Vol. 10 No 7 - Pi Mu Epsilon

578
PI MU EPSILON JOURNAL
PROBLEMS AND SOLUTIONS 579
and
1 2 (2n- Ii- 2 2 (2n- 2) 2 + ••• + (-l)n+ 1 n 2 n 2 = ..!.n[l - (-1rn 3 ].
2
917. Proposed by Murray S. Klamkin, University of Alberta, Edmonton,
Alberta, Canada.
Determine necessary and sufficient conditions on the real numbers w 1
,
w2, ... , wn so that for all vectors V; in Em,
918. Proposed by Murray S. Klamkin, University of Alberta, Edmonton,
Alberta, Canada.
Evaluate the integral
(n/3 Pi
I = lo ln(l + y3 tanx)dx.
919. Proposed by the Editor.
Erect directly similar nondegenerate triangles DBC, ECA, FAB on sides
BC, CA, AB of triangle ABC. At D, E, F center circles of radii k•BC, k•CA,
k•AB respectively for fixed positive k. Let P be the radical center of the three
circles. If P lies on the Euler line of the triangle, show that it always falls on
the same special point.
*920. Proposed by Richard 1 Hess, Rancho Palos Verdes, California.
The sorted Fibonacci sequence is produced by starting with the first two
terms 1 and 1 and defining each succeeding term as the sum of the prior two
terms with the digits sorted into ascending order. Thus we have 1, 1, 2, 3,
5, 8, 13, 12, 25, 37, 26, ... This sequence eventually falls into a repeating
cycle.
a) Are there any two initial terms that produce a diverging sequence
b) How many different repeating cycles can you find
*921. Proposed by Richard l Hess, Rancho Palos Verdes, California.
Place 13 three-digit square numbers in the spaces in the accompanying
grid. (The solution is unique.)
922. Proposed by David Iny, Baltimore, Maryland.
Suppose that j{f{x)) = 0 for all real x. Show that a necessary and
sufficient condition that ensures thatj{x) = 0 for all xis thatfbe infinitely
differentiable on the real line. ·
923. Proposed by A. Stuparu, Valcea, Romania.
Let S(n) denote the Smarandachefunction: ifn is a positive integer, then
S(n) = kif k is the smallest nonnegative integer such that k! is divisible by
n. Thus S(l) = 0, S(2) = 2, S(3) = 3, and S(6) = 3, for example. Prove that
the equation S(x) = p, where p is a given prime number, has just 2p- 1
solutions between p and p!.
*924. Proposed by George Tsapakidis, Agrino, Greece.
Find an interior point of a triangle so that its projections on the sides of
the triangle are the vertices of an equilateral triangle.
925. Proposed by Richard A. Gibbs, Fort Lewis College, Durango,
Colorado.
Given positive integers s and c and any integer k such that 0 ::;; k ~ s,
prove that
(-l)Jct(c+j-l)(j) = E