Problem of the Week 831

Penny Parity

Consider n pennies arranged in a straight line. A move consists of taking a penny and turning it over (from head to tail or visa versa) and of doing the same to each of its neighbors. If the penny happens to be at the end of the line, then it will have only one neighbor. For example, if the sequence is HHTH and you choose the 3rd penny, then your move would result in HTHT. The question is, given any sequence of n pennies, is it possible to find a sequence of moves to bring it to all Heads? Show that the answer is no for n=5 but yes for n=6.

Source: The February 17 St. Olaf College "Math Mess."

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Bonus Problem

The following problem was given to me by Charles R. Johnson (William & Mary) who has an algebraic proof that uses a fairly deep result. Is there an elementary/geometric proof out there?

Let P₁, P₂, P₃, P₄ be points in R³, let d_i,j be the Euclidean distance from P_i to P_j, and let

a = d_1,2 × d_3,4,
b = d_1,3 × d_2,4,
c = d_1,4 × d_2,3.

Show that a, b, c are lengths of sides of a triangle. That is,

a <= b + c,
b <= a + c, and
c <= a + b.

© Copyright 1997 Stan Wagon. Reproduced with permission.