1. Let p1, p2, p3, . . . be the prime numbers listed in increasing order, and let x0 be a real number between 0 and 1. For positive integer k, define
where {x} = x − bxc denotes the fractional part of x. Find, with proof, all x0satisfying 0 < x0< 1 for which the sequence x0, x1, x2, . . . eventually becomes 0.
Solution: The sequence eventually becomes 0 if and only if x0
is a rational number.
First we prove that, for k≥ 1, every rational term xk has a rational predecessor xk−1. Suppose xk is rational. If xk = 0 then either xk−1= 0 or pk/xk−1is a positive integer; either way, xk−1is rational.
If xk is rational and nonzero, then the relation xk =
which shows that xk−1is rational. Since every rational term xk with k≥ 1 has a rational predecessor, it follows by induction that, if xk is rational for some k, then x0is rational. In particular, if the sequence eventually becomes 0, then x0 is rational.
To prove the converse, observe that if xk−1= m/n with 0 < m < n, then xk = r/m, where r is the remainder that results from dividing npk by m. Hence the denominator of each nonzero term is strictly less than the denominator of the term before. In particular, the num-ber of nonzero terms in the sequence cannot exceed the denominator of x0.
Note that the above argument applies to any sequence{pk} of posi-tive integers, not just the sequence of primes.
2. Let ABC be a triangle, and draw isosceles triangles BCD, CAE, ABF externally to ABC, with BC, CA, AB as their respective bases. Prove that the lines through A, B, C perpendicular to the lines EF, F D, DE, respectively, are concurrent.
First Solution: We first show that for any four points W, X, Y, Z in the plane, the lines W X and Y Z are perpendicular if and only if W Y2− W Z2= XY2− XZ2. (∗) To prove this, introduce Cartesian coordinates such that W = (0, 0), X = (1, 0), Y = (x1, y1), and Z = (x2, y2). Then (∗) becomes
x21+ y12− x22− y22= (x1− 1)2+ y21− (x2− 1)2− y22,
which upon cancellation yields x1 = x2. This is true if and only if line Y Z is perpendicular to the x-axis W X.
If P is the intersection of the perpendiculars from B and C to lines F D and DE, respectively, then the fact noted above yields
P F2− P D2= BF2− BD2, and
P D2− P E2= CD2− CE2.
From the given isosceles triangles, we have BF = AF, BD = CD, and CE = AE. Therefore
P F2− P E2= AF2− AE2.
Hence line P A is also perpendicular to line EF , which completes the proof.
Second Solution: Let C1 be the circle with center D and radius BD,C2 the circle with center E and radius CE, andC3 the circle of center F and radius AF . The line through A and perpendicular to EF is the radical axis of circlesC2 and C3, the line through B and perpendicular to DF is the radical axis of circlesC1andC3, and the line through C and perpendicular to DE is the radical axis of circles C1 andC2. The result follows because these three radical axes meet at the radical center of the three circles.
Third Solution: Let A0, B0, C0be points on EF , DF , DE, respec-tively, with AA0 ⊥ EF , BB0 ⊥ DF , and CC0 ⊥ DE. In addition, let D0, E0, F0be points on BC, AC, AB, respectively, with DD0⊥ BC, EE0 ⊥ AC, and F F0 ⊥ AB. Because DD0, EE0, and F F0 are the perpendicular bisectors of the sides of triangle ABC, these three lines are concurrent, meeting at the circumcenter of triangle ABC.
Thus, by the trigonometric form of Ceva’s Theorem applied in tri-angle DEF , Because ∠F DD0 and ∠CBB0 have orthogonal sides, the two angles are congruent. Similarly, ∠EDD0 is congruent to ∠BCC0, ∠DEE0 is congruent to ∠ACC0, ∠F EE0 is congruent to ∠CAA0, ∠EF F0 is congruent to ∠BAA0, and ∠DF F0 is congruent to ∠ABB0. Using these congurences in in (∗) results in
sin ∠BCC0
By the trigonometric form of Ceva’s Theorem applied in triangle ABC, it follows that AA0, BB0, and CC0 are concurrent.
3. Prove that for any integer n, there exists a unique polynomial Q with coefficients in{0, 1, . . . , 9} such that Q(−2) = Q(−5) = n.
Solution: First suppose there exists a polynomial Q with coeffi-cients in{0, 1, . . . , 9} such that Q(−2) = Q(−5) = n. We shall prove that this polynomial is unique. By the Factor Theorem, we can write Q(x) = P (x)R(x) + n where P (x) = (x + 2)(x + 5) = x2+ 7x + 10 and R(x) = r0+ r1x + r2x2+· · · is a polynomial. Then r0, r1, r2, . . . are integers such that
10r0+ n, 10rk+ 7rk−1+ rk−2 ∈ {0, 1, . . . , 9}, k≥ 1 (1) (with the understanding that r−1 = 0). For each k, 1 uniquely determines rk once rj is known for all j < k. Uniqueness of R, and therefore of Q, follows.
Existence will follow from the fact that for the unique sequence{rk} satisfying 1, there exists some N such that rk = 0 for all k ≥ N.
First note that{rk} is bounded, since |r0|, |r1| ≤ B and B ≥ 9 imply
|rk| ≤ B for all k. This follows by induction, using 10|rk| ≤ 7|rk−1|+
|rk−2| + 9 ≤ 10B. More specifically, if ri≤ M for i = k − 1, k − 2, then
rk ≥ −7rk−1
10 −rk−2
10 ≥ −4M 5 , while if ri≥ L for i = k − 1, k − 2, then
rk≤ −7rk−1
10 −rk−2 10 + 9
10 ≤ −4L 10 + 9
10. Since the sequence{rk} is bounded, we can define
Lk = min{rk, rk−1, . . .}, Mk= max{rk, rk+1, . . .}.
Clearly Lk≤ Lk+1 and Mk ≥ Mk+1 for all k.
Since Lk≤ Mkfor all k, the non-decreasing sequence{Lk} must stop increasing eventually, and, similarly, the non-increasing sequence {Mk} must stop decreasing. In other words, there exist L, M, N such that Lk= L and Mk= M for all k≥ N. Certainly L ≤ M, and M ≥ 0, since no three consecutive terms in {rk} can be negative, but the above arguments also imply L≥ −4M/5 and M ≤ −4L/5+9/10.
A quick sketch shows that the set of real pairs (L, M ) satisfying these conditions is a closed triangular region containing no lattice points other than (0, 0). It follows that rk = 0 for all k ≥ N, proving existence.
4. To clip a convex n-gon means to choose a pair of consecutive sides AB, BC and to replace them by the three segments AM, M N , and N C, where M is the midpoint of AB and N is the midpoint of BC.
In other words, one cuts off the triangle M BN to obtain a convex (n + 1)-gon. A regular hexagonP6 of area 1 is clipped to obtain a heptagonP7. ThenP7is clipped (in one of the seven possible ways) to obtain an octagonP8, and so on. Prove that no matter how the clippings are done, the area ofPn is greater than 1/3, for all n≥ 6.
Solution: The key observation is that for any side S of of P6, there is some sub-segment of S that is a side ofPn. (This is easily proved by induction on n.) Thus Pn has a vertex on each side of P6. SincePn is convex, it contains a hexagonQ with (at least) one
vertex on each side ofP6. (The hexagon may be degenerate, as some of its vertices may coincide.)
LetP6= A1A2A3A4A5A6, and letQ = B1B2B3B4B5B6, with Bion AiAi+1 (indices are considered modulo 6). The side BiBi+1 ofQ is entirely contained in triangle AiAi+1Ai+2, soQ encloses the smaller regular hexagon R (shaded in the diagram below) whose sides are the central thirds of the segments AiAi+2, 1≤ i ≤ 6. The area of R is 1/3, as can be seen from the fact that its side length is 1/√
3 times the side length ofP6, or from a dissection argument (count the small equilateral triangles and halves thereof in the diagram below).
Thus Area(Pn) ≥ Area(Q) ≥ Area(R) = 1/3. We obtain strict inequality by observing thatPn is strictly larger that Q: if n = 6, this is obvious; if n > 6, then Pn cannot equal Q because Pn has more sides.
Note. With a little more work, one could improve 1/3 to 1/2. The minimal area of a hexagonQ with one vertex on each side of P6is in fact 1/2, attained when the vertices ofQ coincide in pairs at every other vertex ofP6, so the hexagonQ degenerates into an equilateral triangle. If the conditions of the problem were changed so that the
“cut-points” could be anywhere within adjacent segments instead of just at the midpoints, then the best possible bound would be 1/2.
5. Prove that, for all positive real numbers a, b, c,
(a3+ b3+ abc)−1+ (b3+ c3+ abc)−1+ (c3+ a3+ abc)−1≤ (abc)−1.
Solution: The inequality (a− b)(a2− b2) ≥ 0 implies a3+ b3 ≥ ab(a + b), so
1
a3+ b3+ abc ≤ 1
ab(a + b) + abc= c abc(a + b + c). Similarly
1
b3+ c3+ abc ≤ 1
bc(b + c) + abc = a abc(a + b + c),
and 1
c3+ a3+ abc ≤ 1
ca(c + a) + abc= b abc(a + b + c).
Thus
Solution: Any x that lies in all of the half-open intervals In= an
n , an+ 1 n
, n = 1, 2, . . . , 1997 will have the desired property. Let
L = max [L, U ) has the desired property.
We prove (∗) for all m, n ranging from 1 to 1997 by strong induction.
The base case m = n = 1 is trivial. The induction step splits into three cases. If m = n, then (∗) certainly holds. If m > n, then the induction hypothesis gives (m− n)an < n(am−n+ 1), and adding n(am−n+ an) ≤ nam yields (∗). If m < n, then the induction hypothesis yields man−m < (n− m)(am+ 1), and adding man ≤ m(am+ an−m+ 1) gives (∗).