 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Contestants in a Circle (Posted on 2011-01-31) The eight contestants were seated at a circular table in alphabetic order clockwise: Alice, Bob, Carol, Dave, Eve, Frank, Gisele and Harry. Some points had already been awarded by the judges, in varying amounts to the different competitors, but the competitors didn't know what their totals were. As a last competition, they were to deduce all their scores thus far based upon the announcement of eight totals of three contestants' scores, each total being the sum of three successive people around the table, so for example, the total for Harry, Alice and Bob (the triplet centered on Alice) was given as 38.

The successive triplets centered on Bob through Harry totaled as follows: 49, 45, 48, 52, 44, 39 and 48.

What were in fact the individual scores to that point in the competition?

 See The Solution Submitted by Charlie No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Possible solution | Comment 1 of 3

It seems that Alice wins, as long as she is one of the first three to solve the simultaneous equations:

a+b+c=49, b+c+d=45, c+d+e=48, d+e+f=52, e+f+g=44, f+g+h=39, g+h+a=48, h+a+b=38:

whereby: a = 28,   b = 6,   c = 15,   d = 24,   e = 9,   f = 19,   g = 16,   h = 4.

A total of 121 points has been scored so far, so assuming that the same number of points is allocated for each round, one plausible distribution might be 5,3,2,1 per round for 11 rounds (any more equitable distribution would already ensure Alice the victory), in which case only Dave can only beat Alice in the 12th and final round, and even then only if he both wins the last round and Alice is fourth or worse.

Edited on January 31, 2011, 3:52 pm
 Posted by broll on 2011-01-31 15:20:03 Please log in:

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