Vol. 10 No 5 - Pi Mu Epsilon

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PROBLEMS AND SOLUTIONS 425865. [Fall 19951 Proposed by Miguel Amengwl Covas, Mallorca,Spain.Let ABC be a triangle with sides of lengths a9 b, and c, semiperimeters, and area K. Show that, if Ea(s - a) = 4K, then the three circlescentered at the vertices A, B, and C and of radii s - a9 s - b, and s - c,respectively, are all tangent to the same straight line.Solution by ?Villiam H. Peirce, Delray Beach, Florida.Let rl = s - a, r2 = s - b, and r3 = s - c be the radii of the three circlescentered respectively at A, B, and C. From these defmitions we get a = r2+ r3, b = r3 + rl, c = rl + r2, and s = rl + r2 + r3. ThenSince we havewe may eliminate K between these two expressions to obtainThat is9 F = 0 is equivalent to the statement Za(s - a) = 4K.Consider any two extemally tangent circles of radii rl and r2. By simplegeometry the distance between the two points of tangency along an externaltaugent line is 2 a . Therefore, when the three externally tangent circlesof the problem are tangent to the same line, one and only one of thefollowing three situations will occur:are the four factors of F, so that F = 0 is equivalent to the three circlesbeing tangent to the same straight line. The theorem follows.Also solved by Paul S. Bruckman, David Iny, and the Proposer.866. [Fall 19951 Proposed by J. Rodriguez, Sonora, Mmko.For any nonzero integer n, the Smra~chefincZion is the smallestinteger S(n) such that (S(n))! is divisible by n. Thus S(12) = 4 since 12divides 4! but not 3!.a) Find a strictly increasing idmite sequence of integers such that forany consecutive three of them the Smarandache function is neither increasingnor decreasing.*b) Find the longest increasing sequence of integers on which theSmarandache function is strictly decreasing.I. Solution by David Zny, Baltimore, Maryland.a) Obviously, if p is prime9 then S@) = p. Also, if p is an odd primegreater than or equal to 5, then p + 1 is divisible by distinct integers 2 and... is any@ + 1112, so that S@ + 1) S @ + 1)12. Thus, if p,, p2Âincreasing sequence of primes each greater than or equal to 5, then thesequence of integers pl, pl + 1, p2* p2 + 1, . . is an increasing sequencewhose Smarandache function values alternately increase and decrease.b) We extend the observation of Pa (a) to note that, if p is prime andk Now we fmd primes pl, p2, .. ,pn7 all greater than n, so that SM = Qk. If pl, p;, pi, ..., # is increasingwith pl, 2p2Â ..., npn decreasing, we are done. Recall that asymptoticallythere are mllnm primes less than or equal to m.The construction now follows. Fix n > > 1 and pick primep" > > nn.Suppose we have already picked pa, pn-l . . . , pk+ for k 2 1. Then we picka prime pk so that A < 4:; and Qk > (k + l)pk+ That is,These three expressions, along with the nonzero factorSince nn < < pn < pn-l < ... < pk+17 then we have that(k+lyk 2 (n911kpk+I 2 npk+l since k s n - 1.Pk+l
PROBLEMS AND SOLUTIONS 427Thus it is sufficient to pick pk prime so thatk+l~Pk+l < Pk < nPk+l-We estimate the number of possible primes asul=x+y+z+w,u2 = q' + yz + zw + wx + xz + yw,u3 = xyz + yzw + mx + wxy, andU4 = q'ZW.Let si = 2 + y i + z i + d. Then the system can be written assl = ul = 7, s2 = ISy s3 = 37, and u4 = 6.since n > > 1. This completes the construction.11. Comment by Paul S. Bruckman, Salmya, Kuwait.This problem appeared almost verbatim in Z%e Fibonacci Quarterly,Vol. 32Â No. 1, February 1994, by the same proposer, as Problem H-484.My solution appeared in the same journal, Vol. 33, No, 2, May 1995, pp.189-192.ALSO solved by Charles Ashbacher, James Campbell, Wiiam Chau,Thomas C. Leong, H.-J. Seiffert, Rex H. Wu, and the Proposer.867. pall 19951 Proposed by Seung-Jin Bang, AJOU Universiv,Suwon, Korea.Fid the number of solutions (x, y, z, w) to the systemx+y+z+w=72+f+2+d=152+f+2+d=37xyzw = 6.Solution by Henry S. Liebennan, Waban, Massachusetts.The solutions to the system are all of the permutations of (1, 1,2,3), ofwhich there are 12. The left sides of the equations are rational integralsymmetric functions of x, y, z, and w. We note the following elementaysymmetric hctions of x, y, z, and w:We claim that (x, y, z ,w) solves the system if and only if these values arethe zeros of the polynomial in f, f - ale + u23 - u3s + u4. From formulasin [I] we see thatWe have a, = 7 and u4 = 6, and from (11,which yield u2 = u3 = 17. It is easy to discover that the zeros ofare 1, 1,2, and 3. Hence, (1, 1,2,3) and its permutations solve the system.Reference1. B. L. Van der Waerden, Modem Algebra, vol. 1, Frederick UngarPublishing Company, New York, 1953, page 81.Ako solved by Miguel Amengual Covas, Pad S. Bruchan, WilliamChau, Russell Euler, Mark Evans, Robert C. Gebhardt, S. Gendler, RichardI. Hess* David Iny, Carl Libis, David E. Manes, Yoshinobu Murayoshi,Ke~eth M. Wilke, and the Proposer.
Page 2 and 3: - Another Matching Problem with aMa
Page 4 and 5: Red Light, Green Light:A Model of T
Page 6 and 7: BENNINK. RED LIGHT, GREEN LIGHT 357
Page 8 and 9: BENNINK, RED LIGHT, GREEN LIGHT 361
Page 10 and 11: 365 PI MU EPSILON JOURNALROUSSEAU,
Page 12 and 13: 369 PI MU EPSILON JOURNALwe havec(n
Page 14 and 15: 3 73 PI MU EPSILON JOURNALThis can
Page 16 and 17: VIAR, DIMENSION FORMULA 377subspace
Page 18 and 19: TOMFORDE, SELF-SIMILARITY 38 1m, n
Page 20 and 21: TOMFORDE, SELF-SIMILARITY 385Note t
Page 22 and 23: TOMFORDE, SELF-SIMILARITY 3 896. G.
Page 24 and 25: 393 PI MU EPSILON JOURNALA Particul
Page 26 and 27: 397 PI MU EPSILON JOURNALy = ax - b
Page 28 and 29: 40 1 PI MU EPSILON JOURNALFigure 8P
Page 30 and 31: The Power Means Theorem via the Wei
Page 32 and 33: MISCELLANY 409The 1995 GameIn die F
Page 34 and 35: 0. Co-author of convexity theoremP.
Page 36 and 37: Pi MU EPSILON JOURNALii) JK is para
Page 38 and 39: PROBLEMS AND SOLUTIONS 42 1Henry S.
Page 42 and 43: PROBLEMS AND SOLUTIONS 429868. [Fal
Page 44 and 45: BENNINK, RED LIGHT, GREEN LIGHT 3 6
Page 46 and 47: PROBLEMS AND SOLUTIONS 437Solution
Page 48 and 49: Editor's NoteThe Pi Mu Epsilon Jour

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