/D1P = O1C/P B = r1/P B, and similarly CD2/D2P = r2/P A. Thus CD1/D1P = CD2/D2P and D1= D2, and so AO2, BO1, ` have a common point.
3. Let a, b, c be real numbers and let M be the maximum of the function y =|4x3+ ax2+ bx + c| in the interval [−1, 1]. Show that M ≥ 1 and find all cases where equality occurs.
Solution: For a = 0, b = −3, c = 0, we have M = 1, with the maximum achieved at−1, −1/2, 1/2, 1. On the other hand, if M < 1 for some choice of a, b, c, then
(4x3+ ax2+ bx + c)− (4x3+ 3x)
must be positive at −1, negative at −1/2, positive at 1/2, and negative at 1, which is impossible for a quadratic function. Thus M ≥ 1, and the same argument shows that equality only occurs for (a, b, c) = (0,−3, 0). (Note: this is a particular case of the minimum deviation property of Chebyshev polynomials.)
4. The real numbers a1, a2, . . . , an (n≥ 3) form an arithmetic progres-sion. There exists a permutation ai1, ai2, . . . , ain of a1, a2, . . . , an
which is a geometric progression. Find the numbers a1, a2, . . . , an if they are all different and the largest of them is equal to 1996.
Solution: Let a1< a2<· · · < an= 1996 and let q be the ratio of the geometric progression ai1, . . . ain; clearly q6= 0, ±1. By reversing the geometric progression if needed, we may assume|q| > 1, and so
|ai1| < |ai2| < · · · < |ain|. Note that either all of the terms are positive, or they alternate in sign; in the latter case, the terms of either sign form a geometric progression by themselves.
There cannot be three positive terms, or else we would have a three-term geometric progression a, b, c which is also an arithmetic pro-gression, violating the AM-GM inequality. Similarly, there cannot be three negative terms, so there are at most two terms of each sign and n≤ 4.
If n = 4, we have a1 < a2 < 0 < a3 < a4 and 2a2 = a1+ a3, 2a3= a2+ a4. In this case, q <−1 and the geometric progression is either a3, a2, a4, a1 or a2, a3, a1, a4. Suppose the former occurs (the argument is similar in the latter case); then 2a3q = a3q3+ a3 and 2a3+ a3q + a3q2, giving q = 1, a contradiction.
We deduce n = 3 and consider two possibilities. If a1 < a2 <
0 < a3 = 1996, then 2a2 = a2q2+ a2q, so q2+ q − 2 = 0 and
q = −2, yielding (a1, a2, a3) = (−3992, −998, 1996). If a1 < 0 <
a2 < a3 = 1996, then 2a2 = a2q + a2q2, so again q = −2, yielding (a1, a2, a3) = (−998, 499, 1996).
5. A convex quadrilateral ABC is given for which ∠ABC + ∠BCD <
180◦. The common point of the lines AB and CD is E. Prove that
∠ABC = ∠ADC if and only if
AC2= CD· CE − AB · AE.
Solution: Let C1 be the circumcircle of ADE, and let F be its second intersection with CA. In terms of directed lengths, we have AC2= CD· CE + AB · AE if and only if
AB· AE = AC2− CD · CE = CA2− CA · AF = AC · AF, that is, if and only if B, C, E, F are concyclic. But this happens if and only if ∠EBC = ∠EF C, and
∠EF C = ∠EF A = π − ∠ADE = ∠CDA
(in directed angles modulo π), so B, C, E, F are concyclic if and only if ∠ABC = ∠ADC (as undirected angles), as desired.
6. Find all prime numbers p, q for which pq divides (5p− 2p)(5q− 2q).
Solution: If p|5p− 2p, then p|5 − 2 by Fermat’s theorem, so p = 3.
Suppose p, q 6= 3; then p|5q − 2q and q|5p − 2p. Without loss of generality, assume p > q, so that (p, q− 1) = 1. Then if a is an integer such that 2a≡ 5 (mod q), then the order of a mod q divides p as well as q− 1, a contradiction.
Hence one of p, q is equal to 3. If q6= 3, then q|53− 23 = 9· 13, so q = 13, and similarly p ∈ {3, 13}. Thus the solutions are (p, q) = (3, 3), (3, 13), (13, 3).
7. Find the side length of the smallest equilateral triangle in which three discs of radii 2, 3, 4 can be placed without overlap.
Solution: A short computation shows that discs of radii 3 and 4 can be fit into two corners of an equilateral triangle of side 11√
3 so
as to just touch, and that a disc of radius 2 easily fits into the third corner without overlap. On the other hand, if the discs of radii 3 and 4 fit into an equilateral triangle without overlap, there exists a line separating them (e.g. a tangent to one perpendicular to their line of centers) dividing the triangle into a triangle and a (possibly degenerate) convex quadrilateral. Within each piece, the disc can be moved into one of the corners of the original triangle. Thus the two discs fit into the corners without overlap, so the side length of the triangle must be at least 11√
3.
8. The quadratic polynomials f and g with real coefficients are such that if g(x) is an integer for some x > 0, then so is f (x). Prove that there exist integers m, n such that f (x) = mg(x) + n for all x.
Solution: Let f (x) = ax2+ bx + c and g(x) = px2+ qx + r;
assume without loss of generality p > 0 and q = 0 (by the change of variable x→ x − q/(2p)). Let k be an integer such that k > s This tends to a/p as k increases, so a/p must be an integer; moreover, b must equal 0, or else the above expression will equal a/p plus a small quantity for large k, which cannot be an integer. Now put m = a/p and n = c− ms; then f(x) = mg(x) + n.
Solution: We will show by induction that √
n≤ an ≤ n/√ n− 1 for n ≥ 1, which will imply the claim. These inequalities clearly hold for n = 1, 2, 3. Now assume the inequality for some n. Let
On the other hand, using that an > (n− 1)/√
n− 2 (which we just proved), we get for n≥ 4,
an+1= fn(an) < fn
n− 1
√n− 2
= (n− 1)2+ n2(n− 2) (n− 1)n√
n− 2 <√ n + 2.
10. The quadrilateral ABCD is inscribed in a circle. The lines AB and CD meet at E, while the diagonals AC and BD meet at F . The circumcircles of the triangles AF D and BF C meet again at H.
Prove that ∠EHF = 90◦.
Solution: (We use directed angles modulo π.) Let O be the circumcenter of ABCD; then
∠AHB = ∠AHF +∠F HB = ∠ADF +∠F CB = 2∠ADB = ∠AOB, so O lies on the circumcircle of AHB, and similarly on the circum-circle of CHD. The radical axes of the circumcircum-circles of AHB, CHD and ABCD concur; these lines are AB, CD and HO, so E, H, O are collinear. Now note that
∠OHF = ∠OHC+∠CHF = ∠ODC+∠CBF = π
2−∠CAD+∠CBD, so ∠EHF = ∠OHF = π/2 as desired. (Compare IMO 1985/5.) 11. A 7× 7 chessboard is given with its four corners deleted.
(a) What is the smallest number of squares which can be colored black so that an uncolored 5-square (Greek) cross cannot be found?
(b) Prove that an integer can be written in each square such that the sum of the integers in each 5-square cross is negative while the sum of the numbers in all squares of the board is positive.
Solution:
(a) The 7 squares
(2, 5), (3, 2), (3, 3), (4, 6), (5, 4), (6, 2), (6, 5)
suffice, so we need only show that 6 or fewer will not suffice.
The crosses centered at
(2, 2), (2, 6), (3, 4), (5, 2), (5, 6), (6, 4)
are disjoint, so one square must be colored in each, hence 5 or fewer squares do not suffice. Suppose exactly 6 squares are colored. Then none of the squares (1, 3), (1, 4), (7, 2) can be col-ored; by a series of similar arguments, no square on the perime-ter can be colored. Similarly, (4, 3) and (4, 5) are not covered, and by a similar argument, neither is (3, 4) or (5, 4). Thus the center square (4, 4) must be covered.
Now the crosses centered at
(2, 6), (3, 3), (5, 2), (5, 6), (6, 4)
are disjoint and none contains the center square, so each con-tains one colored square. In particular, (2, 2) and (2, 4) are not colored. Replacing (3, 3) with (2, 3) in the list shows that (3, 2) and (3, 4) are not colored. Similar symmetric arguments now show that no squares besides the center square can be covered, a contradiction. Thus 7 squares are needed.
(b) Write −5 in the 7 squares listed above and 1 in the remaining squares. Then clearly each cross has negative sum, but the total of all of the numbers is 5(−7) + (45 − 7) = 3.