1. Let a be a rational number, b, c, d be real numbers, and f : R → [−1, 1] a function satisfying
f (x + a + b)− f(x + b) = cbx + 2a + bxc − 2bx + ac − bbcc + d for each x∈ R. Show that f is periodic, that is, there exists p > 0 such that f (x + p) = f (x) for all x∈ R.
Solution: Note that for any integer n, f (x + n + a)− f(x + n)
= cbx − b + nc + 2a + bx − b + nc − 2bx − b + n) + ac − bbcc + d
= cb(x − b) + n + 2a + bx − bc + n − 2bx − b + ac − 2n − bbcc + d
= cbx − b) + 2a + bx − bc − 2bx − b) + ac − bbcc + d
= f (x + a)− f(x).
Let m be a positive integer such that am is an integer. Then for all natural number k,
f (x + kam)− f(x)
=
x
X
j=1 m
X
i=1
(f (x + jam + ai)− f(x + jam + a(i − 1)))
= k
m
X
i=1
(f (x + ai)− f(x + a(i − 1)))
= k(f (x + am)− f(x)).
Since f (x)∈ [−1, 1], f(x + kam) − f(x) is bounded, so f(x + am) − f (x) must be 0. This implies f (x + am) = f (x), so f (x) is periodic.
2. Let AB be a given line segment. Find all possible points C in the plane such that in the triangle ABC, the altitude from A and the median from B have the same length.
Solution: Let D be the foot of the altitude from A and E be the foot of the median from B. Let F be the foot of the perpen-dicular from E to line BC. Then EF k AD and E is the midpoint
of AC, so EF = 1/2(AD) = 1/2(BE) and ∠EBC = ±π/6. (All angles will be directed mod π unless otherwise specified.) Now, let P be the point such that B is the midpoint of AP . Then BEk P C, so ∠P CB = ∠EBC is constant. The locus of all points C such that
∠P CB is constant is a circle. Hence the locus of C here consists of two congruent circles through BP (one corresponding to π/6 and the other to−π/6). Using the the particular case of ABC an equilateral triangle, we find that each circle has radius AB and center located so that ABQ = 2π/3 (undirected).
3. [Corrected] Let n≥ 3 be an integer, and suppose that the sequence a1, a2, . . . , anof positive real numbers satisfies ai−1+ ai+1= kiaifor some sequence k1, k2, . . . , kn of positive integers. (Here a0= an and an+1= a1.) Show that
2n≤ k1+ k2+· · · + kn≤ 3n.
Solution: The left inequality follows from AM-GM once we note that
k1+· · · + kn=
n
X
i=1
ai ai+1
+ai+1 ai
.
On the other side, we will actually show that k1+· · · + kn ≤ 3n − 2 for n≥ 2, by induction on n. For n = 2, if a1≥ a2, then 2a2= k1a1, so either a1= a2and k1+ k2= 4 = 3× 2 − 2, or a1= 2a2and again k1+ k2= 4 = 3× 2 − 2. For n > 2, we may assume the ai are not all equal; then there exists some i such that ai≥ ai−1, ai+1 with strict inequality in at least one of the two cases. Then ai−1+ ai+1 < 2ai
and so ki= 1. We conclude that the sequence with ai removed also satisfies the given condition with ki−1and ki+1decreased by 1 and ki
dropped. Since the sum of the resulting kiis at most 3(n− 1) − 2 by assumption, the sum of the original ki is at most 3n− 2, as desired.
4. Let k = 22n+ 1 for some positive integer n. Show that k is a prime if and only if k is a factor of 3(k−1)/2+ 1.
Solution: Suppose k is a factor of 3(k−1)/2+ 1. This is equiv-alent to 3(k−1)/2 ≡ −1 (mod k). Hence 3k−1 ≡ 1 (mod k). Let d be the order of 3 mod k. Then d6 |(k − 1)/2 but d|(k − 1), hence
(k− 1)|d, so d = k − 1 (since d must be smaller than k). Therefore k is prime.
Conversely, suppose k is prime. By quadratic reciprocity,
3
5. Let ABCD be a tetrahedron. Show that
(a) If AB = CD, AD = BC, AC = BD, then the triangles ABC, ACD, ABD, BCD are acute;
(b) If ABC, ACD, ABD, BCD have the same area, then AB = CD, AD = BC, AC = BD.
Solution:
(a) It follows from the hypothesis that the four faces are congruent and that the trihedral angle at each vertex is made up of the three distinct angles of a face. Now, let M be the midpoint of BC. By the triangle inequality,AM + M D > AD = BC = 2M C. Triangles ABC and DCB are congruent, so AM = DM . Thus 2M D > 2M C; that is, M D is greater than the radius of the circle in the plane of BCD with diameter BC. Therefore D lies outside this circle and angle BDC is acute. The same argument applies to every face angle. (This problem is USAMO 1972/2; see Klamkin’s USAMO book for other solutions.) (b) Since AB and CD are not parallel (or the tetrahedron is
pla-nar), we can uniquely choose two parallel planes, one (call this P ) containing AB, and one (call this Q) containing CD. Let the distance between P and Q be d. Also let A0 and B0 be the projections of A and B (respectively) onto Q, and let C0 and D0 be the projections of C and D onto P . Now, since trian-gles ACD and BCD have the same area, and they have the same base CD, they have the same altitude; call this h. Now construct the (infinite) cylinder with axis CD and radius h; ob-viously A and B are on this cylinder. They are also on plane P , which intersects the cylinder in one or two lines parallel to CD.
Now, A and B cannot be on the same one of these lines (or the tetrahedron is planar), so there are two intersection lines between the cylinder and plane P , one containing A and one containing B. They are also parallel to and equidistant from line C0D0, the distance being (h2− d2)1/2. Therefore, line C0D0 bisects line segment AB.
Analogous reasoning shows that line A0B0 bisects line segment CD, or, transferring everything from plane Q to plane P via projection, line AB bisects line segment C0D0. The fact that segments AB and C0D0 bisect each other implies that AC0BD0 is a parallelogram, so AC0 = BD0 (call this distance x) and BC0= AD0 (call this distance y). Then we have that
AC = (AC02+ C0C2)1/2= (x2+ d2)1/2
= (BD02+ D0D2)1/2= BD BC = (BC02+ C0C2)1/2= (y2+ d2)1/2
= (AD02+ D0D2)1/2= AD.
Using a different pair of edges at the outset, we also get AB = CD.
6. [Corrected] Let X be the set of integers of the form a2k102k+ a2k−2102k−2+· · · + a2102+ a0,
where k is a nonnegative integer and a2i ∈ {1, 2, . . . , 9} for i = 0, 1, . . . , k. Show that every integer of the form 2p3q, for p and q nonnegative integers, divides some element of X.
Solution: In fact, every integer that isn’t divisible by 10 divides some element of X. We first note that there exists a multiple of 4p in X with 2p− 1 digits for all nonnegative integer p. This follows by induction on p: it’s obvious for p = 0, 1, and if x is such a multiple for p = k, then we can choose a2k so that x + a2k102k≡ 0 (mod 4k+1) since 102k ≡ 1 (mod 4k).
Now we proceed to show that any integer n that isn’t divisible by 10 divides some element of X. Let n = 2pk, where k is odd. Then by the lemma above one can find a multiple of 2p in X. Let m be the multiple, d be the number of digits of m, and f = 10d+1− 1.
By Euler’s extension of Fermat’s theorem, 10φf k ≡ 1 (mod fk).
Therefore m(10(d+1)φ(f k)− 1)/(10d+1− 1) is divisible by 2pk and lies in X (since it is the concatenation of m’s).
7. Determine all positive integers k for which there exists a function f : N→ Z such that
(a) f (1997) = 1998;
(b) for all a, b∈ N, f(ab) = f(a) + f(b) + kf(gcd(a, b)).
Solution: Such f exists for k = 0 and k =−1. First take a = b in (b) to get f (a2) = (k + 2)f (a). Applying this twice, we get
f (a4) = (k + 2)f (a2) = (k + 2)2f (a).
On the other hand,
f (a4) = f (a) + f (a3) + kf (a) = (k + 1)f (a) + f (a3)
= (k + 1)f (a) + f (a) + f (a2) + kf (a)
= (2k + 2)f (a) + f (a2) = (3k + 4)f (a).
Setting a = 1997 so that f (a)6= 0, we deduce (k + 2)2 = 3k + 4, which has roots k = 0,−1. For k = 0, an example is given by
f (pe11· · · penn) = e1g(p1) +· · · + eng(pn),
where m is a prime factor of 1997, g(m) = 1998 and g(p) = 0 for all primes p6= m. For k = 1, an example is given by
f (pe11· · · penn) = g(p1) +· · · + g(pn).
8. Let ABC be an acute triangle with circumcenter O and circumradius R. Let AO meet the circumcircle of OBC again at D, BO meet the circumcircle of OCA again at E, and CO meet the circumcircle of OAB again at F . Show that OD· OE · OF ≥ 8R3.
Solution: Let D0, E0, F0 be AO∩ BC, BO ∩ CA, CO ∩ AB respec-tively. Then they are images of D, E, F respectively under an inver-sion through the circumcircle of ABC, since the inverinver-sion maps each
of the three circles through the sides of the triangle into the lines con-taining the sides. Therefore OD0·OD = OE0·OE = OF0·OF = R2. Hence the inequality in the problem is equivalent to:
AO i.e., ABC is an equilateral triangle.
9. For n≥ k ≥ 3, let X = {1, 2, . . . , n} and let Fk be a family of k-element subsets of X such that any two subsets in Fk have at most k− 2 common elements. Show that there exists a subset Mk of X with at leastblog2nc + 1 elements containing no subset in Fk. Solution: If k ≥ log2n then we have nothing to prove, so as-sume k < log2n. Let m =blog2nc + 1. Since each (k − 1)-element subset of X lies in at most one subset of Fk, and each element of Fk
contains k (k− 1)-element subsets, we have
#(Fk)≤ 1
On the other hand, for a randomly chosen m-element subset of X, the expected number of elements of Fk that it contains is
m
It suffices to prove that the latter is less than 1, for then some m-element subset must contain no m-element of Fk.
Of course mk ≤ Pi m
i = 2m, but this estimate is not quite suffi-cient for our purposes. Fortunately, one can easily prove the better estimate mk ≤ 3 · 2m−3 for m ≥ k ≥ 3, by induction on m. This gives
1 n− k + 1
m k
≤ 3n
4(n− k + 1) < 1 for n≥ 3, completing the proof.