Vol. 10 No 5 - Pi Mu Epsilon

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 < p is a positive integer, then S(p9 = @. 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