Construction for AIME 1985.14

Filed on Today, at 6:22 pm

AIME 1985.14
In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded 1 point, the loser got 0 points, and each of the two players earned 1/2 point if the game was a tie. After the completion of the tournament, it was found that exactly half of the points earned by each player were earned against the ten players with the least number of points. (In particular, each of the ten lowest scoring players earned half of her/his points against the other nine of the ten). What was the total number of players in the tournament?

Construction
Err this wording sucks but it's late. Looking for symmetry, we may let the lowest ten players each get 4.5 points from each other and also 4.5 points from the other fifteen players, who get 7 points from each other and also 7 points from the lowest ten players. Let the lowest ten players be a. If a_i beats a_{i + 1,\ldots,i + 4\bmod{10}} and ties a_{i + 5\bmod{10}} then each has 4.5 points. Let the other fifteen players be b. If b_i beats b_{i + 1,\ldots,i + 7\bmod{15}} then each has 7 points. Now, consider only relations between the two groups. Tie b_{i,\;1\le i\le5} with a_{2i - 1},a_{2i}, and let edges denote some b defeating some a. Then 6 = \deg b_{i,\;1\le i\le5} and 7 = \deg b_{i,\;6\le i\le15}. Also, \deg a = 15 - 1 - 4 = 10. Connect b_{i,\;1\le i\le5} to a_{j,\;j\ne2i\pm1,(2i + 1)\pm1\pmod{10}}, giving each 10 - 4 = 6 edges and 3 edges to each a thus far. Now for each of the remaining b_i,\;6\le i\le15, connect to a_{i + 1,\ldots,i + 7\mod{10}}, giving each 7 edges and 3 + 7 edges total for all a.

Summarizing, there is at least one possible construction:
    a_i beats a_{i + 1,\ldots,i + 4\bmod{10}} and ties a_{i + 5\bmod{10}}
    b_i beats b_{i + 1,\ldots,i + 7\bmod{15}}
    b_{i,\;1\le i\le5} beats a_{j,\;j\ne2i\pm1,(2i + 1)\pm1\pmod{10}}
    b_{i,\;6\le i\le15} beats a_{i + 1,\ldots,i + 7\mod{10}}
    all unmentioned relationships contain some a_i beating some b_j

Tags: 1985, AIME, Combinatorics

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  • ok done

    By math154, on Mon Nov 23, 2009 8:11 pm

  • dude you should add me as contrib! Wink

    By ghjk, on Sun Nov 22, 2009 9:48 pm

  • hey wait what

    By Poincare, on Fri Oct 16, 2009 2:50 am

  • NOOOOOOOOOOOOOOOOOOOESSSS
    <end randomness>

    By abacadaea, on Tue Aug 11, 2009 5:54 pm

  • contribbbbb.

    By nikeballa96, on Wed Jul 29, 2009 12:46 pm

  • /me wants contrib

    By gauss1181, on Wed Jul 29, 2009 11:22 am

  • wasnt mewto like ur teammate or somethin? u should make him a contributor Mr. Green Mr. Green

    By draops, on Thu Jul 23, 2009 7:11 pm

  • um dude i should contrib????

    By Mewto55555, on Wed Jul 15, 2009 5:15 pm

  • uh that link just led to this blog Shocked

    By gauss1181, on Thu Jul 02, 2009 3:55 pm

  • yes indeed

    By math154, on Wed Jul 01, 2009 9:18 am

  • woah

    Algebra sucks

    By xpmath, on Tue Jun 30, 2009 6:12 am

  • hm 48=2^4\times3

    By gauss1181, on Sat Jun 06, 2009 3:12 pm

  • test

    This is a test

    By dnkywin, on Thu May 28, 2009 1:16 pm

  • actually 1*3*7*13*37

    By Anonymous, on Thu May 21, 2009 1:45 pm

  • lawl@hide tag

    anyway 10101=10^4+10^2+1=(10^2-10+1)(10^2+10+1)=91*111=(3*7)*(13*37) is pretty cool.

    By math154, on Wed May 20, 2009 5:29 pm

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