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trait for a total of 8. Since each woman sion as n approaches infinity; when i is 1, this limit is defined as e, the base Perfect
*Beaudet, Paul R.
ra, three traits; A also cannot be rb and rr, for then she would also be ra, three answer is approaching e times 108.
traits. Therefore, A is ra, rb or ra, rr, 4 The distance of the center of the 1-
Barb (B) and Cleo (C) are ri, rr, and one four traits; and A cannot be ra, rr, for possibilities for this: (1) ra, rb, ri; (2) ra, rb, rr; (3) ra, ri, rr; and (4) rb, ri, rr. 4√3. AC = √(62 -22) = 4√2, and BC = √(52 -12) = 2√6. Let d be the center four traits; (3) can be eliminated since AD = √{52 – (H – 4)2}, BD = √{42 – (H – 3)2} , and CD = √{32 – (H – 2)2}. Use also ra, which leaves rb for B, but then B is also ra, so all three are ra. Thus, C must be ra, rb, ri, and B is ri, rr; so 2 An estimated 18.37 percent of the
Yes,” “Answer No,” and “Have You 5 The young girl should release from
Cheated” cards are selected with equal static position. She will land about 8.7 is 784 for each card. Then 928 – 784 = lum, which the swing is, total energy is difficult regular problem was No. 5 about mg(h – h ), plus kinetic energy, mv2/2, the vertical belt sander. However, of the at any point in its arc. Equating kinetic and potential energy gives v2/2 = gh – gh, so v = √[2g(h – h)] = √[2gL(cosθ 3 After one year, you will owe the
cal and L is the length of the chains. Reader’s entries for the Fall problems will is (1 + i/n)n, where i is the annual inter- trajectory of a projectile, which are x 1 Cleo is not remarkably rich. There
est rate and n is the number of times = x + v t and 0 = y + v t – gt2/2. Now, v are four “remarkable” traits—artistic = vcosθ and v = vsinθ. Solving for t (the (ra), beautiful (rb), intelligent (ri), and time it takes the girl to hit the ground) gives t = [v + √(v 2 +2gy )]/g. Substi- tuting into the equation for x gives x = x + (vcosθ/g)[(vsinθ + √(v2sin2θ + 2gy )], where x = Lsinθ and y = L(1 1 Find the smallest positive integer
– cosθ) + h . Using trial and error (with L = 6 ft and h = 2 ft) on the swing’s 5 A sparrow, flying horizontally in a
straight line, is 20 m directly below an B onus. The punter should aim at the
—Christopher R. Oliver, AL E ’08
2 Five pirates find a chest contain-
tion is given by p = [1/(σ√(2π))]exp(- θ2/2σ2), where θ is the deviation from hawk is flying at 30 m/s, how fast is the the aim angle, θ and varies from -45o to +45o, and σ = 7.5 is the standard devia- vote, and if half or more of the pirates B onus. A regular dodecahedron has
After all, this is football!) Pick a yard- line aim point y ; then the aim angle gons. If some, including none or all, of is given by θ = tan-1[(50-y )/(80/3)]. by y = 50-(80/3)tan(θ +θ), except that when y is negative (indicating that the ball crossed the goal line), y = 20. The expected value of the yard line for spot- -45o to +45o of pydθ. It is easy to write a C omputer Bonus. A, B, and C are
program that varies y to find the mini- 3 My wife and I recently attended a
use the digits 1 through 9 once and only fies the equation, A x B = C2, and where well. Program the expressions for y and p into the spreadsheet (with y as a variable), and calculate y for unit increases in θ from -45 to 45. (This is sufficiently accurate for this problem.) with various values of y until the mini- Tau Beta Pi, P. O. Box 2697, Knoxville,
TN 37901-2697 or email to BrainTick-
lers@tbp.org only as plain text. The C omputer Bonus. The next two pal-
with 16 digits! This is quite a distance 4 Professor Asterisk treated his
to the judges who are Dr. H.G. McIl-
abcddcba where a can equal 1, 4, 5, 6, vried III, PA G ’53; F. J. Tydeman, CA
or 9 and b, c, and d can equal 0 through Δ ’73; J. L. Bradshaw, PA A ’82; and the
—D. A. Dechman, TX A ’57.
6, 8, 10, 12, 14, and, finally, 16-digit in the range of 29 to 39, inclusive. After

Source: http://www.tbp.org/pubs/BTs/W11.pdf