A large number of coins are spread out on a table.  Exactly 199 of them
are heads up.  Your job is to make two groups of coins such that each
group of coins has exactly the same number of heads showing as the other.

The catch is that you are blindfolded: you can't see which coins show
heads and which show tails (and you can't tell the difference by feel or
any other way.)

How can you accomplish this task?

There will be a $1 prize for the best correct solution submitted by a
midshipman.


A solution is "correct" if it answers the question correctly and explains
the answer; a solution is "best" if it includes the clearest correct
explanation.  Solutions are due by noon on Wednesday, February 10, 1999.
Submit solutions to Prof. Hanna at mathprob@nadn.navy.mil, or via the
mailbox in Chauvenet 301.

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No midshipman submitted a correct answer to problem #85, dividing a 3-4-5
triangle into two pieces of equal area.  The shortest line segment which
will do this has length 2, forming the base of an equilateral triangle
with equal sides of length sqrt(10) which lie along the longest two sides
of the original triangle.


Oops. For "equilateral" read "isosceles."