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All use subject to JSTOR Terms and Conditions with 1 < k < n, let [n] denote {1, 2, ... , n}, andlet P be the discrete probabilityspace consisting of the k-element subsets of [n], each equally likely. For S e P, let 7rs be the permutationof [n] obtained by listing S in increasing orderfollowed by [n] - S in increasing order. Let X (S) be the number of inversions in -rs (that is, the number of instances of i, j E [n] such that j < i and 7rs(j) > Trs(i)). Prove that Var(X)/E(X) does not depend on k. 11043. Proposed by Michael Golomb, Purdue University, WestLafayette, IN. Let f and g be nonnegative real-valued L'1-functionson a finite interval I, and suppose that

ff - g = 1. (a) Show that for q E Z with q > 2 there exists a subinterval J of I such that

If

=fga= (b)* What happens if the restrictionto nonnegative functions is dropped? 11044. Proposed by Michel Bataille, Rouen, France. Let N be an odd integer greater than 1. Let Ao = 1, and for k E N let

Ak- reR -)

H(r) i=1l1.2i ' where R is the set of all k-tuples of nonnegative integers such that j= jr = k and a (r) denotes y j= rj. For n E N, show that 1/q.

kri (1)n-1l(N (--2n -2kk--1 1 Ak= N2 (N 11045. Proposed by Manoj Prakash Singh, New Delhi, India. Prove that when n is a sufficiently large positive integer there exists a finite set S of prime numbers such that the sum of Ln/pJ over p E S is equal to n. 11046. Proposed by ChristophSoland, Neuchatel, Switzerland.Let ABC be a triangle, let I be the incircle of ABC, and let r be the radius of I. Let K1, K2, and K3 be the three circles outside I and tangent to I and to two of the three edges of ABC. Let ri be the radius of Ki, for 1 < i < 3. Show that r = rYr2 + r2r3 + ,r3rI.

10913 [2001, 977]. Proposed by Donald E. Knuth,Stanford University,Stanford, CA. Given a positive integer n, let akbe the transposition((k - 1) mod (n + 1), k mod (n + 1)), and let bkbe the transposition (k mod n, n). For example, for n = 3, we have ao, al, a2, a3, a4, . . . = (3, 0), (0, 1), (1, 2), (2, 3), (3, 0), ....

bo, bl, b2, b3, b4, . . . = (0, 3), (1, 3), (2, 3), (0, 3), (1, 3) ....

Prove that aoal ... ak = bk... blbo for every k > 0. 844

@ THEMATHEMATICAALSSOCIATIONOFAMERICA [Monthly 110 Solution by Reiner Martin, New York,NY Let k = rn + s with 0 < s < n. Also, let rt = (0, 1, .. . , s - 1, s, n), and let r = rt,_. We prove by induction on k that both specified products equal rsTr, where k = rn + s with 0 < s < n. For bk... bIbo, the induction step follows from bi ri-1 = ri for 0 < i < n.

Next note that rai-r-j = ai+j for all positive integers i and j. With k = rn + s, this yields sTrak+l= = isar+k+1tr = Tsar(n+l)+s+lTr = tsas+lr.

The last expression equals rs+Ir' if 0 < s < n - 1, while it equals rotr+l if s = n - 1. Now aoal ... ak = t-r follows by induction on k, as desired.

Also solved by G. Body (U. K.), D. Callan,R. Chapman(U. K.), B. Klimszova(CzechRepublic),P.Kordulova (Czech Republic), J. H. Lindsey II, O. P. Lossers (Netherlands),J. Schlosberg (student),R. Stong, F. Szeged (Hungary),L. Zhou, the GCHQ Problem Solving Group (U. K.), the National Security Agency Problems Group,andthe proposer.

A Sequenceof Integer Polynomials 10915 [2002, 77]. Proposed by C. P Rupert, Durham, NC. Given nonzero polynomials p and q in Z[x] satisfying p2 + mq # 0 for 1 < m < 4, define polynomials t, recursively by t,+2 = pt,+4 + qt, with initial conditions to = 0 and t1 = 1. With [ denoting the Mbbius function, prove for n > 1 that the polynomial s E Q[x] defined by s,(x) =

tdg(n/d) actually belongs to Z[x].

Solution byHRdlinchard Stong, Rice University, Houston, TX. Since s1(x) = tl = 1 Z[x], we assume n > 1. The two roots of the characteristicpolynomial X2- p. - q are ri = (p + p2 + 4q)/2 and r2 = (p - p2 +4q)/2. By assumption, p2 + 4q 0, and thus rl : r2. Solving the recurrencefor t, gives n rn - r2 r'n[(rl/r2)n - 1] tn ri - r2 - - p24q

For s, to be defined, we need tn, 0, which holds if r /r2 is not a root of unity. Let f(X.) = + (p2/q + 2)X-+ 1. Note that rj/r2 is a zero of f, and that zeros of f can only be-2roots of unity if (p2/q + 2) E {-2, -1, 0, 1, 2}. This is not the case, since p 0 and p2 + mq A0 for I < m < 4.

Now recall that Ed, n(n/d) = 0 for n > 1. Also,qddn dlt(n/d) -=0(n), where 0 is Euler's totient function, and -dl xd - 1)g(n/d) =n- (x), where Oi is the nth cyclotomic polynomial. Combining these facts shows that s, (x) = r20~) (r1/r2). Also,

Sn(x) is a self-reciprocal polynomial in Z[x] of degree 0(n). Therefore, s,(x) is a symmetric polynomial in rl and r2, and thus sn(x) e Z[r, + r2, rir2] = Z[p, -q] C Z[x].

Also solved by S. Amghibech (France),D. Callan, R. Chapman(U. K.), NSA ProblemsGroup,R. Richberg (Germany),L. Zhou, and the proposer.

Distancesto the Sides 10929 [2002, 298]. Proposed by Lajos Csete, Hungary. Let P be a point in the interior of triangle ABC, and let rl, r2, r3 denote the distances from P to the sides of the triangle with lengths a,, a2, a3, respectively. Let R be the circumradiusof ABC, and let 0 < a < 1 be a real number.Let b = (2a)/(1 - a). Prove that

+ b rNao+vermab2e0rr0O3B]<LEMA(N2RD(S)aSOL+UaTIONb)1-a

PR November 2003]

PROBLEMSAND SOLUTIONS 845 845