1. Determine the smallest integer k for which there exists a graph on 25 vertices such that every vertex is adjacent to exactly k others, and any two nonadjacent vertices are both adjacent to some third vertex.
Solution: The following list includes all of the vertices: some given vertex v, every neighbor of v, and every neighbor of those neighbors other than v itself. The length of this list is 1 + k + k(k− 1), so k2+ 1≥ 25, or k ≥ 5.
Let us prove that k = 5 is impossible, by counting the number of 4-cycles in such a graph. In the list we made above, there must be exactly one repetition; by construction, it cannot be v that appears more than once. Moreover, if two neighbors of v are adjacent, then each appears twice on the list, which cannot happen. Thus two nonneighbors of v must coincide, and so v belongs to exactly one four-cycle. The same reasoning applies to each vertex; however, 25 vertices cannot be partitioned into 4-cycles, a contradiction.
We now exhibit a graph of the desired form with k = 6. Construct five 5-cycles, and between any two 5-cycles, draw five edges joining each vertex on one cycle to one on the other, so that adjacent vertices on one cycle are joined to nonadjacent vertices on the other. (That is, make each pair of cycles into a Petersen graph.) Then any two vertices lie in a copy of the Petersen graph, and so are adjacent or have a common neighbor.
2. Find the largest real number α for which there exists an infinite sequence a1, a2, . . . of positive integers satisfying the following prop-erties.
(a) For each n∈ N, an > 1997n.
(b) For every n ≥ 2, aαn does not exceed the greatest common divisor of the set{ai+ aj: i + j = n}.
Solution: The largest possible value of α is 1/2.
First, suppose (an)∞n=1 is a sequence such that conditions (a) and (b) hold.
Lemma 1 For every > 0, there are infinitely many values of n∈ N for which a2n≥ a2n−.
Proof. Let > 0, and suppose there exists N ∈ N such that for all n > N , a2n < a2n−. Taking logs and dividing by 2n gives
log a2n
2n <2−
2 · log an
n , so
log a2kn
2kn < 2 − 2
k
log an
n → 0 as k → ∞, impossible as an≥ 1997n so (log an)/n≥ log 1997 for all n. 2 Now let n be one of the values given by the Lemma, so that a2n−≤ a2n. Then
a(2n−)α≤ aα2n≤ gcd { ai+ aj | i + j = 2n } ≤ 2an,
so 2≥ a1n−(2−)α ≥ 1997n(1−(2−)α); since this is true for infinitely many values of n∈ N, we must have α ≤ 1/(2 − ). Since > 0 was arbitrary, we must in fact have α≤ 1/2.
Now we give a sequence satisfying conditions a and b with α = 1/2.
Denote the nth Fibonacci number by Fn. Let t be an even integer such that F2tn > 1997n for all n ∈ N, and define the sequence (an)∞n=1 by an = 3F2tn. Then condition (a) clearly holds. I claim that Ftn| F2ti+ F2tj when i + j = n, so gcd{ ai+ aj | i + j = n } ≥ 3Ftn. In fact,
F2ti = Ft(i+j)Ft(i−j)+1+ Ft(i+j)−1Ft(i−j) F2tj = Ft(i+j)Ft(j−i)+1+ Ft(i+j)+1Ft(j−1)
so
F2ti+ F2tj = 2Ft(i+j)Ft(i−j)+1+ (Ft(i+j)+1− Ft(i+j)−1)Ft(j−i)
= Ft(i+j)(2Ft(i−j)+1− Ft(i−j)) and the claim holds. Now,
an= 3F2tn= 3Ftn(Ftn+1+Ftn−1)≤ 9Ftn2 ≤ (gcd{ ai+aj| i+j = n })2, so a1/2n ≤ gcd{ ai+ aj| i + j = n } and the sequence (an)∞n=1satisfies the conditions of the problem with α = 1/2.
3. [Corrected] Let f : N→ Z be the function defined by
f (0) = 2, f (1) = 503, f (n + 2) = 503f (n + 1)− 1996f(n).
For k∈ N, let s1, . . . , sk be integers not less than k, and let pi be a prime divisor of f (2si) for i = 1, . . . , k. Prove that for t = 1, . . . , k,
2t|
k
X
i=1
pi if and only if 2t| k.
Solution: First, one can easily prove by induction that f (n) = 4n+ 499n for n≥ 0.
Lemma 2 If p is an odd prime, m, n∈ N not divisible by p, s ≥ 0, and p| m2s+ n2s, then p≡ 1 (mod 2s+1).
Proof: Let g be a primitive root mod p, and take k such that gk ≡ mn−1 (mod p). As p divides m2s + n2s, m2s ≡ −n2s (mod p) so (mn−1)2s ≡ −1 (mod p); therefore g2sk ≡ −1 (mod p) and 2sk≡ (p − 1)/2 (mod p − 1), so 2sk = l(p− 1)/2 with l odd. Thus 2s divides (p− 1)/2 and 2s+1 divides p− 1. 2
Now let k∈ N, s1, . . . , sk integers not less than k, and pi a prime divisor of f (2si) = 42si+ 4992si for each i. Then pi is not 2 or 499, so by the Lemma 2si+1 | pi − 1; in particular pi ≡ 1 (mod 2k).
Therefore
k
X
i=1
pi≡ k (mod 2k) so 2t|
k
X
i=1
pi if and only if 2t| k.
4. Find all pairs (a, b) of positive reals such that for every n∈ N and every real number x satisfying
4n2x = log2(2n2x + 1), we have ax+ bx≥ 2 + 3x.
Solution: The pairs satisfying this condition are{ (a, b) | ab ≤ e3}.
Let S ={ x ∈ R | ∃n ∈ N : 4n2x = log2(2n2x + 1)}; since 4n2x = log2(2n2x+1) ⇐⇒ 42n2x= 2n2x+1 ⇐⇒ 2n2x = −1
2 or 0,
S ={ −1/4n2| n ∈ N }∪{0}. We want to show that ax+ bx≥ 2+3x for all x∈ S if and only if ab ≤ e3.
First, suppose ab≤ e3. Then a0+ b0= 2 = 2 + 3· 0; suppose x ∈ S, x < 0. By the power mean inequality
ax+ bx 2
1/x
≤√
ab≤ e3/2 so (as x < 0)
ax+ bx≥ 2e3x/4≥ 2(1 +3x
2 ) = 2 + 3x.
Now suppose ax+ bx≥ 2 + 3x for all x ∈ S. Note that
xlim→0
ax− 1
x = log a, lim
x→0
bx− 1
x = log b, so
xlim→0
ax+ bx− 2
x = log ab, but for x∈ S, x 6= 0,
ax+ bx− 2
x ≥ 3.
Since 0 is an accumulation point of S\ {0}, we must have log ab ≥ 3, so ab≥ e3. Therefore ax+ bx≥ 2 + 3x for all x ∈ S if and only if ab≥ e3.
5. [Corrected] Let n, k, p be positive integers such that k≥ 2 and k(p + 1)≤ n. Determine the number of ways to color n labeled points on a circle in blue or red, so that exactly k points are colored blue, and any arc whose endpoints are blue but contains no blue points in its interior contains at least p red points.
Solution: We may fix the location of one blue point as long as we multiply our final answer by n/k. (By symmetry, the number of arrangements in which a given point is blue is the same for each point.) We are distributing n− k red points into the k arcs formed by the blue points so that each arc receives at least p points. Equiv-alently, we are distributing n− (k + 1)p points into the k arcs with
no restrictions. Remembering to multiply by n/k as noted earlier, our answer becomes
n k
n − (k + 1)p + k − 1 k− 1
.
2 1997 Regional Contests: Solutions
where the denominators contain partial sums of the sequence of re-ciprocals of triangular numbers. Prove that S > 1001.Solution: Note that
Solution: Note that 2 divides 2n + 2 for all n. Also, 11 divides 2n+ 2 if and only if n≡ 6 (mod 10), and 43 divides 2n+ 2 if and only if n≡ 8 (mod 14). Since n = 946 = 2 · 11 · 43 satisfies both congruences, n divides 2n+ 2.
3. Let ABC be a triangle and let la= ma
Ma, lb= mb
Mb, lc= mc
Mc,
where ma, mb, mcare the lengths of the internal angle bisectors and Ma, Mb, Mc are the lengths of the extensions of the internal angle bisectors to the circumcircle. Prove that
la
sin2A+ lb
sin2B + lc
sin2C ≥ 3,
with equality if and only if ABC is equilateral.
Solution: Let the bisector of A intersect BC at P and the circle at Q. Then
AP
AB = sin B
sin 180◦− B − A/2, AB
AQ = sin C sin B + A/2. Therefore
la = AP
AQ = sin B sin C (sin B + A/2)2. We compute lb and lc similarly, and deduce that
la
sin2A+ lb
sin2B+ lc
sin2C ≥sin B sin C
sin2A +sin C sin A
sin2B +sin A sin B sin2C ≥ 3 by AM-GM. (Equality only occurs if sin(B + A/2) = 1 and so forth, which forces A = B = C = 60◦.)
4. The triangle A1A2A3 has a right angle at A3. For n ≥ 3, let An+1
be the foot of the perpendicular from An to An−1An−2.
(a) Show that there is a unique point P in the plane interior to the triangles An−2An−1An for all n≥ 3.
(b) For fixed A1 and A3, determine the locus of P as A2 varies.
Solution:
(a) First, notice each triangle is contained in the previous triangle.
Therefore, any point contained in one triangle is contained in the previous ones. Thus such a point P must exist. Moreover, each triangle is similar to the previous one with the same ratio of similarity; hence the maximum distance between two points in the triangle also decreases by this ratio, and so two points inside all of the triangles cannot lie at any positive distance.
Thus P is unique.
(b) Let the first five points be A, B, C, D, E. Then triangle CDE is similar to triangle ABC. Also, the points are in the same order, so we’ll be dropping perpendiculars in the same order for both. Therefore, P is in the same position relative to ABC
as CDE. In particular, triangles AP C and CP E are similar, so
∠AP C = ∠CP E, ∠P AC = ∠P CE.
However, ∠P AC + ∠P CA = ∠P CE + ∠P CA = 90. Therefore
∠AP C = 90 and ∠CP E = ∠AP C = 90, so AP E is a straight line.
Let ∠CAB = A; then
CE = CD cos A = CA cos A sin A = CA/2 sin 2A, so 0≤ CE ≤ CA/2 and both bounds are optimal. Now the lo-cus of the projection of C on AE (which is P ) is an arc of the cir-cle with diameter AC with endpoint C and angle 2 arctan(1/2) (since the maximal value of tan ∠CAE is 1/2 and the central angle is twice ∠CAE).
5. Persons A1, . . . , An (n≥ 3) are seated in a circle in that order, and each person Ai holds a number ai of objects, such that (a1+· · · + an)/n is an integer. It is desired to redistribute the objects so that each person holds the same number; objects may only be passed from one person to either of her two neighbors. How should the redistribution take place so as to minimize the number of passes?
Solution: Note that allowing negative numbers of objects does not matter, since we can rearrange the moves to avoid the negative numbers.
First suppose we make no moves between A1and An. Consider the quantity
x =|b1| + |b1+ b2| + · · · + |b1+ b2+· · · + bn|,
where bi = ai minus the number of objects Ai ends up with. This quantity equals 0 if and only if we have finished. Moreover, a move between aiand ai+1changes|b1+ . . . + bi| by 1 and does not change the other terms. Hence at least x moves are needed.
Finally, note that there always exists a move decreasing x by 1:
move between ai and ai+1, where i is the smallest integer such that
|b1+ b2+· · · + bi| is nonzero. Hence a redistribution exists using only x moves.
Now we allow moves between A1and An, as in the original problem.
Clearly the optimal strategy is to move between A1 and An if such a move decreases x by 2 or more, and otherwise move to decrease x by 1 as above.