Correct solutions from members of the Bradley family came from Chris Mitton, Kenny Albright  Correct solutions from outside the Bradley University community were received from Juan Carlos Marivela, Alexey Vorobyov, Paul Botham, Brice Masselot, Dan Dima, Lou Cairoli, Ahmed AboAdham, CeeAnn Franklin, Aaron Kahn, Paul Lee. 

Label the coins as on the left.  Consider the three pennies 5,8,9 in the middle.  Two of them must have the same side up; without loss, assume there are two heads.  After rotating if necessary, we may assume without loss that 8 and 9 are tails.  In order to avoid a triangle of heads, 5 and 13 must then be tails.  The triangle at 2,3,5 gives that (at least) one of 2 or 3 must be a head;  after refection if necessary, we may assume without loss that 3 is a head.  Heads at 3 and 8 give a tail at 10.  

Suppose that 2 is a head.  Then heads at 2 and 3 give a tail at 1; tails at 1 and 10 give a head at 7; and 2,7,9 are all heads, which is not allowed.  So 2 is a tail.  

Now argue as follows:  2 and 5 give a head at 4; 4 and 8 give a tail at 7; 7 and 10 give a head at 1; 1 and 4 give a tail at 6; 6 and 13 give a head at 15; 1 and 15 give a tail at 11.  But now 7 and 11 give a head at 12, while 3 and 15 give a tail at 12.  

So we must have a triangle of heads or of tails.

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