1. In the plane are given a line ∆ and three circles tangent to ∆ and externally tangent to each other. Prove that the triangle formed by the centers of the circles is obtuse, and find all possible measures of the obtuse angle.
Solution: Let a, b, c be the radii of the circles, A, B, C the centers of the circles and A0, B0, C0 the projections of A, B, C, respectively, which follows by expanding and applying AM-GM.
To find the possible measures of C, we must maximize sin2C/2.
Given x + y = 1, we need to minimize (1 + x2)(1 + y2) = 2− 2xy + (xy)2. Now xy≤ (x+y)2/4 = 1/4 and the function f (z) = z2−2z+2 is decreasing on [0, 1/4]. Thus the C is maximized for x = y = 1/2, in which case C = 2 arcsin 16/25.
2. Determine all sets A of nine positive integers such that for any n≥ 500, there exists a subset B of A, the sum of whose elements is n.
Solution: Note that if A contains x, y, z with x + y = z, then the sets C ∪ {x, y} and C ∪ {z} give the same sum for any C ⊆
A− {x, y, z}, giving at most 511 − 26 = 447 nonzero sums. There-fore no such x, y, z can exist. Similarly there do not exist x, y, z, w in A with x + y + z = w.
In order for 1 and 2 to occur as sums, 1 and 2 must belong to A.
Since then 3 does not belong, 4 must belong. Now 5,6,7 are excluded, so 8 belongs, which in turn excludes the numbers from 9 to 15, which in turn forces 16 to belong, which excludes the numbers from 17 to 30.
If we write the next element of A as 32− a, with 0 ≤ a ≤ 1, we can now write the numbers from 1 to 63− a as subset sums. Thus the next number must be 64− a − b for some b ≥ 0. Likewise, we find that
A ={1, 2, 4, 8, 16, 32−a, 64−a−b, 128−2a−b−c, 256−4a−2b−c−d}
for some a, b, c, d≥ 0, as long as the maximum sum 511 − 8a − 4b − 2c− d is at least 500, that is, 8a + 4b + 2c + d ≤ 11. One now easily enumerates the 74 possible solutions.
3. Let n≥ 4 be an integer and M a set of n points in the plane, no three collinear and not all lying on a circle. Find all functions f : M → R such that for any circle C containing at least three points of M ,
X
P∈M∩C
f (P ) = 0.
Solution: For each two points A, B, the number k of distinct circles through A, B and another point of M is at least 2. Summing the given condition over these circles, we get
0 = X
P∈M
f (P ) + (k− 1)(f(A) + f(B)).
Thus ifP f (P ) is nonzero, it has the opposite sign as f (A)+f (B) for any A, B, but the sum of f (A)+f (B) over all A, B is (n−1)P f (P ), a contradiction Thus P f (P ) = 0, and also f (A) = −f(B) for any two points A, B. Since n≥ 3, we conclude f(P ) = 0 for all P ∈ M.
4. Let ABC be a triangle, D a point on side BC and ω the circum-circle of ABC. Show that the circum-circles tangent to ω, AD, BD and
to ω, AD, DC, respectively, are tangent to each other if and only if
∠BAD = ∠CAD.
Solution: We focus first on the circle ω1 tangent to ω, AD, BD.
Let this circle touch AD at M , BD at N , let r be its radius and let K be its center. Also let O be the center of ω and R its radius; since ω1
is tangent to ω, OK = R− r. If we put ∠ADB = 2α, ∠KDO = β, and x = DN , this becomes OK = R− x tan α. Applying the law of cosines to triangle ODK, one extracts that
x2+ 2λx + DO2− R2= 0, λ = R sin α− DO cos β
cos α .
Likewise, if y is the length of the tangent from D to the other circle, we get that
y2+ 2µy + DO2− R2= 0, µ = R cos α− DO sin β
sin α .
The circles are tangent if and only if these circles have a common root, which occurs if and only if λ = µ. This in turn is equivalent to R cos 2α = DO sin(β− α) = DO sin ∠ADO. By the Law of Sines in triangle ADO, this in turn is equivalent to R cos 2α = R sin DAO.
Now let H be the foot of the altitude from A to BC; then the con-dition becomes cos 2α = sin HAD, which is equivalent to ∠HAD =
∠DAO. Since H and O are isogonal conjugates, this occurs if and only if AD is the angle bisector of A.
5. Let V A1· · · An be a pyramid with n≥ 4. A plane Π intersects the edges V A1, . . . , V An at B1, . . . , Bn, respectively. Suppose that the polygons A1· · · Anand B1· · · Bnare similar. Prove that Π is parallel to the base of the pyramid.
Solution: By projecting to a pyramid with perpendicular lat-eral edges, one sees that if A0, B0, C0 are points on the lateral edges of triangular pyramid V ABC (with vertex V ), then
[V A0B0C0] [V ABC] = V A0
V A V B0
V B V C0
V C.
Now let x = V Bi/V Ai, and let S be the constant of similitude from A1· · · An to B1· · · Bn. On the one hand, by the previous assertion,
[V BiBjBk]/[V AiAjAk] = xixjxk. On the other hand, if h and H are the lengths of the altitudes from V to B1· · · Bn and A1· · · An, respectively, then the ratio of volumes is also S2h/H. We deduce from this (using that n≥ 4) that x1=· · · = xn = S2/3(h/H)1/3. 6. Let A be the set of positive integers representable in the form a2+2b2
for integers a, b with b6= 0. Show that if p2∈ A for a prime p, then p∈ A.
Solution: The case p = 2 is easy, so assume p > 2. Note that if p2= a2+ 2b2, then 2b2= (p− a)(p + a). In particular, a is odd, and since a cannot be divisible by p, gcd(p−a, p+a) = gcd(p−a, 2p) = 2.
By changing the sign of a, we may assume p− a is not divisible by 4, and so
|p + a| = m2,|p − a| = 2n2
Since|a| < |p|, both p + a and p − a are actually positive, so we have 2p = m2+ 2n2, so p = n2+ 2(m/2)2.
7. Let p≥ 5 be a prime and choose k ∈ {0, . . . , p − 1}. Find the max-imum length of an arithmetic progression, none of whose elements contain the digit k when written in base p.
Solution: We show that the maximum length is p− 1 if k 6= 0 and p if k = 0. In a p-term arithmetic progression, the lowest noncon-stant digit takes all values from 0 to p− 1. This proves the upper bound for k 6= 0, which is also a lower bound because of the se-quence 1, . . . , p− 1. However, for k = 0, it is possible that when 0 occurs, it is not actually a digit in the expansion but rather a leading zero. This can only occur for the first term in the progression, so extending the progression to p + 1 terms would cause an honest zero to appear. Thus the upper bound for k = 0 is p, and the sequence 1, p+1, . . . , (p−1)p+1 shows that it is also a lower bound. (Compare with USAMO 1995/1.)
8. Let p, q, r be distinct prime numbers and let A be the set A ={paqbrc: 0≤ a, b, c ≤ 5}.
Find the smallest integer n such that any n-element subset of A contains two distinct elements x, y such that x divides y.
Solution: Of course n is one more than the size of the largest subset of A no two of whose elements divide one another. The set of paqbrc has this property and contains 27 elements. On the other hand, we can partition A into 27 sequences such that each element of a sequence divides the next element. To see this, identify A with the points (a, b, c) in 3-space. We partition the points in the plane c = 0 into six chains, running from (a, 0, 0) to (a, 5, 0), and we might as well continue to (a, 5, 5). This leaves a 5×6 rectangle in the plane c = 1, which we split into five chains which continue upward. Now c = 2 is left with a 5× 5 rectangle which splits into five chains, and so on. We end up with 6 + 5 + 5 + 4 + 4 + 3 = 27 chains, as desired.
Thus n = 28.
9. Let ABCDEF be a convex hexagon. Let P, Q, R be the intersections of the lines AB and EF , EF and CD, CD and AB, respectively.
Let S, T, U be the intersections of the lines BC and DE, DE and F A, F A and BC, respectively. Show that if AB/P R = CD/RQ = EF/QP , then BC/U S = DE/ST = F A/T U .
Solution: Both sets of equalities are equivalent to the vector equal-ity
A− B + C − D + E − F = 0.
10. Let P be the set of points in the plane and D the set of lines in the plane. Determine whether there exists a bijective function f : P → D such that for any three collinear points A, B, C, the lines f (A), f (B), f (C) are either parallel or concurrent.
Solution: No such function exists. We first note that the inverse images A, B, C of three concurrent or parallel lines l1, l2, l3must be collinear; otherwise, any point on the lines AB, BC, CA would map to a line concurrent with or parallel to these, as then would any point on a line through two such points, i.e. any point in the plane, a contradiction.
In particular, given two pencils P1, P2 of parallel lines, the inverse images of the lines in Piare the points on a line li. Note that l1 and l2 must be parallel, since no line lies in both P1 and P2. In other
words, the lines li for pencils Pi of parallel lines are all parallel to each other.
Pick a pencil Q of concurrent lines, which corresponds to a line m necessarily not parallel to l1. Any line m0 parallel to m also cor-responds to a pencil of lines through a different point (if m0 cor-responded to parallel lines, so would m). However, there is a line through the points corresponding to m and m0, whose inverse image would be a point on both m and m0, a contradiction.
11. [Corrected] Find all continuous functions f : R → [0, ∞) such that for all x, y∈ R,
f (x2+ y2) = f (x2− y2) + f (2xy).
Solution: From x = y = 0 we get f (0) = 0; then putting x = 0 we get f (t) = f (−t) for all t ∈ R. Note that for any a, b > 0, the equations x2− y2= a and 2xy = b have a real solution, and so
f (a) + f (b) = f (p
a2+ b2).
Putting g(x) = f (√
x), we find that g(a + b) = g(a) + g(b) for all a, b ≥ 0. Since g is continuous, a standard argument shows that g(a) = ca for some constant c≥ 0, so f(x) = cx2
12. Let n≥ 2 be an integer and P (x) = xn+ an−1xn−1+· · · + a1x + 1 be a polynomial with positive integer coefficients. Suppose that ak = an−k for k = 1, 2, . . . , n− 1. Prove that there exist infinitely many pairs x, y of positive integers such that x|P (y) and y|P (x).
Solution: There exists at least one such pair, namely (1, P (1)).
If there were only finitely many such pairs, we could choose a pair (x, y) with y maximal. However, we claim that for any pair (x, y) with x|P (y) and y|P (x), the pair (y, P (y)/x) also has this property, or equivalently, that y|P (P (y)/x). Indeed, the given conditions im-ply P (P (y)/x) = (x/P (y))nP (x/P (y)). Since P (y)≡ 1 (mod y), we conclude that P (P (y)/x)≡ xnP (x)≡ 0 (mod y).
Moreover, P (y)/x > y2/x > y, so the pair (y, P (y)/x) has larger second member than does (x, y), a contradiction. Thus infinitely many pairs exist.
13. Let P (x), Q(x) be monic irreducible polynomials over the rational numbers. Suppose P and Q have respective roots α and β such that α + β is rational. Prove that the polynomial P (x)2− Q(x)2 has a rational root.
Solution: Let α + β = q. Then P (x) and Q(q− x) are irreducible polynomials with rational coefficients sharing a root; it follows that P (x) = cQ(q− x) for some constant c, and it is evident that in fact c = (−1)deg Q. In particular, P (x)2= Q(q− x)2, and so q/2 is a root of P (x)2− Q(x)2.
14. Let a > 1 be an integer. Show that the set {a2+ a− 1, a3+ a2− 1, . . .}
contains an infinite subset, any two members of which are relatively prime.
Solution: We show that any set of n elements of the set which are pairwise coprime can be extended to a set of n + 1 elements.
For n = 1, note that any two consecutive terms in the sequence are relatively prime. For n > 1, let N be the product of the numbers in the set so far; then aφ(N )+1+ aφ(N )− 1 ≡ a (mod N), and so can be added (since every element of the sequence is coprime to a, N is as well).
15. Find the number of ways to color the vertices of a regular dodecagon in two colors so that no set of vertices of a single color form a regular polygon.
Solution: Call the colors red and blue. Obviously we need only keep track of equilateral triangles and squares. The vertices of the dodecagon form four triangles, each of which can be colored in 6 nonmonochromatic ways. Thus we have 64= 1296 colorings with no monochromatic triangles.
We now consider how many of these colorings have one, two or three monochromatic squares . The number with a given square monochromatic is 2× 34 = 162, since if the square is colored red, the other two vertices in each triangles can be red-blue, blue-red, or
blue-blue. For two given squares, we have one choice in each triangle if the squares are the same color, and two otherwise, for a total of 2 + 2× 24 = 34. For all three squares, we need that the squares themselves are not all the same color, for a total of 6 colorings.
By inclusion-exclusion, the number of colorings with no monochro-matic triangles or squares is
1296− 3 × 162 + 3 × 34 − 6 = 906.
16. Let Γ be a circle and AB a line not meeting Γ. For any point P ∈ Γ, let P0 be the second intersection of the line AP with Γ and let f (P ) be the second intersection of the line BP0with Γ. Given a point P0, define the sequence Pn+1= f (Pn) for n≥ 0. Show that if a positive integer k satisfies P0 = Pk for a single choice of P0, then P0 = Pk for all choices of P0.
Solution: Perform a projective transformation taking AB to infin-ity. This takes Γ to an ellipse, and a suitable affine transformation takes this ellipse to a circle while keeping AB at infinity. Now the map P → P0 is a reflection across the diameter through the point A (at infinity), while P0 → f(P ) is a reflection across the diameter through B. Thus P → f(P ) is a rotation, so if P0= Pk holds for a single P0, it holds for all P0.